Phy5646/Non-degenerate Perturbation Theory - Problem 3

(Submitted by Team 1)

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

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).

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

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

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

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

The only contributions come from k = n — 1 and k = n + 1, so that \{k\A + A+\n)\2 _ |V^|2 | |V^TI|2 _ i ten n-k 1 -1 and thus 9¾2 £(2) = 2mu> The result is independent of n. We can check for its correctness by noting that the total potential energy is 1 n co 1 J ? 24% \ I i( <&> Y 92¾2 — rruo x + q%x = ■= m<o\ xr H x I = -z ma>\ x H 2 H 2 \ tm? I 2 V mco2/ 2rm>2 Thus the perturbation shifts the center of the potential by —q%/mco2 and lowers the energy by q2%2l2rmp-, which agrees with our second-order result.