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John Baez: Messing around with the vacuum energy, eh?

Michael Weiss complained:
Oh professor, I think you took too much off!
Oh, sorry. So you replaced the Hamiltonian

(p2 + q2)/2 = (a a* + a* a)/2

by its normal-ordered form, where all annihilation operators are pushed to the right:

p2 + q2 - 1/2 = a* a

Somehow I hadn't noticed that. This will come in handy in full-fledged quantum field theory, but it's not necessary here, and it's sort of enlightening not to do it.

Hmm, maybe you're saying the formulas are trying to tell me something-- that I shouldn't monkey around with the zero-point?
Messing around with the vacuum energy, eh? You may unleash powerful forces -- forces that mankind was never meant to meddle with! For example, you have a perfectly nice representation of the Lie algebra of the symplectic group on your hands; getting rid of the vacuum energy will turn it into a nasty projective representation. But never mind, go ahead, just don't blame me for what happens...

In simpler terms: there are lots of interesting classical observables built using quadratic expressions in the p's and q's, of which the Hamiltonian is one. When we replace the classical p's and q's by operators, we'd like Poisson brackets to go over to commutators. If we try to do this for general polynomials in the p's and q's, it doesn't work very well. However, for quadratic expressions in the p's and q's it does, if we don't mess with them by normal-ordering.

As for the rest of your post...

Great. You got a very nice formula:

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

implying

Coh(z) = exp(-|z|2/4) SUMn (z/sq2)n/sqrt(n!) |n>

which is a very precise way of stating what you noted quite a while ago: the number of quanta in a coherent state is given by a Poisson distribution.

But now let's see what happens if we evolve our coherent state in time. We'll see something nice, a cute relation between the harmonic oscillator and the spin-1/2 particle, which we discussed once upon a time...


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Michael Weiss

3/10/1998