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Michael Weiss: Okay, thanks, Baker, Campbell, and Hausdorff!

Just stopping by for a sec, to drop off some homework. I'll be by again later for a longer chat.

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

This corrects a little mistake of Michael's where he claimed that

e-itH Coh(z) = Coh(e-it z).

It's not quite so simple and nice. Hint: vacuum energy.

But I explicitly said I was redefining the zero-point of energy to get rid of those pesky factors of e-it/2. Hmm, maybe you're saying the formulas are trying to tell me something-- that I shouldn't monkey around with the zero-point? (Ominous music wells up on the soundtrack). (Flash to the final scene in a 50s sci-fi movie, as the grey-haired senior scientist portentously intones, ``There are aspects of Nature that we change at our peril. Let this be a Lesson To Us All...'')

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


A) Use the formula

Coh(z) = e(za* - z*a)/ sq2 |0>

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 f(a,a*) is a power series with complex coefficients. Since [a,a]=0 and [a,a*]=1, we'll get the right result for [a,f(a,a*)] with this recipe: compute (d/dx)f(a,x) formally, treating a like a constant; then replace x with a* in the final result.

Using this rule on eza* - z*a, we get

[a, e(za* - z*a)/ sq2] = [a, (za*-z*a)/sq2] = (z/sq2) e(za* - z*a)/ sq2

Now put |0> on the right, we get

a Coh(z) = a e(za*-z*a)/ sq2 |0> = (z/ sq2) e(za*-z*a)/ sq2 |0>

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

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

Coh(z) = C0 (|0> + (z/sq2 |1> + (z/sq2)n sqrt(n!) |n> + ...)

where C0 is the coefficient of |0>.

 We can evaluate |C0| pretty easily. The norm squared of Coh(z)is |C0|2 exp(|z|2/2), from the formula we just got. But Coh(z) has norm 1. How do I know that? Well, e(za*-z*a)/ sq2 is unitary. How do I know that? Well, (za*-z*a)/ sq2 is i times a self-adjoint operator (just take the adjoint and see what you get), so by some theorem or other its exponential is unitary.

So |C0| =  exp(-|z|2/4).  So we've determined e(za*-z*a)/sq2 |0> up to a phase (let's call the phase iota):

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

Hmmm, now for a new twist. The professor asked for the answer in terms of a*. Well, a*n |0> = sqrt(n!) |n> --hey, this works out nicely:

e(za*-z*a)/ sq2 |0> = iota exp(-|z|2/4) SUMn  (z/sq2)n /sqrt(n!) a*n |0>

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

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 eza*-z*a = eza*e-z*a (it isn't), and since a annihilates |0>, e-z*a|0> = |0> (just expand out e-z*a in a Taylor series). So we'd have:

eza*-z*a|0> = eza*|0>      (NOT!!)

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

eza*-z*a = enumber eza* e-z*a

Let's ask. Oh, professor!

The Baker-Campbell-Hausdorff formula says that when [A,B] commutes with everything

eA+B = e-[A,B]/2 eAeB
Hey, keen! How do I prove that?
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:

A = za*/ sq2

B = -z*a/ sq2

[A,B] = -(1/2) zz* [a*,a] = zz*/2

which is a number and so commutes with everything, so

eA+B = e-zz*/4 eAeB


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

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.

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Michael Weiss