Quantum Mechanics | Part IB, 2002

Consider a quantum mechanical particle moving in an upside-down harmonic oscillator potential. Its wavefunction Ψ(x,t)\Psi(x, t) evolves according to the time-dependent Schrödinger equation,

iΨt=222Ψx212x2Ψi \hbar \frac{\partial \Psi}{\partial t}=-\frac{\hbar^{2}}{2} \frac{\partial^{2} \Psi}{\partial x^{2}}-\frac{1}{2} x^{2} \Psi

(a) Verify that

Ψ(x,t)=A(t)eB(t)x2\Psi(x, t)=A(t) e^{-B(t) x^{2}}

is a solution of equation (1), provided that

dAdt=iAB\frac{d A}{d t}=-i \hbar A B


dBdt=i22iB2\frac{d B}{d t}=-\frac{i}{2 \hbar}-2 i \hbar B^{2}

(b) Verify that B=12tan(ϕit)B=\frac{1}{2 \hbar} \tan (\phi-i t) provides a solution to (3), where ϕ\phi is an arbitrary real constant.

(c) The expectation value of an operator O\mathcal{O} at time tt is

O(t)dxΨ(x,t)OΨ(x,t),\langle\mathcal{O}\rangle(t) \equiv \int_{-\infty}^{\infty} d x \Psi^{*}(x, t) \mathcal{O} \Psi(x, t),

where Ψ(x,t)\Psi(x, t) is the normalised wave function. Show that for Ψ(x,t)\Psi(x, t) given by (2),

x2=14Re(B),p2=42B2x2\left\langle x^{2}\right\rangle=\frac{1}{4 \operatorname{Re}(B)}, \quad\left\langle p^{2}\right\rangle=4 \hbar^{2}|B|^{2}\left\langle x^{2}\right\rangle

Hence show that as tt \rightarrow \infty,

x2p24sin2ϕe2t\left\langle x^{2}\right\rangle \approx\left\langle p^{2}\right\rangle \approx \frac{\hbar}{4 \sin 2 \phi} e^{2 t}

[Hint: You may use

dxeCx2x2dxeCx2=12C\frac{\int_{-\infty}^{\infty} d x e^{-C x^{2}} x^{2}}{\int_{-\infty}^{\infty} d x e^{-C x^{2}}}=\frac{1}{2 C}

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