| course |
course_year |
question_number |
tags |
title |
year |
Quantum Mechanics |
IB |
71 |
IB |
2015 |
Quantum Mechanics |
|
Paper 2, Section II, D |
2015 |
A quantum-mechanical harmonic oscillator has Hamiltonian
$$\hat{H}=\frac{\hat{p}^{2}}{2}+\frac{1}{2} k^{2} \hat{x}^{2}$$
where $k$ is a positive real constant. Show that $\hat{x}=x$ and $\hat{p}=-i \hbar \frac{\partial}{\partial x}$ are Hermitian operators.
The eigenfunctions of $(*)$ can be written as
$$\psi_{n}(x)=h_{n}(x \sqrt{k / \hbar}) \exp \left(-\frac{k x^{2}}{2 \hbar}\right),$$
where $h_{n}$ is a polynomial of degree $n$ with even (odd) parity for even (odd) $n$ and $n=0,1,2, \ldots$. Show that $\langle\hat{x}\rangle=\langle\hat{p}\rangle=0$ for all of the states $\psi_{n}$.
State the Heisenberg uncertainty principle and verify it for the state $\psi_{0}$ by computing $(\Delta x)$ and $(\Delta p)$. [Hint: You should properly normalise the state.]
The oscillator is in its ground state $\psi_{0}$ when the potential is suddenly changed so that $k \rightarrow 4 k$. If the wavefunction is expanded in terms of the energy eigenfunctions of the new Hamiltonian, $\phi_{n}$, what can be said about the coefficient of $\phi_{n}$ for odd $n$ ? What is the probability that the particle is in the new ground state just after the change?
[Hint: You may assume that if $I_{n}=\int_{-\infty}^{\infty} e^{-a x^{2}} x^{n} d x$ then $I_{0}=\sqrt{\frac{\pi}{a}}$ and $I_{2}=\frac{1}{2 a} \sqrt{\frac{\pi}{a}}$.]