Paper 2, Section II, D

Quantum Mechanics | Part IB, 2015

A quantum-mechanical harmonic oscillator has Hamiltonian

H^=p^22+12k2x^2\hat{H}=\frac{\hat{p}^{2}}{2}+\frac{1}{2} k^{2} \hat{x}^{2}

where kk is a positive real constant. Show that x^=x\hat{x}=x and p^=ix\hat{p}=-i \hbar \frac{\partial}{\partial x} are Hermitian operators.

The eigenfunctions of ()(*) can be written as

ψn(x)=hn(xk/)exp(kx22),\psi_{n}(x)=h_{n}(x \sqrt{k / \hbar}) \exp \left(-\frac{k x^{2}}{2 \hbar}\right),

where hnh_{n} is a polynomial of degree nn with even (odd) parity for even (odd) nn and n=0,1,2,n=0,1,2, \ldots. Show that x^=p^=0\langle\hat{x}\rangle=\langle\hat{p}\rangle=0 for all of the states ψn\psi_{n}.

State the Heisenberg uncertainty principle and verify it for the state ψ0\psi_{0} by computing (Δx)(\Delta x) and (Δp)(\Delta p). [Hint: You should properly normalise the state.]

The oscillator is in its ground state ψ0\psi_{0} when the potential is suddenly changed so that k4kk \rightarrow 4 k. If the wavefunction is expanded in terms of the energy eigenfunctions of the new Hamiltonian, ϕn\phi_{n}, what can be said about the coefficient of ϕn\phi_{n} for odd nn ? What is the probability that the particle is in the new ground state just after the change?

[Hint: You may assume that if In=eax2xndxI_{n}=\int_{-\infty}^{\infty} e^{-a x^{2}} x^{n} d x then I0=πaI_{0}=\sqrt{\frac{\pi}{a}} and I2=12aπaI_{2}=\frac{1}{2 a} \sqrt{\frac{\pi}{a}}.]

Typos? Please submit corrections to this page on GitHub.