Quantum Gravity Seminar

Week 11, Track 2

John Baez

January 8, 2001

"So, now the track 1 students should leave...."

"I think I'll stay" said Alissa.

"Great!" said the Wizard. "Joining Track 2, eh? Okay, now I can let you in on a secret... everything you learnt in Track 1 was false."

He chuckled. "Just kidding. Actually, you're joining at a good time, because last quarter we studied the Lagrangian approach to classical field theories, but now we're switching gears: we're going to quantize some of theories. And we'll start with an introduction to quantization.

In Track 1 we just covered some basic principles of quantum theory. Now let's bring time into the game. In ordinary quantum mechanics, a crucial assumption is that moments of time correspond to real numbers, and that we have a clock unaffected by the system we're studying, which we can use to measure this real number at any moment. One also assumes that the laws governing our system don't change with the passage of time. Another assumption is that states of our system are described by unit vectors in a single fixed Hilbert space H. These assumptions are idealizations, and they may break down when we get to quantum gravity -- but for now, let's see what we can get from them.

First, we expect that for each real number t there's a linear operator

U(t): H -> H

which describes how states change as time passes: if we start with any state vector x, after time t it becomes the state vector U(t)x. Since t can be negative, we can ask what the state was as well as what it will be.

Second, we expect that waiting no time at all does nothing at all:

U(0) = 1H

Third, we expect that waiting a time t and then waiting a time t' is the same as waiting a time t+t':

U(t')U(t) = U(t+t')

We can already conclude some interesting stuff from these assumptions! The operator U(-t) must be the inverse of U(t):

U(t)U(-t) = U(-t)U(t) = 1H.

Since the operator U(t) maps state vectors to state vectors, it must satisfy

||x|| = 1 implies ||U(t)x|| = 1

which by linearity implies

||U(t)x|| = ||x||

for all vectors in H, not just unit vectors. From this, we can use a trick called "polarization" to get

<U(t)x, U(t)y> = <x,y>

for all vectors x and y. By the definition of adjoint, this means

U(t)* U(t) = 1H

Now if H is finite-dimensional this says U(t) is unitary! If H is infinite-dimensional, we also need to use the fact that U(t) is invertible. Luckily we saw this is true. Multiplying both sides on the right by the inverse of U(t) we get

U(t)* = U(t)-1

and then multiplying on the left by U(t) we get

U(t) U(t)* = 1H

Now that we know

U(t)* U(t) = U(t) U(t)* = 1H,

we know U(t) is unitary.

So far, we can summarize everything we know about these operators U(t) by saying they form a "1-parameter unitary group". In other words: U is a homomorphism from the real numbers to the unitary operators on our Hilbert space. Or in still other words: U is a unitary representation of R on our Hilbert space. These all mean the same thing. The more ways you can say it, the smarter you'll feel.

Can anyone think of yet another way to say the same thing? I just taught you about categories. How can you show off your newfound knowledge?"

The Acolyte Richard, who had been studying categories on the side, said: "A group is a category with one object and with all morphisms being invertible, so we can think of the real numbers R as a category. Then U is a just a functor from R to Hilb!"

"Right! And this puts TQFTs into proper perspective. Time evolution in ordinary quantum mechanics is given by a functor from R to Hilb, while time evolution in topological quantum field theory is given by a functor from nCob to Hilb. So TQFTs amount to replacing the "time line" -- the real numbers -- by the category of all n-dimensional spacetimes!"

Here the Wiz let out a gale of crazed laughter and danced a little jig. Some of the class nodded sagely; others looked a bit scared.

"So far, so good. But in quantum mechanics, people also throw in an assumption of a different kind. They assume for any state x, U(t)x varies continuously with t. This comes from our experience that the world doesn't change discontinously with time. Perhaps more importantly, it lets us use a big theorem to say what U(t) is like!

First, some more jargon: if we have a 1-parameter unitary group U where U(t)x is continuous as a function of t, we call it "strongly continuous" - since analysts know about millions of other flavors of continuity.

Okay, here's the theorem:

Stone's Theorem - If U is a strongly continuous 1-parameter unitary group, then there exists a unique self-adjoint operator H such that

U(t) = exp(-itH)

for all t. Conversely, if H is self-adjoint, exp(-itH) defines a strongly continuous 1-parameter unitary group.

At this point Toby smiled wickedly and interjected, "Hey! You already used H to stand for the Hilbert space. Now you're using it for a self-adjoint operator on that Hilbert space."

The wizard stared aghast at the blackboard. "Damn! Damn! Everyone uses H for this operator, because it's called the Hamiltonian. How did Hilbert and Hamilton get the dumb idea of starting their last names with the same letter? We could use a fancy script H for the Hilbert space, but I hate that. We could call the Hamiltonian "E", since that what it really is: it's the energy..."

He turned aside to the rest of the class and explained: "You see, in quantum theory self-adjoint operators correspond to observables. Now we've seen that starting from some plausible assumptions about the nature of time, every quantum-mechanical system comes equipped with a god-given observable! This observable turns out to be none other than the energy! We only call it the "Hamiltonian" to make it sound impressive."

Turning back to the blackboard, he said "Ugh! I guess I'll use the letter "A" for the Hamiltonian..." With a wave of his wand, all the H's turned into A's.

"Anyway, Stone's theorem is great, but it's not use if you don't know what "exp(-itA)" means. If our Hilbert space is finite-dimensional, we can define this using the usual power series for the exponential:

exp(-itA) = sum (-itA)n / n!
If our Hilbert space is infinite-dimensional, things get trickier: our self-adjoint operator A can be unbounded, which implies it's only defined on a dense subspace, and so on.... but it all works out. You gotta take an analysis class to learn this stuff; I'm not gonna explain it here.

Okay, now for the final punchline. Suppose we start with a state vector x, and let x(t) be what this state vector becomes after time t. In other words:

x(t) = U(t)x

Then if x(t) is differentiable as a function of t, as is always the case when our Hilbert space is finite dimensional, we get

dx(t)/dt = d/dt (exp(-itA)x)
         = -iA exp(-itA)x 
         = -iA x(t)
This is called Schroedinger's equation! People usually write the state as "psi" rather than x(t), and they usually write the Hamiltonian as "H" instead of A, so it looks like this:

d psi /dt = -iH psi

Often people start with this formula when they do quantum mechanics, but here I wanted to derive it from some assumptions, so we can see where it comes from. Doing ordinary quantum mechanics is like riding a big fat Cadillac -- but we need to look under the hood and figure out what all the parts do, so we can build our own car.

Next time we'll look at some examples. Class dismissed!"

baez@math.ucr.edu © 2001 John Baez