next up previous
Next:  In Boston... Up: Schmotons Previous: John Baez: To quote

John Baez: A lower bound on slickness

Michael Weiss wrote:
Speaking of Baker-Campbell-Hausdorff: is there a slick proof of their formula, or does one just have to fight it out with Taylor series?
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 of A and B, and it says [see the postscript version of these notes. --ed.]

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 that eA+Be-Be-A is a number ... Next we appeal to Schur's lemma.
Watch 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?

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

eA+B = eAeB e-[A,B]/2

it's obviously enough to show

et(A+B) = etAetBe-[tA,tB]/2

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

(d/dt) etAetBe-[tA,tB]/2 = (A+B) etAetBe-[tA,tB]/2

(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:

H = (p2 + q2)/2

the kinetic energy operator:

K = p2/2

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

S = qp

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 sl(2,R), which is a 3-dimensional Lie algebra. So it's natural to guess that we've got sl(2,R) on our hands.

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.

next up previous
Next: In Boston... Up: Schmotons Previous: John Baez: To quote 
Michael Weiss