But first--oh professor, I think you took too much off! You say:

This corrects a little mistake of Michael's where he claimed thatBut I explicitly said I was redefining the zero-point of energy to get rid of those pesky factors ofeCoh^{-itH}(z) =Coh(e^{-it}z).It's not

quiteso simple and nice. Hint: vacuum energy.

(Actually, the grey-haired senior scientist has already said something about this zero-point stuff.)

Okay.

A) Use the formula

to get a curiously similar formula involving an exponential of only *creation*
operators, applied to the vacuum.

Here *z*=*c*+*ib*.

I did almost this computation once before, but let's do a quick recap.

First: we want to show that Coh*(z)* is an eigenvector of the annihilation
operator *a*. For this we need to compute some commutators, and the
slick way is to notice that [*a,-*] acts like a derivative on many
operators. At least this works for power series in *a* and *a ^{*}*.
Say

Using this rule on *e ^{za*}*

Now put |0> on the right, we get

since *a* annihilates |0>. So *a *Coh*(z) = z/*sq2*
*Coh*(z)*.

Next we expand Coh*(z)* in the basis |0>, |1>* ...*.From
the eigenvalue equation, we get immediately:

where *C _{0}* is the coefficient of |0>.

We can evaluate |*C _{0}*| pretty easily. The norm
squared of Coh

So |*C _{0}*| = exp(-|

Hmmm, now for a new twist. The professor asked for the answer in terms
of *a ^{*}*. Well,

What are we going to do about that phase *iota*?

Hmmm, let's take another approach. If life was *really* simple,
we could just say that *e ^{za*}*

But skimming back over the thread, we get strong hints that

Let's ask. Oh, professor!

The Baker-Campbell-Hausdorff formula says that when [Hey, keen! How do I prove that?A,B] commutes with everythinge^{A}^{+B}=e^{-[A,B]/2}e^{A}e^{B}

You don't. You thank Baker, Campbell, and Hausdorff for proving it.Okay, thanks! (They all read the newgroups? I've seen a post from Galileo, so maybe.)

Well, *that* makes short work of this half of the problem. Let's
set:

which is a number and so commutes with everything, so

so

so the factor *iota* is 1.

Whew! Heavy firepower, just to determine that measly little phase factor
*iota*! But then, rumor has it that Gauss spent two years of Sundays
just trying to determine the sign of a certain square root.