Phy5646/Non-degenerate Perturbation Theory - Problem 3: Difference between revisions

Latest revision as of 22:13, 24 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}}\left(a+a^{\dagger}\right) }

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)}= \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 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 \!} 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

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 \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

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 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\!} . 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}\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 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\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.

@@ Line 2: / Line 2: @@
 This example taken from "Quantum Physics" 3rd ed., Stephen Gasiorowicz, p. 177.
+-----
 '''Problem:'''
 A charged particle in a simple harmonic oscillator, for which <math>{H}_0 = \frac{p^{2}}{2m} + \frac{mw^{2}}{2}</math>,
 subject to a constant electric field so that <math>{H}^' = q\mathcal{E} x</math>. Calculate the energy shift
-for the <math>n^{th}</math> level to first and second order in <math>(q\mathcal{E})</math>.
+for the <math>n^{th}\!</math> level to first and second order in <math>(q\mathcal{E})</math>.
 (Hint: Use the operators <math>a</math> and <math>a^{\dagger}</math> for the evaluation of the matrix elements).
+-----
 '''Solution:'''
-(a) To first order we need to calculate <math>{H}^' = q\mathcal{E} \langle n|x|n\rangle</math>. It is easy to show that (n[x|n) = 0. One way is to
-use the relation (7-4), from which we get
+(a) To first order we need to calculate <math>{H}^' = q\mathcal{E}\langle n|x|n\rangle</math>. It is easy to show that <math>\langle n|x|n\rangle = 0</math>. One way is to use the relation
-h (A + A+)
-mio
+:<math>
-and since A\n) = \fn\n — 1) and A+\n) = Vn + \\n + 1) we see that («|jc|«) = 0. Another way of
+x=\sqrt{\frac{\hbar}{2m\omega}}\left(a+a^{\dagger}\right)
-seeing this is that the matrix element involves the integral
+</math>
-CO CO
-J 0*(*W>„(jc) = J dxx\$n(x)\2
+and since <math>A|n\rangle = \sqrt{n}|n-1\rangle</math> and <math>A^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle</math> we see that <math>\langle n|x|n\rangle = 0</math>.
-Since |</>„(jt)|2 is always an even function of x (</>„(— x) = (— l)"<£„(jt)), the integral from — °= to +»
-involving an odd integrand is zero.
 (b) The second-order term involves
-a^y \{k\x\n)\2 _q2%2 h \(k\A+A+\n)\2
-ten hm(n — k) fuo 2mio ^n n — k
+:<math>
-The only contributions come from k = n — 1 and k = n + 1, so that
+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}
-\{k\A + A+\n)\2 _ |V^|2 | |V^TI|2 _ i
+</math>
-ten n-k 1 -1
+The only contributions come from <math>k=n-1 \!</math> and <math>k=n+1 \!</math>, so that
+:<math>
+\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
+</math>
 and thus
-¾2
-£(2) =
+:<math>
-mu>
+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
+</math>
+The result is independent of <math>n\!</math>. We can check for its correctness by noting that the total potential
 energy is
-n co 1 J ? 24% \ I i( <&> Y 92¾2
-— rruo x + q%x = ■= m<o\ xr H x I = -z ma>\ x H
+:<math>
-H 2 \ tm? I 2 V mco2/ 2rm>2
+\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 —q%/mco2 and lowers the energy by
+</math>
-q2%2l2rmp-, which agrees with our second-order result.
+Thus the perturbation shifts the center of the potential by <math> -\frac{q\mathcal{E}}{m\omega^{2}}</math> and lowers the energy by
+<math>\frac{q^{2}\mathcal{E}^{2}}{2m\omega^{2}}</math>, which agrees with our second-order result.

Phy5646/Non-degenerate Perturbation Theory - Problem 3: Difference between revisions

Latest revision as of 22:13, 24 April 2010

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