A Modern Approach to Quantum Mechanics · Problem 7.19

Path Integrals — John S. Townsend

Physics — Quantum Mechanics · operator-exponential-identity

Problem

Prove that

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

when the operators

\hat{A}

and

\hat{B}

each commute with their commutator

[\hat{A},\hat{B}]

, that is,

[\hat{A},[\hat{B},\hat{A}]]=0

and

[\hat{B},[\hat{B},\hat{A}]]=0

. (a) First use induction to show that

[\hat{B},\hat{A}^n] = n\hat{A}^{n-1}[\hat{B},\hat{A}]

Recall from Problem 3.1 that in general

[\hat{A},\hat{B}\hat{C}] = \hat{B}[\hat{A},\hat{C}] + [\hat{A},\hat{B}]\hat{C}

(b) Use the result of (a) to show that

[\hat{B},F(\hat{A})] = F'(\hat{A})[\hat{B},\hat{A}]

where

F'(x) = dF/dx

. Suggestion: Think of

F(x)

in terms of a Taylor series expansion. (c) Define

f(\lambda) = e^{\lambda\hat{A}}e^{\lambda\hat{B}}

. Show that

\frac{df}{d\lambda} = (\hat{A}+\hat{B}+\lambda[\hat{A},\hat{B}])f(\lambda)

Finally, integrate this equation to obtain

f(\lambda) = e^{\lambda(\hat{A}+\hat{B})}e^{[\hat{A},\hat{B}]\lambda^2/2}