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# John Baez: To quote a student of Segal...

Michael Weiss wrote:
So after 2 pi seconds, we've got minus the original state vector.
Right! Just like a spin-1/2 particle, the harmonic oscillator picks up a phase of -1 when it goes all the way around. Coincidence?

Before I answer that, let me describe another way to do the computation Michael just did, which takes advantage of the relation between the Heisenberg picture and the Schrödinger picture.

In the Heisenberg picture, operators evolve in time while states stand still. After a time t, an operator A evolves to

A(t) = eiHt A e-iHt

Measuring an observable A(t) just means ``measuring the observable A at time t''.

If we differentiate this equation with respect to t we see that operators change at a rate given by their commutator with the Hamiltonian:

(d/dt) A(t) = i [H,A(t)]      (Heisenberg)

Okay, so how does the creation operator evolve in time in our problem? I claim it evolves in time in the simplest possible way: just like the classical harmonic oscillator, it goes round and round!

a*(t) = eit a*

How do we show this? Well, it's clearly true for t = 0, so let's just check that it satisfies (Heisenberg). First note that

[H, a*] = [a*a + 1/2, a*] = [a*a, a*] = a*

Its commutator with the Hamiltonian is itself! Thus we have

i [H, a*(t)] = i [H, eit a*]

= ieit a*

= (d/dt) eit a*

= (d/dt) a*(t)

as desired.

So:

eiHt a* e-iHt = eit a*

raise both sides to the n and you get:

eiHt a*n e-iHt = (eit)n a*n

since adjacent factors of e-iHt and eiHt cancel out on the left hand side. Thus

eiHt eza*/ sq2 e-iHt = SUMn (z/sq2)n/ sqrt(n!)  (eit)n a*n = exp(zeita*/sq2)

Now replace t with -t and move a factor to the other side, getting

e-iHt eza*/ sq2 = exp(ze-ita*/sq2)e-iHt

and thus

e-iHt Coh(z) = e-iHt exp(-|z|2/4) eza*/ sq2 |0>

= exp(-|z|2/4) exp(e-itza*/sq2) e-iHt |0>

= exp(-|z|2/4) exp(e-itza*/sq2) e-it/2 |0>

=e-it/2Coh(e-itz)

The coherent state Coh(z) is the quantum analog of a particle at the point z in phase space. Our dot races around clockwise (with our sign conventions).

The moral is clear: when we quantize the harmonic oscillator, the creation operator evolves in a way that completely mimics the evolution of classical solutions. Since coherent states are built from the vacuum by hitting it with exponentiated creation operators, it's also true that coherent states evolve in a way which completely mimics the evolution of the corresponding classical solutions! Except for a phase, coming from the vacuum energy.

All this will apply to quantum field theory as well, which is why it's worthwhile going over it in such painstaking detail.

I think we are done with the harmonic oscillator! When we turn to quantum field theory, we'll find that we've already done most of the hard work.

Now, on that analogy between the harmonic oscillator and the spin-1/2 particle...to quote a student of Segal:

This funny extra 1/2 in the eigenvalues of the harmonic oscillator Hamiltonian can be thought of as the ``zero-point energy'' or ``vacuum energy'' due to the uncertainty principle. However, we've seen that the fact that it's exactly 1/2 is no coincidence! Just as you need to give a particle of half-integer spin two rotations of 360 degrees for it to get back to the way it was, with no funny phase factor of -1, so you need to let the harmonic oscillator wait two classical periods for it to get back to exactly the way it was. In the first case we are using the fact that SO(3) has a double cover SU(2) -- or more generally, SO(n) has a double cover Spin(n). In the second case we are using the fact that Sp(2) has a double cover Mp(2) -- or more generally, Sp(2n) has a double cover Mp(2n).

As I've said before, SO(n) is to fermions as Sp(2n) is to bosons. The first has to do with the canonical anticommutation relations, and Clifford algebras, while the second has to do with canonical commutation relations, and Weyl algebras. So there is a big beautiful pattern here.

Next: John Baez: A lower Up: Schmotons Previous: Michael Weiss: Vacuum energy,
Michael Weiss

3/10/1998