When he heard the Wiz was talking about quantum mechanics in Track 2, Oz decided to attend these lectures instead of leaving after Track 1. But he knew the Wiz wouldn't like this, so he hid himself under a desk during the break.

Pretty soon the rest of the class returned. Then the Wiz walked in and said:

"Last time I gave an outline of quantum mechanics. Now I want to tackle some specific examples. Eventually I want to quantize a bunch of the field theories we talked about last quarter. For the sake of the nonphysicists here, I'll start with something simpler: a free point particle on the line. If you've already taken a course on quantum mechanics this will be pretty boring, so you might as well go to sleep right now."

"Oh good," thought Oz. "Maybe I'll understand this lecture!"

"However," the Wiz continued, "next time we'll see that this example is isomorphic to one of the field theories we're interested in: the vacuum Maxwell equations on a 2d spacetime shaped like a cylinder! So once we quantize one, we can instantly quantize the other. And from there, it's a hop and a jump to 2d Yang-Mills theory, and then 2d BF theory.

So let's get going! We'll start with a classical point particle on the line. This is just a boring old dot whose position at any time is some real number q."

Oz couldn't restrain himself from asking a question here, but he knew the Wizard would punish him if he were discovered. So he did his best to mimic Toby's voice, and he asked: "Why do you use the letter q for position?"

The Wiz turned to Toby and replied, "Because p is already taken --
it stands for momentum! Anyway, the space of all possible positions is
called the "configuration space". In this example it's just
**R**, but in general a configuration space in classical mechanics can be
any manifold whatsoever.

Now let's bring in time. How does our particle move as a function of time? It follows Newton's famous law

F = mawhere m is the particle's mass, F is the force on it, and a is its acceleration. If the force depends only on the particle's position, and its position as a function of time is q(t), this law says

F(q(t)) = m q''(t)This is a second-order differential equation, so with any luck we can solve it given q(t) and q'(t) at some particular time. In physics lingo, the position and velocity at a given time determine the particle's "state". The idea is that knowing the state of a system now determines its state at any other time.

It's often good to work with the particle's momentum

p = m q'instead of its velocity."

Again Oz was overcome with curiosity, so mimicking Toby's voice, he asked, "Why do we call the momentum p?"

The Wiz glared at Toby. "Because m is already taken -- it stands for mass! Seriously, I don't know why people call position q and momentum p. All I know is that if you use any other letters, people can tell you're not a physicist. So I urge you to follow tradition on this point.

Anyway, let's describe the particle's state using a position-momentum
pair (q,p). The space of these position-momentum pairs is called the
"phase space" of our particle. In this example it's just the
plane **R**^{2}. In fancier examples, where the
configuration space is some manifold M, the phase space will be the
cotangent bundle of M. The position is a point in M, the velocity is a
tangent vector, and the momentum is a cotangent vector.

Using position and momentum, we can rewrite Newton's law as a pair of first-order differential equations

q'(t) = p(t)/m p'(t) = F(q(t))called Hamilton's equations. Actually, Hamilton wrote these in a clever way using the concept of *energy*. I hope you're familiar with this idea, because I'm feeling too lazy to explain it very well now! I'll just remind you....

The energy of our point particle consists of two parts: its kinetic energy and its potential energy. The kinetic energy depends only on the particle's momentum; it equals

pThe potential energy depends only on the particle's position; it's^{2}/2m

V(q)where V is some function with

V'(q) = -F(q)Of course we can add any constant to V without spoiling this equation, so potential energy is only determined up to a constant."

Oz asked in Toby's voice: "Why is the potential called V?"

The Wiz turned to Toby and shouted, "Because p is already taken -- it stands for momentum!!! Now listen: just because you're a fictional character whose only role is to push the exposition forwards, doesn't mean you have to ask so many bone-headed questions!"

"But I'm *not*," said Toby.

"Yeah, right," said the Wiz, and flung a fireball at him, which exploded on impact, blackening him to a crisp.

"No fair!" cried Toby. But being a fictional character, he just shook himself off and was as good as new.

Unperturbed, the Wiz continued. "Okay. Once we know the kinetic and potential energy, we add them to get the total energy. The total energy is called the "Hamiltonian", and it's

H(q,p) = pThe great thing is that as our particle moves along, q and p change with time, but the Hamiltonian does not:^{2}/2m + V(q).

dH(q(t),p(t))/dt = 0Energy is conserved!

But Hamilton's really clever idea was this: once you know the energy as a function of momentum and position, it tells you how momentum and position change with time. We've already seen that

q' = p/m p' = F(q),but these are equivalent to

q' = dH/dp p' = -dH/dqwhich are what people usually call Hamilton's equations. You'd better check this for yourself if you haven't yet! Otherwise your life as a physicist will be nasty, brutish and short.

So: the Hamiltonian tells us how the state of a system evolves in time. We've seen this for a point particle on the line, but this principle is much more general. It's true for all classical systems... and it's true for quantum ones too!

The details are very different in the quantum case, though:

In classical mechanics the state is a point in some "phase
space"... but in quantum mechanics we've seen that it's a unit
vector in some *Hilbert space* -- or really an equivalence class
of unit vectors, since multiplying the unit vector x by a phase doesn't
change any transition probabilities |<y,x>|^{2}.

In classical mechanics the Hamiltonian is a function on our phase space... but in quantum mechanics it's a self-adjoint operator on our Hilbert space.

In classical mechanics the Hamiltonian describes how states evolve in time, via Hamilton's equations... but in quantum mechanics it describes how states evolve in time by means of Schroedinger's equation:

d psi / dt = - iH psi.We talked about this last time.

