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\begin{document}
\begin{center}
{\bf Wick Powers \\}
{\small John\ C.\ Baez, April 13, 2004\\}
\vspace{0.3cm}
\end{center}
\vskip 1em
\noindent
{\it To get the combinatorics of Feynman diagrams to work nicely,
it'll be best to use the {\bf Wick} or {\bf normal-ordered} powers
of the position operator instead of its
ordinary powers. Roughly speaking, the Wick power $\:q^n\:$ is
designed to be as much as possible like the ordinary power $q^n$,
{\rm except} for the fact that
\[ \langle 1 , \:q^n\: 1 \rangle = 0 .\]
This will greatly simply calculations of transition
amplitudes when we study perturbed versions of the harmonic
oscillator --- which counts for about 80\% of what people
actually do in quantum field theory.
Your goal in this homework is to learn what these
Wick powers are. But first, let me briefly remind you of everything
you need to know to tackle this!
\vskip 1em
\noindent
Recall that the {\bf Weyl algebra}, $\W,$ is the associative
algebra over $\C$ generated by elements $q,p$ satisfying this
relation:
\[ pq - qp = -i , \]
or
\[ [p,q] = -i \]
for short. The Weyl algebra has a number of interesting
representations, but for now let us work in the Fock representation.
\vskip 1em
\noindent
In the {\bf Fock representation}, we think of $p$ and $q$
as operators on the space $\C[z]$ consisting of complex polynomial
in one variable $z$. To do this, we
first define the {\bf annihilation operator} $a$
and {\bf creation operator} $a^\ast$, as follows:
\[ (a\psi)(z) = \psi'(z) \]
\[ (a^*\psi)(z) = z \psi(z) .\]
These satisfy
\[ [a,a^\ast] = 1 . \]
Then we define the {\bf position operator} $q$ and
{\bf momentum operator} $p$ in terms of these:
\[ q = {a + a^* \over \sqrt{2}} \]
\[ p = {a - a^* \over \sqrt{2}i } . \]
These satisfy $[p,q] = -i$, as desired.
\vskip 1em
\noindent
In the homework {\bf $k$-Colorings as Categorified Coherent States},
we saw that $\C[z]$ has a unique inner product such that $1 \in \C[z]$
is a unit vector and
\[ \langle a^\ast \phi, \psi \rangle = \langle \phi, a\psi \rangle .\]
This inner product is given explicitly by
\[ \langle z^n,z^m \rangle = n! \, \delta_{nm} .\]
\vskip 1em
\noindent
The vector $z^n$ has an important physical significance! When
we normalize it, we get a state in which the
quantum harmonic oscillator has $n$ quanta of energy. More precisely,
if we define the {\bf harmonic oscillator Hamiltonian} by
\[ H_0 = \frac{1}{2}(p^2 + q^2 - 1) = a^\ast a ,\]
then we have
\[ H_0 z^n = n z^n .\]
The state $z^0 = 1$ is especially important since it's the
{\bf ground state} of the harmonic oscillator: that is, the state
with the least energy.
\vskip 1em
\noindent
But enough reminders... on to something new!
\vskip 1em
\noindent
It's easy to check that that the powers of $q$ satisfy
\[ [p,q^n] = -inq^{n-1} , \qquad \qquad [q,q^n] = 0 .\]
The {\bf Wick powers} of $q$, denoted $\:q^n\:$, are elements
of $\W$ satisfying
\[ [p,\:q^n\:] = -in\:q^{n-1}\: , \qquad \qquad [q,\:q^n\:] = 0 \]
but also
\[ \langle 1, \:q^n\: 1\rangle = 0 .\]
This says that the expectation value of $:q^n:$ in the ground
state of the harmonic oscillator is zero. By definition we have
\[ \:q^0\: = 1 , \]
but the higher Wick powers of $q$ are, in general,
different from the ordinary powers!
\vskip 1em
\noindent
Your job is to figure out what they are.
}
\vskip 1em
\noindent
1. Show that an element of the Weyl algebra is determined up
to an additive constant by its commutators with $p$ and $q$.
\vskip 1em
\noindent
In other words,
show that if $w,w'$ are two elements
of $W$ with
\[ [p,w] = [p,w'] ,\]
\[ [q,w] = [q,w'] , \]
then $w - w' = c1$ for some $c \in \C$.
\vskip 1em
\noindent
{\it Hint: I'll let you use without proof the fact that any
element $w \in \W$ can be written as a `polynomial'
\[ w = \sum_{m,n \in \N} w_{mn} p^m q^n \]
for a unique choice of the coefficients $w_{mn} \in \C$. Use
this to work out $[ip,w] = f$ and $[-iq,w] = g$, and check that
$f$ and $g$ determine $w$ up to an additive constant.
