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\magnification=1200
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\centerline {\Big\bf QUANTUM THEORY AND ANALYSIS}
\medskip
\centerline {John C. Baez 1989}
\section{1. States, Observables, and Symmetries}

The basic ingredients of a theory of a physical system are mathematical objects describing 1) a set $\S$ of {\bf states} of
the system, 2) the set $\O$ of {\bf observables} of a system, that is,
real-valued properties of the system which can be experimentally
determined, and 3) the group $\G$ of {\bf symmetries} of the system.
The difference between classical physics and quantum theory arises from
very different choices of mathematical objects used to describe states,
observables, and symmetries.  But in both classical and quantum physics
certain mathematical structures appear that are associated with
$(\S,\O,\G)$ .  Three of the most important are: \medskip 
1. Given a state $\psi\in\S$ and an observable $A\in\O$ there is a 
probability measure $\mu_{A,\psi}$ on $\R$ which represents the
probability distribution of values that the observable $A$ may be measured to have when the system is in the state $\psi$.
(We are not assuming that the observable $A$ has a unique numerical value in a given state $\psi$; repeated measurements of
the same observable a system in the same state may have {\it different} values, and these values are
distributed according to a probability distribution.) 
\aside{A {\bf probability measure} $\mu$ on a measure space $X$ is a finite measure such that $\mu(X) = 1$.  Measures
on a topological space, e.g. $\R$, will always be assumed to be defined on the $\sigma$-algebra of Borel sets.}
2.  The group $\G$ has an action $U$ on the set $\S$ of states. 
\aside{An {\bf action} $U$ of a group $G$ on a set $S$ is a mapping $U\colon G\times S \rightarrow S$, usually
written as $(g,x) \mapsto U(g)x$, such that: 1) $U(e)x = x$ where $e$ is 
the identity in $G$; 2) $U(gh)x = U(g)U(h)x$.  These imply $U(g^{-1}) = U(g)^{-1}$.} 
The most important and simple example is where $\G = \R$ and $U$ is {\bf time evolution}; i.e. given a state
$\psi\in\S$ and $t\in\R$, the state $U(t)\psi$ is obtained by simply waiting $t$ units of time, letting the
state evolve according to the physical laws the system obeys.  The requirement $U(t+s) = U(t)U(s)$ amounts to saying
that the physical laws don't change with time, i.e. there are no time-dependent external influences on the system.
\medskip
3. The group $\G$ is a {\bf topological group}, that is, a group with a topology such multiplication 
is continuous from $\G\times\G$ to $\G$ and the map $g\mapsto g^{-1}$ is continuous from $\G$ to $\G$.  
A continuous homomorphism $t \mapsto g_t$ from $\R$ to $\G$ 
is called a {\bf one-parameter subgroup}, though strictly speaking its {\it range} is a subgroup of $\G$.  
Given a one-parameter subgroup let $U(t) = U(g_t)$.  Then associated to the one-parameter subgroup there is an observable 
$A\in\O$ called the
{\bf associated conserved quantity}, which has the property that $\mu_{A, U(t)\psi}$ is
independent of $t$.  
For example, in physics the conserved quantity associated with time evolution is called {\bf energy} or the {\bf Hamiltonian}.  In
systems which have symmetry under spatial translations, the associated conserved quantities are called {\bf momenta} (in the   
$x,y$, and $z$ directions, for example).  In systems which have rotational symmetry the associated conserved
quantities are called {\bf angular momenta} (about the $x,y$ and $z$ axes, for example).  In electromagnetism there is
a one-parameter group of symmetries called {\bf global gauge transformations} which corresponds to the conserved 
quantity {\bf electric charge}.
\aside{This relation between states, observables, and symmetries is less obvious and more interesting
than the first two; it often goes by the name of {\bf N\"other's Theorem}, after Emmy N\"other, the student of Hilbert,
since a version of it applicable to problems in variational calculus was
proved in her `Invariante Variationsprobleme' article in 1918, available
in English in {\it Transport Theory and Stat. Phys.}{\bf 1}, 186-207 (1971).} 
We now will say what mathematical objects fill the roles of $(\S,\O,\G)$
in classical mechanics and  
quantum theory.  In the case of quantum theory we will immediately be
led to the spectral theorem and Stone's theorem, which we will then prove.  

\section{2. States, Observables, and Symmetries in Classical Mechanics}
In classical mechanics, the set of states $\S$ is $\R^{2n} = \R^n\oplus\R^n$, called {\bf phase space}, and
the first summand $\R^n$ is called {\bf configuration space}.  One uses coordinates $\{q_i\}_{i=1}^n$ for configuration space;
these coordinates are called {\bf positions}, and one uses coodinates $\{q_i,p_i\}_{i=1}^n$ for phase space; the
$p_i$ being called {\bf momenta}.  \aside{The relation between these `momenta' and the `momenta' mentioned above is
indirect.   A typical example of phase space appears in the description of a
system of $n$ point particles in ordinary 3-dimensional space; each particle's
state is described by its position $\vec q_i = (q_{1i},q_{2i},q_{3i})$ and
its momentum, or mass times velocity vector, $\vec p_i = (p_{1i},p_{2i},p_{3i})$, and the whole assemblage is described by the following vector in the phase space $\S = 
\R^{3n}\oplus\R^{3n}$: $(\vec q_1,\dots,\vec q_n,\vec p_1,\dots,\vec p_n)$.}
The set of observables $\O$ is the space $C(\R^{2n})$ of continuous real-valued functions on phase space.  Given a state
$\psi\in\R^{2n}$ and an observable $A\in C(\R^{2n})$, the probabilty measure $\mu_{A,\psi}$ mentioned above is
just $\delta_{A(\psi)}$, the Dirac delta measure at $A(\psi)\in\R$.  In other words, if the state is $\psi$ the value of any
observable $A$ is {\it always} measured to be $A(\psi)$; there's no `uncertainty' in the value of $A$. 

We will consider only the simplest case, in which the group $\G$ of symmetries is just $\R$.  
  Given a smooth function $H\in C^\infty(\R^{2n})$, we define
the associated vector field $v_H$ on $\R^{2n}$ by
$$v_H = {\partial H\over\partial p_i}{\partial\over\partial q_i} - {\partial H\over\partial q_i}{\partial \over\partial p_i}\>.$$
A vector field $v$ on $R^{2n}$ is {\bf integrable} if there is an action $U$ of $\R$ on $\R^{2n}$ such that
$${d\over dt}U(t)\psi = v(U(t)\psi)\>.$$
If $v_H$ is integrable and $U$ is an action of $\R$ on $\R^{2n}$ such that ${d\over dt}U(t)\psi = v_H(U(t)\psi)$, and we write in coordinates $U(t)\psi
 = (q_i,p_i)\in \R^{2n}$, then {\bf Hamilton's equations}
 hold:
$${dq_i\over dt} = {\partial H\over\partial p_i}\quad;\quad
{dp_i\over dt} = -{\partial H\over\partial q_i}\>.$$
 In classical mechanics the action $U$ of the group $\R$
 on states is required to be one
arising from some $H\in C^\infty(\R^{2n})$ in this way, and $H$ is the conserved quantity associated to $U$.  
\medskip Exercise:  Show that Hamilton's equations are indeed equivalent to
$${d\over dt}U(t)\psi = v_H(U(t)\psi)\>.$$ Show that if $H\in C^\infty(\R^{2n})$ and $v_H$ is integrable, the directional derviative
$v_H H$ equals zero.  Show that this implies that the measure $\mu_{H, U(t)x}$ is independent of $t$ for any state $x\in \R^{2n}$.
(This is a baby version of N\"other's theorem.)
\medskip
Exercise:  Show that if $H ={1\over 2} \sum_{i=1}^n q_i^2 + p_i^2 $ (the {\bf harmonic oscillator Hamiltonian})
the vector field $v_H$ is integrable by explicitly solving Hamilton's equations.  
Show that if $H = \sum _{i=1}^n {1\over2}(q_i^2 + p_i^2 ) + {1\over
4}\lambda_i p_i^4$ (the {\bf anharmonic oscillator Hamiltonian}) the vector field $v_H$ is integrable if and
only if all the {\bf coupling constants} $\lambda_i$ are non-negative.
\medskip
\aside {The setup for classical mechanics as we have described it can be substantially generalized.  The three
most important directions of generalization are: 1) The group of symmetries $\G$ can be an arbitrary {\bf Lie group} 
(a group which is also a smooth manifold, such that the group operations are smooth maps).  
2) The phase space $\S$ can be an arbitrary {\bf symplectic manifold} (a manifold with a non-degenerate closed 2-form).
3)  The group $\G$ and the phase space $\S$ can be infinite-dimensional; this occurs in the study of classical
field theory, i.e. the study of wave equations.}

\vfill\eject
\section{3. States, Observables and Symmetries in Quantum Theory}
In quantum mechanics the set of states $\S$ is 
the unit sphere in a Hilbert space $\H$; i.e. $$\S = \{\psi\in\H\colon\Vert\psi
\Vert = 1\}\>.$$
\aside {As we shall see, for many purposes it's better to take the state space to be the quotient of this sphere by  
the equivalence relation: $\psi\sim\phi$ if $\psi = c\phi$ for some $c\in \C$, i.e. our present description is `redundant', but
this is unimportant for introductory purposes.}   

The set of observables $\O$ is then the set of self-adjoint operators on $\H$.
Recall that a {\bf  operator on} $\H$ is a linear map $A$
from a linear subspace of $\H$ to $\H$.   This subspace, the domain
of $A$, is denoted $D(A)$, and we are {\it not} requiring even that it
be dense in $\H$.  If $D(A)$ is dense in $\H$ we say that $A$ is
{\bf densely defined}.  An operator with domain equal to all of $\H$, satisfying
$$\exists c\ge 0\;\forall \psi\in\H\quad \Vert A\psi\Vert \le c\Vert\psi\Vert$$
is called a {\bf bounded} operator, and the space of all bounded operators on $\H$
is denoted by $\L(\H)$.  The {\bf adjoint} of an operator $A$ is supposed to have the
property $\langle\phi,A\psi\rangle = \langle A\adj\phi,\psi\rangle$, where we use
$\langle\cdot,\cdot\rangle$ for the inner product in $\H$. 
 (Also, we follow the 
physicists' convention that the inner product is linear in the {\it 
second} variable.)  But for operators that are not bounded
this requires some care in formulation.  Thus we define the {\bf adjoint} of
a densely defined operator $A$ to be the operator $A\adj$ with domain $$D(A\adj) = \{\phi\in\H\colon
\exists\phi'\in\H\;\forall\psi\in D(A)\quad
\langle\phi,A\psi\rangle = \langle \phi',\psi\rangle\}$$
and noting that if $\phi'$ with the above property exists it is {\it unique}
(because $A$ is densely defined), we define $A\adj\phi$ for $\phi\in D(A\adj)$
to be the unique $\phi'$ such that
$$\forall\psi\in D(A)\quad
\langle\phi,A\psi\rangle = \langle \phi',\psi\rangle\>.$$
An operator $A$ on a Hilbert space is said to be {\bf self-adjoint} if
it is densely defined and equal to its adjoint (which in particular
means that $D(A) = D(A\adj)$.)  \aside {The definition of adjoint may
look complicated, but it is the most natural possible definition for
densely defined operators.  It foreshadows the fact that densely defined
(unbounded) operators are altogether more subtle than bounded operators.
Physicists typically ignore these nuances, and indeed the fastest way to
tell if a physicist is a `mathematical physicist' is to ask him to
define `adjoint'.} Given a state $\psi$ in $\S$ and a self-adjoint
operator $A$ on $\H$ there is a probability measure $\mu_{A,\psi}$ on
$\R$, called the `spectral measure', which describes the probability
distribution of measured values for $A$ when the system is in the state
$\psi$.  This measure arises as a consequence of the spectral theorem,
which we shall soon state and prove.

A bounded operator $T$ on a Hilbert space $\H$ is {\bf unitary} if $TT\adj =
 T\adj T = I$, or equivalently if $T$ is onto and $\Vert T\psi\Vert = \Vert\psi
\Vert$ for all $\psi\in\H$.  
Recall that an action $U$ of a group $G$ on a vector space $V$ 
 is a {\bf representation} if $U(g)$ is linear for all $g\in G$; if
in addition $S$ is a Hilbert space and $U(g)$ is unitary for all $g\in G$ we say $U$ is a {\bf unitary representation}.
If $\G$ is a topological group we say that a representation of $\G$ on a Hilbert space is
{\bf strongly continuous} if $g_\alpha\goesto g$ in $\G$ implies $\Vert U(g_\alpha)\psi - U(g)\psi\Vert
\rightarrow 0$ for all $\psi \in \H$.  
In quantum theory one requires that the group $\G$ of symmetries be a topological group and that
the action $U$ of $\G$ on the states $\S$ come from a strongly continuous unitary
representation of $\R$ on $\H$.  Stone's theorem then associates to  
any one-parameter subgroup $g_t$ of $\G$ a conserved quantity $A$, a self-adjoint operator on $\H$ also called
the `generator' of $U(t) = U(g_t)$.

\aside {We recall here the three most important topologies on $\L(\H)$.  The {\bf norm} topology is defined by:
$T_\alpha \goesto T\;$ if $\Vert T_\alpha - T \Vert \goesto 0$.  The {\bf strong} topology is defined by:
$T_\alpha \goesto T\;$ if $\Vert T_\alpha \psi - T \psi\Vert \goesto 0$ for all $\psi\in\H$.  And the {\bf weak}
topology is defined by: $T_\alpha\goesto T\;$ if $\langle\phi, (T_\alpha - T)\psi\rangle \goesto 0$
for all $\psi,\phi\in\H$.  A function from or to $\L(\H)$ is said to be norm-continuous, strongly continuous,
or weakly continuous depending on which topology is used for $\L(\H)$.}
\medskip

Exercise - Let $\ell^2$ denote the Hilbert space of complex sequences
$\psi = (\psi_1,\psi_2,\dots)$ with finite norm $\Vert \psi\Vert = (\sum |\psi_i|^2)^{1/2} $, given the inner product $\langle\psi,\phi\rangle = \sum 
\overline\psi_i\phi_i$.  
Let $T$ be the {\bf left shift} operator, given by 
$T(\psi_1,\psi_2,\dots) = (\psi_2,\psi_3,\dots)$.  Calculate $T\adj(\psi_1,\psi_2,\dots)$ and
explain why $T\adj$ is called the {\bf right shift} operator.  These shift operators
illustrate some of the things that bounded linear operators can do in infinite
dimensions that they cannot in finite dimensions.  For example, show that
$T\adj$ is an {\bf isometry}, e.g. $\Vert T\adj\psi \Vert = \Vert \psi \Vert$,
but not unitary.  Show that an isometry of a finite-dimensional Hilbert space is
 unitary.  Show that $T$ and $T\adj$ do not commute.  

Exercise - Show that norm convergence implies strong convergence, which implies weak convergence, but that
the converses do not hold.  (For counterexamples, consider the sequences
$\{T^n\}$ and $\{T\adj^n\}$, where $T$ is the left shift.)  Show that converses
hold in a finite-dimensional Hilbert space.  Show also that if $U_\alpha$ are {\it unitary} and $U_\alpha \goesto U$
weakly then $U_\alpha\goesto U$ strongly; thus a weakly continuous unitary representation is automatically
strongly continuous.

Exercise - Show that for bounded operators $S,T$ we have $T\adj\adj = T$, $\Vert T\adj \Vert = \Vert T \Vert$,
$\Vert T\adj T\Vert = \Vert T\Vert^2$, $(cT)\adj = \overline c T\adj$,
$(S+T)\adj = S\adj + T\adj$, and $(ST)\adj = T\adj S\adj$.  Show that the map $T\mapsto T\adj$ is
norm-continuous and weakly continuous but not strongly continuous. (For
the counterexample consider shift operators.)

\section{4.  The Spectral Theorem - Bounded Case} 

We begin by proving the spectral theorem for bounded self-adjoint operators and then deal with the unbounded case.
First we define the {\bf spectrum} of a bounded operator (not necessarily self-adjoint) $T\in\L(\H)$ to be the
set of $\lambda\in \C$ such that $T-\lambda I$ is not invertible, i.e. is not a bijection with a
bounded inverse.  The spectrum of $T$ is denoted $\sigma(T)$.
The complement of the spectrum of $T$ is called the {\bf resolvent set} of $T$ and denoted $\rho(T)$.
The exercises below show that the spectrum is a natural generalization of the set of eigenvalues of a matrix.
\medskip
Exercise - Show that if $\H = \C^n$ the spectrum of $T\in\L(H)$ is just the set of $\lambda\in\C$ such that
$\det(T - \lambda I) = 0$.  Show {\it in this case} that the spectrum of $T$ is the set of
eigenvalues of $T$ (for self-adjoint $T$ this is easy; for general $T$ one can use the Jordan canonical form -
see any advanced book on linear algebra, e.g. Birkhoff and MacLane, {\sl A Survey of Modern Algebra}.)

Exercise - Given a bounded operator $T$ on $\H$ define $\lambda\in\C$ to be an {\bf approximate
eigenvalue} of $T$ if for any $\epsilon > 0$
 there exists
$\psi\in\H$ with $\Vert\psi\Vert = 1$ and $\Vert T\psi - \lambda\psi\Vert \le \epsilon$.  Show that if
$\lambda$ is an approximate eigenvalue for $T$ then $\lambda\in\sigma(T)$.    For a partial converse, show that 
if $A$ is a bounded self-adjoint operator on a Hilbert space and $\lambda\in\sigma(A)$, then $\lambda$ is
an approximate eigenvalue of $A$.  Hint for the second part: show that 
 for any densely defined operator $T$ on $L(\H)$,
$$Ran(T)^\perp = Ker(T\adj)\>,$$
so that for bounded self-adjoint $A$, $(A -\lambda I)$ is invertible unless $\lambda$ is an approximate eigenvalue of $A$.

Exercise - Show that $\lambda\in\sigma(T)$ implies $\overline\lambda\in\sigma(T\adj)$.  Show that if $A$ is bounded and self-adjoint $\sigma(A)\subseteq\R$.
\vskip 2ex 
\Proclaim The Spectral Mapping Theorem (for polynomials).  If $T$ is a
bounded operator and $P$ is a nonconstant polynomial
then $\sigma(P(T)) = \{P(\lambda)\colon\lambda\in\sigma(T)\}$.
(For short we simply write $\sigma(P(T)) = P(\sigma(T))$.)

Proof - Let $\lambda\in\sigma(T)$.  Since $x = \lambda$ is a root of
$P(x) - P(\lambda)$, we can factor: 
$$P(x) - P(\lambda) = (x - \lambda)Q(x)\>,$$
so $P(T) - P(\lambda) = (T - \lambda I)Q(T)$.  Since the operator  $T -
\lambda I$ is not invertible neither is $P(T) - P(\lambda)$, i.e.
$P(\lambda)\in \sigma(P(T))$. 

Conversely suppose $\mu\in \sigma(P(T))$.  Factor $P(x) - \mu$ as $a(x -
\lambda_1)\cdots(x - \lambda_n)$.   Then 
$P(T) - \mu I = a(T - \lambda_1)\cdots(T - \lambda_n)$.
If all the $\lambda_i$ are {\it not} in $\sigma(T)$ then we have 
$$(P(T) - \mu I)^{-1}  = a^{-1}(T - \lambda_1)^{-1}\cdots(T -
\lambda_n)^{-1}\>,$$ so $\mu$ is {\it not} in $\sigma( P(T))$, a
contradiction.  Thus there must be some $\lambda_i \in \sigma(T)$, and 
$$P(\lambda_i) - \mu = a(\lambda_i - \lambda_1)\cdots(\lambda_i -
\lambda_n) = 0$$ 
so $\mu = P(\lambda_i)$. \qed
\medskip

\Proclaim Theorem.
The {\bf spectral radius} of a bounded operator $T$ is defined to be
$r(T) = \sup \{\vert\lambda\vert\colon \lambda\in \sigma(T)\}\>.$ 
 If $A$ is a bounded self-adjoint operator on $\H$ then $r(A) = \Vert A
\Vert$.  

Proof - If $\lambda\in \sigma(A)$ then $\lambda$ is an approximate eigenvalue 
for $A$ so there exists $\psi_n \in \H$
with $\Vert \psi_n \Vert = 1 $ and $\Vert A \psi_n \Vert \goesto \vert \lambda \vert$.  Thus $\Vert A \Vert \ge \vert \lambda \vert$,
so $\Vert A \Vert \ge r(A)$.  

Now choose $\psi_n$ such that $\Vert \psi_n \Vert = 1$ and $\Vert A\psi_n\Vert \goesto \Vert 
A\Vert$.  Let $\lambda = \Vert A \Vert$.
If $A$ is self-adjoint then $$\eqalign{\Vert A^2 \psi_n - \lambda^2 \psi_n\Vert^2 &= 
\langle A^2 \psi_n - \lambda^2 \psi_n, A^2 \psi_n - \lambda^2 \psi_n\rangle \cr
&= \Vert A^2 \psi_n \Vert ^2 - 2\lambda^2 \Vert A\psi_n\Vert^2 + \lambda^4\cr
&\le (\Vert A \Vert\> \Vert A\psi_n\Vert )^2 - 2\lambda^2 \Vert A\psi_n\Vert^2 + \lambda^4\cr
&= -\lambda^2\Vert A\psi_n\Vert^2 + \lambda^4 \goesto 0\;.\cr}$$
Thus $\lambda^2$ is an approximate eigenvalue for $A^2$, so $\lambda^2 \in \sigma(A^2) = \sigma(A)^2$.
 It follows that $\Vert A \Vert \le r(A)$ .  Thus $\Vert A \Vert = r(A)$.  
\qed

\aside{The remarkable thing about this theorem is that it relates a topological concept, the norm, to a concept which is defined
purely algebraically, the spectral radius.  This interplay of topology 
and algebra is behind the  beautiful theory of 
$C^\ast$-algebras, a kind of algebra over the complex numbers which has a norm
 and $\ast$ operator
satisfying axioms like those which hold for $\L(\H)$.  $C^\ast$-algebras
are important in the mathematics of quantum field theory, and in understanding
the relation between classical and quantum physics.}
\vskip 2ex

Exercise - Fill out the following proof of a generalization of the theorem
above: if $T$ is a bounded operator then $r(T) \le \|T\|$ and  
 $$r(T) = \limsup_{n\goesto\infty}\Vert T^n \Vert^{1/n}\>.$$
The idea is to use the Laurent series for $(T - \lambda I)^{-1}$, which is modeled after the 
series for $(z - \lambda)^{-1}$:
$$(T - \lambda I )^{-1} = -\sum_{n=0}^\infty \lambda^{-n-1} T^n\quad.$$ 
In the region in $\C$ where the series is norm-convergent, term-by-term multiplication is
justified in showing that $(T - \lambda I)$ multiplied by $ -\sum_{n=0}^\infty \lambda^{-n-1} T^n$, in either
order, is the identity.  
 Recall however Hadamard's theorem that 
the radius of convergence of $\sum a_n z^n$ is $R = 1/\limsup_{n\goesto\infty} |a_n|^{1/n}$, and that
$\vert\sum a_n z^n\vert \goesto \infty$ for some sequence of $z$'s with $\vert z \vert \goesto R$. The same argument
(fortified with the uniform boundedness theorem) shows that the radius of convergence of $\sum T^n z^n$ is
$r = 1/\limsup\Vert T^n \Vert^{1/n}$, and that $\Vert \sum T^n z^n \Vert \goesto \infty$ for some
sequence of $z$'s with $\vert z \vert \goesto R$.   
Taking $\lambda = 1/z$, it follows that
$(T - \lambda)^{-1}$  exists for $\lambda \ge R = 
\limsup \Vert T^n \Vert ^{1/n}$, and
$$\Vert( T - \lambda I)^{-1} \Vert \goesto \infty $$ for some sequence of $\lambda$'s with $\vert \lambda \vert $ converging
to $R$.  This immediately implies that $r(T) \le R$, and to show that it also implies $r(T) = R$ it suffices
to show that $\Vert (T - \lambda I)^{-1} \Vert$ is continuous on $\rho(T)$, since then $r(T) < R$ would imply
have $\Vert (T - \lambda I)^{-1} \Vert$ bounded near $\vert\lambda\vert = R$, a contradiction.
The continuity follows from a similar power series argument.

Exercise - Let $T$ denote the left shift operator on $\ell^2$.  Compute
the spectral radius of $T$, show that the whole unit circle in $\C$
consists of approximate eigenvalues of $T$, and show that $0$ is an
eigenvalue of $T$.  Show that the whole unit circle in $\C$ consists
of approximate eigenvalues of $T\adj$ and that $0$ is in the spectrum of
$T\adj$, but $0$ is not an approximate eigenvalue of $T\adj$.
\medskip

\Proclaim Theorem.  If $T$ is a bounded operator its spectrum is compact.