Now, "quantization" is the game of guessing the quantum
description of a physical system from the classical description. It's a
bit of a hit-or-miss business, but there are some tricks that often
work. To get the Hilbert space of the quantum system, we often take
L^{2}(X), where X is the configuration space of the
corresponding classical system. And to get the Hamiltonian of the
quantum system, we often take the Hamiltonian for the classical system,
wave a magic wand over it... and turn it into an operator!

But instead of describing all this in general, let me just do the case
of a point particle. The configuration space of the classical point
particle is **R**, so the Hilbert space of the quantum point particle is
L^{2}(**R**). A state is thus a function psi with

\int |psi(x)|Physicists call psi the "wavefunction". The quantity |psi(x)|^{2}dx = 1

The Hamiltonian for the classical point particle is

H = pTo reinterpret this as an operator on L^{2}/2m + V(q)

(q psi)(x) = x psi(x) (p psi)(x) = -i psi'(x)Note we're using the same letters q and p to mean completely different things now that we're in quantum-land! q is the operator of multiplication by x, and p is the operator of differentiation... times -i, to make it self-adjoint.

This was Schroedinger's great idea, this trick for reinterpreting q and p as operators. Let's see if we can use it to reinterpret H as an operator. The kinetic energy part is easy, since everyone knows how to square an operator:

((pBut for the potential energy, we need to apply the function V to the operator q, and get an operator V(q). How do you apply a function to an operator? You can do it using something called the "functional calculus", which you'd learn in a heavy-duty analysis class. I'll just tell you the answer:^{2}/2m) psi) (x) = -(1/2m) psi''(x).

(V(q) psi) (x) = V(x) psi(x).In other words, since q is the operator of multiplying by x, V(q) is the operator of multiplying by V(x)! Now we're all set:

(H psi) (x) = -(1/2m) psi''(x) + V(x) psi(x).The black magic is now over. We have a Hilbert space and a Hamiltonian, so we can go ahead and see how states evolve in time. Let's do it for a particle with no forces at all acting on it -- a "free" particle. For zero force, we should take V to be constant, so we might as well take it to be zero. We get

(H psi) (x) = -(1/2m) psi''(x).Given this, we want to solve Schroedinger's equation:

d psi / dt = - iH psior in other words

(d/dt) psi(t,x) = (i/2m) (dTo solve a partial differential equation like this -- linear, with constant coefficients -- the usual trick is to try an exponential:^{2}/dx^{2}) psi(t,x)

psi(t,x) = exp(i(-wt + kx))Don't worry about the minus sign; that's just a convention that physicists like. w is called the "frequency" since it says how fast psi wiggles as a function of time. k is called the "wavenumber" since it says how fast psi waves as a function of space. Differentiating an exponential is easy:

(d/dt) psi(t,x) = -iw psi(t,x) (dso we get a solution of our equation if^{2}/dx^{2}) psi(t,x) = -k^{2}psi(t,x)

w = kDoes this formula remind you of anything? It's just the relation between momentum and kinetic energy!^{2}/2m

Now, the solution we just got isn't square-integrable as a function of x, so it doesn't reallly lie in our Hilbert space. But we can cure this problem by taking linear combinations like this:

psi(t,x) = (1/sqrt(2 pi) \int exp(i(-wt + kx)) f(k) dkwhere f is square-integrable and w is related to k as above. This trick is called a Fourier transform. We can undo it to find f if we know psi; when t = 0, we have

psi(x) = (1/sqrt(2 pi) \int exp(ikx) f(k) dkso taking an inverse Fourier transform we get

f(k) = (1/sqrt(2 pi) \int exp(-ikx) psi(x) dx.The factors of 1/sqrt(2 pi) are just there to make sure the L

The upshot is that given any L^{2} function on the real line
representing the state of a quantum particle at some time, we know how
to solve Schroedinger's equation to figure out the state at all other
times."

At this point Richard raised his hand and asked, "But what does this
have to do with the *classical* point particle?"

"Good question. Let's compare the classical and quantum version of
the free point particle. As you'd expect, the *classical* free point
particle just moves along at constant velocity. If you solve Hamilton's
equations for a free particle, you get

q(t) = q + tp/m p(t) = pwhere (q,p) are the position and momentum at time 0. What about the quantum version? Well, if we start with a wavefunction such that |psi(x)|

\int x |psi(t,x)|will move along at constant velocity. Even better, if we take the mass m to be very large, the rate at which the wavefunction spreads out will very slow. So for massive particles, our classical picture of a point particle as a mere dot is a pretty good approximation to the quantum picture, where it's described by a wavefunction.^{2}dx = <psi(t), q psi(t)>

We can see this clearly in the Heisenberg picture. We've been working in the Schroedinger picture, where we evolve states like this:

psi |-> psi(t) = exp(-itH) psiIn the Heisenberg picture, we evolve observables in time rather than states. It goes like this:

A |-> A(t) = exp(itH) A exp(-itH)If the system starts off in the state psi, the expectation value of the observable A measured at time t is

<psi(t) A, psi(t)> = <psi, A(t) psi>In the case of the free particle on the line, we have

q(t) = q + tp/m p(t) = pSee? The same formulas as in the classical case! So we get

<psi(t) q, psi(t)> = <psi, q(t) psi> = <psi, q psi> + <psi, p psi> (t/m)meaning that the expected value of the position marches along at a constant rate as time goes by.

Hmm, I bet nobody who didn't already know quantum mechanics could follow what I just said, so I'll stop here."

At this point Miguel mentioned coherent states, which are wavefunctions
that do their best at the impossible task of having a well-defined position
*and* momentum. The discussion drifted off into technicalities that
made Oz's head spin, so he very quietly crawled out from under the desk,
over to the door, and out into the hallway. Then he ran away as fast as
he could.

baez@math.ucr.edu © 2001 John Baez