John Baez: To quote  Schmotons  Michael Weiss: Dangerous signs

Michael Weiss: Vacuum energy, source of all of humanity's future energy needs

OK, time evolution. We have:

Coh(z) = exp(-|z|²/4) e^{za*/ sq2} |0>

and

H = (aa^* + a^*a)/2 = a^*a + 1/2

including the vacuum energy this time, source of all of humanity's future energy needs. (Right.)

We want to know how Coh(z) evolves (in the Schrödinger picture, since it doesn't evolve in the Heisenberg picture). I.e., we want to compute

e^-iHt Coh(z)

Now it would be nice if we could just plug in the formulas we just obtained. Alas, we'd need to compute e^a*a e^za* (give or take an annoying factor), and the Baker-Campbell-Hausdorff formula we have only helps with e^Ae^B when [A,B] commutes with A and B. Is that the case here?

[a^*a, a^*] = [a^*,a^*]a + a^*[a,a^*] = 0 + a^* = a^*.

No such luck, the commutator does not commute with a^*a.

Probably we could do something slick anyway, but at least the dull way is quick.

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

e^-iHt |n> = e^-it/2 e^-int |n>

so

e^-iHt Coh(z) =  exp(-|z|²/4) e^-it/2 SUM_n (z/sq2)ⁿ/ sqrt(n!) e^-int |n>

but since e^-int = (e^-it)ⁿ, we can fold this into the (z/sq2)ⁿ factor and get:

e^-iHt Coh(z) = e^-it/2 Coh(ze^-it)

So after 2 pi seconds, we've got minus the original state vector.

3/10/1998