Speaking of Baker-Campbell-Hausdorff:The Baker-Campbell-Hausdorff formula is complicated enough to set a certain lower bound on the slickness of any proof thereof. It doesn't make strong assumptions on the special nature ofisthere a slick proof of their formula, or does one just have to fight it out with Taylor series?

It comes in handy you want to define the product in a Lie group knowing the bracket in the Lie algebra: it says that the latter uniquely determines the former, and says in painstaking detail exactly how.

I've never really understood the proof; I see it before me on pages
114-120 of Varadarajan's *Lie Groups, Lie Algebras, and their Representations*,
but it looks like black magic.

However, we don't need the full-fledged version here, since in our example
[*A*,*B*] commutes with everything and all the higher terms go
away. This watered-down version is a lot easier to prove.

I did notice one thing: we can prove pretty easily thatWatch it! Someday Schur's lemma is going to get sick of all these appeals...compassion fatigue will set in, and then where will we be?e^{A}^{+B}e^{-B}e^{-A}is anumber... Next we appeal to Schur's lemma.

If you could compute that number you'd be done. But here's a closely
related approach that should also work. Assume [*A*,*B*] commutes
with *A* and *B*. Then to show

it's obviously enough to show

for all real *t*. And for this, it suffices to show that the right-hand
side satisfies the differential equation that defines the left-hand side.
In other words, we just need to check that

(If you are concerned about rigor, wave your hands and mutter the phrase ``Stone's theorem'' here.)

But this is just a straightforward computation! [To see this
computation, get
the Postscript version of these notes. --*ed*.]

Let me decode my somewhat cryptic remarks about the drawbacks of normal-ordering.
Consider *homogeneous* quadratic expressions in the classical *p*'s
and *q*'s. If we have just one *p* and one *q*, a basis
of these is given by the harmonic oscillator Hamiltonian:

the kinetic energy operator:

and the generator of scale transformations or ``dilations'':

The Poisson brackets of any two of these guys is a linear combination of
these guys, so what we've got on our hands is a little 3-dimensional Lie
algebra. Now, the group of symplectic transformations of a 2d phase space
is *SL(2, R)*, so its Lie algebra is

Exercise: work out the Poisson brackets of *H*, *S*, and *K*
and show they form a Lie algebra isomorphic to *sl(2, R)*.

Now, it turns out that if we replace the classical *p*'s
and *q*'s in these expressions by quantum *p*'s and *q*'s,
and pick the right factor ordering for *S* -- carefully, because *pq*
isn't the same as *qp* in quantum-land -- we get operators whose commutators
perfectly mimic the classical brackets. In other words, we get a representation
of *sl(2, R)*!

This is quantization like physicists always dreamt it would be: the
classical Lie algebra of symmetries is now the quantum one. Ah, were it
always so simple! However, it only works if we use the above formula for
*H*. It *doesn't* work if we use the normal-ordered version of
the harmonic oscillator Hamiltonian, where we subtract off the vacuum energy.
Then our commutation relations only mimic the classical Poisson brackets
*up to constants*.

Exercise: show that's true.

And this is a pity, because in quantum field theory we *have to*
use the normal-ordered version of the Hamiltonian. This leads to ``anomalies''
and other scary things.