Phy5646/Non-degenerate Perturbation Theory - Problem 3: Difference between revisions
No edit summary |
No edit summary |
||
Line 21: | Line 21: | ||
(b) The second-order term involves | (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)} | |||
</math> | |||
The only contributions come from k = n — 1 and k = n + 1, so that | 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 | \{k\A + A+\n)\2 _ |V^|2 | |V^TI|2 _ i |
Revision as of 07:02, 3 April 2010
(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 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}_0 = \frac{p^{2}}{2m} + \frac{mw^{2}}{2}} , subject to a constant electric field so 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 {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 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|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 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}
.
(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)} }
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.