\vskip 1em
\noindent
In doing this,
you should see that the commutators $[ip,w]$ and $[-iq,w]$
are like the partial derivatives of $w$ with respect to $q$ and $p$,
respectively. The above result is thus a
quantum (i.e.\ noncommutative) version of the following
fact: any polynomial $w \in \C[x,y]$ is uniquely determined up
to an additive constant by its partial derivatives
\[ {\partial w\over \partial x} = f, \qquad \qquad
{\partial w \over \partial y} = g .\]
}
\vskip 1em
\noindent
2. Using Problem 1,
show that there is at most one sequence of elements
$\:q^n\: \in \W$, the {\bf Wick powers} of $q$, such that
\[ [ip,\:q^n\:] = n\:q^{n-1}\: , \qquad \qquad [q,\:q^n\:] = 0 , \]
\[ \langle 1, \:q^n\: 1\rangle = 0 \]
for $n > 0$, and
\[ \:q^0\: = 1 . \]
\vskip 1em
\noindent
{\it This proves the Wick powers are {\rm unique.} But why do they
{\rm exist?} For this, its easiest to give a
nice {\rm formula} for them...}
\vskip 1em
\noindent
3. Show that if we define
\[ \:q^n\: = ({1 \over \sqrt{2}})^n \sum_{k = 0}^n {n \choose k}
{a^\ast}^k a^{n - k} \]
then
\[ [ip,\:q^n\:] = n\:q^{n-1}\: , \qquad \qquad [q,\:q^n\:] = 0 , \]
\[ \langle 1, \:q^n\: 1\rangle = 0 \]
for $n > 0$, and
\[ \:q^0\: = 1 . \]
By Problem 2, the Wick powers of $q$ must be given by this formula.
\vskip1em
\noindent
{\it
This formula for $\:q^n\:$ should remind you of the binomial formula!
We would get this formula if we used
\[ q = {a + a^* \over \sqrt{2}} \]
to expand $q^n$ in terms of $a$ and $a^*$ and then {\rm brutally
moved all the $a$'s to the right of the $a^*$'s,
as if they commuted!} This is why people also call $\:q^n\:$
the {\bf normal-ordered} $n$th power of $q$: {\bf normal ordering}
means putting all the annihilation operators on the right.
\vskip1em
\noindent
Of course the annihilation and creation operators {\rm don't}
commute, so $:q^n:$ is not equal to $q^n$. But, they're close
relatives.
We make this more precise in the next exercise by showing
that $:q^n:$ is a polynomial of degree $n$ in $q$.
}
\vskip1em
\noindent
4. Show inductively that there for each $n$
there exists a polynomial $P_n$ with
\[ P'_n(q) = nP_{n-1}(q) ,\]
\[ \langle 1, P_n(q) 1\rangle = 0 \]
for $n > 0$, and
\[ P_0(q) = 1. \]
Show that $\:q^n\: = P_n(q)$.
\vskip 1em
\noindent
{\it Hint: once you know $P_{n-1}$,
the equation $P'_n = nP_{n-1}$ determines everything
about $P_n$ except its constant term, which we can choose
to obtain $\langle 1, P_n(q) 1\rangle = 0$. Once you have polynomials
$P_n$ satisfying these equations, you can check using Problem
2 that $P_n(q) = \, \:q^n\:$. Remember that $[ip, \cdot]$ acts
like the partial derivative ${\partial \over \partial q}$.
}
\vskip 1em
\noindent
It's still a bit of work to figure out explicitly what these
polynomials $P_n$ are. Up to some normalization factors, they
turn out to be Hermite polynomials. Here are some examples:
\[
\begin{array}{ccl}
\:q^0\: &=& 1 \\
\\
\:q^1\: &=& q \\
\\
\:q^2\: &=& q^2 - {1\over 2} \\
\\
\:q^3\: &=& q^3 - {3\over 2} q \\
\\
\:q^4\: &=& q^4 - 3 q^2 + {3\over 4}
\end{array}
\]
\vskip 1em
\noindent
To see that these are correct, first note that the polynomials
$P_n(q)$ on the right-hand side satisfy $P'_n = nP_{n-1}$. This
determines each polynomial in terms of the previous one up to an
additive constant. To check that we have the constant right we
need to check $\langle 1, P_n(q) 1\rangle = 0$ for $n > 0$.
For this, we use this formula, which I'll prove in class:
\[ \langle 1, q^n 1 \rangle = \left\{
\begin{array}{cl}
0 & {\it n} \; {{\rm odd}} \\
{(n-1)!! \over (\sqrt{2})^n} & {\it n} \; {{\rm even }}
\end{array}
\right.
\]
for $n > 0$.
\vskip 1em
\noindent
So, for example, to check the constant is right in the formula
for $\:q^4\:$, we note
\[
\begin{array}{ccl}
\langle 1, (q^4 - 3 q^2 + {3\over 4} ) 1 \rangle &=&
{3!! \over \sqrt{2}^4} - 3 \cdot {1!! \over \sqrt{2}^2} + {3\over 4} \\
\\
&=& {3\over 4} - {3 \over 2} + {3 \over 4} \\
\\
&=& 0 .
\end{array}
\]
\vskip1em
\noindent
\end{document}