View source for Phy5646/Non-degenerate Perturbation Theory - Problem 3

(Submitted by Team 1)

This example taken from "Quantum Physics" 3rd ed., Stephen Gasiorowicz, p. 177.

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'''Problem:'''

A charged particle in a simple harmonic oscillator, for which <math>{H}_0 = \frac{p^{2}}{2m} + \frac{mw^{2}}{2}</math>, 
subject to a constant electric field so that <math>{H}^' = q\mathcal{E} x</math>. Calculate the energy shift 
for the <math>n^{th}\!</math> level to first and second order in <math>(q\mathcal{E})</math>. 
(Hint: Use the operators <math>a</math> and <math>a^{\dagger}</math> for the evaluation of the matrix elements).

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'''Solution:'''

(a) To first order we need to calculate <math>{H}^' = q\mathcal{E}\langle n|x|n\rangle</math>. It is easy to show that <math>\langle n|x|n\rangle = 0</math>. One way is to use the relation 

:<math>
x=\sqrt{\frac{\hbar}{2m\omega}}\left(a+a^{\dagger}\right)
</math>

and since <math>A|n\rangle = \sqrt{n}|n-1\rangle</math> and <math>A^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle</math> we see that <math>\langle n|x|n\rangle = 0</math>.

(b) The second-order term involves 

:<math>
q^{2}\mathcal{E}^{2}\sum_{k\neq n}\frac{|\langle k|x|n\rangle|^{2}}{\hbar \omega (n-k)}= \frac{q^{2}\mathcal{E}^{2}}{\hbar\omega}\frac{\hbar}{2m\omega}\sum_{k\neq n} \frac{\left|\left\langle k\left|a+a^{\dagger}\right|n\right\rangle\right|^{2}}{n-k}
</math>

The only contributions come from <math>k=n-1 \!</math> and <math>k=n+1 \!</math>, so that 

:<math>
\sum_{k\neq n} \frac{\left|\left\langle k\left|a+a^{\dagger}\right|n\right\rangle\right|^{2}}{n-k}=\left|\sqrt{n}\right|^{2}-\left|\sqrt{n+1}\right|^{2}=-1
</math>

and thus 

:<math>
E_{n}^{(2)}=-\frac{q^{2}\mathcal{E}^{2}}{2m\omega}
</math>

The result is independent of <math>n\!</math>. We can check for its correctness by noting that the total potential  
energy is 

:<math>
\frac{1}{2}m\omega^{2}x^{2}+q\mathcal{E}x=\frac{1}{2}m\omega^{2}\left(x^{2}+\frac{2q\mathcal{E}}{m\omega^{2}}x\right)=\frac{1}{2}m\omega^{2}\left(x+\frac{q\mathcal{E}}{m\omega^{2}}\right)^{2}-\frac{q^{2}\mathcal{E}^{2}}{2m\omega^{2}}
</math>

Thus the perturbation shifts the center of the potential by <math> -\frac{q\mathcal{E}}{m\omega^{2}}</math> and lowers the energy by 
<math>\frac{q^{2}\mathcal{E}^{2}}{2m\omega^{2}}</math>, which agrees with our second-order result.