Proof - In the previous theorem we showed  $r(T) \le \Vert T\Vert$, so  $\sigma(T)$ is bounded
and it suffices to show that $\sigma(T)$ is closed or equivalently that the resolvent set $\rho(T)$ is
open.  Suppose $\lambda_0\in\rho(T)$ and let $R = (T - \lambda I)^{-1}$.  Then for $\lambda$ sufficently
close to $\lambda_0$ the power series
$$\sum _{n=0}^\infty (\lambda - \lambda_0)^n R^{n+1}$$
is norm-convergent and 
$$(T - \lambda I) 
(\sum _{n=0}^\infty (\lambda - \lambda_0)^n R^{n+1}) = $$
$$\sum _{n=0}^\infty[(T - \lambda_0 I)(\lambda - \lambda_0)^n R^{n+1} + 
(\lambda_0 - \lambda)(\lambda - \lambda_0)^n R^{n+1}] = $$
$$\sum _{n=0}^\infty[(\lambda - \lambda_0)^n R^{n} - 
(\lambda - \lambda_0)^{n+1} R^{n+1}] = I\>.$$
The same argument shows that
 $$[(\sum _{n=0}^\infty (\lambda - \lambda_0)^n R^{n+1})](T - \lambda I) = I \>,$$
so we have found an explicit inverse for $(T - \lambda I)$.  Thus $\lambda \in \rho(T)$.\qed

\aside {Note the use of power series techniques in both the proof of this theorem and in
the proof that $r(T) = \limsup_{n\goesto\infty}\Vert T^n \Vert ^{1/n}$ in the exercise
above.  Such techniques from complex analysis are very useful in operator theory; $(T - \lambda I)^{-1}$ is
called an analytic operator-valued function on $\rho(T)$ because it admits such local
power series expansions in $\lambda$.}

\Proclaim Spectral Theorem - continuous functional calculus.
If $A$ is bounded self-adjoint operator on the Hilbert space $\H$ there is a unique map $\pi\colon C(\sigma(A))\goesto \L(\H)$
such that:
\medskip 1. $\pi$ is a {\bf $\ast$-homomorphism}, i.e.
$$\pi(\lambda f) = \lambda\pi(f)\quad, \quad\pi(f+g) = \pi(f)+\pi(g)\quad,\quad \pi(fg) = \pi(f)\pi(g)\>,$$
$$\pi(\overline f) = \pi(f)\adj \quad,\quad \pi(1) = I\>,$$
\medskip
2. $\pi$ is continuous, i.e., $\Vert\pi(f)\Vert \le C\Vert f\Vert_\infty$,
\medskip
3. If $h$ is the function $h(\lambda) = \lambda$, then $\pi(h) = A$.
\medskip\noindent
We shall usually write $f(A)$ for $\pi(f)$. In this notation, $\pi$ has the following properties:
\medskip
4.  If $f\ge 0$ then $f(A) \ge 0$.
\medskip
5.  $\Vert f(A)\Vert = \Vert f\Vert_\infty $.
\medskip
6. $\sigma(f(A)) = f(\sigma((A))$. (The {\bf spectral mapping theorem}.)
\medskip
7.  If $A\psi = \lambda\psi$ then $f(A)\psi = f(\lambda)\psi$.

\medskip
Proof - We first define $\pi$ on polynomials by $\pi(p) = p(A)$.  The idea is to show that the 
definition of a {\it polynomial function} of a self-adjoint operator extends to {\it continuous functions}, obtaining
a `continuous functional calculus'.  The trick is to use Weierstrass' theorem, which says that the
set of polynomials is dense in $C(\sigma(A))$, since $\sigma(A)$ is compact.  If we can show 2 for polynomial
$f$, it will then follow
that $\pi$ extends uniquely from the polynomials to all of $C(\sigma(A))$, and satisfies 2 for all $f$.
(Here we are using the easy but all-important {\bf B.L.T. theorem} (for `bounded linear transformation'), which
says that if $X$ and $Y$ are Banach spaces, $D$ is a dense subspace of $X$, and $f\colon D\goesto Y$ satisfies
$\Vert f(x)\Vert \le C\Vert x\Vert$ for all $x\in D$,
then $f$ extends uniquely to a bounded linear function from $X$ to $Y$.)
Thus we first verify 2 for a polynomial $p$.  We have
$$\eqalign{\Vert \pi (f) \Vert &= \Vert p(A)\Vert\cr
&=(\Vert p(A)\adj p(A) \Vert)^{1/2}  \cr
&= \Vert (p\overline p)(A) \Vert^{1/2} \cr
&= r((p\overline p)(A))^{1/2} \>,\cr}$$
the latter since  $p(A)\adj p(A) = (p\overline p)(A)$ is self-adjoint. 
By the spectral mapping theorem for polynomials we have
$$\eqalign{ r((p\overline p)(A)) &=  \sup\{p(\lambda)\overline p(\lambda)\colon \lambda\in\sigma(A)\} \cr
&= \Vert p\Vert_ \infty^2\;
,\cr}$$
Thus $$ \Vert \pi(p)\Vert = \Vert p \Vert _\infty$$
if $p$ is the restriction of a polynomial to $\sigma(A)$, verifying 2 for polynomials.

It follows from the B.L.T. theorem that we may uniquely extend $\pi$ to continuous functions, and that 2 and 7 
hold.  Clearly 3 holds, and since 1 holds for polynomial $f,g$ its easy to see that it extends by
continuity to hold for all $f,g\in C(\sigma(A))$.  It also follows from the B.L.T. theorem that $\pi$
as we've defined it is the {\it unique} function satisfying 1 - 3.  

Property 4 clearly holds for polynomial $f$ and extends by continuity to all $f\in C(\sigma(A))$.
We've shown that 5 holds.  7 holds for polynomials and extends by continuity to all $f\in C(\sigma(A))$.

It remains to prove 6, which we know holds for polynomials.  We show that 
$f(\sigma(A))\subseteq \sigma(f(A))$ as follows.   Let $\lambda\in \sigma(A)$; $\lambda$ is
an approximate eigenvalue of $A$ so choose $\psi_n \in \H$ with $\Vert\psi_n\Vert = 1$ and
$\Vert (A - \lambda I)\psi_n\Vert \le 1/n$.  Choose polynomials $P_m$ converging uniformly to
$f$ on $\sigma(A)$, with $\Vert P_m - f \Vert \le 1/m$.  Then for any $m$, 
$$\eqalign{\Vert (f(A) - f(\lambda) I)\psi_n\Vert &\le 
 \Vert (f(A) - P_m(A))\psi_n \Vert + \Vert (P_m(A) - P_m(\lambda) I)\psi_n \Vert\cr &\quad +
  \Vert (P_m(\lambda) - f(\lambda) \psi_n\Vert \cr
&\le \Vert f(A) - P_m(A) \Vert  + \|(P_m(A) - P_m(\lambda))\psi_n\| + \Vert P_m - f \Vert_{L^\infty (\sigma(A))} \cr
&\le \|(P_m(A) - P_m(\lambda))\psi_n\| + 2/m\>.} $$ 
Since $\|\psi_n\| = 1$ and $\|A\psi - \lambda\psi\| \le 1/n$, we have, for any 
$k \ge 0$,
$$ \eqalign{\|A^k\psi_n - \lambda^k\psi_n\| &= \|(A^{k-1} + \lambda A^{k-2} + 
\cdots +\lambda^{k-1} I)(A -\lambda I)\psi_n\| \cr
\le \|A^{k-1} + \lambda A^{k-2} + \cdots + \lambda^{k-1}\|/n\>,}$$
which goes to zero as $n\to \infty$, so 
$$ \lim_{n\to\infty} \|(P_m(A) - P_m(\lambda))\psi_n\| = 0\>,$$
so for $n$ large enough this implies that  
$$\|(f(A) - f(\lambda)I)\psi_n \| \le 3/m \>.$$
Since $m$ is arbitrary this implies that 
$$\|(f(A) - f(\lambda)I)\psi_n \| \to 0 \>.$$
Thus $f(\lambda)$ is an approximate eigenvalue for $f(A)$, so $f(\lambda) \in 
\sigma(f(A))$. 

To show that $\sigma(f(A)) \subseteq f(\sigma(A))$, suppose that $\lambda\notin 
f(\sigma(A))$.  Then the function $g(x) = (f(x) - \lambda)^{-1}$ is continuous 
on $\sigma(A)$, so we obtain $g(A)$ by the functional calculus, and 
$$g(A)(f(A) - \lambda I) = I = (f(A) - \lambda I)g(A) $$
by the $\ast$-homomorphism property.  Thus $f(A) - \lambda I$ is invertible, 
so $\lambda \notin \sigma(f(A))$.  
 \qed

\aside{The spectral theorem has many versions, and a long and tangled history.
There are three main forms of the spectral theorem: the functional calculus 
form, the projection-valued measure form (see the remarks on spectral 
projections in the next section), and the following form: if $A$ is a 
self-adjoint operator on a Hilbert space $\H$, there is a measure space 
$(X,\mu)$, a real-valued measurable function $f$ on $X$, and a unitary 
transformation $U:\H \to L^2(X)$ such that $A = U^{-1}M_fU$, where $M_f$ denotes multiplication by $f$.  That is, every self-adjoint
operator is unitarily equivalent to a multiplication operator.}
\vskip 3em

Exercise - Prove the B.L.T. theorem.
\vskip 3em

Note that if $\psi\in\H$ the above theorem implies that $f\mapsto \langle\psi,f(A)\psi\rangle$ is a continuous
linear functional from $C(\sigma(A))$ to $\C$, which moreover is positive by property 4.  Thus by the Riesz-Markov
 theorem there is a unique measure $\mu_{A,\psi}$ on $\sigma(A)$,
 the {\bf spectral measure}, such that 
$$ \langle\psi,f(A)\psi\rangle = \int_{\sigma(A)} f(\lambda) d\mu_{A,\psi}(\lambda)\quad.$$
If $\Vert\psi\Vert = 1$ we have $\int_{\sigma(A)} d\mu_{A,\psi}(\lambda) = 1$, so the spectral measure is
a probability measure.  This makes good on our promise that in quantum theory the spectral theorem would
yield a probability measure given a state and an observable - though so far we've only shown this for bounded
$A$.

The spectral measure also immediately allows us to strengthen the above theorem:

\Proclaim Spectral Theorem - measurable functional calculus.
Let $A$ be a bounded self-adjoint operator on $\H$.  Let $\B(\sigma(A))$ denote the bounded Borel functions on $\sigma(A)$ - {\rm not} identifying those which are equal a.e..
Given $A$ on $\H$ and $f \in \B(\sigma(A))$, there is a unique
bounded operator $\pi(f) = f(A)$ such that 
$$\langle\psi ,f(A)\psi\rangle =  \int_{\sigma(A)} f(\lambda) d\mu_{A,\psi}(\lambda)\quad.$$
Then $\pi: \B(\R)\goesto\L(\H)$  is the unique map satisfying:
\medskip 1. $\pi$ is a $\ast$-homomorphism,
\medskip
2.  $\Vert f(A)\Vert = \Vert f\Vert_\infty^\prime$, where the $L^{\infty\prime}$ norm is defined with the following modified
notion of `almost everywhere': we say a property holds a.e.$^\prime$ if for each $\psi\in\H$ the property holds except
for a set of measure zero relative to $\mu_{A,\psi}$.
\medskip
3. If $h$ is the function $h(\lambda) = \lambda$, then $h(A) = A$.
\medskip\noindent
Moreover 
4.  If $f\ge 0$ then $f(A) \ge 0$.
\medskip
5. $\sigma(f(A)) = f(\sigma((A))$. 
\medskip
6.  If $A\psi = \lambda\psi$ then $f(A)\psi = f(\lambda)\psi$.
\medskip
7.  If $AT = TA$ for some $T\in\L(\H)$, then $f(A)T = Tf(A)$.
\medskip
8.  Suppose that $f_n \goesto f$ pointwise a.e.$^\prime$ and $\Vert f_n\Vert_\infty^\prime$ is bounded.  Then $f_n(A)\goesto f(A)$
strongly.

Sketch of proof - For the uniqueness one uses the polarization identity (see 
below), which implies that a bounded operator is
determined by the inner products $\langle\psi,T\psi\rangle$.  
  Property 8 follows from the dominated convergence
theorem and the rest of the properties follow by limiting arguments and the previous version of the spectral
theorem, using Lusin's theorem that given any Borel measure $\mu$ on $\R$, any  $f\in\B(\R)$ 
is a pointwise $\mu-$a.e. limit of a uniformly bounded sequence of
 continuous functions.\qed


Exercise - Prove the {\bf polarization identity}  
for any $T\in\L(\H)$:

$$\eqalign{\langle\phi,T\psi\rangle 
&= {1\over 4} \bigg[\,\{\,\langle\psi+\phi,T(\psi+\phi)\rangle  
- \langle \psi-\phi,T(\psi-\phi)\rangle\, \}\cr
&-i\{\,\langle\psi+i\phi,T(\psi+i\phi)\rangle 
 -\langle\psi-i\phi,T(\psi-i\phi)\rangle\,\}\,\bigg]
\>.}$$

Exercise - Work out the details of the proof of the above theorem. 
\medskip
The following exercises contain important generalizations 
of the spectral theorem, and the results will be used in the next section:
\medskip

Exercise - Prove the following: 
\Proclaim Spectral Theorem for Collections.  
Let $A_1,\dots,A_n$ be a collection of pairwise commuting bounded self-adjoint
operators on $\H$.  Let $$X = \sigma(A_1)\times\cdots\times\sigma(A_n)\subseteq
\R^n\>.$$  Let $\B(X)$ denote the bounded Borel functions on $X$, again {\rm not} identifying those which are equal a.e..
There is a map $\pi: \B(X)\goesto\L(\H)$ and
for each $\psi\in\H$ a measure $\mu_{\psi}$ on $X$ such that 
 $$\langle\psi,\pi(f)\psi\rangle = \int_X f(x) d\mu_\psi(x)$$
and:
\medskip 1. $\pi$ is a $\ast$-homomorphism,
\smallskip
2.  $\Vert\pi(f)\Vert \le \Vert f\Vert_\infty^\prime$, where the $L^{\infty\prime}$ norm is defined with the following modified
notion of `almost everywhere': we say a property holds a.e.$^\prime$ if for each $\psi\in\H$ the property holds except
for a set of measure zero relative to $\mu_{\psi}$.
\smallskip
3. If $h\colon X\to \R$ is the function $h(\lambda_1,\dots,\lambda_n) = \lambda_i$, then $\pi(h) = A_i$.
\smallskip
4.  If $f\ge 0$ then $\pi(f) \ge 0$.
\smallskip
5.  If some $T\in\L(\H)$ has $A_iT = TA_i$ for all $i$,
 then $\pi(f)T = T\pi(f)$.
\smallskip
6.  Suppose that $f_n \goesto f$ pointwise a.e.$^\prime$ and $\Vert f_n\Vert_\infty^\prime$ is bounded.  Then $f_n(A)\goesto f(A)$
strongly.

\medskip 
\smallskip One usually writes $f(A_1,\dots,A_n)$ for $\pi(f)$.
\medskip
To prove this one can essentially copy the proof for
a single operator.  However to  
 get the B.L.T. theorem to work don't use polynomials in the $A_i$,
but instead linear combinations of `rectangle functions'.  Namely, define
a {\bf rectangle} in $\R^n$ to be a set of the form $S_1\times\cdots\times S_n$
with each $S_i$ a Borel subset of $\R$.  Define $\pi$ on characteristic
functions of rectangles by:
$$\pi(\chi_R) = \chi_{S_1}(A_1)\cdots\chi_{S_n}(A_n)$$
where each $\chi_{S_i}(A_i)$ is defined by the spectral theorem.
Using the spectral theorem for a single bounded operator, and
in particular statement 8 of the `measurable functional calculus' version, 
show that if $R_i$ is a finite collection of disjoint rectangles then
$$\Vert \sum a_i\pi(\chi_{R_i}) \Vert \le \Vert \sum a_i\chi_{R_i}\Vert_\infty
\>.$$
Define $\pi$ on finite linear combinations of rectangle functions by
$\pi(\sum a_i\chi_{R_i}) = \sum a_i\pi(\chi_{R_i})$, and use the B.L.T. theorem
to extend $\pi$ to $\B(X)$. 
\medskip
Exercise - A bounded operator is {\bf normal} if it commutes with
its adjoint.  E.g., unitary and bounded self-adjoint operators are
normal.  Show that if $T$ is normal then ${\rm Re}T = {1\over 2}(T+T\adj)$
and ${\rm Im}T = {1\over 2i}(T-T\adj)$ are bounded self-adjoint and commute.
Use the spectral theorem for collections 
to prove the following:

\Proclaim Spectral Theorem for Normal Operators.
Let $T$ be a normal operator on $\H$.  Let $\B(\sigma(T))$ denote the bounded Borel functions on $\sigma(T)$.
There is a map $\pi: \B(\sigma(T))\goesto\L(\H)$ 
and for each $\psi\in\H$ a measure $\mu_{T,\psi}$ on $X$  such that
$$\langle\psi ,f(T)\psi\rangle =  \int_{\sigma(T)} f(\lambda) d\mu_{T,\psi}(\lambda)\quad.$$
and:
\medskip 1. $\pi$ is a $\ast$-homomorphism,
\smallskip
2.  $\Vert f(A)\Vert = \Vert f\Vert_\infty^\prime$, where the $L^{\infty\prime}$ norm is defined with the following modified
notion of `almost everywhere': we say a property holds a.e.$^\prime$ if for each $\psi\in\H$ the property holds except
for a set of measure zero relative to $\mu_{\psi}$.
\smallskip
3. If $h$ is the function $h(\lambda) = \lambda$, then $h(T) = T$.
\smallskip
4.  If $f\ge 0$ then $f(T) \ge 0$.
\smallskip
5. $\sigma(f(T)) = f(\sigma((T))$. 
\smallskip
6.  If $T\psi = \lambda\psi$ then $f(T)\psi = f(\lambda)\psi$.
\smallskip
7.  If $ST = TS$ for some $S\in\L(\H)$, then $f(S)T = Tf(S)$.
\smallskip
8.  Suppose that $f_n \goesto f$ pointwise a.e.$^\prime$ and
 $\Vert f_n\Vert_\infty^\prime$ is bounded.  Then $f_n(T)\goesto f(T)$
strongly.

\vskip 3ex

Exercise - Show using the theorem above that if $T$ is normal, the spectrum of 
$T$ is the closure of $\bigcup_{\psi\in\H} {\rm supp}\mu_{T,\psi}$.  
(Hint: show that if $\lambda$ is not in the closure of $\bigcup {\rm supp}\mu_{T,\psi}$, one can construct $(T - \lambda I)^{-1}$ using the functional calculus.)

Exercise - Use the exercise above to show that the spectrum of a normal operator
consists of approximate eigenvalues.  Conclude that if $U$ is unitary, $\sigma
(U)$ is a subset of the unit circle in $\C$.  

Exercise - Assume $A\in \L(\H)$.  Use the polarization identity to show that 
if $A\ge 0$, 
i.e.~$\langle\psi, A\psi\rangle \ge 0$ for all $\psi\in\H$, then 
$A$ is self-adjoint.

Exercise - Show that $A\ge 0$ if and only if $\sigma(A)$
is a subset of $[0,\infty)$ (for one direction use the previous exercise).  
Show that if $A\ge 0$ there exists $B\in\L(\H)$ 
with $B\ge 0$ and $B^2 = A$. 

\section{5. The Spectral Theorem - Unbounded Case}
The spectral theorem for unbounded self-adjoint operators proceeds by reduction to the bounded case.
We begin by establishing a useful property of self-adjoint operators.
We define an operator $T$ on $\H$ to be {\bf symmetric} if it
is densely defined and for all $\psi,
\phi\in D(T)$ we have 
$$\langle\phi,T\psi\rangle = \langle T\phi,\psi\rangle\>.$$
Note that this is equivalent to saying that $T\subseteq T\adj$, since
it says that for every $\phi\in D(T)$ there exists $\phi'\in\H$ (namely $T\phi$)
 such that 
$$\langle\phi,T\psi\rangle = \langle \phi',\psi\rangle\>,$$
so that $D(T)\subseteq D(T\adj)$, and $T\adj\phi =\phi' = T\phi$,
so that $T\subseteq T\adj$.
Thus every self-adjoint operator is symmetric, but as we'll see the converse
does {\it not} hold.
\aside{Beware: physicists often say self-adjoint when they mean symmetric.
This is a result of insufficient education and the following facts: 
1) `sufficiently nice' symmetric operators are self-adjoint,
and 2) it's much easier to check if an operator is symmetric than if its
self-adjoint - in other words, it can be very difficult to show that a symmetric
operator is `sufficiently nice'.  To check that an operator $T$ is symmetric
one must show that
$\langle\phi,T\psi\rangle = \langle T\phi,\psi\rangle$ for every
$\phi,\psi\in D(T)$; this is often easy to do, and if $T$ is a differential
operator it usually amounts to integration by parts.  To check that $T$
is self-adjoint one must {\it also} show that if $\phi\in\H$ has
a $\phi'\in\H$ such that
$\langle\phi,T\psi\rangle = \langle \phi',\psi\rangle$, then $\phi$ lies in 
$D(T)$.  This is usually done indirectly using
various theorems, some of which we will discuss later, and 
many problems in mathematical physics consist of proving that
symmetric operators are self-adjoint.  This is because the spectral theorem 
only holds
for self-adjoint operators, so self-adjoint operators are much better than
merely symmetric operators. 
\smallskip We also note term `hermitian' is also used by
 mathematicians
to mean symmetric, and by physicists to mean self-adjoint, by which they mean
symmetric\dots we will avoid this term.}
We define an operator $T$ to be {\bf closed} if given $\psi_i\in D(T)$
such that $\psi_i\goesto\psi\in\H$ and $T\psi_i\goesto\psi'$ then
$\psi\in D(T)$ and $T\psi = \psi'$.  If $T$ is an operator
and $T\subseteq\S$ for some
operator $S$, we say $S$ is an {\bf extension} of $T$.  If
$T$ is an operator with a closed extension we say $T$ is
{\bf closable}.  If $T$ is closable the intersection of all
closed extensions of $T$ is a closed extension of $T$ called the
{\bf closure} of $T$, written $\overline T$.

Exercise - Prove the claim above the intersection of a set of
closed extensions is a closed extension.

Exercise - Show that if $T$ is a densely defined operator satisfying
$\Vert T\psi\Vert \le C\Vert\psi\Vert$ then $T$ has a unique
closed extension, and that this extension is a bounded operator.

Exercise - An operator that's not closable.  Let $\psi_n$ be an 
orthonormal basis for a Hilbert space $\H$ and let $\phi\in\H$ be a vector 
that is not a linear combination of finitely many $\psi_n$.  Let $D$ be the 
set of finite linear combinations of $\phi$ and the $\psi_n$, and let 
$T$ be the operator with domain $D$ defined by:
$$ T\left(a\phi + \sum_{n=1}^N b_n \psi_n\right) = a\phi \>.$$
Show that $T$ is a densely defined linear operator.  Show that any 
closed extension $S$ of $T$ would have $S\phi = \phi$ and $S\phi = 0$,
 obtaining a contradiction.

Exercise - A nonzero operator whose adjoint is zero (roughly speaking).  
Let $\psi_n$ be an orthonormal basis for a Hilbert space $\H$ and let 
$D$ be the set of all finite linear combinations of the $\psi_n$.
Let $T$ be the operator with domain $D$ defined by:
$$ T\left(\sum_{n=1}^N a_n \psi_n\right) = (\sum_{n=1}^N a_n)\psi_1 \>.$$
Show that $T$ is a densely defined linear operator.
Show that the domain of $T\adj$ consists of all vectors orthogonal 
to $\psi_1$, and that $T\adj$ is zero on this domain.

Note that in matrix language $$T = \pmatrix{1&1&1&\cdots\cr
0&0&0&\cdots\cr
0&0&0&\cdots\cr
\cdot&\cdot&\cdot&\cdots\cr}$$
So why is $T\adj$ zero?

\Proclaim The Basic Criterion for Self-adjointness.
Let $T$ be a symmetric operator on a Hilbert space $\H$.  Then the following
three statements are equivalent:
\medskip
1.  $T$ is self-adjoint.
\medskip
2.  $T$ is closed and $Ker(T\adj \pm i) = \{0\}$.
\medskip
3.  $Ran(T\pm i) = \H$.

Proof.  $1\Rightarrow 2$: Assume $T$ is self-adjoint.  
To show that $T$ is closed we need to show that if
$\psi_i\goesto\psi\in\H$ and $T\psi_i\goesto\psi'$ then
$\psi\in D(T)$ and $T\psi = \psi'$.  Since $T = T\adj$
we only need to show that $\psi\in D(T\adj)$ and
$T\adj\psi = \psi'$.  In other words, we need to show that
for all $\phi\in D(T)$,
$$\langle\psi,T\phi\rangle = \langle \psi',\phi\rangle\>.$$
Since $T$ is self-adjoint and $\psi_i\in D(T)$ we have:
$$\eqalign{\langle\psi,T\phi\rangle &= \lim_{i\to\infty} \langle\psi_i,T\phi\rangle \cr
&= \lim_{i\to\infty} \langle T\psi_i,\phi\rangle \cr
&= \langle \psi',\phi\rangle \cr
}$$
as desired.  

