Physics Problems & Solutions: Quantum Mechanics

The raising and lowering operators for the quantum harmonic oscillator satisfy

$\begin{aligned} a^\dagger |n\rangle&= \sqrt{(n+1)}|n+1\rangle\\ a |n\rangle &= \sqrt{n}|n-1\rangle \end{aligned}$

for energy eigenstates $|n\rangle$ with energy E_n

. Which of the following gives the first-order shift in the n=2

energy level due to the perturbation

$\Delta H=V(a+a^\dagger )^2$

where

is constant?

A.

C. $\sqrt 2 V$
D. $2\sqrt 2 V$
E. 5 V

(GR0177 #94)

Solution:

The energy

$\Delta E=\langle n| \Delta H |n\rangle$

with n=2

and $\Delta H=V(a+a^\dagger )^2$

$\begin{aligned} \Delta E&=\langle 2| V(a+a^\dagger )^2 |2\rangle\\ &=V\langle 2| (aa+a^\dagger a^\dagger + aa^\dagger+a^\dagger a|2\rangle \\ &=V\langle 2|aa|2\rangle+V\langle 2|a^\dagger a^\dagger|2\rangle+V\langle 2|aa^\dagger|2\rangle+V\langle 2|a^\dagger a|2\rangle \end{aligned}$

Raising and lowering operators:

$\begin{aligned} a^\dagger |1\rangle &= \sqrt{1+1}|1+1\rangle=\sqrt{2}|2\rangle\\ a^\dagger |2\rangle &= \sqrt{2+1}|2+1\rangle=\sqrt{3}|3\rangle\\ a^\dagger |3\rangle &= \sqrt{3+1}|3+1\rangle=\sqrt{4}|4\rangle\\ \end{aligned}$

$\begin{aligned} a |1\rangle &= \sqrt{1}|1-1\rangle=\sqrt{1}|0\rangle\\ a |2\rangle &= \sqrt{2}|2-1\rangle=\sqrt{2}|1\rangle\\ a |3\rangle &= \sqrt{3}|3-1\rangle=\sqrt{3}|2\rangle\\ \end{aligned}$

$\langle 2|aa|2\rangle = \langle 2|a\sqrt{2}|1\rangle=\sqrt{2}\langle 2|a|1\rangle=\sqrt{2}\langle 2|\sqrt{1}|0\rangle= \sqrt{2}\sqrt{1}\langle 2|0\rangle=0$

$\langle 2|a^\dagger a^\dagger|2\rangle = \langle 2|a^\dagger\sqrt{3}|3\rangle=\sqrt{3}\langle 2|a^\dagger|3\rangle=\sqrt{3}\langle 2|\sqrt{4}|4\rangle= \sqrt{3}\sqrt{4}\langle 2|4\rangle=0$

$\langle 2|aa^\dagger|2\rangle = \langle 2|a\sqrt{3}|3\rangle=\sqrt{3}\langle 2|a|3\rangle=\sqrt{3}\langle 2|\sqrt{3}|2\rangle= \sqrt{3}\sqrt{3}\langle 2|2\rangle=3$

$\langle 2|a^\dagger a|2\rangle = \langle 2|a^\dagger\sqrt{2}|1\rangle=\sqrt{2}\langle 2|a^\dagger|1\rangle=\sqrt{2}\langle 2|\sqrt{2}|2\rangle= \sqrt{2}\sqrt{2}\langle 2|2\rangle=2$

$\Delta E=3V+2V=5V$

Answer: E

Physics Problems & Solutions

Quantum Mechanics - Perturbation Theory

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