Let gooby do teh hoemwerk: Baker-Campbell-Hausdorff identities


Let A and B be linear operators, which commute with the commutator of A and B, i.e.

\bigl[A,[A,B]\bigr]= \bigl[B,[A,B]\bigr]=0\,.

Show that the following formulae hold:

e^A e^B = e^B e^A e^{[A,B]}\,,\\  e^{A+B} = e^A e^B e^{-[A,B]/2}\,,

which are known as the Baker-Campbell-Hausdorff identities. They play an important rule, e.g. in quantum mechanics, whenever one deals with exponentiation of the position and the momentum operator.


We will start by proving the first equation. To this end we introduce an auxiliary operator O_1(\lambda), which depends on a real parameter \lambda:

O_1(\lambda) = e^{\lambda B}e^Ae^{[A,\lambda B]}e^{-\lambda B}e^{-A}\,.

We will show, that O_1(\lambda)= 1, which will prove the claim. Clearly this statement is true for \lambda=0. The derivative of O_1(\lambda) with respect to \lambda is

\frac{d}{d\lambda}O_1(\lambda) = e^{\lambda B}\Bigl(Be^A e^{[A,\lambda B]}e^{-\lambda B}+e^A[A,B]e^{[A,\lambda,B]}e^{-\lambda B}-e^Ae^{[A,\lambda B]}Be^{-\lambda B}\Bigr)e^{-A}\notag\\  \phantom{\frac{d}{d\lambda}O_1(\lambda)} = BO_1(\lambda)+[A,B]O_1(\lambda)-e^{\lambda B}e^Ae^{[A,\lambda B]}Be^{-\lambda B}e^{-A}\notag\\  \phantom{\frac{d}{d\lambda}O_1(\lambda)} =[A,B]O_1(\lambda)-e^{\lambda B}[e^A,B]e^{[A,\lambda B]}e^{-\lambda B}e^{-A}

It is easy to see by induction, that under the conditions imposed on A and B,

[A^n,B]= n[A,B]A^{n-1}\,,
which implies
[e^A,B]= [A,B]e^A

and thus \frac{d}{d\lambda}O_1(\lambda)=0 and consequently validates the first identity.

In order to prove the second equation, we will do a similar trick and define

O_2(\lambda)=e^{-\lambda(A+B)}e^{\lambda A}e^{\lambda B}e^{-\lambda^2[A,B]/2}\,.

Obviously, O_2(0)= 1, and the derivative

\frac{d}{d\lambda}O_2(\lambda)= -(A+B)O_2(\lambda)+e^{-\lambda(A+B)}Ae^{\lambda A}e^{\lambda B}e^{-\lambda^2[A,B]/2}\notag\\  \phantom{\frac{d}{d\lambda}O_2(\lambda)=}+e^{-\lambda(A+B)}e^{\lambda A}Be^{\lambda B}e^{-\lambda^2[A,B]/2}-\lambda[A,B]O_2(\lambda)\notag\\  \phantom{\frac{d}{d\lambda}O_2(\lambda)}=-BO_2(\lambda)+[e^{-\lambda(A+B)},A]e^{\lambda A}e^{\lambda B}e^{-\lambda^2[A,B]/2}+e^{-\lambda(A+B)}Be^{\lambda A}e^{\lambda B}e^{-\lambda^2[A,B]/2}\notag\\  \phantom{\frac{d}{d\lambda}O_2(\lambda)}=0\,.

In the second step I have interchanged B with e^{\lambda A} in the third term. The commutation rest cancels the fourth term.

This completes the proof.


About goobypl5

pizza baker, autodidact, particle physicist
This entry was posted in Math and tagged , , . Bookmark the permalink.

One Response to Let gooby do teh hoemwerk: Baker-Campbell-Hausdorff identities

  1. I remember I had to prove this theorem in a “Quantum Mechanics II” course homework exercise a few years ago, and it took forever to find clues for the solution online, until I found out some page where this was proven, but it was called “Glauber’s formula” there.

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