Next we need to show that if $T$ is self-adjoint and
 $(T + i) \psi = 0$ or   $(T - i) \psi = 0$ for some      
$\psi\in D(T)$, then $\psi = 0$.  If $(T-i)\psi = 0$, 
$$\eqalign{\langle \psi, i\psi\rangle &= \langle \psi, T\psi\rangle\cr
 &= \langle T \psi, \psi\rangle\cr
 &= \langle i \psi, \psi\rangle\cr
 &= -\langle  \psi, i\psi\rangle\>,\cr}$$
so $\psi = 0$, and the other case works similarly.

$2\Rightarrow 3$:  Assume $T$ is symmetric and closed and $Ker(T\adj\pm i) = 0$.
We shall show $Ran(T\pm i)$ is closed and dense in $\H$.
Since $Ker(T\adj\pm i) = Ran(T\mp i)^\perp$, and the orthogonal complement of 
a subspace is empty if and only if the subspace is dense, clearly
$Ran(T\pm i)$ is dense.

To show that $Ran(T+i)$ is closed we naturally use the hypothesis
that $T$ is closed (the proof that $Ran(T-i)$ is closed is similar).
Suppose we have $(T+i)\psi_i \to \phi$ in $\H$.  We need to show that there 
exists $\psi \in D(T)$ such that $\phi = (T+i)\psi$.
We have
$$0\leftarrow \Vert (T+i)\psi_i - (T+i)\psi_j\Vert^2$$
$$=\Vert T(\psi_i - \psi_j)\Vert^2  + \Vert \psi_i - \psi_j \Vert^2$$
$$ + \langle i(\psi_i - \psi_j), T(\psi_i - \psi_j)\rangle
+\langle T(\psi_i - \psi_j), i(\psi_i - \psi_j)\rangle\>,$$
and the last two terms cancel because $T$ is symmetric.  Thus $\psi_i \to \psi$
 and
$T\psi_i \to T\psi$ for some $\psi \in D(T)$, since $T$ is closed.  Thus we have
$\phi = \lim (T+i)\psi_i = (T+i)\psi$.

$3\Rightarrow 1$: Assume that $Ran(T \pm i) = \H$.
Given $\phi\in D(T\adj)$ we want to show that
$\phi\in D(T)$, since this will imply
$D(T) = D(T\adj)$ which together with $T\subseteq T\adj$
implies $T = T\adj$.  

Since $Ran(T - i) =  \H$ there exists $\psi\in D(T)$
such that $(T\adj - i)\phi = (T - i)\psi$.  Since
$T\subseteq T\adj$ this implies $\phi - \psi \in D(T\adj)$
and $$(T\adj - i)(\phi - \psi) = 0\>.$$  Since
$Ran(T+i) = \H$, $Ker(T\adj-i) = 0$ so $\phi = \psi \in D(T)$,
as was to be shown.\qed



Exercise - Let $X$ be a measure space, $\mu$ a measure on $X$, and 
$\H = L^2(X,\mu)$.  Let $f$ be a real-valued measurable function
on $X$ and let $M_f$ be the operator with domain equal to $D = \{\psi\in
\H\colon \; f\psi\in\H\}$ and for $\psi\in D$ let $M_f\psi = f\psi$.
$M_f$ is called a {\bf multiplication operator}.  Show that: 
1) $M_f$ is densely defined, 2) $M_f$ is symmetric, and 3)
$M_f$ is self-adjoint.  
One can prove 3) directly from the definitions
or using the criterion above.

\Proclaim Spectral Theorem - measurable functional calculus.  
Suppose that $A$ is a self-adjoint operator on $\H$.
Then for each $\psi\in\H$ there is a {\bf spectral measure} $\mu_{A,\psi}$, and 
for each $f\in\B(\R)$ there is a unique bounded operator $\pi(f) = f(A)$
such that
$$\langle\psi,f(A)\psi = \int_\R\, f(\lambda)d\mu_{A,\psi}(\lambda)$$
and:
\medskip 1. $\pi$ is a $\ast$-homomorphism,
\medskip
2.   $\Vert f(A)\Vert =  \Vert f\Vert_\infty^\prime$,
where the $L^{\infty\prime}$ norm is defined with the modified notion of
`almost everywhere' in which we say a property holds a.e.$^\prime$ if for each $\psi\in\H$
the property holds a.e. except for a set of measure zero relative to $\mu_{A,\psi}$. 
\medskip
3. If $f_n\in\B(\R)$ are functions such that $f_n(\lambda) \to \lambda$ pointwise and
$\vert f_n(\lambda)\vert \le \vert\lambda\vert$, then $f_n(x)\psi \to A\psi$ for any
$\psi\in D(A)$.
\medskip
4.  If $f\ge 0$ then $f(A) \ge 0$.
\medskip
5.  If $A\psi = \lambda\psi$ then $f(A)\psi = f(\lambda)\psi$.
\medskip
6.  Suppose that $f_n \goesto f$ pointwise a.e.$^\prime$ and $\Vert f_n\Vert_\infty^\prime$ is bounded.  Then $f_n(A)\goesto f(A)$
strongly.

Sketch of proof -  The idea is to reduce to the bounded case, so we need a 
function that
`compresses' $A$ down to a normal operator without losing any information.
For example, the operator consisting of multiplication by $x$ on $L^2(\R)$ is self-adjoint and unbounded, and our
procedure will `compress' it to the operator consisting of multiplication by 
$(x+i)(x-i)^{-1}$,
which is unitary. 
Let $\alpha(x) = (x+i)(x-i)^{-1}$.  
Our idea is to construct $f(A)$ by constructing
$(f\circ\alpha^{-1})(\widetilde A)$, where $\widetilde A = (A+i)(A-i)^{-1}$ will be a well-defined unitary operator, called the {\it Cayley transform} of $A$.  
Composing $f$ with $\alpha^{-1}$ `undoes' the effect of the Cayley transform 
$A\mapsto \widetilde A$.
We begin with some lemmas:

\proclaim Lemma 1.  If $A$ is self-adjoint then $\widetilde A = (A+i)(A-i)^{-1}$
is a well-defined unitary operator on $\H$.

Proof - If $A$ is self-adjoint the maps
$(A \pm i)\colon D(A)\to\H$ are 1-1 and onto. Thus the composite
  $(A+i)(A-i)^{-1}$
is a well-defined 1-1 and onto map from $\H$ to $\H$, and to show that it is 
unitary it suffices to show that it is norm-preserving.

We need to show that $\Vert (A+i)(A-i)^{-1}\phi \Vert = \Vert \phi \Vert$ for 
all $\phi \in\H$.   We can use the criterion for self-adjointness to write $\phi = (A-i)\psi$ for some $\psi\in D(A)$.  Thus it suffices to show that 
$\Vert (A+i) \psi \Vert = \Vert (A-i)\psi\Vert$ for all $\psi\in D(A)$.
This follows from:
$$\Vert(A \pm i)\psi\Vert^2 = \Vert A\psi\Vert^2 + \Vert \psi \Vert^2\>,$$
shown earlier in the proof of the criterion for self-adjointness.\qed


\proclaim Lemma 2.  Let $S^1$ denote the unit circle in the complex plane.  
Given $f\in \B(\R)$ define $\widetilde f\in\B(S^1)$
by: $\widetilde f(x) = f(\alpha^{-1}(x))$ on the range of $\alpha$, and 
$\widetilde f = 0$ elsewhere.
Then the map $f\mapsto\widetilde f$ defines a $\ast$-homomorphism from 
$\B(\R)$ to $\B(S^1)$ modulo functions that are zero a.e.\ $\prime$,
 where $f = 0$ a.e.$\prime$ means that $f = 0$ a.e.\ with repsect to all 
the measures $\mu_{\widetilde A, \psi}$.  

Proof - 
An easy calculation shows that $\alpha\colon \R\to S^1$, so that
$\widetilde f\in\B(S^1)$.  The range of $\alpha$ is just $S^1 - \{1\}$.
 We prove one of the $\ast$-homomorphism properties and leave most of
 the rest as 
easy exercises: 
$$(\widetilde f + \widetilde g)(x) = f(\alpha^{-1}(x)) + g(\alpha^{-1}(x)) = (f+g)(\alpha^{-1}(x)) = \widetilde{(f+g)}(x)\;.$$
The only tricky point is that we need to show that $\widetilde 1$, that is 
the function equal to $1$ on $S^1 - \{1\}$ and $0$ at $\{1\}$, is equal to 
$1$ a.e.$\prime$.  It suffices to show that $\mu_{\widetilde A,\psi}(1) = 0$ 
for all $\psi\in\H$.  We use the:

\proclaim Sublemma.  If $T$ is normal, then $\lambda\in \sigma(T)$ has nonzero 
measure with respect to some $\mu_{T,\psi}$ if and only if $T\phi = 
\lambda\phi$ for some nonzero $\phi\in\H$.


Proof - Suppose $\mu_{T,\psi}(\lambda) \ne 0$.  Let $\chi_\lambda$ 
be the characteristic function of $\{\lambda\}$, and let $\phi 
= \chi_\lambda(T)\psi$.    We claim that $\phi\ne 0$ and $T\phi = 
\lambda\phi$.  First note that by the spectral theorem  
$$\eqalign{ \|\phi\|^2 &= \langle \psi, \chi_\lambda(T)^2 \psi\rangle \cr
&= \langle \psi, \chi_\lambda(T) \psi \rangle \cr
&= \int \chi_\lambda (\lambda') d\mu_{T\psi}(\lambda) \cr 
& = \,u_{T,\psi} (\{\lambda\})  > 0\>,}$$
so $\phi\ne 0$.  Similarly,  
$$\eqalign{ \|(T - \lambda I)\phi\|^2 &= \langle \psi, \chi_\lambda(T)^2 
(T - \lambda)\adj(T - \lambda) \psi \rangle \cr
&= \int \chi_\lambda(\lambda')^2 |\lambda' - \lambda|^2 d\mu_{T,\psi} \cr
&= 0 \>,}$$
so $T\phi = \lambda\phi$.  

The converse uses similar ideas and is left as an exercise. \qed
 
By this sublemma it suffices to show that there is no nonzero $\phi\in\H$ 
with $\widetilde A \phi = \phi$.  Suppose there was, i.e.\ suppose that 
$$\phi = (A+i)(A-i)^{-1}\phi\>.$$
Let $\psi = (A-i)^{-1}\phi$.  Then 
$$(A-i)\psi = (A+ i)\psi$$
so $\psi = 0$.  It follows that $\phi = 0$ because $Ker(A-i)^{-1} = 0$; 
this is a contradiction.  \qed 
 
Note that $\sigma(\widetilde A) \subseteq S^1$ since 
$\widetilde A$ is unitary.
Thus $\widetilde f$ restricts to a bounded Borel function on 
$\sigma(A)$, and given $f\in \B(\R)$ we may define
 $f(A) = \widetilde f (\widetilde A)$ 
using the spectral theorem for bounded operators.  Most of the statements of 
the theorem now follow rather simply.

Statement 1 follows from Lemma 2 and the functional calculus for 
normal operators, e.g.:
$(fg)(A) = \widetilde{(fg)}(\widetilde A) = \widetilde f \widetilde g (\widetilde A) = 
\widetilde f (\widetilde A) \widetilde g (\widetilde A) = f(A)g(A)$.

We define the spectral measures $\mu_{A,\psi}$ using the formula given in the statement of the theorem.
Note that if $\mu = \mu_{A,\psi}$ and $\widetilde\mu = \mu_{\widetilde A,\psi}$ then unraveling
the definitions implies
$$\int f d\mu = \langle\psi,f(A)\psi\rangle = \langle\psi,\widetilde f(\widetilde A)\psi\rangle =
\int \widetilde f d\widetilde\mu$$
Thus the $\mu$-measure of a set $S$ equals the $\widetilde\mu$-measure of the set $\alpha (S)$,
and statement 2 follows from the corresponding statement for bounded operators.

Statements 4-6 also follow easily from the spectral theorem for bounded operators, and we leave them
as exercises.  Statement 3 
is more touchy since the functions $\{f_n\}$ are not uniformly bounded.
We need a lemma:



\Proclaim Lemma 3.  Let $g(x) = (x+i)^{-1}$.  Then $g(A) = (A+i)^{-1}$.  

Proof - By definition, $g(A) = \widetilde g(\widetilde A)$, and it is easily 
computed that $\widetilde g(x) = {x-1\over 2ix}$.  Thus by the functional 
calculus for normal operators:
$$\eqalign{g(A) &= \widetilde g(\widetilde A) = 
(\widetilde A - 1)(2i\widetilde A)^{-1}\cr
&= {1\over 2i}(1 - {\widetilde A}^{-1})\cr 
&= {1\over 2i}(1 - (A-i)(A+i)^{-1})\cr
&= {1\over 2i}((A+i) - (A-i))(A+i)^{-1}\cr
&= (A+i)^{-1}\>;\cr}$$
here as always one must take care to justify all the algebraic manipulations 
involving unbounded operators, but we leave this as an exercise. \qed

Now suppose $f_n\in\B(\R)$ are functions such that $f_n(\lambda) \to \lambda$ 
pointwise and $\vert f_n(\lambda)\vert \le \vert\lambda\vert$. 
Let $g$ be defined as in the lemma; then the sequence $(f_n+i)g\in\B(\R)$
 is uniformly 
bounded and converges to $1$ pointwise, so by statement 6 and the lemma 
we have:
$$\lim_{n\to\infty} (f_n(A) + i)(A+i)^{-1}\phi = 
\lim_{n\to\infty} (f_n(A) + i)g(A)\phi = \phi$$
for all $\phi\in\H$.  Given any $\psi\in D(A)$, we may write 
$\psi = (A+i)^{-1}\phi$, and conclude that 
$\lim_{n\to\infty} (f_n(A) + i)\psi = (A+i)\psi$, or
$$\lim_{n\to\infty} f_n(A)\psi = A\psi\>,$$
as desired.\qed

\aside{The criterion for self-adjointness 
is due to von Neumann.  The extension of the spectral 
theorem to unbounded operators is due to von Neumann, `Allgemeine 
Eigenwerttheorie Hermitescher Functionaloperatoren,' {\it Math. Ann.} {\bf 102},49-131 (1929-1930), and Stone, `Linear Transformations in Hilbert Spaces and 
their Applications to Analysis,' {\it A.~M.~S. Colloq. Publ.} {\bf 15} (1932). 
Carleman in 1923 had pointed out that the spectral theorem would not work for 
arbitrary symmetric operators.}

The spectral
 theorem has many uses; our first big application of it will be to prove 
Stone's theorem in the next section.  Note that it fulfills our 
promise to associate to each state $\psi$ and observable $A$ of a quantum 
system a probability measure $\mu_{A,\psi}$.  We shall now describe some 
basic consequences of this.       

It follows from the the theorem above that for all states 
 $\psi\in D(A)$ we have 
$$\langle\psi , A\psi\rangle = \int  \lambda \,
d\mu_{A,\psi}(\lambda)\>.$$ 
Since $\mu_{A,\psi}$ represents the probability distribution of values that 
$A$ can be measured to have in the state $\psi$, the quantity $\langle\psi,
 A\psi\rangle$ naturally corresponds to the mean value, or {\bf expectation 
value}, of the observable $A$ in the state $\psi$.  Note that this only 
makes sense for $\psi\in D(A)$.

\vskip 3ex
Exercise - A {\bf projection} on a Hilbert space $\H$ is a 
bounded self-adjoint operator  $P$ with  $P^2 = P$.  (Sometimes one considers 
projections that are not self-adjoint, but in quantum theory one typically 
assumes self-adjointness, so we build it into the definition.)  Show that there is a 1-1 correspondence between projections and closed subspaces of $\H$ given 
by associating to each projection its range, or conversely to each closed 
subspace the orthogonal projection onto that subspace.  Given a self-adjoint 
operator $A$ on $\H$ and a Borel set $S\subseteq \R$, define the 
{\bf spectral projection} onto $S$ to be the operator $\chi_S (A)$ given by 
the functional calculus, 
where $\chi_S$ is the characteristic function of $S$.
Show that the spectral projections are indeed projections and satisfy 
$\chi_{S\cup T}(A) = \chi_S(A) + \chi_T(A)$ if the sets $S$ and $T$ are 
disjoint. 
\vskip 3ex

Note that $\langle \psi, \chi_S(A)\psi\rangle = \int _S d\mu_{A,\psi}(\lambda)$,
 so that   $\langle \psi, \chi_S(A)\psi\rangle$ corresponds to the probability 
that the observable $A$ is measured to have a value in the set $S$.  Spectral 
projections are thus very important for computing probabilities.  

\vskip 3ex
Exercise - Show that if $\R$ is the disjoint union of a countable collection of 
Borel sets $S_i$ that the sum of the probabilities $p_i$ that $A$ will be 
measured to have a value lying in $S_i$ is one.
\vskip 3ex

 This is 
only the beginning of detailed correspondence between concepts from
 probability theory and from quantum theory.  Indeed quantum theory may be 
regarded as a `noncommutative generalization' of probability theory - the 
noncommutativity of operator multiplication gives rise to numerous strange 
effects such as the `uncertainty principle', which we will discuss later.

\section{6. Stone's Theorem}

Stone's theorem gives a 1-1 correspondence between one-parameter subgroups of 
unitary operators on $\H$ and self-adjoint operators on $\H$, which in 
quantum mechanics connects symmetries with their associated conserved
 quantities.  To prove it we need the notion of an `essentially self-adjoint' 
operator.  

Recall that a densely defined operator is closable if it has closed extensions, in which case the closure of the operator is its smallest closed 
extension.  The following exercise shows that symmetric operators are always closable.

\vskip 3ex
Exercise - If $T$ is densely defined then $T\adj$ is closed.  If $T$ is symmetric it is closable, since $T\adj$ is a closed extension of $T$. 
\vskip 3ex

A symmetric operator $A$ on a Hilbert space $\H$ is {\bf 
essentially self-adjoint} if its closure is self-adjoint.
The importance of this is that it will be easier to show that 
operators are essentially self-adjoint than to show that they're 
self-adjoint, and for some purposes essential self-adjointness is 
good enough.  One has: self-adjoint$\Rightarrow$essentially self-adjoint$\Rightarrow$symmetric. 
 There are a lot of useful relationships between these concepts, but for we'll just mention a couple that we'll need for Stone's theorem: 

\vskip 3ex 
Exercise - If $S$ and $T$ 
are densely defined and $S\subseteq T$, then $T\adj\subseteq S\adj$.  
If $A$ is self-adjoint and $B$ is a symmetric 
extension of $A$ then $B = A$.  (Hint: show $A\subseteq B$ 
and $B\subseteq B\adj\subseteq A\adj = A$.)
\vskip 3ex

Exercise - Copy the proof of the basic criterion for self-adjointness
 to show the following: 
\vskip 2em
\Proclaim Basic Criterion for Essential Self-Adjointness. 
Let $T$ be a symmetric operator on a Hilbert space $\H$.  Then the following
three statements are equivalent:
\medskip
1.  $T$ is essentially self-adjoint.
\medskip
2.  $Ker(T\adj \pm i) = \{0\}$.
\medskip
3.  $Ran(T\pm i)$ are dense in $\H$.


\Proclaim Stone's Theorem.  Let $A$ be a self-adjoint operator and let $U(t) = 
e^{itA}$.  Then $U(t)$ is a strongly continuous one-parameter unitary group, 
i.~e.,  for all $t$ $U(t)$ is unitary, $U(t+s) = U(t)U(s)$, and $t_\alpha \to t$
implies $U_\alpha(t)\psi\to U(t)\psi$ for all $\psi\in\H$.  Moreover 
$$D(A) = \{\psi\in\H\colon\;\lim_{t\to 0}{U(t)\psi - \psi \over t} \>{\rm exists}\}$$
and for all $\psi\in D(A)$
$$iA\psi = \lim_{t\to 0}{U(t)\psi - \psi \over t} \>.$$
Conversely, if $U(t)$ is a strongly continuous one-parameter unitary group, 
there is a unique self-adjoint operator $A$ such that $U(t) = e^{itA}$.  $A$ is 
called the {\bf infinitesimal generator} of $U(t)$.

Proof - If $A$ is a self-adjoint operator it is an easy exercise using the spectral theorem to show that $e^{itA}$ is a strongly continuous unitary group.  If 
$\psi\in D(A)$ then since $(e^{it\lambda} -1)/t$ converges pointwise to $\lambda$ as $t\to 0$, with $|(e^{it\lambda} -1)/t| \le \lambda$ for $|t| \le 1$, statement 3 of the spectral theorem for unbounded operators implies that 
$$iA\psi = \lim_{t\to 0}{U(t)\psi - \psi \over t} \>.$$
Defining $B$ by 
$$D(B) = \{\psi\in\H\colon\;\lim_{t\to 0}{U(t)\psi - \psi \over t} \>{\rm exists}\}$$
and 
$$iB\psi = \lim_{t\to 0}{U(t)\psi - \psi \over t} \>,$$
it follows from the above that $A\subseteq B$, and $B$ is symmetric since 
$$\eqalign{\langle B\psi,\phi\rangle &= \lim_{t\to 0} \langle (it)^{-1}
(U(t) - I)\psi , \phi\rangle \cr
&= \lim_{t\to 0} \langle \psi ,-(it)^{-1}(U(-t) - I)\phi\rangle\cr
&= \lim_{t\to 0} \langle \psi ,(it)^{-1}(U(t) - I)\phi\rangle\cr
&= \langle \psi, B\phi\rangle\>.\cr}$$
Since $B$ is a symmetric extension of the self-adjoint oerator $A$, the exercise above implies $B = A$.  Thus 
$$D(A) = \{\psi\in\H\colon\;\lim_{t\to 0}{U(t)\psi - \psi \over t} \>{\rm exists}\}\>.$$

Conversely, suppose that $U(t)$ is a strongly continuous one-parameter 
unitary group.  We will obtain an essentially self-adjoint operator by 
differentiating $U(t)\psi$ for $\psi$ 
in a domain of `smooth' vectors, and the closure of this operator will be our 
self-adjoint $A$.

We use an important trick of G\aa rding's to obtain vectors for which $U(t)\psi$
 is differentiable.  Given $f\in C_0^\infty(\R)$ define $$\phi_f = 
\int f(t)U(t)\phi \,dt\>,$$
where the integral can be taken in the sense of Riemann (one can develop a 
full-blown theory of Lebesgue integration for Hilbert-space-valued functions, but for 
continuous functions like we have here the Riemann integral will do).  
Note that 
$$\eqalign{\lim_{t\to 0} \left( {U(t) - I\over t}\right) \phi_f &= 
\lim_{t\to 0}\int f(s) \left({U(t+s) - U(s)\over t}\right) \phi \,ds\cr
&=\lim_{t\to 0} \int {f(\tau - t) - f(\tau)\over t}U(\tau)\phi\,d\tau\cr
&=-\int f'(\tau)U(\tau)\phi\, d\tau\cr
&= \phi_{-f'}\>,\cr}$$
leaving as an exercise the justification for passing the limit through the integral!

We define the {\bf G\aa rding domain} by $D = \{\phi_f:\>f\in C_0^\infty(\R)\}$ and define the operator 
$B$ on $D$ by $$B\phi = \lim_{t\to 0}\left({U(t) - I\over it}\right)\phi\>.$$
We claim: 1) $D$ is dense, 2) $B$ is essentially self-adjoint, 3) if $A = \overline 
B$ then $U(t) = e^{itA}$.  

For 1), given $\phi\in\H$ choose a sequence $f_n\in C_0^\infty(\R)$ with 
${\rm supp}f_n \subset [-1/n, 1/n]$ and $\int f_n = 1$.  Then each $\phi_{f_n}$ is 
in $D$ and 
$$\eqalign{\Vert \phi_{f_n} - \phi \Vert 
&= \Vert\int f_n(t)(U(t)\phi - \phi) \,dt\>\Vert\cr
&\le \sup_{t\in [-1/n,1/n]} \Vert U(t)\phi - \phi\Vert\>;\cr}$$
and since $U(t)$ is strongly continuous it follows that $\phi_{f_n} \to \phi$, so $D$ is dense.

