Introduction to Quantum Mechanics · Problem 3.29

Formalism — David J. Griffiths

Physics — Quantum Mechanics · baker_campbell_hausdorff

Problem

Consider operators A^\hat{A} and B^\hat{B} that do not commute with each other (C^=[A^,B^])\left(\hat{C} = \left[\hat{A}, \hat{B}\right]\right) but do commute with their commutator: [A^,C^]=[B^,C^]=0\left[\hat{A}, \hat{C}\right] = \left[\hat{B}, \hat{C}\right] = 0 (for instance, x^\hat{x} and p^\hat{p}). (a) Show that [A^n,B^]=nA^n−1C^.\left[\hat{A}^n, \hat{B}\right] = n\hat{A}^{n-1}\hat{C}. *Hint:* You can prove this by induction on nn, using Equation 3.65. (b) Show that [eλA^,B^]=λeλA^C^,\left[e^{\lambda \hat{A}}, \hat{B}\right] = \lambda e^{\lambda \hat{A}} \hat{C}, where λ\lambda is any complex number. *Hint:* Express eλA^e^{\lambda \hat{A}} as a power series. (c) Derive the Baker–Campbell–Hausdorff formula: eA^+B^=eA^eB^e−C^/2.e^{\hat{A}+\hat{B}} = e^{\hat{A}} e^{\hat{B}} e^{-\hat{C}/2}. *Hint:* Define the functions f^(λ)=eλ(A^+B^),g^(λ)=eλA^eλB^e−λ2C^/2.\hat{f}(\lambda) = e^{\lambda\left(\hat{A}+\hat{B}\right)}, \quad \hat{g}(\lambda) = e^{\lambda \hat{A}} e^{\lambda \hat{B}} e^{-\lambda^2 \hat{C}/2}. Note that these functions are equal at λ=0\lambda = 0, and show that they satisfy the same differential equation: df^/dλ=(A^+B^)f^d\hat{f}/d\lambda = \left(\hat{A} + \hat{B}\right)\hat{f} and dg^/dλ=(A^+B^)g^d\hat{g}/d\lambda = \left(\hat{A} + \hat{B}\right)\hat{g}. Therefore, the functions are themselves equal for all λ\lambda.
Intuitive explanation

The key to this problem is recognizing that non-commuting operators can still be controlled if they satisfy special commutation relations with their own commutator. You'll build up from a simple power rule for commutators (part a), extend it to exponentials using power series (part b), and finally use a clever differential equation argument to show that exponentials of non-commuting operators nearly factorize—with a correction term capturing their non-commutativity. This is the deep structure behind why eA^+B^≠eA^eB^e^{\hat{A}+\hat{B}} \neq e^{\hat{A}}e^{\hat{B}} in general.

Intuitive explanation

The key to this problem is recognizing that non-commuting operators can still be controlled if they satisfy special commutation relations with their own commutator. You'll build up from a simple power rule for commutators (part a), extend it to exponentials using power series (part b), and finally use a clever differential equation argument to show that exponentials of non-commuting operators nearly factorize—with a correction term capturing their non-commutativity. This is the deep structure behind why eA^+B^≠eA^eB^e^{\hat{A}+\hat{B}} \neq e^{\hat{A}}e^{\hat{B}} in general.

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