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 ${H}_{0}={\frac {p^{2}}{2m}}+{\frac {mw^{2}}{2}}$, subject to a constant electric field so that ${H}^{'}=q{\mathcal {E}}x$. Calculate the energy shift for the $n^{th}\!$ level to first and second order in $(q{\mathcal {E}})$. (Hint: Use the operators $a$ and $a^{\dagger }$ for the evaluation of the matrix elements).

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Solution:

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

$x={\sqrt {\frac {\hbar }{2m\omega }}}\left(a+a^{\dagger }\right)$

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

(b) The second-order term involves

$q^{2}{\mathcal {E}}^{2}\sum _{k\neq n}{\frac {|\langle k|x|n\rangle |^{2}}{\hbar \omega (n-k)}}={\frac {q^{2}{\mathcal {E}}^{2}}{\hbar \omega }}{\frac {\hbar }{2m\omega }}\sum _{k\neq n}{\frac {\left|\left\langle k\left|a+a^{\dagger }\right|n\right\rangle \right|^{2}}{n-k}}$

The only contributions come from $k=n-1\!$ and $k=n+1\!$, so that

$\sum _{k\neq n}{\frac {\left|\left\langle k\left|a+a^{\dagger }\right|n\right\rangle \right|^{2}}{n-k}}=\left|{\sqrt {n}}\right|^{2}-\left|{\sqrt {n+1}}\right|^{2}=-1$

and thus

$E_{n}^{(2)}=-{\frac {q^{2}{\mathcal {E}}^{2}}{2m\omega }}$

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

${\frac {1}{2}}m\omega ^{2}x^{2}+q{\mathcal {E}}x={\frac {1}{2}}m\omega ^{2}\left(x^{2}+{\frac {2q{\mathcal {E}}}{m\omega ^{2}}}x\right)={\frac {1}{2}}m\omega ^{2}\left(x+{\frac {q{\mathcal {E}}}{m\omega ^{2}}}\right)^{2}-{\frac {q^{2}{\mathcal {E}}^{2}}{2m\omega ^{2}}}$

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