For 2), first note that $B$ can be proved symmetric in the same way that 
the operator $B$ 
appearing earlier in the proof was shown to be symmetric.
Thus to show that $B$ is essentially self-adjoint we can use the basic criterion and 
show that $Ker(B\adj\pm i) = \{0\}$.  We treat $(B\adj-i)$, the case of $(B\adj+i)$ being 
analogous.  Suppose that $\psi\in D(B^\adj)$ has $B\adj \psi = i\psi$.  Then for 
each $\phi\in D(B) = D$ we have 
$$\eqalign{{d\over dt} \langle U(t)\phi,\psi\rangle 
&= \langle iBU(t)\phi,\psi\rangle\cr
&= -i\langle U(t)\phi, B\adj \psi\rangle\cr
&= -i\langle U(t)\phi, i\psi\rangle\cr
&= \langle U(t)\phi,  \psi\rangle\>,\cr}$$
where in the first line we use the fact that $U(t)\psi\in D$, so that ${d\over dt}U(t)\psi = iBU(t)\psi$.  
Thus the function $f(t) = \langle U(t)\phi,\psi\rangle$ satisfies 
$f' = f$, so $f(t) = ce^t$.  But $f(t)$ is bounded, since $U(t)$ is unitary, 
so we must have $c= 0$, and in particular $\langle\phi,\psi\rangle = 0$ for all 
$\phi\in D$.  Since $D$ is dense this means that $\psi = 0$ as desired.

For 3) we compare the two unitary groups $U(t)$ and $e^{itA}$.  Given 
$\phi\in D$ let $w(t) = U(t)\phi - e^{itA}\phi$.  Then $w(t)$ is differentiable and 
$$w'(t) = iBU(t)\phi - iAe^{itA}\phi = iAU(t)\phi - iAe^{itA}\phi = iAw(t)$$
since $A|_D = B$.  Thus
$$\eqalign{ {d\over dt} \Vert w(t)\Vert^2 &= \langle w'(t),w(t)\rangle + 
\langle w(t), w'(t)\rangle \cr
&= \langle iAw(t),w(t)\rangle + \langle w(t), iAw(t)\rangle\cr
&= 0\cr}$$
so $w(t) = 0$ for all $t$, since $w(0) = 0$.  Thus for all $\phi\in D$ we have
$U(t)\phi = e^{itA}\phi$.  Since $D$ is dense we must have $U(t) = e^{itA}$
(two bounded operators which agree on a dense set are equal).\qed

Note that if $U(t)$ is a strongly continuous one-parameter 
unitary group with infinitesimal generator $A$, for all $t$ and 
all $f\in\B(\R)$ we have
$$\eqalign{\int f(\lambda) d\mu_{A,\psi}(\lambda) &= 
\langle \psi, f(A)\psi\rangle\cr  &= 
\langle e^{itA}\psi, f(A)e^{itA}\psi\rangle\cr &=
\int f(\lambda) d\mu_{A,U(t)\psi}(\lambda) \cr}$$
so
$$\mu_{A,U(t)\psi} = \mu_{A,\psi}.$$

This makes good on our promise that Stone's theorem would 
associate to each one-parameter subgroup of symmetries an 
`associated conserved quantity' - the infinitesimal generator. 
In the most basic case $U(t)\psi$ represents the state obtained 
from $\psi$ by time translation - that is, wating $t$ units of 
time.  Then the infinitesimal generator of $U(t)$ is called
 the Hamiltonian, and is often denote $H$.  The fact that
the probability distribution of measured values of $H$ doesn't 
change with time, i.e.
$\mu_{H,U(t)\psi} = \mu_{H,\psi}$, is called {\bf conservation of
energy}.  

Also note that if $\psi\in D(H)$ then 
$${d\over dt}U(t)\psi = HU(t)\psi\>.$$
This is called (the abstract) {\bf Schr\"odinger's equation} and is the basic equation of quantum mechanics, just as Hamilton's 
equations are fundamental in classical mechanics.  Note also how much Schr\"odinger's equation resembles the modern version of Hamilton's equations:
$${d\over dt}U(t)\psi = v_HU(t)\psi\>.$$
This is naturally no coincidence; there is a subject of mathematics called `geometric quantization' which arose from investigations 
into the relationship between these two equations.

Note that one can't {\it do} anything with Schr\"odinger's equation until one knows what the Hamiltonian $H$ is!
In what follows we will describe the Hamiltonians for various 
important physical systems, for example atoms and molecules.
We will begin, however, with simpler examples designed to get a 
firmer grip on the meaning of the spectral theorem and Stone's theorem.
\vfill


\section{7. Self-adjoint Matrices and the Spin ${1\over 2}$-Particle}
So far our only examples of self-adjoint operators are multiplication operators.
If $X$ is a measure space, $f$ is a real-valued measurable function on $X$,
 and $M_f$ is the operator on $L^2(X)$ with domain
$D = \{\psi\in L^2(X):\, f\psi\in L^2(X)\}$ defined by:
$$M_f\psi = f\psi\>,\; \psi\in D$$
 then $M_f$ is self-adjoint.

Another sort of example comes from finite-dimensional matrices.  
If $\H$ is finite-dimensional,
a dense subspace must be the whole space so a densely defined operator $T$ on
$\H$ must have $D(T) = \H$, and any such operator must be bounded.  Thus all 
self-adjoint operators on finite-dimensional $\H$ are bounded.  For any  
self-adjoint operator $A$ on $\H$ there is a unitary $U$ such that $U\!AU^{-1}$ 
is a diagonal matrix with real coefficients, the eigenvalues of $A$, and 
$\sigma(A)$ is the set of eigenvalues of $A$, or the set of numbers along 
the diagonal of $U\!AU^{-1}$.   The following exercise is 
crucial for getting a feeling for spectral measures: 
\medskip
Exercise - Let $A$ be a self-adjoint operator on $\C^n$ and let 
$\{\psi_i\}_{i=1}^n$ be an orthonormal basis of eigenvectors with 
$A\psi_i = \lambda_i\psi_i$.  Let $\psi$ be the state (= unit vector)
with $\psi = \sum_{i=1}^n a_i\psi_i$ .
Show that the spectral measure  $\mu_{A,\psi}$ is given by:
$$  \mu_{A,\psi} = \sum |a_i|^2 \delta_{\lambda_i}\>.$$
\vskip 1em

The simplest and also most important use of finite-dimensional matrices 
in quantum theory is in the study of `spin-${1\over 2}$' particles.
During 1921-1924, Stern and Gerlach did experiments in which they 
sent a beam of silver atoms through an inhomogeneous magnetic field.
To their surprise it split into two components.  The same effect occurs
in a simpler context if one uses electrons rather than silver atoms.  
Goudsmit and Uhlenbeck hypothesized in 1925 that this strange effect was  
due to an intrinsic angular momentum of the electron, called `spin'.  
The electrons are either `spin up' or 
`spin down' (relative to the gradient of the magnetic 
field).  Those with `spin up' are bent one way by the magnetic field, 
and those with `spin down' are bent the other.  But this happens 
{\it no matter from which angle the magnetic field is applied}!
It's as if one had a spinning tennis ball whose axis of rotation was 
vertical {\it in any rotated coordinate system}, which is clearly 
impossible. 

Wolfgang Pauli, Enrico Fermi, and Paul Dirac developed a mathematical 
theory of spin-${1\over 2}$ 
particles (which include protons and neutrons as well as electrons), and we present 
part of it here in a summarized form.  In what follows we ignore the 
position and momentum aspects of the 
electron's state and concentrate on the description of its `spin'.  

The spin of the electron about {\it any} axis 
is always measured to be $\pm{1\over 2}\hbar$,
where $\hbar= h/2\pi$ and $h$ is {\bf Planck's constant}, a 
physical constant equal to $\,6.626196\times10^{-27}\,$ erg sec in the metric 
system.  This constant is so important that mathematical 
physicists typically work in a modified system of units in which $\hbar = 1$,
and we shall do so from now on.   

How can the spin of something {\it always} be measured to be 
 $\pm {1\over 2}$ about {\it any} axis?  This does not occur in classical
 mechanics, where the angular 
momenta (of, say, a spinning tennis ball) about the $x$,$y$, and $z$ axes 
form a vector $\vec J = (J_1,J_2,J_3)$  whose components can be {\it 
arbitrary} real numbers, and transform in the usual way under 
rotations.    
We need to specify mathematical descriptions of the states, observables, and 
symmetry group for the spinning electron.  We begin with the states 
and observables, and discuss the symmetry group later. 

In quantum mechanics the states of an 
electron (ignoring its position and momentum) are described by the unit vectors in a 2-dimensional Hilbert space 
$\H = \C^2$, and the spin angular momenta about the $x$, $y$, and $z$ axes are 
observables described by $2\times 2$ self-adjoint matrices,
$$J_i = {1\over 2}\sigma_i\>,$$
where the $\sigma_i$ are the {\bf Pauli matrices}: 
$$\sigma_1 = \left(\matrix{0&1\cr 1&0\cr}\right)$$
$$\sigma_2 = \left(\matrix{0&-i\cr i&0\cr}\right)$$
$$\sigma_3 = \left(\matrix{1&0\cr 0&-1\cr}\right)\>.$$

We develop the properties of these matrices in the following exercises:
\vskip 3em

Exercise - Show that the each of the matrices $\sigma_i$ has eigenvalues 
$\pm 1$ (thus the matrices $J_i$ have eigenvalues $\pm {1\over 2}$).  
Determine the eigenvectors of the $\sigma_i$.  Note that no two $\sigma_i$
have a (nonzero) eigenvector in common.
\medskip
Exercise - Show that the operator $J$ defined by 
$$J = (J_1^2 + J_2^2 + J_3^2)^{1/2}$$ has $J = \sqrt{3/4}I$.  
(In general 
if $\vec J$ are the angular momentum operators for a `spin-s particle' 
one has $J = \sqrt{s(s+1)}I$.)  
\medskip
Exercise - The {\it commutator} $[S,T]$ of $S,T\in \L(\H)$ is defined to 
be $ST - TS$.  Show that if $A,B\in \L(\H)$ are self-adjoint then $i[A,B]$ 
is self-adjoint.  Show that $$\sigma_i^2 = I$$ and
$$\sigma_i\sigma_j = -\sigma_j\sigma_i = i\sigma_k$$
if $(i,j,k)$ is a cyclic permutation of $(1,2,3)$, hence $[J_i,J_j] = iJ_k$. 
\aside {In particular, this implies that real linear combinations of 
the matrices $i\sigma_1$, $i\sigma_2$, $i\sigma_3$ and the identity $I$ 
form an algebra over $\R$ that is isomorphic to Hamilton's {\bf 
quaternions}, which recall is an algebra generated by elements $i,j,k$ satisfying $$i^2 = j^2 = k^2 = -1\>,$$
$$ij = -ji = k\>,\;jk = -kj = i\>,\; ki = -ik = j\>.$$
In the late 1800's Hamilton's quaternions (invented 1843) were in strong 
competition with Gibbs' vector notation as a tool for the mathematics of  
3-dimensional space.  Vector notation, being more elegant for the problems at 
hand, eventually won out.  But it is now clear that quaternions (now 
often represented as matrices) are necessary for describing certain features of 
the geometry of $\R^3$.  The appropriate generalization to higher dimensions 
are the `Clifford algebras', known in the four-dimensionial case by physicists 
as `Dirac matrices'.} 
\medskip
Exercise - Let $\psi\in\H$ be the unit vector $(a,b)$ , $|a|^2 + |b|^2 = 1$ .  
Determine the spectral measures $\mu_{\sigma_i,\psi}$, using the exercise 
above on spectral measures.
\vskip 3em

In particular, the `spin-up' and `spin-down' states
$$ \uparrow\, = \left(\matrix{1\cr 0\cr}\right)\>,\;
\downarrow\, = \left(\matrix{0\cr 1\cr}\right)\>,$$
are the eigenvectors of the operator $J_3$ : 
 $$J_3\!\uparrow\, ={1\over 2}\!\uparrow\>,\;\;
  J_3\!\downarrow\, =
-{1\over 2}\!\downarrow\>,$$
and they correspond to an electron with spin pointing up or down, i.e., 
rotating counterclockwise or clockwise along the $z$ axis.  An electron in 
any state can be written as a linear combination or, in physics 
language, {\bf superposition} of these states:
$$\left( \matrix{a\cr b\cr}\right) = a\!\uparrow +\, b\!\downarrow\>,$$
and has the spectral measure 
$$\mu_{J_3,\psi} = |a|^2\delta_{-1/2} + |b|^2\delta_{1/2}\>.$$
Thus it has probability $|a|^2$ of being measured to have spin 
${1\over 2}$ and probability $|b|^2$ of being measured to have spin 
$-{1\over 2}$, {\it if one measures its angular momentum along the z axis}.
If one measures its spin along the $x$ axis,
 one will measure it to be $\pm{1\over 2}$, distributed 
according to the probability distribution $\mu_{J_1,\psi}$.   
    
Of course there's nothing special about the 
$x$,$y$, and $z$ axes.  Let $\vec v\in\R^3$ be the unit vector 
$(v_1,v_2,v_3)$.  Then the self-adjoint operator corresponding to spin 
in the $\vec v$ direction is $$\vec v \cdot \vec J = \sum_{i=1}^3 v_iJ_i\>.$$

Now we introduce the symmetry group.  Recall that the {\bf special orthogonal group} $SO(n)$ is the group of $n\times n$ real orthogonal matrices
 with determinant one, and the {\bf special unitary group} $SU(n)$ is 
the group of $n\times n$ complex unitary matrices with determinant one.   $SO(3)$ is the group of rotations 
of 3-dimensional space, and is the 
obvious candidate for being the symmetry group of the spinning electron, but
in fact the group of symmetries is $SU(2)$, which 
we shall show is a {\bf double cover} of $SO(3)$. That is, there is an onto,  
two-to-one homomorphism $\alpha\colon SU(2)\to SO(3)$.  

Note first that the group $SU(2)$ has a natural strongly continuous unitary 
representation $U$ on $\H$ given by $$U(g)\psi = g\psi\>.$$ (Here as always 
we give groups of finite-dimensional matrices the topology such that 
$T_\alpha \to T$ iff $(T_\alpha)_{ij} \to T_{ij}$ for all $i, j$.)  
Next note that if $A$ is a self-adjoint operator on $\H$ 
which is {\bf traceless}, i.e. ${\rm tr}A = 0$ 
(where ${\rm tr}A = \sum A_{ii}$), then $ e^{itA} \in SU(2)$, by the following:
\medskip
Exercise - If $\H$ is an arbitary Hilbert space, show that for self-adjoint 
$A\in \L(\H)$, $$e^{itA} = \sum_{n=0}^\infty {(itA)^n\over n!}\>.$$ 
Show that for any self-adjoint $A\in \L(\C^n)$,
$$\det e^{itA} = e^{it\,{\rm tr}A}\>.$$
\medskip
For any unit vector $\vec v\in\R^3$ the one-parameter subgroup 
$e^{i\theta\vec v\cdot \vec J}$ ($\theta\in\R$) corresponds 
to rotations about the $\vec v$ axis. 
In other words, given an electron with state described by $\psi$,
if one rotates it by an angle $\theta$ about the $\vec v$ axis one obtains 
an electron with state $e^{i\theta \vec v\cdot\vec J}\psi$.  
Note however that
$$ e^{i\theta J_3} = \left(\matrix{ e^{i\theta /2} &0\cr 0&e^{-i\theta/2}\cr}
\right)\>,$$
so rotating the state $\psi$ by $2\pi$ gives, {\it not} $\psi$, but 
$-\psi$!  Rotating by $4\pi$, we get back to the state $\psi$! 
This is one reason why it makes sense to call such particles 
`spin-${1\over 2}$'.  And it's a physical fact that if you leave an 
electron lying on the table and someone sneaks in when you're not looking 
and rotates it by 360 degrees, when you come back you can 
tell - if you have the right apparatus!  But if he it turns around twice
you can't.  
\aside{Here we are glossing over an important subtlety.  For every observable 
$A$ on the Hilbert space $\H$ we have $\mu_{A,\psi} = \mu_{A,-\psi}$, so 
there is no obervable property of an electron in {\it isolation} that 
changes upon rotating it 360 degrees.  But its interactions with 
other parts of a larger system may differ.  
\medskip
The spin of a particle can be $0, 1/2, 1, 3/2,$ etc. 
Mathematically these various spins correspond to the different irreducible 
strongly continuous unitary representations of $SU(2)$.  Various mesons such as 
pions, kaons, etc. are spin-zero; electons, protons, neutrons, 
neutrinos and various baryons are spin-$1/2$; and photons are spin-one.  
Assemblages of particles can act like particles with various spins, 
thus atomic nuclei may be said to have spin $5/2$ and so on.}
\vskip 1em

We have:

\Proclaim Theorem.  There is an onto, two-to-one homomorphism 
$\alpha\colon SU(2)\to SO(3)$ such that the element  
$e^{i\theta\vec v \cdot \vec J}$, where $\vec v$ is a unit vector,
 is mapped onto the matrix representing a rotation of angle $\theta$ 
about the $\vec v$ axis.  

Sketch of Proof - We construct $\alpha$ as follows.  There is an isomorphism 
between $\R^3$ and the $2\times2$ self-adjoint traceless matrices 
given by:
$$\vec v \mapsto \vec v\cdot \vec \sigma = 
\left(\matrix{ v_3 & v_1 - iv_2 \cr
v_1 + iv_2 & v_2 \cr}\right)\;.$$
Note that 
$$\|\vec v\|^2 = -\det (\vec v\cdot \vec J)\>.$$

The group $SU(2)$ has a representation on the $2\times 2$ traceless
 self-adjoint matrices given by 
$$(g,A) \mapsto gAg^{-1}\>;$$
we leave it as an exercise to verify this.  Via the above 1-1 correspondence this
 gives rise to a representation of $SU(2)$ on $\R^3$, which we call 
$\alpha$.  Note that 
$$\eqalign{\|V(g)\vec v\|^2 &= 
-\det g(\vec v\cdot \vec J)g^{-1} \cr
&= -\det (\vec v\cdot\vec J)  \cr
&= \|\vec v\|^2 \>,}$$
so that $\alpha(g)$ is an orthogonal $3\times 3$ matrix.  
 
It's clear that $\alpha(I) = I$, and $\alpha(gh) = \alpha(g)\alpha(h)$ 
because
$$(gh)A(gh)^{-1} = g(hAh^{-1})g^{-1}\>.$$ 
Thus $\alpha$ is a homomorphism from $SU(2)$ to $O(3)$, the group 
of $3\times 3$ orthogonal matrices.  It remains to show that:
1)  $Ran(\alpha) = SO(3)$, 2) $\alpha$ is two-to-one, 3) $\alpha$
maps 
$e^{i\theta\vec v \cdot \vec J}$ to the rotation of angle $\theta$ 
about $\vec v$.

1) can be shown by a calculation, but it's easier to 
use a few facts about these groups.  First, since $SU(2)$ is 
connected and $\alpha$ is continuous, $\alpha$ must map $SU(2)$ into 
the connected component of $O(3)$ that contains the identity.  This 
is just $SO(3)$; the other connected component consists of matrices with determinant 
-1.  Since every element of $SO(3)$ is a rotation of some angle about 
some axis, if we show 3) it will follow that $\alpha$ is onto $SO(3)$.

2) is equivalent to saying that the kernel of $\alpha$ consists of two 
elements.  Unraveling the definitions, $g\in SU(2)$ is in $Ker\,\alpha$
 iff for all traceless self-adjoint matrices $A$ we have $gAg^{-1} = A$.
It is easily seen that the only matrices which commute with all traceless 
self-adjoint matrices are those of the form $cI$.  The only ones of these 
which lie in $SU(2)$ are ${\pm I}$.  Thus $Ker\,\alpha = \pm I$.

To show 3) it suffices to show - differentiating with respect to 
$\theta$ - that:
$${d\over d\theta}\alpha(e^{i\theta\vec v \cdot \vec J})\vec w|_{\theta = 0} = 
-\vec v\times \vec w\>.$$
(Here the minus sign is an unimportant consequence of our sign conventions; 
the main point is that when one rotates the vector $\vec w$ about the 
$\vec v$ axis 
the derivative is $\pm \vec v\times \vec w$, with the sign depending on whether 
one rotates clockwise or counterclockwise.)   
Unraveling the definitions, this is equivalent to:
$${d\over d\theta}e^{i\theta\vec v \cdot \vec J}(\vec w\cdot\vec J) 
e^{-i\theta\vec v \cdot \vec J}|_{\theta=0} 
 = -(\vec v\times \vec w)\cdot \vec J\>.$$
Doing the differentiation on the left side, it suffices to show:
$$[i \vec v \cdot \vec J,\>\vec w\cdot\vec J]   
 = -(\vec v\times \vec w)\cdot \vec J\>.$$
This follows from the result of an exercise above, that 
$$[\sigma_i,\sigma_j] = 2i\sigma_k\>.$$
\hskip 30 em\qed


Note also that by general principles the operator $\vec v\cdot \vec J$ is 
the conserved quantity corresponding to the one-parameter group $e^{i\theta 
\vec v\cdot\vec J}$ .  This means that the probability distribution of 
measured spins in the $\vec v$ direction doesn't change if you rotate the 
electron by a fixed angle about the $\vec v$ axis - which makes sense.
\medskip
Rather than meditating further on the oddity of the quantum theory 
of spin-${1\over 2}$ particles, let us turn to infinite-dimensional 
diagonal matrices.  Recall that $\ell^2$ is the usual Hilbert space of complex 
sequences
$\psi = (\psi_i)_{i=1}^\infty$ such that $\|\psi\|^2 = \sum |\psi_i|^2$ is 
finite, with the inner product $\langle \psi,\phi\rangle = \sum \overline\psi_i
\phi_i$.

\Proclaim Theorem.  Let $(a_i)_{i=1}^\infty$ be a sequence of real numbers. 
 Let $A$ be the operator on $\ell^2$ with domain
$$D(A) = \{\psi\in\ell^2\colon\> \sum |a_i\psi_i|^2 < \infty\}$$
given on this domain by 
$$(A\psi)_i = a_i\psi_i\>.$$
Then $A$ is self-adjoint.  The restriction of $A$ to the domain 
$D_0 = \{\psi\in\ell^2\colon\>$ {\rm only finitely many} $\psi_i$ 
{\rm are nonzero}$\}$ 
is essentially self-adjoint.

Proof -  We can think of $\ell^2$ as $L^2(N)$, where $N = {1,2,3,\dots}$ with 
the `counting measure' that assigns to each point the measure 1.  Then 
$A$ is a multiplication operator and is self-adjoint by the general 
theory of such.  To show that $A_0 = A|_{D_0}$ is essentially self-adjoint, 
we use the basic criterion: it is easily seen that $A_0$ is symmetric, and 
to show that $Ran(A\pm i)$ are dense in $\H$ it suffices to show that they 
contain $D_0$, which is dense.  Suppose $\psi\in D_0$, and define 
$\phi$ by $\phi_i = (a_i \pm i)^{-1}\psi_i$.  Then $\phi\in D_0$, and 
$(A_0\pm i)\phi = \psi$.  Thus $Ran(A\pm i)\supseteq D_0$.
\hskip 2 em\hskip 30 em  \qed

Exercise - Show that if $A$ is the self-adjoint operator on $\ell^2$ given 
above and $\psi\in\ell^2$ then $\mu_{A,\psi} = \sum |\psi_i|^2\delta_{a_i}$ .

\vskip 2em

The spectrum of an unbounded self-adjoint operator may be defined in more 
or less the same way as for bounded operators.
We give two definitions and prove that they are equivalent, after giving an 
exercise used in the proof:
\medskip
Exercise - Show that if $T$ is a densely defined operator on $\H$ then 
 $(T -\lambda I)\adj = T\adj - \overline\lambda I$, where $T - \lambda I$ 
is the operator with domain $D(T)$, defined in the obvious way.

\Proclaim Theorem.  Let $A$ be a self-adjoint operator on a Hilbert 
space $\H$.  Then the following are equivalent:
\medskip
1) $\lambda$ is in the closure of $\,\bigcup{\rm supp}\mu_{A,\psi}$.
\medskip
2) There does not exist $R\in\L(\H)$ such that $R(A-\lambda I)\psi = \psi$ 
for all $\psi\in D(A)$ and $(A - \lambda I)R\psi = \psi$ for all $\psi\in \H$.
(Note that for the latter equation to make sense we need to have 
$Ran(R)\subseteq D(A)$.)
\medskip
If either of these holds we say $\lambda$ is in the {\bf spectrum} of $A$,
denoted $\sigma(A)$.

Proof - $1)\Rightarrow 2)$:  Suppose 1) holds.  Then given any $\epsilon > 0$
there exists $\psi\in\H$ such that $\lambda$ is distance $\le \epsilon$
from supp$\mu_{A,\psi}$.  This implies that the intersection of the 
interval  $ I = [\lambda -\epsilon,\lambda + \epsilon]$ with 
supp$\mu_{A,\psi}$ is nonzero, so by the spectral theorem
$$\eqalign {\|\chi_I(A)\psi\|^2 &= \langle \psi,\chi_I(A)\psi\rangle  \cr
&= \int_I \mu_{a,\psi} > 0\>.}$$
Let $\phi = \chi_I(A)\psi\,$; by the above $\phi\ne 0$, and clearly 
$\phi = \chi_I(A)\phi\,$.  It is easily seen using the spectral theorem 
that $\phi\in D(A)$.  Moreover 
$$\eqalign{\|(A-\lambda)\phi\|^2 
&= \langle \phi, (A-\lambda)^2\chi_I(A)\phi\rangle    \cr
&= \int_I (x-\lambda)^2 d\mu_{A,\phi}(\lambda)  \cr
&\le \epsilon^2 \int_I d\mu_{A,\phi} 
= \epsilon^2 \|\phi\|^2  \>.\cr}$$

Thus for any $\epsilon > 0$ there exists nonzero  $\phi\in D(A)$ such 
that $\|(A-\lambda I)\phi\| \le \epsilon \|\phi\|$.  It is an 
easy exercise to show that no bounded $R$ can have $R(A-\lambda I)\phi 
=\phi $ for all such $\phi$. 

