Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 32D (Principles of Quantum Mechanics)

Write $D=[A,B]$, so $[A,D]=[B,D]=0$. Repeated use of the product rule for the commutator gives $[A,B^n]=nD{B}^{n-1}$. Summing the exponential series yieldsThese manipulations hold for bounded operators, or as identities on a common invariant domain where the series and derivatives are justified.

The same commutation relation implies $e^{\lambda A}B{e}^{-\lambda A}=B+\lambda D$: its derivative is $e^{\lambda A}[A,B]e^{-\lambda A}=D$ and its initial value is $B$. Put $G(\lambda )=e^{\lambda A}{e}^{\lambda B}$. ThenFor the specified $F(\lambda )=G(\lambda )e^{-\lambda (A+B)}$, the operator $G$ commutes with $A+B$. Indeed conjugation by $e^{\lambda B}$ sends $A+B$ to $A+B-\lambda D$, and subsequent conjugation by $e^{\lambda A}$ restores $A+B$. Thereforeso $F(\lambda )=e^{{\lambda }^{2}D/2}$. At $\lambda =1$ this proves the central-commutator case of the Baker--Campbell--Hausdorff formula:The last equality holds because $D$ commutes with $A+B$.

The spatial translation operator is $U(\alpha )=e^{-i\alpha \hat{p}/\hslash }$. From $[\hat{x},\hat{p}]=i\hslash$ and the preceding central-commutator result,Thus $\hat{x}U|x⟩=U\hat{x}|x⟩+\alpha U|x⟩=(x+\alpha )U|x⟩$, consistent with $U|x⟩=|x+\alpha ⟩$. Since $⟨x|U=⟨x-\alpha |$, the translated wavefunction is $\boxed{\psi (x-\alpha )}$.

For the quantum harmonic oscillator, the creation and annihilation operators give $\hat{p}=i\sqrt{m\hslash \omega /2}(a^{†}-a)$. Put $s=\alpha \sqrt{m\omega /(2\hslash )}$. The central commutator of $s{a}^{†}$ and $-sa$ is $s^2$, soAs $a|0⟩=0$ and $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$,Taking position wavefunctions provesThe coefficient squares sum to $e^{-s^2}{\sum }_n{s}^{2n}/n!=1$, checking normalization of this coherent state.