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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n^{th}} level to first and second order in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (q\mathcal{E})} . (Hint: Use the operators Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a^{\dagger}} for the evaluation of the matrix elements).

Solution: (a) To first order we need to calculate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {H}^' = q\mathcal{E}\langle n|x|n\rangle} . It is easy to show that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n|x|n\rangle = 0} . One way is to use the relation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

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

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

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

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 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.