  $2)\Rightarrow 1)$:  If 1) does not hold, then the function 
$(x-\lambda)^{-1}$ has finite $L^{\infty\prime}$ norm, 
and we can define $R_\lambda$ 
by the functional calculus to be $(A-\lambda I)^{-1}$.  Note that by 
the functional calculus we have $R_{\overline\lambda} = R_\lambda\adj\,$. 
If we let 
$f_n$ be a sequence of functions with $f_n(x) \to x$ for all $x\in\R$ and 
and $|f_n(x)|\le x$, then by the functional calculus we have 
$f_n(A)\psi \to A\psi$ for all $\psi\in D(A)$, and 
$R_\lambda (f_n(A) - \lambda I) \to I$ strongly, hence 
for all $\psi\in D(A)$
$$\eqalign{R_\lambda(A -\lambda I)\psi
 &= R_\lambda \lim_{n\to\infty} (f_n(A) - \lambda I) \psi\cr 
&=  \lim_{n\to\infty} R_\lambda(f_n(A) - \lambda I)\psi \cr
&= \psi\>.}$$
Furthermore, if $\psi\in\H$, for all $\phi\in D(A-\lambda I)$ we have
$$\eqalign{\langle \psi,\phi \rangle
 &= \langle\psi,  R_{\overline\lambda}
(A - \overline\lambda I)\phi\rangle\cr
&= \langle R_\lambda \psi, (A - \overline\lambda I)\phi\rangle\>.}$$
hence by the definition of adjoint,
 $R_\lambda \psi\in D((A-\overline\lambda I)\adj),$
 and $$(A-\lambda I)R_\lambda\psi = (A-\overline\lambda I)\adj R_\lambda\psi
 = \psi\>.$$  
Thus 2) does not hold.
 \qed

We apply this to infinite-dimensional diagonal matrices in the following 
exercises:
\medskip
Exercise - Show that a self-adjoint operator $A$ on $\H$ is bounded if and 
only if $\sigma(A)$ is a bounded set, in which case $\|A\| = r(A)$.
\medskip
Exercise - If $A$ is the self-adjoint operator on $\ell^2$ given by 
$$(A\psi)_i = a_i\psi_i\>,$$
show that $\sigma(A) = \{a_i\}$.  Show that $A$ is bounded if and only if
$a_i$ is a bounded sequence, in which case $\|A\| = \sup |a_i|$.
\vskip 3em
The theorems above have many applications to 
differential equations.  For example:
\medskip
Exercise - Show that there 
is an orthonormal basis of $L^2[0,1]$ consisting of $\psi_n\in C^\infty[0,1]$ 
such that 
$$-{d^2\over dx^2}\psi_n = \lambda_n\psi_n$$ 
satisfying the {\bf Dirichlet boundary conditions}:  
$$\psi(0) = \psi(1) = 0\>.$$ 
Let $D$ be the space of 
finite linear combinations of the $\psi_n$.  Show that $-{d^2\over dx^2}$ is 
essentially self-adjoint on $D$, and let $\Delta_D$ be the closure of 
this operator.   Determine the spectrum of $\Delta_D$.
\medskip
Exercise - Show that there 
is an orthonormal basis of $L^2[0,1]$ consisting of $\psi_n\in C^\infty[0,1]$ 
such that 
$$-{d^2\over dx^2}\psi_n = \lambda_n\psi_n$$ 
satisfying the {\bf Neumann boundary condtions}:
$$\psi'(0) = \psi'(1) = 0\>.$$ 
Let $N$ be the space of 
finite linear combinations of the $\psi_n$.  Show that $-{d^2\over dx^2}$ is 
essentially self-adjoint on $N$, and let $\Delta_{N}$ be the closure of 
this operator.   Determine the spectrum of $\Delta_N$.
\medskip
One can then use the functional calculus to define operators such as 
$e^{-t\Delta}$ (the `heat kernel'), which are very useful in mathematics 
and physics.
\vfill 
\section{8.  Translation and Differentiation}

Another simple and important self-adjoint operator arises from translation:

\Proclaim Theorem.  Let $U(t)\colon L^2(\R) \to L^2(\R)$ defined for $t\in\R$ 
by $$U(t)\psi(x) = \psi(x + t)\>.$$  
Then $U(t)$ is a strongly continuous one-parameter 
unitary group, and the infinitesimal generator $p$ of $U(t)$ is a self-adjoint 
operator such that 
for all $\psi\in D(p)$, $$p\psi = i^{-1}{d\psi\over dx}$$
where the derivative is taken in the distributional sense.

Proof - Clearly $U(t)$ is a 1-parameter unitary group.  
Note that to show that a 
unitary group is strongly continuous it suffices to show that for all 
$\psi\in H$,
$U(t)\psi \to \psi$ 
as $t\to 0$.  Given $\psi\in L^2(\R)$, for all $\epsilon > 0$  there exists 
$\phi\in C_0^\infty(\R)$ such that $\Vert \psi-\phi\Vert \le \epsilon$. 
Let $M^2 = \int_\R |d\phi/dx|^2$.  Then 
$$\eqalign{\|U(t)\psi - \psi\| & \le \|U(t)\psi - U(t)\phi\| + \|U(t)\phi -  \phi\| +
\|\phi - \psi\| \cr                     
&\le 2\epsilon + (\int |\phi(x+t) - \phi(x)|^2 dx)^{1\over 2}\cr
&\le 2\epsilon + t|M|\cr }$$
so if $|t|M \le \epsilon$ then 
$$\|U(t)\psi - \psi\| \le 3\epsilon\>.$$
Thus $U(t)$ is strongly continuous.

By Stone's theorem $U(t) = e^{itp}$ for some self-adjoint $p$ and
$$D(p) = \{\psi\in L^2(\R):\; \lim_{t\to 0} {U(t)\psi - \psi\over t}\;
{\rm exists \>in\;} L^2(\R)\,\}$$
and $$ip\psi = \lim_{t\to 0} {U(t)\psi - \psi\over t}\>.$$
Thus we only need to show that if the limit 
$$\lim_{t\to 0} t^{-1}(U(t)\psi - \psi)$$
exists in $L^2(\R)$ then it equals the distributional derivative $d\psi/dx$.

Suppose the limit exists in $L^2(\R)$.  Then  
for all test functions $\phi\in C_0^\infty(\R)$
$$\eqalign{\int {d\overline\psi\over dx}\phi \,dx&=
 -\int \overline\psi {d\phi\over dx}\,dx \cr 
&= \lim_{t\to 0} \int t^{-1}\overline\psi(x) \left
(\phi(x-t) - \phi(x)\right)\,dx\cr
&= \lim_{t\to 0} \int t^{-1}\left(\overline\psi(x+t) - \overline\psi(x)\right)
\phi(x)\,dx\cr
&= \lim_{t\to 0}\langle t^{-1}\left(U(t)\psi - \psi\right),\phi\rangle\cr
&= \langle ip\psi,\phi\rangle\;,\cr}$$
so the distributional derivative $d\psi\over dx$ equals $ip\psi$.
\qed

Translation is of course an important physical symmetry, and in the 
context of physics the infinitesimal generator $p$ is 
called `momentum'.  Of course real physics happens in $\R^3$, not $\R$.  
Ignoring spin now, the state of 
a point particle in $\R^3$ 
is described by a unit vector in the Hilbert space $\H = L^2(\R^3)$, 
and the group $\R^3$ of translations has a strongly continuous unitary representation 
on $\H$ given by $$(U(g)\psi)(x) = \psi(x+g)\;,\>g\in\R^3,\,\psi\in\H\>.$$
(The proof of the latter is exactly analogous to that of the theorem above.)
The infinitesimal generators of translation in the $x$, $y$, and $z$ directions 
respectively are self-adjoint operators, the $x$, $y$, and $z$ {\bf momenta}, 
denoted $p_1$, $p_2$, $p_3$.    

We can also obtain the momentum operators, as well as other self-adjoint 
differential operators, in terms of the Fourier transform.  Recall that the 
Fourier transform is a unitary operator on $L^2(\R^n)$ given by 
$$\widehat f(\vec k) = (2\pi)^{-n/2}\int f(\vec x)e^{-i\vec k\cdot \vec x} d^nx\>,$$
where the integral is absolutely convergent for 
$f\in L^2(\R^n)\cap L^1(R^n)$ but is defined by a limit for arbitrary 
$f\in L^2$.

\Proclaim Theorem.  Let $p$ be the self-adjoint generator of translations on 
$L^2(\R)$.
Then the domain of $p$ is equal to $\{f\in L^2(\R)\colon\; k\widehat 
f\in L^2(\R)\}$.

Proof - Let $U(t)$ be the one-parameter group of translations on 
$\H= L^2(\R)$, and let $F\colon\H\to\H$ be the Fourier transform.  We have 
$$\eqalign{\widehat{(U(t)f)}(k) &= 
 (2\pi)^{-1/2}\int f(x+t)e^{-ikx} dx\cr
&=  (2\pi)^{-1/2}\int f(x)e^{-ik(x-t)} dx\cr
&= e^{ikt}\widehat f(k)\>,}$$
or, in short, 
$$FU(t) = M_{e^{ikt}}F\>,$$
or 
$$U(t) = F^{-1}M_{e^{ikt}}F\>,$$
where $M_{e^{ikt}}$ denotes the operator of multiplication by $e^{ikt}$ 
(which is a unitary operator on $\H$).  
Thus $\lim_{t\to 0}t^{-1}(U(t) -  I)\psi$ exists in $\H$  if and and only 
if $\lim_{t\to 0}t^{-1}(e^{ikt} - I)\widehat\psi$ exists in $\H$, which is 
is easily seen to hold if and only if $k\widehat\psi\in \H$.\qed

Exercise - As a spot-check on your real analysis skills, prove the `easily seen' claim above.

Exercise - Determine the spectrum of the operator $p$.  Find approximate 
eigenvectors.
\medskip
We can use the following theorem to show that $i^{-1}{d\over dx}$ is 
essentially self-adjoint on $C_0^\infty(\R)$.  More precisely, if $A$ is 
a self-adjoint operator on $\H$ we say $D\subseteq D(A)$ is a {\bf 
domain of essential self-adjointness} for $A$ or a {\bf core} for $A$ if 
the closure of $A|_D$ is $A$.  (In particular this implies that $A|_D$ is 
essentially self-adjoint.) 

\Proclaim Theorem.  Let $e^{itA} = U(t)$ be a strongly continuous 1-parameter 
unitary group on $\H$ and suppose $D\subset\H$ be a dense space of 
{\bf differentiable vectors} for $U(t)$, i.e., if $\psi\in D$ then the 
limit $$\lim_{t\to 0} {U(t)\psi - \psi\over t}$$
exists in $\H$.  Suppose also that $D$ is {\bf invariant under $U(t)$}, i.e.,
$U(t)D \subseteq D$ for all $t$.  Then $D$ is a core for $A$.

Proof - The proof is much like part of that of Stone's theorem.   
Note that by Stone's theorem $D\subseteq D(A)$.  
Next note that $B = A|_D$ is symmetric since $\langle A\psi,\phi\rangle = 
\langle \psi, A\phi\rangle$ for all $\psi,\phi\in D(A)$, hence for all 
$\psi,\phi\in D$.   

Next we show that $B$ is essentially self-adjoint by the basic criterion, 
by showing that $Ker(B\adj\pm i) = \{0\}$.  We treat $(B\adj-i)$, the case of $(B\adj+i)$ being 
analogous.  Suppose that $\psi\in D(B^\adj)$ has $B\adj \psi = i\psi$.  
Then for each $\phi\in D(B) = D$ we have 
$$\eqalign{{d\over dt} \langle U(t)\phi,\psi\rangle 
&= \langle iBU(t)\phi,\psi\rangle\cr
&= -i\langle U(t)\phi, B\adj \psi\rangle\cr
&= -i\langle U(t)\phi, i\psi\rangle\cr
&= \langle U(t)\phi,  \psi\rangle\>,\cr}$$
where in the first line we use the fact that $U(t)\psi\in D$, 
so that ${d\over dt}U(t)\psi = iBU(t)\psi$.  
Thus the function $f(t) = \langle U(t)\phi,\psi\rangle$ satisfies 
$f' = f$, so $f(t) = ce^t$.  But $f(t)$ is bounded, since $U(t)$ is unitary, 
so we must have $c= 0$, and in particular $\langle\phi,\psi\rangle = 0$ for all 
$\phi\in D$.  Since $D$ is dense this means that $\psi = 0$ as desired.

Thus we have the following situation: $A$ is self-adjoint, $B\subseteq A$ and 
$B$ is essentially self-adjoint.  We'll be done if we show the closure of $B$ is $A$.  This follows from:

\Proclaim Lemma.  Suppose $A$ is a self-adjoint operator, $B$ is an essentially 
self-adjoint operator and $B\subseteq A$.  Then $\overline B = A$.  

Proof - Since $A$ is self-adjoint it is closed, and it is easily seen that 
in general 
$$S\subseteq T \;\Longrightarrow\; \overline S\subseteq\overline T\;,$$
so $\overline B\subseteq \overline A = A$ .  To show the opposite inclusion, 
note that $\overline B\subseteq A$ implies $A\adj\subseteq\overline B\adj$, 
but $A$ and $\overline B$ are self-adjoint, so $A\subseteq \overline B$. \qed
 
\proclaim Corollary.  The space $C_0^\infty(\R)$ is a core for the operator $p$.

Proof - This follows immediately from the theorem. \qed

Differentiation on the unit interval $[0,1]$ is a more subtle matter.  
Let $\H = L^2[0,1]$.  By Stone's theorem we know any self-adjoint operator 
corresponding to $i^{-1}{d\over dx}$ has to generate a one-parameter group 
of unitaries, and by analogy with the $\R$ case we expect this group 
to consist of translations, but if one translates a function on $[0,1]$ it  
will `fall off the edge'! 

Suppose one considers, by analogy with the case of $\R$, an operator $B$ 
by:
$$D(B) = \{\psi \in\H\colon\; \psi\in C^\infty[0,1]\;\;,\; \phi(0) = \phi(1) = 0\}$$
$$B\psi = i^{-1}{d\psi \over dx}\>.$$
Is $B$ essentially self-adjoint? 
An integration by parts shows that it is symmetric: 
$$\eqalign{
\langle \psi, B\phi\rangle &= \int_0^1 \overline\psi(x) i^{-1}\phi'(x)\, dx \cr 
&= \int_0^1 \overline{i^{-1}\psi'(x)} \phi(x) \,dx\cr
&= \langle B\psi,\phi\rangle\>;}$$
note that no boundary terms appear in the integration by parts since 
$\phi(0) = \phi(1) = 0$.  
Unfortunately $B$ is not essentially self-adjoint because
 the kernels $Ker(B\adj\pm i)$ are nonzero. 

Consider the case of $B\adj + i$ (the other is similar).  We claim that 
the function $e^x\in\H$ is in $D(B\adj)$ and $B\adj e^x = -ie^x$, which 
implies $e^x \in Ker(B\adj + i)$.  To prove our claim it suffices to show 
that 
$$\langle e^x, B\phi\rangle = \langle -ie^x,\phi\rangle$$
for all $\phi\in D(B)$.  This follows from integration by parts:
$$\eqalign{
\langle e^x, B\phi\rangle &= \int_0^1 e^x i^{-1}\phi'(x) dx \cr 
&= \int_0^1 \overline{i^{-1}e^x} \phi(x) dx\cr
&= \langle -ie^x,\phi\rangle\>.}$$
The moral here is that the domain $D(B)$ is too small, so the function 
$e^x$ is orthogonal to $(B-i)\phi$ for all $\phi\in\H$, giving $B\adj + i$
a nonzero kernel. 

So why not try a bigger domain?  Define $B$ instead 
by:  
$$D(B) = \{\psi \in\H\colon\; \psi\in C^\infty[0,1]\,\}$$
$$B\psi = i^{-1}{d\psi \over dx}\>.$$
This domain is unfortunately too big: now $B$ isn't symmetric!  For if 
$\psi,\phi\in D(B)$, integration by parts gives: 
$$\eqalign{\langle \psi, B\phi\rangle 
&= \int_0^1 \overline\psi(x)i^{-1}\phi'(x)\,dx\cr
&= \int_0^1  \overline{i^{-1}\psi(x)}\phi(x)dx + \overline\psi\phi\big|_0^1\cr
&\ne \langle B\psi,\phi\rangle\>.}$$

Don't despair, however!  Suppose we define $B$ by 
$$D(B) = \{\psi \in\H\colon\; \psi\in C^\infty[0,1]\;\;,\; 
\psi(1) = \alpha\psi(0)\;\}\;,$$
$$B\psi = i^{-1}{d\psi \over dx}\>,$$
where $\alpha$ is a complex number with $|\alpha| = 1$.  
Now $B$ is symmetric because, integrating by parts: 
$$\eqalign{\langle \psi, B\phi\rangle 
&= \int_0^1 \overline\psi(x)i^{-1}\phi'(x)\,dx\cr
&= \int_0^1  \overline{i^{-1}\psi(x)} \phi(x)dx + \overline\psi\phi\big|_0^1\cr
&=  \langle B\psi,\phi\rangle\>,}$$
since 
$$\overline\psi(1)\phi(1) = |\alpha|^2 \overline\psi(0)\phi(0) = 
 \overline\psi(0)\phi(0)\>.$$

Let $\theta$ be any real number with $e^{i\theta} = \alpha$.  Note that the functions $\psi_n$ defined by 
$$\psi_n(x) = e^{i(2\pi n+ \theta)x}$$
are in $D(B)$, and
$$B\psi_n = (2\pi n + \theta)\psi_n\>.$$
By Fourier theory it's easy to see that the functions $\psi_n$ form an 
orthonormal basis of $\H$, so finite linear combinations of the $\psi_n$,
which lie in $D(B)$, are dense in $\H$.  In particular, given such a 
combination, say $\sum a_n \psi_n$, we can write it as 
$$\sum a_n\psi_n = (B \pm i) \left(\sum a_n(2\pi n + \theta\pm i)^{-1}
\psi_n\right)\>,$$
where the sum on the right side is in $D(B)$, so we see that $Ran(B\pm i)$ 
are dense.  Thus $B$ is essentially self-adjoint.  As one would expect from 
the above calculations, we have:

\medskip
Exercise - $\overline B$ is a self-adjoint operator with spectrum 
equal to $\{2\pi n + \theta\}_{-\infty <n<\infty}$.
\medskip

Thus we see that in this example the expression $i^{-1}{d\over dx}$ may be
given many different interpretations as a self-adjoint operator, depending 
on what domain we choose, and that the spectrum depends on the domain we choose!
 Note also that the choice of domain is closely related to boundary conditions.
  This is typical in elliptic operator theory; an elliptic operator on a 
bounded open set is self-adjoint if we choose the right domain, but its 
spectrum depends on the domain.  
\medskip
Exercise - Describe the unitary group $e^{it\overline B}$ generated by the 
operator $\overline B$ above, and explain how it depends on $\alpha$.  Hint: 
evaluate $e^{it\overline B}\psi_n$.   


\section{9. The Canonical Commutation Relations}

Ignoring spin, the state of a particle in $\R^3$ is described by a unit vector in
$L^2(\R^3)$.  The function $\psi$ on $\R^3$ is often called the particle's 
{\bf wave function}.  As mentioned above, the momenta are the observables 
$p_1,p_2,p_3$ generating translations in the $x$, $y$, and $z$ directions.   
Other fundamental observables are 
the {\bf position} operators, that is, multiplication operators by 
the coordinate functions $x$, $y$, and $z$.  These self-adjoint operators 
are denoted $q_1$, $q_2$, and $q_3$.  (The generalization to $\R^n$ should 
be obvious.) 

Recall that $\psi\in L^2(\R^n)$ is said to be in the {\bf Schwartz space} 
${\cal S}(\R^n)$ if for all $n,m\ge 0$ we have 
$$ (1 + r^2)^n (1 + \Delta)^m \psi \in L^2(\R^n)\>.  $$
where 
$$\Delta = -\sum _{i=1}^n {\partial^2 \over \partial x_i^2}$$
is taken in the distributional sense.  Another equivalent definition is 
commonly used: 
\medskip
Exercise - Show that $\psi\in {\cal S}(\R^n)$ if and only if for all 
$n, m \ge 0$ we have 
$$ (1 + r^2)^n (1 + \Delta)^m \psi \in L^\infty (\R^n)\>.  $$
\medskip 
Much of the importance of the Schwartz space in operator theory 
comes from the following facts:
\medskip
Exercise - Show that if $\psi \in {\cal S}(\R^n)$ then $p_i\psi\in
{\cal S}(\R^n)$ and $q_i\psi\in{\cal S}(\R^n)$ for all $i$, $1 \le i \le n$.  
\medskip
Exercise - Show that ${\cal S}(\R^n)$ is a core for the operators $p_i, q_i$.  
(Hint: use the theorem on cores in the previous section.)
\medskip
Exercise - Let $F$ denote the Fourier transform as a unitary operator 
on $L^2(\R^n)$.  Show that $F$ maps ${\cal S}(\R^n)$ to itself in a 1-1 and 
onto manner. 
\vskip 2em
Suppose that $\psi\in {\cal S}(\R^n)$.  Then note that for all $i, j$,
$$\eqalign{(p_iq_j - q_jp_i)\psi &= i^{-1}\left ({\partial \over\partial x_i}x_j
\psi - x_j{\partial\over\partial x_i}\psi\right)\cr
&= i^{-1} {\partial x_j\over \partial x_i} \psi\cr
&= i^{-1}\delta_{ij}\psi\>.}$$
We also have
$$(p_ip_j - p_jp_i)\psi = 0\;\;,\;\> (q_iq_j - q_jq_i)\psi = 0\>.$$
These relations, often summarized as 
$$[p_i,q_j] = i^{-1}\delta_{ij}\;\;,\;\>[p_i,p_j] = [q_i,q_j] = 0\;,$$
are called the {\bf canonical commutation relations}.
They were discovered by Werner Heisenberg's supervisor, Max Born, while 
studying a paper of Heisenberg's.  At the time (1925) Born was one of the few 
physicists who had a good knowledge of matrices.  He interpreted some quantum-
mechanical calculations of Heisenberg's in terms of 
infinite-dimensional matrices, 
explaining in this way the odd non-commutativity of multiplication that 
Heisenberg had gotten himself into.  This made physicists learn about 
matrices and develop the new `matrix mechanics', as an alternative to 
Schr\"odinger's `wave mechanics'.   
Only later were Heisenberg's position and momentum operators correlated 
by Pauli to the 
operators of multiplication and differentiation by coordinates $x_i$, in 
which form they had appeared in Schr\"odinger's equation for the hydrogen 
atom (more on which later).   Heisenberg had, in effect, been describing 
differential operators in terms of an orthonormal basis for $L^2(\R^n)$.  


\vfill
\section{10. The Uncertainty Principle}

We begin by pointing out how to multiply and add densely defined operators.
Given two densely defined operators $A$ and $B$, we define the product 
$AB$ by:
$$D(AB) = \{\psi\in D(B)\colon\;\;B\psi\in D(A)\}\>,$$
$$AB\psi = A(B\psi)\>.$$
It is not necessarily the case that $D(AB)$ is dense!  This needs to be 
proved in each case; for example, properties of the Schwartz space make 
it clear that products of position and momentum operators such as 
$pq$, $p^2$, $qp^2q$ and so on are all 
densely defined operators.

