Physics Problems & Solutions: Quantum Mechanics

Let $|n\rangle$ represent the normalized n^th energy eigenstate of the one-dimensional harmonic oscillator,

$H|n\rangle=\hbar\omega\left(n+\frac{1}{2}\right)|n\rangle$

If $|\psi\rangle$ is a normalized ensemble state that can be expanded as a linear combination

$|\psi\rangle=\frac{1}{\sqrt{14}}|1\rangle-\frac{2}{\sqrt{14}}|2\rangle+\frac{3}{\sqrt{14}}|3\rangle$

of the eigenstates, what is the expectation value of the energy operator in this ensemble state?

A. $\frac{101}{14} \hbar \omega$

B. $\frac{43}{14} \hbar \omega$

C. $\frac{23}{14} \hbar \omega$

D. $\frac{17}{\sqrt{14}} \hbar \omega$

E. $\frac{7}{\sqrt{14}} \hbar \omega$

(GR0177 #45)

Solution:

The energy eigenstates:

$\begin{aligned} H|1\rangle&=\hbar\omega\left(1+\frac{1}{2}\right)|1\rangle = \frac{3}{2}\hbar\omega|1\rangle \\H|2\rangle&=\frac{5}{2}\hbar\omega|2\rangle \\H|3\rangle&=\frac{7}{2}\hbar\omega|3\rangle \end{aligned}$

The expectation value:

$\begin{aligned} \langle H\rangle &=\langle\psi|H|\psi\rangle \\&=\frac{3}{2}\hbar\omega\left( \frac{1}{\sqrt{14}}\right )^2\langle1|1\rangle+\frac{5}{2}\hbar\omega\left(- \frac{2}{\sqrt{14}}\right )^2\langle2|2\rangle+\frac{7}{2}\hbar\omega\left( \frac{3}{\sqrt{14}}\right )^2\langle3|3\rangle \\&=\left( \frac{3}{28} +\frac{20}{28}+\frac{63}{28}\right)\hbar\omega \\&=\frac{86}{28}\hbar\omega =\frac{43}{14}\hbar\omega \end{aligned}$

Answer: B

Physics Problems & Solutions

Quantum Mechanics - Harmonic Oscillator

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