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 $a$ 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 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}|n\rangle = \sqrt{n+1}|n+1\rangle} we see that $\langle n|x|n\rangle =0$.

(b) The second-order term involves

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^{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 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 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

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 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

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 \frac{1}{2}m\omega^{2}x^{2}+q\mathcal{E}x=\frac{1}{2}m\omega^{2}(x^{2}+\frac{2q\mathcal{E}}{m\omega^{2}}x)=\frac{1}{2}m\omega^{2}(x+\frac{q\mathcal{E}}{m\omega^{2}})^{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 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 \frac{q^{2}\mathcal{E}^{2}}{2m\omega^{2}}} , which agrees with our second-order result.