Similarly we define the sum $A+B$ by:
$$D(A+B) = D(A)\cap D(B)\>,$$
$$(A+B)\psi = A\psi+B\psi\>.$$
Again it needs to be proved that D(A+B) is dense - two dense subspaces of 
a Hilbert space can have {0} as their intersection!  The properties of 
the Schwartz space imply that all polynomials in $p$ and $q$ are densely 
defined operators.
\medskip
Exercise - Show that if $A$ is a self-adjoint operator then any polynomial 
in $A$ is densely defined, and any real polynomial in $A$ is essentially 
self-adjoint.
\medskip
Exercise - Define $\widetilde q$ by 
$$D(\widetilde q) = C_0^\infty(\R^n)\>,$$
$$(\widetilde q\psi)(x) = x\psi(x)\>.$$
Define $\widetilde p$ by 
$$D(\widetilde p) = \{\psi\in L^2(\R)\colon\;\widehat\psi 
\in C_0^\infty(\R)\,\}$$ 
$$(\widetilde p\psi)(x) = i^{-1}{d\over dx}\psi(x)\>.$$
Show that $\widetilde p$ and $\widetilde q$ are essentially self-adjoint and 
have as closures the usual $p$ and $q$.  Show however that
$$D(\widetilde p)\cap D(\widetilde q) = \{0\}$$
so $\widetilde p + \widetilde q$ has as domain only ${0}$!  (Hint: 
for the last part use the fact that if $\widehat \psi\in L^2(\R)$ 
is compactly supported 
then $\psi$ is analytic, which follows from the Paley-Wiener theorem.)
\medskip
Exercise - Show that the operator $\Delta = p_1^2 + p_2^2 + p_3^2$ is 
essentially self-adjoint.  (Hint: show using the Fourier transform 
that it's even essentially self-adjoint when restricted to the smaller domain 
${\cal S}(\R^3)$.)  
\vskip 2em
Given an obervable $A$ on $\H$ and a state $\psi\in D(A^2)$, we can define 
the {\bf standard deviation} of $A$ in the state $\psi$, usually written 
$\Delta A$ even though it depends on $\psi$, by 
$$(\Delta A)^2 = \langle \psi, A^2 \psi \rangle - \langle \psi, A\psi\rangle^2 
\>.$$
Note that this corresponds to the usual definition of standard deviation 
because $\langle \psi, A\psi\rangle$ is the mean value of the observable 
$A$ in the state $\psi$, as described in section 5.

The {\bf uncertainty principle} gives a lower bound on the product 
$\Delta A\Delta B$ for non-commuting observables $A$ and $B$:

\Proclaim Uncertainty Principle.  Let $A$ and $B$ be self-adjoint 
operators on the Hilbert space $\H$, and suppose $\psi \in D(A^2)\cap
D(B^2)\cap D(AB)\cap D(BA)$ is a unit vector.  Then 
$$\Delta A\Delta B \ge {1\over 2}\left|\langle\psi, [A,B]\psi\rangle\right|\>.$$

Proof - First we note that it suffices to show this for 
$$A' = A - \langle \psi, A\psi\rangle$$
and
$$B' = B - \langle \psi, B\psi\rangle$$
instead.  This is because $[A', B'] = [A,B]$ and 
$$\langle\psi, A' \psi\rangle = \langle \psi, A\psi\rangle - 
\langle \psi, \langle\psi, A\psi\rangle\,\psi\rangle = 0\>,$$
hence 
$$\eqalign{ (\Delta A')^2 &= \langle \psi, A'^2 \psi\rangle  \cr
&= \langle \psi, (A^2 - 2\langle \psi, A\psi\rangle A + \langle \psi, A\psi\rangle^2)\psi\rangle\cr
&= \langle\psi, A^2\psi\rangle - \langle\psi,A\psi\rangle^2\cr
&=(\Delta A)^2\>,}$$
and similarly for $B$. 

We have 
$$ \eqalign{\left|\langle \psi, [A',B']\psi\rangle\right| &=
\left|\langle A'\psi,B'\psi\rangle - \langle B'\psi, A'\psi\rangle \right|\cr
&=2 \left| {\rm Im} \langle A'\psi, B'\psi\rangle\right|\cr
&\le 2\left |  \langle  A'\psi,B'\psi\rangle\right|\cr
&\le 2\|A'\psi\|\|B'\psi\|\>,}$$
the last step using Cauchy-Schwarz; but 
$$\|A'\psi\| = \langle \psi, A'^2\psi\rangle^{1/2} = \Delta A'$$
and similarly for $B'$, so 
$$ \left|\langle \psi, [A',B']\psi\rangle\right| \le 2 \Delta A\Delta B\>,$$
as was to be shown. \qed

Recall that the momentum and position operators have $[p,q]\psi = i^{-1}\psi$ 
for all $\psi\in\cal S(\R)$.  Starting from this one can show:
\medskip
Exercise - For all $\psi\in D(p^2)\cap D(q^2)\cap D(pq)\cap D(qp)$,
$[p,q]\psi = i^{-1}\psi\>.$ 
\medskip
This implies the famous `Heisenberg uncertainty principle':
$$ \Delta p \Delta q \ge {1\over 2}\>,$$
for all states $\psi$ for which both sides make sense.  Putting the units back 
in for once:
$$\Delta p \Delta q \ge {\hbar\over 2}\>.$$
The consequence of this is that one can never measure {\it both} the position 
and momentum of a particle to arbitrary accuracy; accuracy in one must be 
obtained at the expense of accuracy in the other past a certain point, which 
is measured by Planck's constant. 

\medskip
Exercise -  Show that the Heisenberg uncertainty principle is sharp: 
if $$\psi(x) = {1\over \sqrt{\pi c}} e^{-x^2/c}$$
show that $\|\psi\| = 1$ and $\Delta p\Delta q = {1\over 2}$.
\medskip
Exercise - Given one of the functions $\psi$ as in the previous 
exercise, show that for any $t\in\R$ the states $e^{itp}\psi$ and 
$e^{itq}\psi$ also 
have $\Delta p\Delta q = {1\over 2}$.  These are translated and 
`boosted' versions of the state $\psi$, respectively.
Such states are often called `wave packets' since they are states for 
which the wave function $\psi$ has minimal uncertainty in position and 
momentum, hence resembles as closely as possible a classical particle.  







\vfill
\section{11. Quantization, Schr\"odinger's Equation, and the Hydrogen Atom}
To describe the dynamics of a quantum system, i.e.\ to specify how states
evolve in time, it suffices to specify the Hamiltonian, a self-adjoint 
operator $H$ on the Hilbert space $\H$ whose unit vectors correspond to 
states: given the state $\psi\in\H$, $e^{itH}\psi$ describes the state 
$t$ units of time later.  

How does one guess the Hamiltonian of a quantum system?  I say `guess' 
because only the real world {\it knows} the Hamiltonian.  Many quantum 
systems correspond to classical systems for which one knows a Hamiltonian 
(see section 2).  For example, if one has a {\it classical} particle in $\R^3$ 
in 
a {\bf potential} $V\colon\R^3\to\R$, the Hamiltonian or energy from classical 
mechanics is a function $H\colon\R^6\to\R$ given by 
$$H_{class} = {1\over 2m}(p_1^2 + p_2^2 + p_3^2) + V(q_1,q_2,q_3)$$
where the coordinate functions on  the state space $\R^6$ are $(q_1,q_2,
q_3,p_1,p_2,p_3)$.  The first term in $H_{class}$ is called the {\bf 
kinetic energy}; $m$ is a positive number called the {\bf mass} of 
the particle.  The second term is called the {\bf potential energy}.  
An important example is an electron in the electric field of a stationary 
proton; here the energy is
$$H_{class} = {1\over 2m_e}(p_1^2 + p_2^2 + p_3^2) - {e^2\over  r} $$
where $m_e$ is the mass of the electron, $e$ is the electric charge of
the electron (which is minus that of the proton), and $r = (q_1^2 + q_2^2 +q_3^2)^{1/2}$.  A hydrogen 
atom consists of an electron and a proton, so classically it would be 
described by this Hamiltonian.  One can show that the solutions of 
Hamilton's equations with this Hamiltonian correspond to circular, 
elliptical, parabolic and hyperbolic motions of the electron about the 
proton: this is essentially what Newton did in the study of a mathematically 
isomorphic problem, that of the planets (and comets) moving in the 
gravitational potential of the sun.  (Newton did not, of course, use 
Hamilton's formalism!)   

\aside{As always when speaking of real physical systems, we are 
glossing over some subtleties.  In classical mechanics the earth 
does not revolve about a {\it stationary} sun; rather, the earth 
and sun revolve about their common center of mass.  Since the sun is 
much more massive than the earth the stationary sun is a good first 
approximation; similarly, the proton is about 1836 times as massive as
the electron.  One can solve the problem exactly using a mathematical 
trick called separation of variables, and it turns out the only effect is 
to replace the mass of the electron by the {\bf reduced mass} 
$m = m_em_p/(m_e + m_p)$, where $m_p$ is the mass of the proton.
\medskip Also, we are temporarily ignoring effects due to the electron's 
spin: the proton produces not only an electric field but also a magnetic 
field, and the latter interacts with the spin of the electron.  
In general, physical systems are described by a hierarchy of increasingly 
accurate models which take into account more and more effects, so when someone 
says `the Hamiltonian is...' this must always be taken with a grain of salt.  
}

  
One guesses the Hamiltonian for the corresponding {\it quantum} system 
in the most simple-minded manner possible: one formally 
substitutes the position and momentum {\it operators} for the $q$'s 
and $p$'s in the expression $H_{class}$.   Thus for the hydrogen atom
the states of the electron are unit vectors in $\H = L^2(\R^3)$, and 
the Hamiltonian is 
$$H = {1\over 2m}(p_1^2 + p_2^2 + p_3^2) - {e^2\over  r}\>, $$
where in this equation $p_i$ stands for the momentum {\it operator} 
($i^{-1}\partial/\partial x_i$) and $r^{-1}$ stands for the {\it operator} of 
multiplication by $r^{-1}$.  Thus $H$ is a differential operator on 
$L^2(\R^3)$.   As Schr\"odinger wrote it, keeping  
in Planck's constant (which we normally set equal to 1):
$$H = -{\hbar^2\over 2m} \nabla^2 - {e^2\over r}\>.$$
The original {\bf Schr\"odinger equation} for the hydrogen atom is:
$${1\over i}{d\psi\over dt} = -{\hbar^2\over 2m}\nabla^2\psi - {e^2\over r}\psi\>,$$
where  $\psi\in D(H)$ and $\psi(t,\vec x) = (e^{itH}\psi)(\vec x)$.  

This process of substituting operator $p$'s and $q$'s for 
functions $p$ and $q$ is called {\bf quantization}.  Historically it 
started as a trick with little justification other than that it seemed
to work, but now there is a well-developed mathematical theory of it,
one aspect of which goes under the name of {\bf geometric quantization}.
(For a mathematically sophisticated treatment see, for example, J\c edrzej 
\'Sniatycki, {\sl Geometric Quantization and Quantum Mechanics}.)
Note that this process is ambiguous in general, because while $pq = qp$
classically, in quantum mechanics $pq \ne qp$, leading to {\bf 
operator ordering problems} in the quantization of expressions involving 
products of $p$'s and $q$'s.  These ambiguities can sometimes be resolved 
through other criteria; note however that in expressions of the form 
$$H = {1\over 2m} (p_1^2 + p_2^2 + p_3^2) + V(q_1,q_2,q_3)$$
they do not occur!  Indeed, one wonders what the history of quantum mechanics 
would have been like if there has been operator ordering problems in the 
Hamiltonian of the hydrogen atom (the first system to which quantum theory 
was successfully applied).  

After deciding upon this strategy for quantization, the first problem we must 
address is: is the quantized Hamiltonian a self-adjoint operator?  
The next problem is:
what is its spectrum?  These questions can be answered by computations 
in the case of the hydrogen atom Hamiltonian, 
explicitly determining eigenvectors and approximate eigenvectors.  For 
more complicated systems this `explicit' approach is hardly ever possible.  
So rather than wasting time solving the hydrogen atom, which 
can be found in any quantum mechanics text, we simply describe the answer. 

The answer is: the operator $H_0$ given by 
$$D(H_0) = C_0^\infty(\R^3)\>;$$
$$H_0\psi = (-{\hbar^2\over 2m} \nabla^2 - {e^2\over r})\psi \>,$$
is essentially self-adjoint.  Note that because of the singularity of 
the potential $e^2/r$ at $r=0$, some work is needed to show that 
$H_0$ is really well-defined:

\vskip 1em

Exercise - Show that if $\psi\in D(H_0)$ then $H_0\psi\in L^2(\R^3)$.  
Show that $H_0$ is symmetric.  
\vskip 1em

Let $\overline H_0 = H$.  Then the spectrum of the self-adjoint operator $H$ is 
$$\{-{me^4\over 2\hbar^2 n^2}\}_{n=1}^\infty\, \cup\, [0,\infty)\>.$$
For $n = 1,2,3\dots$ there are $n^2$ linearly independent vectors $\psi$ 
such that 
$$H\psi = -{me^4\over 2\hbar^2 n^2}\psi\;;$$
the integer $n$ is called the 
{\bf principal quantum number}.
Eigenvectors of the Hamiltonian are also called {\bf bound states}.  
The $n=1$ and $n=2$ states are given as follows:
\vskip 1em

Exercise - Show that the function 
$$\psi(\vec x) = e^{-r/a_0}\>,$$ 
where $r = \|\vec x\|$  and $a_0 = \hbar^2/m e^2$, satisfies 
$$H\psi = - {me^4\over 2\hbar^2} \psi\>.$$
Thus $\psi$ is an eigenvector of $H$ with $n = 1$.   Show that the functions 
$$\psi(\vec x) = xe^{-r/2a_0}\>,\>ye^{-r/2a_0}\>,\>ze^{-r/2a_0}\>,{\rm\  and} \;\,
(2 - {r\over a_0})e^{-r/2a_0}$$
are eigenvectors of $H$ with $n = 2$.  The constant $a$ is called the 
{\bf Bohr radius}  of the hydrogen atom and is about $.529$ angstroms, that is, $.529\times 10^{-8}$ centimeters.  
\vskip 1em

In general the states with principal quantum number $n$ are of the form: 
$e^{-r/na_0}$ multiplied by a polynomial of degree $n$ in $x,y,z$ and $r$.
Thus the probability of finding the electron in one of these states at a 
distance greater than $r$ 
from the proton decreases rapidly with $r$.   These states correspond 
physically to states 
in which the electron is `in orbit' around the proton, so they are also called, 
especially in chemistry, 
{\bf orbitals}.  
The `orbit' idea shouldn't be taken to seriously, however,  
because {\it none} of the bound states is an eigenvector of the position 
operators $q_i$ (which have no eigenvectors!), so the electron 
does not follow a curve as time goes on: rather, at any time its position is
described by the probability density $|\psi(\vec x)|^2$ (this follows from 
the probability interpretation of the spectral measure).   If at time $t$ the state is an eigenfunction $\psi$ with $H\psi = \lambda\psi$, then at time $t$ the 
state is given 
by $$\psi(t) = e^{itH}\psi = e^{it\lambda}\psi\>,$$
so the probability distribution $|\psi(t,\vec x)|^2$ is independent of time!  
Thus thinking of the electron as a little ball moving around the proton is 
misleading for such states.  

For $\lambda \ge 0$ 
there is {\it no} eigenvector $\psi\in D(H)$ such that $H\psi = \lambda \psi$, 
rather, these $\lambda$  are only {\it approximate} eigenvalues of $H$.  
The states which are orthogonal to the bound states are called {\bf scattering 
states} and correspond physically to states in which the electron is {\bf 
ionized}, that is, has enough energy to move arbitrarily far from 
the proton, rather than orbit it.  Approximate eigenvectors for $H$ 
corresponding to the approximate eigenvalues $\lambda\ge 0$ are given as 
follows:  
 
\vskip 1em
Exercise - Let $f\in C_0^\infty(\R)$ be a decreasing function with 
$f(r) = 1$ for $0\le r \le 1/2$ and $f(r) = 0$ for $r \ge 1$.  Given 
$\vec k\in \R^3$ let 
$$\psi_n(\vec x) = cf(\|\vec x - \vec x_0\|/n) e^{i \vec k\cdot\vec x/\hbar} $$
where $\vec x_0$ is any point with $\|\vec x_0\| = 2n$ and $c$ is chosen 
such that the $L^2$ norm of $\psi_n$ equals $1$. 
Show that $\psi_n\in D(H)$ and that 
$$\lim_{n\to\infty} \|H\psi_n - \lambda\psi_n\| = 0\>,$$
where $\lambda = \|\vec k\|^2/2m$.  
Conclude that $[0,\infty)$ is in the spectrum of $H$.  
 

 

\vfill
\section{12. The Kato-Rellich Theorem}
In this section we describe a general method for showing that 
`sufficiently small' perturbations of self-adjoint operators 
are self-adjoint.  Using it we will show that the hydrogen 
atom Hamiltonian of the previous section is self-adjoint.  

Suppose that $A$ and $B$ are densely defined operators on a 
Hilbert space $\H$ and that: 1) $D(B) \supseteq D(A)$; 2) 
for some $a,b\in\R$, 
$$\|B\phi\| \le a\|A\phi\| + b\|\phi\|$$
for all $\phi\in D(A)$.  Then $B$ is said to be {\bf {\it A}-bounded}.  
The infimum of such $a$ is called the {\bf relative bound} of $B$ with 
respect to $A$.  
Typically $b$ will have to be chosen larger as $a$ is chosen smaller.

We begin with two exercises that we'll need:

\vskip 1ex
Exercise - Show that if $A$ is a self-adjoint operator then 
so is $cA$ for any real $c\ne 0$ (where $D(cA) = D(A)$).   
Conclude that if $A$ is densely 
defined and symmetric then $A$ is self-adjoint if and only if
$Ran(A\pm is)  = \H$ for some $s>0$, in which case it holds for
all $s>0$.   

Exercise -  Show that if $A$ is densely defined and symmetric then 
$A$ is essentially self-adjoint if and only if $Ran(A\pm is)$ is 
dense in $\H$ for some $s>0$, in which case it holds for all $s>0$.   
\vskip 1ex

\Proclaim Kato-Rellich Theorem.  Suppose that $A$ is 
self-adjoint and $B$ is $A$-bounded with relative bound $< 1$.  
Then $A+B$ is self-adjoint on $D(A)$ and essentially self-adjoint 
on any core for $A$.  

Proof - To show that $A+B$ is self-adjoint we will  
show that $Ran(A+B+ is)=\H$ for all $s\in\R$ with $|s|$ large enough.    
Since $A$ is self-adjoint, $(A+ is)\colon D(A)\to \H$ is onto and
$$(A+B+ is)\phi = (I + B(A+ is)^{-1})(A+ is)\phi$$
for all $\phi\in D(A)$. 
Thus to show that $Ran(A+B+ is)= \H$ 
we need to show that $Ran(I + B(A+is)^{-1}) = \H$.  For this it 
suffices to show that $-1 \notin \sigma(B(A+is)^{-1})$, and since the 
spectral radius is less than or equal to
 the norm it suffices to show that $\|B(A+is)^{-1}\|< 1$.   

For $\phi\in D(A)$ we have $$\|(A+is)\phi\|^2 = \|A\phi\|^2 + s^2\|\phi\|^2\>.$$
Letting $\phi = (A+is)^{-1}\psi$, it follows that $\|A(A+is)^{-1}\|\le 1$
 and $\|(A+is)^{-1}\|\le |s|^{-1}$.  Since $B$ is $A$-bounded, we have
$$\eqalign{\|B(A+is)^{-1}\psi\| &\le a\|A(A+is)^{-1}\psi\| + b\|(A+is)^{-1}\|\cr
&\le (a + b|s|^{-1})\|\psi\|\>.}$$
Thus for $|s|$ sufficiently large, $B(A + is)^{-1}$ has norm less than one, 
since we can choose $a<1$.  

Suppose $D_0$ is a core for $A$. 
Then $(A+ is)D_0$ is dense in $\H$ if $s\ne 0$, $s\in\R$.   
Choosing $|s|$ large enough, $Ran(I + B(A+ is)^{-1})=\H$ by the above.  It follows from the exercise below that 
$Ran(A+B+ is) = (I + B(A+ is)^{-1})(A+ is)D_0$ is dense in $\H$.  
Thus $D_0$ is a core for $A+B$.  \qed

Exercise - Show that if 
$T\in\L(\H)$, $Ran T = \H$, and $D\subseteq\H$ is dense in $\H$, then 
$TD$ is dense in $\H$.   
\vskip 2ex
To apply the Kato-Rellich theorem to the hydrogen atom we take 
$$A = -{1\over 2m}\nabla^2\;,\;B = -{e^2\over r}\;.$$
The work will consist of showing that $B$ is $A$-bounded.  
More precisely, recall that $-\nabla^2 = p_1^2 + p_2^2 + p_3^2$ is essentially 
self-adjoint (an exercise in section 10).   
We let $\Delta$ denote the closure, 
a self-adjoint operator on $L^2(\R^3)$.  By the following exercise $C_0^\infty$ is a core for $\Delta$: 

\vskip 1em
Exercise - Use the fact that $C_0^\infty(\R^3)$ is dense in $\S(\R^3)$ to show 
that $\Delta$ is essentially self-adjoint when restricted 
to $C_0^\infty(\R^3)$.  (Hint: express $\Delta$ in terms of the Fourier 
transform.)   
\vskip 1em

\Proclaim Theorem.  
 Suppose $V\in L^\infty(\R^3) + L^2(\R^3)$, that is, $V= V_1 + V_2$ 
with $V_1\in L^\infty(\R^3)$ and $V_2 \in L^2(\R^3)$.  Then the operator 
$\Delta  + V$, where here $V$ denotes the operator of multiplication by 
$V$, is essentially self-adjoint on $C_0^\infty(\R^3)\subset L^2(\R^3)$ and 
is self-adjoint on $D(\Delta)$.  

\Proclaim Lemma.  For all $a>0$ there exists $b$ such that 
$$\|\psi\|_\infty \le a\|\Delta\psi\|_2 + b\|\psi\|_2$$
for all $\psi\in D(\Delta)\subseteq L^2(\R^3)$.  

Proof - If $f$ is the Fourier transform of $\psi$, we have  
$$\|\psi\|_\infty\le \|f\|_1$$
by the Riemann-Lebesgue lemma and 
$$\|\psi\|_2 = \|f\|_2\;,\;\|\Delta\psi\|_2 = \|k^2f\|_2$$ 
by the Plancherel theorem, so it suffices to show that for all $a>0$ there 
exists $b$ such that 
$$\|f\|_1 \le a\|k^2f\|_2 + b\|f\|_2\>.$$
Since $\psi\in D(\Delta)$, $k^2f\in L^2(\R^3)$, so for any $c>0$,
$((ck)^2+1)f\in L^2(\R^3)$.  It is easily seen 
that $((ck)^2 + 1)^{-1} \in L^2(\R^3)$, so 
 by the Cauchy-Schwartz inequality
$$\|f\|_1   \le \|((ck)^2+1)^{-1}\|_2 \,\|((ck)^2+1)f\|_2$$
Note that by a change of variables 
$$\eqalign{ \|((ck)^2+1)^{-1}\|_2 &= (\int {d^3k\over ((ck)^2 + 1)^2})^{1/2}\cr
&= (\int {c^{-3}d^3k\over (k^2 + 1)^2})^{1/2}\cr
&= c^{-3/2}\|(k^2+1)^{-1}\|_2\>,}$$
so letting $d = \|(k^2+1)^{-1}\|_2$, we have 
$$\eqalign{\|f\|_1 &\le c^{-3/2}d \|((ck)^2 + 1)f\|_2 \cr 
&\le dc^{1/2}\|k^2f\|_2 + dc^{-3/2}\|f\|_2\>.}$$
Choosing $d$ such that $dc^{1/2} = a$, 
$$\|f\|_1 \le a\|k^2f\|_2 + b\|f\|_2\>.\hskip 5em\Box$$

\vskip 1ex
Proof of Theorem - It suffices to show that $V$ is $\Delta$-bounded with 
relative bound less than one.   
We have 
$$\|V\psi\|_2 = \|(V_1 + V_2)\psi\|_2 \le \|V_1\|_2\|\psi\|_\infty 
 + \|V_2\|_\infty\|\psi\|_2\>,$$
so by the lemma, for any $a>0$ there exists $b$ such that 
$$\|V\psi\|_2   \le \|V_1\|_2(a\|\Delta\psi\|_2 + b\|\psi\|_2)  
 + \|V_2\|_\infty\|\psi\|_2\>,$$
so $V$ is $\Delta$-bounded with relative bound zero.  \qed
 
\Proclaim Corollary. The hydrogen atom Hamiltonian, 
$${1\over 2m}\Delta - {e^2\over r}\>,$$
is essentially self-adjoint on $C_0^\infty(\R^3)$ and self-adjoint on 
$D(\Delta)$.  

