Paper 1, Section II, A

Principles of Quantum Mechanics
Part II, 2016

A particle in one dimension has position and momentum operators x^\hat{x} and p^\hat{p} whose eigenstates obey

xx=δ(xx),pp=δ(pp),xp=(2π)1/2eixp/\left\langle x \mid x^{\prime}\right\rangle=\delta\left(x-x^{\prime}\right), \quad\left\langle p \mid p^{\prime}\right\rangle=\delta\left(p-p^{\prime}\right), \quad\langle x \mid p\rangle=(2 \pi \hbar)^{-1 / 2} e^{i x p / \hbar}

For a state ψ|\psi\rangle, define the position-space and momentum-space wavefunctions ψ(x)\psi(x) and ψ~(p)\tilde{\psi}(p) and show how each of these can be expressed in terms of the other.

Write down the translation operator U(α)U(\alpha) and check that your expression is consistent with the property U(α)x=x+αU(\alpha)|x\rangle=|x+\alpha\rangle. For a state ψ|\psi\rangle, relate the position-space and momentum-space wavefunctions for U(α)ψU(\alpha)|\psi\rangle to ψ(x)\psi(x) and ψ~(p)\tilde{\psi}(p) respectively.

Now consider a harmonic oscillator with mass mm, frequency ω\omega, and annihilation and creation operators

a=(mω2)1/2(x^+imωp^),a=(mω2)1/2(x^imωp^)a=\left(\frac{m \omega}{2 \hbar}\right)^{1 / 2}\left(\hat{x}+\frac{i}{m \omega} \hat{p}\right), \quad a^{\dagger}=\left(\frac{m \omega}{2 \hbar}\right)^{1 / 2}\left(\hat{x}-\frac{i}{m \omega} \hat{p}\right)

Let ψn(x)\psi_{n}(x) and ψ~n(p)\tilde{\psi}_{n}(p) be the wavefunctions corresponding to the normalised energy eigenstates n|n\rangle, where n=0,1,2,n=0,1,2, \ldots.

(i) Express ψ0(xα)\psi_{0}(x-\alpha) explicitly in terms of the wavefunctions ψn(x)\psi_{n}(x).

(ii) Given that ψ~n(p)=fn(u)ψ~0(p)\tilde{\psi}_{n}(p)=f_{n}(u) \tilde{\psi}_{0}(p), where the fnf_{n} are polynomials and u=(2/mω)1/2pu=(2 / \hbar m \omega)^{1 / 2} p, show that

eiγu=eγ2/2n=0γnn!fn(u) for any real γe^{-i \gamma u}=e^{-\gamma^{2} / 2} \sum_{n=0}^{\infty} \frac{\gamma^{n}}{\sqrt{n !}} f_{n}(u) \text { for any real } \gamma

[You may quote standard results for a harmonic oscillator. You may also use, without proof, eA+B=eAeBe12[A,B]e^{A+B}=e^{A} e^{B} e^{-\frac{1}{2}[A, B]} for operators AA and BB which each commute with [A,B].]\left.[A, B] .\right]