Proof - It suffices to show that $\Delta - 2me^2r^{-1}$ has these properties, 
so by the Kato-Rellich theorem it suffices to show that $r^{-1}\in L^2(\R^3)+
L^\infty(\R^3)$.  We may write $$r^{-1} = \chi_{\{r\le 1\}}r^{-1} + 
\chi_{\{r>  1\}}r^{-1} \>,$$
and the first term is in $L^2$ while the second is in $L^\infty$.  \qed

{\bf Kato's theorem} is an extension of the corollary above 
to Hamiltonians describing arbitrary atoms and molecules,  
or more generally, collections of charged particles.  To describe a system 
of $n$ particles in $\R^3$ one uses the Hilbert space $L^2(\R^{3n})$, where 
we give $\R^{3n}$ the coordinates $(x_{1i},x_{2i},x_{3i})_{i=1}^n$.  The
 position and momentum operators for the $i$th particle are 
$$q_{1i} = M_{x_{1i}}\;,\;q_{2i} = M_{x_{3i}}\;,\;q_{3i} = M_{x_{3i}}$$
and 
$$p_{1i} = i^{-1}{\partial\over\partial x_{1i}}\;,\;
p_{2i} = i^{-1}{\partial\over\partial x_{2i}}\;,\;
p_{3i} = i^{-1}{\partial\over\partial x_{3i}}\>.$$
If the $i$th particle has mass $m_i$ and charge $e_i$, the Hamiltonian is 
$$H = \sum_{i=1}^n  {1\over 2m_i}(p_{1i}^2 +p_{2i}^2 +p_{3i}^2) \;-\; 
\sum_{1\le i<j\le n} {e_ie_j\over \|\vec q_i - \vec q_j\|}\>.$$
\vskip 1ex 
Exercise - Prove {\bf Kato's Theorem}: the Hamiltonian $H$ above is 
essentially self-adjoint on $C_0^\infty(\R^{3n})$.  (Hint: copy the proof 
for the hydrogen atom). 
\vskip 1ex

The kinetic energy term,
$H_0 = \sum_{i=1}^n  {1\over 2m_i}(p_{1i}^2 +p_{2i}^2 +p_{3i}^2)\>,$
 is often called the {\bf free Hamiltonian}, since it is the Hamiltonian 
for $n$ `free' or non-interacting particles: 
\vskip 1ex
     
Exercise - Let the {\it classical} free Hamiltonian be given by
$$H_{class} = \sum_{i=1}^n  {1\over 2m_i}(p_{1i}^2 +p_{2i}^2 +p_{3i}^2)\>,$$
where here the $p$'s are functions on $\R^{6n}$ as in section 2.  Show that 
the solutions of Hamilton's equations correspond to particles moving 
with constant velocity along straight lines.
\vskip 1ex
Showing that the potential energy term or {\bf interaction Hamiltonian}
 $$H_{int} = -\sum_{1\le i<j\le n} {e_ie_j\over \|\vec q_i - \vec q_j\|}$$
is $H_0$-bounded is a way of making precise the idea that it is in  
some sense a `small perturbation' of the free Hamiltonian.  
This situation 
is typical of mathematical physics: one tries to model a complicated 
`interacting' system as a perturbation of a simpler `free' system.   

This approach has failed so far to prove self-adjointness of 
the quantum-field-theoretic Hamiltonians describing the basic laws of physics,
i.e., quantum electrodynamics, or more comprehensively, 
the `standard model' of the known particles, interacting via the 
electromagnetic, weak, and strong forces.  
These Hamiltonians are not even known to be well-defined operators on 
a dense domain in some Hilbert space!   
This is probably {\it the} problem of mathematical physics today, and 
the subject of `constructive field theory' is the hardest testing-ground 
for new techniques in analysis.  
There is no reason (any more) to think that in quantum field theory 
the interaction Hamiltonian is 
a `small' perturbation of the free Hamiltonian, but there is no 
other clear way to begin an attack on the problem.  Recently, however, 
I. Segal has proposed a method which amounts to considering the   
free Hamiltonian as a small perturbation of the interacting Hamiltonian! 
While this idea is unlikely to crack the problem without a lot of 
work being put into it, it is an amusing twist.  

\aside{The Kato-Rellich theorem is originally due to F. Rellich, `St\"orungstheorie der Spektralzerlegung, II', {\it Math. Ann.} {\bf 116} (1939) 555-570. 
Kato's application of this result to atomic Hamiltonians appeared in 
`Fundamental properties of Hamiltonian operators of Schr\"odinger type', 
{\it Trans. Amer. Math. Soc.} {\bf 70} (1951), 195-211.  
For introductions to mathematically rigorous quantum field theory, try 
{\it Introduction to Axiomatic Quantum Field Theory} by Bogolubov, Logonov and 
Todorov, or {\it Quantum Physics: A Functional Integral Point of View} by 
Glimm and Jaffe, or - please forgive the advertising - 
{\it Introduction to Algebraic and Constructive Quantum Field Theory} by 
Baez, Segal and Zhou.  }






\vfill
\eject
\section {13. Singular Perturbations and Deficiency Indices}     

Let $H_{p,c}$ be the following generalization of the hydrogen atom Hamiltonian:
$$H_{p,c} = \Delta + cr^{-p}\;,\; p \ge 0\>.$$
Of course, this is not an honest operator until we also specify its domain.  
We have seen that for $p = 1$, $H_{p,c}$ is essentially self-adjoint on
$C_0^\infty(\R^3)$.  We will now study the question: for which $c, p$ is 
$\Delta + cr^{-p}$ essentially
 self-adjoint on some reasonable domain?  Note that as $p$ increases, the 
potential $r^{-p}$ becomes more singular at the origin, so 
our question is an example of the general problem of  
 `singular perturbations': how `big' can the 
interaction Hamiltonian $H_{int}$ be compared to the free Hamiltonian $H_0$ and 
still have $H_0+H_{int}$ be self-adjoint? 
 Mathematically the situation turns out to be rather complicated, but there 
are good physical reasons for this complexity.  

The important thing is not the particular operator $H_{p,c}$ 
we'll be studying but certain {\it general methods} that have been 
developed to study such situations.  In this section we consider 
situations where a symmetric differential operator admits many 
self-adjoint extensions due to different choices of boundary conditions.  
We will introduce the `deficiency indices' as a way of measuring 
how many self-adjoint extensions a symmetric operator has.   
\vskip 2em 
First, note the following:
\vskip1em
Exercise - Show using the Kato-Rellich theorem that $H_{p,c}$ 
is essentially self-adjoint on $C_0^\infty(\R^3)$ for $0\le p < 3/2$. 

Exercise - Show that $r^{-p}\psi$ need not be in $L^2(\R^3)$ 
for $\psi\in C_0^\infty(\R^3)$ if $p \ge 3/2$.   
\vskip 1em 
Thus $H_{p,c}$ is only defined on $C_0^\infty(\R^3)$ for $p\le 3/2$, in which 
case it is essentially self-adjoint.  One might hope that some other domain 
would work for larger $p$.  In a rather misleading sense this is true for 
all $p$.  First, we note that we can always choose a dense domain on which 
$H_{p,c}$ is symmetric:
\vskip 1em
Exercise - Show that the operator $H_{p,c}^0$ defined by 
$$D(H_{p,c}^0) = C_0^\infty(\R^3 - 0)\>;$$
$$H_{p,c}^0\psi = (\Delta + cr^{-1})\psi\>,$$
is densely defined and symmetric.  (Recall that $C_0^\infty(\R^3-0)$ is the 
space of all compactly supported $C^\infty$ functions on $\R^3$ 
that vanish in a neighborhood of the origin.) 
\vskip 1em

The next step is to ask: can we find a dense domain on which $H_{p,c}$ is 
{\it closed} and symmetric?  The answer is yes, because the closure of 
a symmetric operator is symmetric:
\vskip 1em 

Exercise - Show that if $T$ is a closable operator, then the graph of the 
closure of $T$ is the closure of the graph of $T$:
$$\{(\psi,\overline T\psi)\colon\, \psi\in D(\overline T)\} = 
\overline{ \{(\psi,T\psi)\colon\, \psi\in D(T)\}}\>.$$
(Hint: showing $\supseteq$ is easy from the definitions; to show 
$\subseteq$ show that the right side is the graph of a closed extension 
of $T$, hence contains the graph of $\overline T$.)

Exercise - Recall the result of an exercise in section 6: every symmetric 
operator is closable.  Show that if $T$ is symmetric then $\overline T$ is
symmetric, using the exercise above.
\vskip 1em 

Thus we can take the closure of $H_{p,c}^0$ and get a closed
 symmetric operator.  But is this closure self-adjoint, or if not, does 
it have self-adjoint extensions?     
Given a symmetric operator $T$ on $\H$, define the 
{\bf deficiency subspaces} $\H_\pm$ of $T$ by:
$$\H_\pm = Ran(T\pm i)^\perp = Ker(T\adj\mp i) \>.$$
Define the {\bf deficiency indices} of $T$ to be $n_\pm = \dim \H_\pm$.  
The basic criterion for self-adjointness says that a closed symmetric operator 
$T$ is essentially self-adjoint if and only if $n_+ = n_- = 0$.  
The following generalization is due 
to von Neumann:

\Proclaim Theorem.  Let $T$ be a symmetric operator with deficiency 
indices $n_+$ and $n_-$.  Then $T$ has self-adjoint extensions if and only 
if $n_+ = n_-$.  There is a one-to-one correspondence between self-adjoint 
extensions of $T$ and unitary maps from $\H_+$ to $\H_-$.  

Proof - We omit the proof; see for example Reed and Simon, 
{\it Methods of Modern Mathematical Physics}, 
vol. II, Theorem X.2 and corollary thereof. \qed

We can use this to show that $H_{p,c}^0$ has self-adjoint extensions.  
Define a {\bf conjugation} on a 
Hilbert space $\H$ to be a map $K\colon\H\to\H$ such that:
$$K(\psi + \phi) = K\psi + K\phi\;\;,\; K(c\psi) = \overline c K(\psi)\;,$$
$$ K^2 = I\;\;,\; \|K\phi\| = \|\phi\| \;,$$
for all $c\in\C$ and $\phi,\psi\in \H$.  
These properties abstract those of the usual complex conjugate.    
\vskip 1em
Exercise - Show that if $K$ is a conjugation, then for all $\psi,\phi\in\H$
$$\langle K\psi,K\phi \rangle = \langle \psi,\phi \rangle\>.$$
(hint: use the polarization identity - but $K$ is conjugate-linear, not linear, so watch out!)

\Proclaim Theorem.  If $T$ is a closed symmetric operator on $\H$ and 
$K$ is a conjugation on $\H$ such that 
$K\colon D(T) \to D(T)\,$ and $KT = TK$, then the deficiency indices $n_\pm$
of $T$ are equal, so $T$ admits self-adjoint extensions.

Proof -  Suppose that $\psi\in \H_+$.   Then for all $\phi\in D(T)$,
$$\langle \psi, (T+i)\phi\rangle = 0\>,$$ 
so by the exercise above,
$$\langle K\psi, K(T+i)\phi\rangle = \langle K\psi, (T-i)K\phi \rangle = 0\>.$$
We thus can conclude that $K\psi\in \H_-$ if we can show that 
every vector in $D(T)$ 
is of the form $K\phi$ for some $\phi\in D(T)$.  By our hypotheses $K\colon
D(T)\to D(T)$, and since $K^2 = I$, we have $K^{-1} = K$, so $K$ is one-to-one 
and onto from $D(T)$ to itself.  Thus every vector in $D(T)$ is of the form 
$K\phi$ for some $\phi\in D(T)$.  

Thus $K\colon\H_+ \to \H_-$.  The same argument shows that $K\colon\H_- \to 
\H_+$.  Since $K = K^{-1}$ this implies that the map $K\colon\H_+\to\H_-$ is 
1-1 and onto, so that the dimensions of $\H_\pm$ are equal, i.e. $n_+ = n_-$. 
\hskip 5em \Box 

\Proclaim Corollary.  If $V$ is a real-valued measurable function on $\R^3$ such that 
$\Delta + V$ is densely defined, then $\Delta + V$ admits self-adjoint extensions.

Proof - It is easily checked that if $\Delta + V$ is densely defined then it is 
symmetric, since $\Delta$ and $V$ are.  Let $K\colon\H\to\H$ be given by:
$$(K\psi)(x) = \overline\psi(x)\>.$$
Then $K$ is a conjugation, and since $V$ is real and differentiation 
commutes with $K$, $K$ commutes with $\Delta + V$.  It follows
from the theorem above that $\Delta + V$ admits self-adjoint extensions. \hskip 
3em\Box

\Proclaim Corollary.  The operator $H_{p,c}^0$ admits self-adjoint extensions.

Proof - $H_{p,c}^0$ is symmetric and commutes with $K$ as in the previous 
corollary. \qed

One might think that the corollary above solves our problems: can't one simply
take one of the self-adjoint extensions of $H_{p,c}^o$ to be the Hamiltonian?  
Unfortunately, if there is more than one self-adjoint extension, different 
choices of self-adjoint extensions generate different one-parameter 
unitary groups,
corresponding to different rules for time evolution.  
As Reed and Simon put it, `{\it different} self-adjoint extensions correspond 
to {\it different physics}.' 

For example, by analogy with the Hamiltonians 
we've considered so far, the `free particle on a half-line' should be described 
by the Hilbert space $L^2[0,\infty)$ and the Hamiltonian:  
$$-{1\over 2m} {d^2\over dx^2}\>.$$ 
But the interpretation of $-d^2/d^2x$ as a self-adjoint operator 
depends on what domain we choose.  If we define $T$ by:
$$D(T) = C_0^\infty(0,\infty)\>;$$
$$T\psi = -{d^2\psi\over dx^2}\>,$$
clearly $T$ is densely defined, and it is easily seen using integration by parts
 that $T$ is symmetric:
$$\int_0^\infty \overline\psi T\phi 
= \int_0^\infty \overline\psi\,' \phi' 
= \int_0^\infty \overline T\psi \phi $$
since $\psi,\phi\in C_0^\infty(0,\infty)$.   
We can determine the deficiency indices $n_\pm$ of $T$ as follows.  
If $\psi\in\H_+$, then 
$$\int_0^\infty \overline\psi(T+i)\phi = 0$$
for all $\phi\in D(T) = C_0^\infty(0,\infty)$.  Integrating by parts,  
$$\int_0^\infty \overline{(\psi'' - i\psi)}\phi = 0\>,$$
and since is this is true for all $\phi\in C_0^\infty(0,\infty)$, we must 
have $\psi'' = i\psi$.  Thus 
$\psi\in\H_+$ must be a linear combination of the solutions 
$$\psi_1(x) = e^{\sqrt i x} \;,\; \psi_2(x) = e^{-\sqrt i x}\>.$$
But only $\psi_2\in L^2(0,\infty)$, since $\psi_1(x)$ grows exponentially 
as $x\to +\infty$.  Therefore $\H_+$ is spanned by $\psi_2$, so $n_+ = 1$. 
Similarly one can show that $\H_-$ is spanned by 
the complex conjugate 
$\overline\psi_2$; this also follows from the theorem on conjugations.  
Thus $n_- = 1$ as well.  

It follows that $T$ has self-adjoint extensions.  These are described 
as follows:
\vskip 1em

Exercise - Let $T_a$, $a\in\R$, be the operator given by: 
$$D(T_a) = \{\psi\in C_0^\infty[0,\infty)\;\colon\;\psi'(0) + a\psi(0) = 0\}
\>,$$
$$T_a\psi = -{d^2\psi\over dx^2}\>.$$
Let $T_\infty$ be given by:
$$D(T_\infty) = \{\psi\in C_0^\infty[0,\infty)\;\colon\;\psi'(0)  = 0 \} \>,$$
$$T_\infty\psi = -{d^2\psi\over dx^2}\>.$$
Show that $T_a$ and $T_\infty$ are essentially self-adjoint by determining 
$H_\pm$ as above.  Conclude that $\overline T_a$ and $\overline T_\infty$ are 
self-adjoint extensions of $T$.  
\vskip 1em
 
Thus the choice of self-adjoint extension of $T$ corresponds to a choice 
of boundary conditions.  
The choice of boundary conditions determines what happens when 
a wave function $\psi\in C_0^\infty(0,\infty)$ is evolved in time according 
to the 1-parameter group generated by the Hamiltonian 
until it `hits' the `wall' at $0$.  
Dirichlet boundary conditions correspond to the operator $\overline T_\infty$; they
 produce a situation like 
that which occurs when one wiggles one end of a rope up and down and the other 
end is 
held fixed, so the displacement at that end is zero. 
If instead the the other end is free to move up and down in a slot, one 
has Neumann boundary conditions, which correspond to the operator $\overline T_0$.  
(A real rope satisfies 
the wave equation, not Schr\"odinger's equation, but the similarities 
rather than the differences concern us here.)  If one shakes the rope once,
producing a pulse which travels to the other end and bounces back, the pulse 
will come back upside down given Dirichlet boundary conditions, but right-side 
up given Neumann boundary conditions!   

\aside{
To verify this, note that $i^{-1}{d\over dx}e^{ikx} = ke^{ikx}$, so that, 
roughly speaking,  
 $e^{ikx}$ is an eigenvector of $p$ with eigenvalue $k$, or a state with 
momentum exactly $k$.  (Of course $e^{ikx}$ is not in $L^2[0,\infty)$, but 
one can obtain approximate eigenvectors by `cutting it off' at some 
large value of $x$.)  
Show that $\psi = e^{-ikx} + \alpha e^{ikx}$ (suitably cut off) is in the 
domain of $T_a$ where $\alpha = (ik - a)/(ik + a)$ (where possibly $a=\infty$).
Show that $\psi$ is an eigenvector of $T_a$ (ignoring the cutoff).  Physically, 
one interprets $\psi$ as a `plane wave' of momentum $k$ moving to the {\it 
left} superposed with a `plane wave' of momentum $k$ moving to the {\it 
right}, i.e.\ `reflecting', with the reflected wave having a `phase shift' of 
$\alpha$.  Note that Dirichlet boundary conditions correspond to a phase shift 
of $-1$ for all momenta, while Neumann boundary conditions correspond to a 
phase shift of $1$ for all momenta.}

Something similar happens with $H_{p,c}^0$.  If 
it has more than one self-adjoint extension, different self-adjoint 
extensions say different things about how the particle `pops out' 
when it hits the singularity at $r = 0$.  In other words, the dynamics is 
not uniquely determined by $H_{p,c}^0$.  The following theorem says 
when there are such different choices.  

\Proclaim Theorem.   $H_{c,p}^0$ is essentially self-adjoint  
 if $cr^{-p} \ge {3\over 4}r^{-2}$.  $H_{c,p}^0$ has more than one self-adjoint 
extension if $ cr^{-p} < {3\over 4}r^{-2}$.     

Proof - The proof is similar to our study of the free particle on 
the half-line; one obtains a differential equation $\psi\in\H_\pm$ 
must satisfy and sees which solutions are in $L^2(\R^3)$.  
Unfortunately it's rather lengthy, so we omit it.  For a proof that $H_{p,c}^0$ is essentially self-adjoint for $p = 2$, $c \ge 3/4$, see Barry Simon's paper 
`Essential self-adjointness of Schr\"odinger operators with singular potentials' in {\sl Arch.\ Rational Mech.\ Analysis} {\bf 52} (1973), 44--48.  Also see Theorem XIII.24, 
Dunford and Schwartz, {\it Theory of Linear 
Operators}, vol. II.   For even more detail on this problem see D.\ M.\ Gitman, I.\ V.\ Tyutin and B.\ L.\ Voronov's paper `Self-adjoint extensions and spectral analysis in the Calogero problem' in {\sl Jour.\ Phys.\ A} {\bf 43} (2010), 145205.  This paper
is also available for free on the arXiv. \qed 

Thus we only have essential self-adjointness on $C_0^\infty(\R^3 - 0)$ 
if  $p> 2$ and $c> 0$ or $r = 2$ and $c \ge 3/4$!  These conditions are 
saying, roughly, that the potential $cr^{-p}$ is sufficiently 
positive, i.e.\ {\it  repulsive}, to make sure that the particle doesn't 
blunder into the origin, so that no extra rules are needed to describe 
how it pops out.  

\vskip 2em 
Exercise - Let $T$ be the operator on $L^2[0,1]$ given by:
$$D(T) = C_0^\infty(0,1) \>,$$
$$T\psi = -{d^2\over dx^2} \psi\>.$$
Show that the deficiency indices of $T$ are 
$n_+ = n_- = 2$ by explicitly determining $H_\pm$.  (Hint: write down the 
definition $$\H_\pm = \{\psi\in \H\colon \,\forall \phi\in D(T)\;
 \int_0^1 \overline\psi(T \pm i) \phi  = 0 \, \}\>,$$ 
and integrate by parts to get a differential equation $\psi$ must satisfy.)
\vskip1em
Exercise - Let $T$ be the operator on $L^2[0,\infty)$ given by:
$$D(T) = C_0^\infty(0,\infty)\>,$$
$$T\psi = i^{-1}{d\psi\over dx}\>.$$
Show that the deficiency indices of $T$ are $n_+ = 1$, $n_- = 0$.  Show that 
$T$ has no self-adjoint extensions.  Show that $\overline T$ is a closed 
symmetric operator that is not self-adjoint!   

\vfill 
\eject



\section {14. Singular Perturbations and Positivity} 

Recall that an operator $T$ is {\bf bounded below} if for some 
$c$, $T \ge c I$, that is, for all $\psi\in D(T)$
$$\langle \psi,T\psi\rangle \ge   c \|\phi\|^2\>.$$
The operator $T$ is {\bf nonnegative} if $T \ge 0$, that is,  
$$\langle \psi,T\psi\rangle \ge 0$$ 
for all $\psi\in D(T)$.  The operator $T$ is ${\bf positive}$ if 
$\langle \psi,T\psi\rangle > 0$ for 
all nonzero $\psi\in D(T)$.  In physics, `well-behaved' Hamiltonians are
bounded below; roughly speaking, systems with Hamiltonians not bounded below 
would be `unstable', being able to emit an infinite amount of energy. 
 This idea is made precise in thermodynamics, where it is seen that 
at low temperatures a system in thermal equilibrium with its surroundings 
is in the state of least possible energy (roughly), 
and that `equilibrium' is only well-defined for Hamiltonians that 
are bounded below. 
Note that the free Hamiltonian for a particle in $\R^3$, 
${1\over 2m}\Delta$, is bounded below, in fact positive, 
 for $m > 0$, since 
$$\langle \psi,\Delta\psi\rangle = \int |\nabla\psi|^2 = 
\int k^2 |\widehat \psi|^2 d^3k > 0$$
for all nonzero $\psi\in D(\Delta)$.  (Here the gradient is taken in the 
distributional sense.)  This is one reason why Schr\"odinger's equation is 
only a reasonable description of particles with {\it positive} mass.  

\aside{Photons and neutrinos are thought to have zero mass, although present 
experimental evidence does not rule out a small mass for neutrinos. Any 
reasonable description of massless particles require special relativity to be 
taken into account; such particle always move at the speed of light, and 
at speeds near that of light the classical Hamiltonian is seen to be only 
an approximation to a `relativistic' Hamiltonian.  Electrons in heavy 
atoms move at speeds near enough to that of light for these corrections 
to become important; in the hydrogen atom relativistic effects are quite small 
(the speed of the electron is about 1/137 times that of light) but very  
well measured, and provide important confirmation of quantum 
electrodynamics, the theory of electromagnetism which takes quantum theory 
and special relativity into account.} 

Mathematically, bounded below operators are nice in many ways.  First of all, 
it follows from polarization that if $T$ is positive then $T$ is symmetric.  
As we will show in the next section, there is a nice self-adjoint extension of 
any positive operator, called the Friedrichs extension.  

We will apply these ideas to our Hamiltonian 
$$D(H_{p,c}^0) = C_0^\infty(\R^3 - 0)\>,$$
$$H_{p,c}^0\psi = (\Delta + cr^{-p})\psi\>.$$ 

\Proclaim Theorem.  $H_{p,c}^0$ is not bounded below 
if $c< 0$ and $p> 2$. 

Proof -  Let $\psi\in C_0^\infty(\R^3 - 0)$ be nonzero.  Then 
clearly 
$\langle \psi,\Delta\psi \rangle > 0$
while if $c < 0$, 
$\langle \psi,cr^{-p}\psi \rangle < 0\>.$
Thus for some $\epsilon > 0$, 
$$\langle \psi,\Delta\psi   \rangle = \epsilon\|\phi\|^2$$
and for some $\delta > 0$, 
$$\langle \psi,\psi \rangle = -\delta\|\phi\|^2$$

Now let $\psi_a$ be the `dilation' of $\psi$ given by 
$$\psi_a(x) = \psi(ax)\>.$$
Note that by a change of variables 
$$\|\psi_a\|^2 = \int |\psi(ax)|^2 d^3x  =  \int|\psi(y)|^2\,a^{-3} d^3y = a^{-3}\|\psi\|^2 \>.$$
On the other hand, 
$$\eqalign{\langle \psi_a,\Delta\psi_a\rangle &= 
\int\overline\psi_a\,\Delta\psi_a\cr 
&= a^2\int \overline\psi(ax) \,(\Delta\psi)(ax) d^3x \cr
&= a^2\int \overline\psi(y)\,\Delta\psi(y)\, a^{-3}d^3 y\cr
&= a^{-1} \langle \psi,\Delta\psi\rangle\cr
&= a^{-1}\epsilon \|\psi\|^2 \>,\cr}$$
while
 $$\eqalign{\langle \psi_a,c^{-p}\psi_a\rangle &= 
\int c\|x\|^{-p}\overline\psi(ax) \psi(ax) d^3 x\cr
&= \int c\|y/a\|^{-p} \overline\psi(y) \psi(y) a^{-3} d^3 y\cr
&= a^{p-3}\langle\psi,cr^{-p}\psi\rangle \cr
& = -a^{p-3}\delta\|\psi\|^2\>.\cr}$$
It follows that for any $a> 0$, $\psi_a$ satisfies
$$\eqalign{ \langle\psi_a, H_{p,c}^0 \psi_a\rangle &= 
\langle\psi_a, (\Delta + cr^{-p})\psi_a\rangle \cr
& = (a^{-1}\epsilon - a^{p-3}\delta)\|\psi\|^2 \cr
&= (a^2\epsilon - a^{p}\delta) \|\psi_a\|^2 \>.}$$ 
It follows that if $p > 2$, no matter how large $M$ is we can choose $a$ large 
enough that 
$$\langle\psi_a,H_{p,c}^0\psi_a\rangle \le -M \|\psi_a\|^2 \>.$$
Thus $H_{p,c}^0$ cannot be bounded below if $c < 0$ and $p > 2$. \qed

\aside{This is a typical example of a {\bf scaling} argument: many counterexamples 
can be produced by considering the family of all dilations $\psi_a$ of 
a given function.  Here the proof worked by showing that 
for functions concentrated sufficiently near the origin, namely $\psi_a$ with $a$ 
large, $cr^{-p}\psi_a$ is bigger than $\Delta\psi_a$ if $p> 2$. 
The exponent $p=2$ is an example of a {\bf critical exponent}, at which 
the proof breaks down because $\Delta$ and $r^{-2}$ have the same 
behavior under scaling.  In physics terminology, $\Delta$ and $r^{-2}$ both 
have dimensions of length$^{-2}$.  Many subtle and interesting problems 
in mathematics and physics arise from critical exponents.  
} 
 
\Proclaim Theorem.  $H_{p,c}^0$ is bounded below if $c > 0$ or $p < 2$. 
If $p = 2$, $H_{p,c}^0$ is nonnegative if $c \ge -1/4$, and unbounded 
below if $c < 1/4$.    

Proof - $H_{p,c}^0$ is clearly positive if $c > 0$, since 
$\Delta$ is positive and $cr^{-p}$ is nonnegative.  
For the rest, we use the following beautiful 
inequality:

\Proclaim Lemma.  Let $\psi\in C_0^\infty(\R^3)$.  Then 
$$\int_{\R^3} {1\over 4r^2} |\psi(x)|^2 \,d^3 x 
\le \int_{\R^3} |\nabla \psi(x)|^2 \,d^3 x\>.$$

Proof - We may suppose without loss of generality that $\psi$ is 
real-valued.  Note that if $r > 0$ 
$$\nabla(r^{1/2}\psi) = r^{1/2}\nabla\psi + {1\over 2}r^{-3/2}\vec x \psi$$ 
so 
$$\eqalign{ |\nabla\psi |^2 
&= (r^{-1/2}\nabla(r^{1/2}\psi) - {1\over 2}r^{-2}\vec x\psi)^2\cr
&\ge - r^{-5/2}\psi\, \vec x\cdot \nabla(r^{1/2}\psi) + 
 {1\over 4}r^{-2}|\psi|^2\>, }$$  
where we are using the fact that $(\vec v -\vec  w)^2 \ge \vec v\,^2 - 2\vec v\cdot\vec w$.  
Recalling that $\vec x \cdot\nabla f = r{\partial f\over \partial r}$, we obtain
$$ |\nabla\psi |^2 
\ge - r^{-3/2}\psi {\partial\over\partial r }(r^{1/2}\psi) +  
{1\over 4}r^{-2}|\psi|^2\>. $$   
and using the fact that
$$\eqalign{  r^{-3/2}\psi {\partial\over\partial r }(r^{1/2}\psi) &= 
r^{-1}\psi{\partial\psi\over\partial r} + {1\over 2}r^{-2}\psi^2 \cr
&= {1\over 2}r^{-2}{\partial \over\partial r} (r\psi^2)}$$
we obtain 
$$ |\nabla\psi |^2 
\ge - {1\over 2r^2}{\partial\over \partial r}(r\psi^2)  +  
{1\over 4r^2}|\psi|^2\>. $$   
Integrating, this yields 
$$\int |\nabla\psi|^2 \ge \int {1\over 4r^2}|\psi|^2 - 
{1\over 2} \int {\partial\over  \partial r}(r\psi^2) \,
\sin^2 \theta \,drd\theta d\phi\>,$$
and doing the second integral by parts we get zero, so the lemma is proved. \qed

Proof of Theorem, continued - If $p = 2$, the lemma above immediately 
implies that 
$H_{p,c}^0$ is nonnegative if $c \ge -1/4$.  We treat the case $c < 1/4$ in the exercise below.   
We treat the case $p < 2, c < 0$ as 
follows.  Given $a > 0$, choose $\epsilon$ such that $r^{-p} \le ar^{-2}$ for 
all $r \le \epsilon$.   Then if $\psi\in C_0^\infty(\R^3)$, we use the lemma 
to show that:  
$$\eqalign{ \int_{\R^3} r^{-p} |\psi|^2 
&= \int_{r\le \epsilon} r^{-p} |\psi|^2 +
 \int_{r>\epsilon} r^{-p} |\psi|^2 \cr 
&\le \int_{r\le \epsilon} ar^{-2} |\psi|^2 +
\epsilon ^{-p} \int_{r>\epsilon}  |\psi|^2 \cr 
&\le a\int_{\R^3} r^{-2} |\psi|^2 +
\epsilon ^{-p} \int_{\R^3}  |\psi|^2 \cr 
&\le 4a\int_{\R^3}  |\nabla\psi|^2 +
\epsilon ^{-p} \int_{\R^3}  |\psi|^2  
}$$

In particular, if $\psi\in C_0^\infty(\R^3 - 0)$, we have 
$$ \langle \psi, H_{p,c}^0 \psi\rangle = 
\int cr^{-p}|\psi|^2 + |\nabla\psi|^2$$ 
$$\ge (1 + 4ac)\int |\nabla\psi|^2 + c\epsilon^{-p} \int |\psi|^2\>, $$
so if we choose $a$ small enough that $1 + 4ac \ge 0$, 
$$\langle \psi, H_{p,c}^0 \psi\rangle  
\ge   c\epsilon^{-p} \int |\psi|^2 = c\epsilon^{-p}\|\psi\|^2\>, $$
which shows that $H_{p,c}^0$ is bounded below.  \qed


Exercise - Show that if $p = 2$ and $c < -1/4$ then $H_{p,c}^0$ is 
not bounded below, as follows.  Assuming $0 < a < 1/2$, show that 
the function $\psi(r) = r^{-a}$ satisfies 
 $$|\nabla \psi| = {a\over r}|\psi|\>,$$  
and the integrals $$\int_{\R^3} |\nabla \psi|^2 \;,\; \int_{\R^3} {1\over r^2} |\psi|^2 $$
converge and satisfy
  $$\int |\nabla \psi|^2 = \int {a^2\over r^2} |\psi|^2\>. $$
Show that for any $\epsilon > 0$ there exists  $\phi\in C_0^\infty(\R^3 - 0)$, 
obtained by making $\psi$ smooth and zero near the origin, such that 
   $$\int |\nabla \psi|^2 = \int ({a^2 \over r^2}+\epsilon) |\psi|^2\>. $$
Show that if $c < -1/4$, for any $M > 0$ we can choose $a$ near $1/2$ and 
$\epsilon$ small, so that 
$$ \langle \phi,H_{p,c}^0\phi \rangle < -M\|\phi\|^2 \>.$$

 




\vfill
\section{15. Singular Perturbations and Sesquilinear Forms}

Fundamental problems in mathematical physics often require for their 
solution the generalization of existing concepts.  Often these 
generalizations arise from the need to deal with more singular objects 
than the existing framework is capable of handling.  For example, 
distributions arose as a generalization of functions in order to make 
sense of objects such as the Dirac delta `function'.  Densely 
defined operators arose as a  
generalization of bounded operators because most of the Hamiltonians arising 
in physics are not bounded.  It turns out that 
there is a very useful further generalization of densely defined operators, 
called `sesquilinear forms', which are useful in dealing with 
singular perturbations of self-adjoint operators.  In this section 
we will develop a bit of 
the theory of sesquilinear forms and apply it to the Hamiltonian $H_{p,c}$ 
we have been discussing.   

A {\bf sesquilinear form} on a Hilbert space $\H$ is 
a map $q \colon Q(q)\times Q(q) \to \C$, linear in the second argument 
and conjugate-linear in the first, where $Q(q)$ is a dense subspace of $\H$, 
called the {\bf form domain} of $q$.  (`Sesqui' is Greek for `one and a 
half'.)   

Given a densely defined operator $T$ there is a sesquilinear form $q$ with 
$Q(q) = D(T)$ given by:  $$q(\psi,\phi) = \langle \psi, T\phi \rangle\>.$$
Not all sesquilinear forms arise in this way, however; a different 
sort of example is the sesquilinear form $q$ with $Q(q) = C_0^\infty(\R)
\subset L^2(\R)$ given by 
$$q(\psi,\phi) = \overline\psi(0)\phi(0)\>.$$
Roughly speaking, $q$ corresponds to `multiplication by the delta function 
at zero', since 
$$q(\psi,\phi)= \int \overline\psi(x) \delta(x) \phi(x) dx\>,$$ 
though of course there is no such multiplication operator, since $\delta$ is 
a distribution but not a measurable function.  Thus 
sesquilinear forms are able to describe more singular situations than operators 
can.   

The product of a sesquilinear form $q$ by a complex number $c$ is defined 
to have form domain $Q(cq) = Q(q)$ and on that domain is given by:
$$ (cq)(\psi,\phi) = c\,q(\psi,\phi)\>.$$
The adjoint of a sesquilinear form $q$ is defined to have form 
domain $Q(q\adj) = Q(q)$ and on that domain is given by:
$$ q\adj(\psi,\phi) = q(\phi,\psi) \>.$$
Note that we {\it always} have $q\adj\adj = q$, contrary to the situation for 
densely defined operators, which is a relief.
The sum of two sesquilinear forms $q$ and $q'$ is defined to have 
form domain $Q(q+q') = Q(q)\cap Q(q')$ and on that domain is given by:
$$(q+q')(\psi,\phi) = q(\psi,\phi) + q'(\psi,\phi)\>.$$
The sesquilinear form $I$ is defined by $Q(I) = \H$ and 
$$I(\psi,\phi) = \langle \psi,\phi \rangle\>.$$ 

\aside{ 
One problem with sesquilinear forms is that there is no good general notion of 
a product of two forms, just as there is no general way to multiply 
distrubutions; in certain circumstances, however, a useful product exists.  
See John Baez, `Wick Products of the Free Bose Field', to appear in 
{\it Jour. Funct. Anal.}  For an detailed introduction to sesquilinear 
forms (also sometimes called quadratic forms) see Barry Simon, {\it Quantum 
Mechanics for Hamiltonians Defined as Quadratic Forms}.  } 

A very important fact is that if $A$ is an unbounded self-adjoint operator and 
$q$ is the sesquilinear form with $Q(q) = D(A)$ and $q(\psi,\phi) =
\langle \psi,A\phi \rangle $, the form $q$ can be extended to a larger domain. 

\Proclaim Theorem.  Let $A$ be a self-adjoint operator on a Hilbert space 
$\H$.  There is a unique sesquilinear form $q_A$ such that 
$$Q(q_A) = \{\psi\in\H\,\colon \,\int_{\sigma(A)} |\lambda| d\mu_{A,\psi}(\lambda)
< \infty \,\}$$
and 
$$q_A(\psi,\psi) = \int_{\sigma(A)} \lambda\, d\mu_{A,\psi}(\lambda) $$
for all $\psi\in Q(q_A)$.  $D(A) \subseteq Q(q_A)$ and 
$q_A(\psi,\phi) = \langle \psi,A\phi \rangle $  
if $\psi,\phi\in D(A)$.  
$q_A$ is called the sesquilinear form {\bf associated to} $A$, and $Q(q_A)$ 
is called the {\bf form domain of} $A$.  

Proof - It's not so clear that $Q(q_A)$ is a vector space; 
we show this as follows. 
   Suppose $f \in \B(\R)$.  
If $\psi,\phi\in Q(q_A)$, 
$$ \|f(A)(\psi+\phi)\| \le \|f(A)\psi\| + \|f(A)\phi\|$$
so using $(a+b)^2 \le 2(a^2 + b^2)$, 
$$ \langle (\psi + \phi),f(A)^2 (\psi + \phi)\rangle \le 
2\langle \psi,f(A)^2 \psi \rangle +  
2\langle \phi,f(A)^2 \phi \rangle \>.$$
By the spectral theorem this implies 
$$\int f^2 d\mu_{A,\psi+\phi} \le 2\int f^2 d\mu_{A,\psi} + 
2\int f^2 d\mu_{A,\phi}\>.$$
Since $f^2$ can be any positive function in $\B(\R)$, this implies   
that  
$$ d\mu_{A,\psi+\phi} \le 2d\mu_{A,\psi} + 2 d\mu_{A,\phi}\>.$$
In particular 
$$ \int |\lambda| d\mu_{A,\psi+\phi} \le 
2\int |\lambda| d\mu_{A,\psi}+ 2 \int |\lambda| d\mu_{A,\phi} <\infty$$
so $\psi+\phi\in Q(q_A)$.  

Given $\psi \in Q(q_A)$, we define $q_A(\psi,\psi)$ as above:
$$q_A(\psi,\psi) = \int_{\sigma(A)} \lambda d\mu_{A,\psi}(\lambda)\>. $$
Given $\psi,\phi \in Q(q_A)$, we define $q_A(\psi,\phi)$ by the polarization 
identity; it's straightforward to verify that this 
makes $q_A$ really a sesqilinear form.  Similarly the polarization identity
implies that $q_A$ is the {\it unique} form satisfying the conditions of 
the theorem.  

Suppose $\psi\in D(A)$.  Then $\psi = (A+ i)^{-1}\phi$ 
for some $\phi \in\H$.  For all $f\in \B(\R)$, 
$$\eqalign{\int f \,d\mu_{A,\psi} &= \langle \psi,f(A)\psi\rangle \cr
&= \langle\phi, (A - i)^{-1} f(A) ( A+ i)^{-1} \phi\rangle \cr
&= \int {f(\lambda) \over \lambda^2 + 1} \, d\mu_{A,\phi}(\lambda) \cr
}$$
so $$d\mu_{A,\psi}(\lambda)  = (\lambda^2 + 1)^{-1} d\mu_{A,\phi}(\lambda)\>.$$
In particular,
$$ \int (\lambda^2 + 1) d\mu_{A,\psi} = \int 1\, d\mu_{A,\phi} < \infty\>,$$
so 
$$D(A)\subseteq  
\{\psi\in\H\,\colon \,\int_{\sigma(A)} \lambda^2\, d\mu_{A,\psi}(\lambda)\}
\subseteq Q(q)\>.$$
Also, for $\psi\in D(A)$ 
$$\langle \psi,A\psi\rangle = \int_{\sigma(A)} \lambda\, d\mu_{A,\psi} = 
q_A(\psi,\psi)$$
so polarization implies that 
$q_A(\psi,\phi) = \langle \psi,A\phi \rangle $  
if $\psi,\phi\in D(A)$.  
\qed

To get a feeling for what's going on in the the theorem, try the following:

\vskip 1em 
Exercise - Show that 
$$D(\Delta) = \{\psi\in\H\colon\, \int |\Delta\psi|^2 < \infty\}$$
while 
$$Q(q_\Delta) =  \{\psi\in\H\colon\, \int |\nabla\psi|^2 < \infty\}\>.$$
\vskip 1em 

Given a sesquilinear forms $q$ and $q'$ we write $q \ge q'$ if 
$$q(\psi,\psi) \ge q'(\psi,\psi)$$ 
for all $\psi \in Q(q)\cap Q(q')$.  We define {\bf bounded below}, 
{\bf nonnegative}, and {\bf positive} sesquilinear forms by copying the 
definitions for operators; in particular, $q$ is bounded below if $q$ 
satisfies $q \ge -cI$ for some $c$.    
Sesquilinear forms that are bounded below have many nice properties, and 
for the rest of this section we will discuss forms that are bounded 
below. 
  
Suppose the sesquilinear form $q$ satisfies $q \ge -cI$.  Then there is an 
inner product on $Q(q)$ given by:
$$\langle \psi,\phi\rangle _{1} = q(\psi,\phi) + 
(c+1)\langle \psi,\phi\rangle\>.$$
Note that for topological purposes it doesn't matter which $c$ we pick here: 

\vskip 1ex 
Exercise - Show that if $q \ge -cI$ and $q \ge -c'I$, the two norms 
on $Q(q)$: 
$$\|\psi\|^2 _{1} = q(\psi,\psi) + (c+1)\langle \psi,\psi\rangle$$
and 
$$\|\psi\|^2 _{1'} = q(\psi,\psi) + (c'+1)\langle \psi,\psi\rangle$$
are equivalent, i.e.\ for some constant $a > 0$,
$$\|\psi\|_1 \le a \|\psi\|_{1'}\;,\;\|\psi\|_{1'} \le a \|\psi\|_{1}\>.$$
\vskip 2ex

Of course $Q(q)$ may not be a Hilbert space under the inner product 
$\langle\cdot,\cdot\rangle_1$ because it may not be complete in the 
norm $\|\cdot\|_1$.  Given a sesquilinear form $q$ that is bounded below, we 
define it to be {\bf closed} if $Q(q)$ is complete in the norm $\|\cdot\|_1$.  
We say that a bounded-below form $q$ is {\bf closable} if  
there is closed bounded-below form $\overline q$ such that $Q(q)$ is dense 
in $Q(\overline q)$, the latter given the topology coming from its 
$\|\cdot\|_1$ norm.  In this case we say $\overline q$ is the 
{\bf closure} of $q$:  

\vskip 1ex
Exercise - Show that the closure of a closable bounded-below sesquilinear 
form is {\it unique}.
\vskip 1ex
 
There is a one-to-one correspondence between closed bounded-below 
sesquilinear forms and bounded below self-adjoint operators:

\Proclaim Theorem.
 If $A$ is a self-adjoint operator, bounded below, the 
associated sesquilinear form $q_A$ is closed and bounded below.  
If $q$ is a sesquilinear form that is closed and bounded below, there 
is a unique bounded-below self-adjoint operator such that $q = q_A$.  

Proof - We omit this proof; it can found, for example, in Reed and 
Simon, {\it Methods of Modern Mathematical Physics}, vol. I, Theorem 
VIII.15.  
\qed

This theorem is rather remarkable because while not all closed {\it operators} 
that are bounded below are self-adjoint, it says that all closed {\it forms} 
that are bounded below correspond to self-adjoint operators.  As a consequence, 
there is a distinguished self-adjoint extension of any operator that is 
bounded below:

\Proclaim Friedrichs Extension Theorem.  Let $A$ be an operator that is 
 bounded below and let $q(\psi,\phi) = \langle \psi,A\phi\rangle$ for 
$\psi,\phi\in D(A)$.  Then $q$ is closable, and the closure 
$\overline q$ equals $q_{\hat A}$ for a unique bounded-below 
self-adjoint operator $\widehat A$.  $\widehat A$ is an extension of 
$A$, called the {\bf Friedrichs extension} of $A$.  

Proof - If $A\ge cI$, let $\langle\cdot,\cdot\rangle_1$ be the inner product on 
$Q(q) = D(A)$ given by:   
$$\langle \psi,\phi\rangle _{1} = q(\psi,\phi) + 
(c+1)\langle \psi,\phi\rangle\>.$$
Let $\H_1$ be the Hilbert space completion of $Q(q)$ in the norm 
$\|\cdot\|_1$.  It is easy to see that $q_1$ extends by continuity to a unique 
form $q_1$ on $H_1$.
 
Let $\iota\colon Q(q)\to \H$ be the inclusion map.  Since $Q(q)$ is dense in 
$\H_1$ and $\|\iota\psi\| \le \|\psi\|_1$ for $\psi\in Q(q)$, the B.L.T. theorem 
implies that $\iota$ extends to a unique continuous linear map 
$\iota_1 \colon\H_1\to\H$.  
We show that $\iota_1$ is one-to-one as follows: if 
$i_1(\psi) = 0$ for some $\psi\in \H_1$, then since $Q(q)$ is dense in 
$\H_1$, we can choose $\psi_i\in Q(q)$  such that $$\|\psi_i - \psi\|_1\to 0\>,$$
hence $$\|\psi_i\| = \|\iota_1\psi_i\| \to \|\iota_1\psi\| = 0\>.$$ 
It follows that 
$$\eqalign{ \|\psi\|_1^2 &= \lim_{j\to\infty} \lim_{i\to\infty} 
\langle \psi_i,\psi_j\rangle_1
 \cr & =   \lim_{j\to\infty} \lim_{i\to\infty} 
 \langle \psi_i, A\psi_j\rangle + (c+1)\langle \psi_i,\psi_j\rangle
\cr &= 0\>,}$$
so $\psi = 0$.  

Since $\iota_1$
is one-to-one, it is an isomorphism between $H_1$ and $Ran\,\iota_1\subseteq \H$, 
allowing us to identify $\H_1$ with $Ran \,\iota_1$  and $\widehat q$ with a 
form on $Ran\,\iota_1$, which is clearly the closure $\overline q$ since 
$Q(q)$ is dense in $\H_1$ and $\H_1$ is complete.   
Since $\overline q$ is closed and bounded below, by the theorem above it is of 
the form $q_{\hat A}$
for a unique bounded-below self-adjoint operator $\widehat A$.  Since for all $\psi,\phi
\in D(A)$, 
$$\langle \psi, A\phi\rangle = q(\psi,\phi) = \overline q(\psi,\phi) = 
\langle \psi, \widehat A\phi\rangle\>, $$ 
and $D(A)$ is dense in $\H$, we must have $A\phi = \widehat A\phi$.  Thus 
$\widehat A$ is an extension of $A$.  \qed

An important good feature of the Friedrichs extension is simply its 
{\it naturality}, that is, it takes no {\it extra information} to specify 
which self-adjoint extension of a bounded-below symmetric operator to 
pick.  Thus in the cases where one is trying to get the correct Hamiltonian 
out of an operator that is bounded below but not 
essentially self-adjoint, mathematical physicists take its Friedrichs 
extension.  The following are other nice properties of the Friedrichs extension:

\vskip 1ex 
Exercise -  Show that if $A$ is a symmetric operator with $A \ge cI$, then 
$\widehat A \ge cI$.

Exercise - Show that if $A$ is a symmetric and bounded-below operator then 
$\widehat A$ is the only self-adjoint extension of $A$ whose domain is 
contained in $Q(\overline q)$, where $q$ is as in the theorem.    

Exercise -  Show that if $A$ is self-adjoint and bounded below then 
$\widehat A = A$.  
\vskip 1ex

As an example of how the Friedrichs extension can be used on very 
singular perturbations, we have:

\Proclaim Corollary.  If $V\colon \R^3\to\R$ is continuous except at 
finitely many points and satisfies $V\ge c$ for some $c\in\R$, then 
the operator $\Delta + V$ is densely defined and bounded below, hence 
admits a Friedrichs extension.  

Proof - Let $S$ be the set of points on which $V$ is not continuous.  
Then for $\psi\in C_0^\infty(\R^3 -S)$, $V\psi\in L^2(\R^3)$ and $\Delta\psi
\in L^2(\R^3)$.  Thus $C_0^\infty(\R^3-S) \subseteq D(\Delta)\cap D(V)$, so 
$\Delta + V$ is densely defined.  It's clear also that $\Delta + V$ is 
bounded below, so the Friedrichs extension exists. \qed 

The above theorem applies to operators of the form $\Delta + cr^{-p}$ with 
$c\ge 0$.   

\vskip 1em 
Here, then, is a summary of results on the Hamiltonian $H_{p,c}$.  We haven't proved
all these results, but we've proved most of them:   
\vskip 3em
Positive potential - $c> 0$ : $H_{p,c}^0$ is positive hence Friedrichs 
extension exists.
\vskip 0.5em
$0 \le p < 3/2$ : $H_{p,c}$ essentially self-adjoint on $C_0^\infty(\R^3)$ by Kato-Rellich. 

$p = 2$, $0 < c < 3/4$ : $H_{p,c}^0$ is not essentially self-adjoint.  

$p = 2$, $c \ge 3/4$ : $H_{p,c}^0$ is essentially self-adjoint.  

$p > 2$ : $H_{p,c}^0$ is essentially self-adjoint. 

\vskip 1em
Negative potential - $c < 0$ : $H_{p,c}^0$ is not essentially self-adjoint, but admits self-adjoint extensions.  
\vskip 0.5em 
$0 \le p < 3/2$ : $H_{p,c}$ essentially self-adjoint on $C_0^\infty(\R^3)$ by Kato-Rellich. 

$0 \le p < 2$ : $H_{p,c}^0$ is bounded below hence Friedrichs extension exists.  

$p = 2$, $c \ge -1/4$ : $H_{p,c}^0$ is bounded below hence Friedrichs extension  exists.  

$p = 2$, $c < -1/4$ : $H_{p,c}^0$ is not bounded below.  

$p > 2$ : $H_{p,c}^0$ is not bounded below.  


\vfill\end



