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Finding expectation values

  1. Sep 19, 2009 #1
    I want to find <x> and ,<x^2>, <p>, and <p^2> of a particle in an infinite well where:
    [tex]V(x)=0, \frac{-a}{4}<x<\frac{3a}{4}[/tex]

    Using the usual method, I found the wavefunction to be:
    [tex]\psi(x)=\sqrt{\frac{2}{a}}sin[\frac{n\pi}{a}(x+\frac{a}{4})][/tex]

    I also found:
    [tex]<x>=\frac{2}{a}\int_{\frac{-a}{4}}^{\frac{3a}{4}}xsin^2[\frac{n\pi}{a}(x+\frac{a}{4})]dx=\frac{a}{4}[/tex]

    <x^2> is where I encounter the 1st problem. Here's my attempt:
    [tex]<x^2>=\frac{2}{a}\int_{\frac{-a}{4}}^{\frac{3a}{4}}x^2sin^2[\frac{n\pi}{a}(x+\frac{a}{4})]dx[/tex]
    Using by parts I let:
    [tex]u=x^2[/tex]
    [tex] du=2xdx[/tex]
    [tex]dv=sin^2[\frac{n\pi}{a}(x+\frac{a}{4})]dx[/tex]
    [tex]v=\frac{x}{2}-\frac{a}{2n\pi}sin[\frac{n\pi}{a}(x+\frac{a}{4})]cos[\frac{n\pi}{a}(x+\frac{a}{4})]=\frac{x}{2}-\frac{a}{4n\pi}sin[\frac{2n\pi}{a}(x+\frac{a}{4})][/tex]
    (double angle formula used)

    [tex]<x^2>=uv-\int vdu=\frac{2}{a}\left [ \left ((x^2)(\frac{x}{2}-\frac{a}{4n\pi}sin[\frac{2n\pi}{a}(x+\frac{a}{4})]) \right )_{\frac{-a}{4}}^{\frac{3a}{4}} -\int_{\frac{-a}{4}}^{\frac{3a}{4}}2x(\frac{x}{2}-\frac{a}{4n\pi}sin[\frac{2n\pi}{a}(x+\frac{a}{4})])dx\right ][/tex]
    [tex]=\frac{2}{a}\left [ \frac{27a^3}{128}-\frac{-a^3}{128}-\int_{\frac{-a}{4}}^{\frac{3a}{4}}x^3dx+\frac{a}{4n\pi}\int_{\frac{-a}{4}}^{\frac{3a}{4}}2xsin[\frac{2n\pi}{a}(x+\frac{a}{4})])dx\right ][/tex]
    [tex]=\frac{2}{a}\left [ \frac{7a^3}{32}-\frac{5a^4}{64}+\frac{a}{2n\pi}\left ( \frac{a^2}{4n^2\pi^2}sin[\frac{2n\pi}{a}(x+\frac{a}{4})])-\frac{a}{2n\pi}xcos[\frac{2n\pi}{a}(x+\frac{a}{4})])\right )_{\frac{-a}{4}}^{\frac{3a}{4}}\right ][/tex]
    Sine term vanishes at both limits while cosine term as 1 at both limits:
    [tex]=\frac{2}{a}\left [ \frac{7a^3}{32}-\frac{5a^4}{64}+\frac{a}{2n\pi}\left ( \frac{-3a^2}{8n\pi}+\frac{a^2}{8n\pi} \right )\right ]=\frac{7a^2}{16}-\frac{a^2}{4n^2\pi^2}-\frac{5a^3}{32}=\frac{a^2(7n^2\pi^2-4)}{16n^2\pi^2}-\frac{5a^3}{32}[/tex]

    This is obviously wrong since the units don't add up/ How can there be a^2 and a^3 at the same time? I spend hours checking and still can't find my mistake.

    I plug the equation into a graphic calculator, I think the answer should be:
    [tex]\frac{a^2(7n^2\pi^2-24)}{64n^2\pi^2}[/tex]


    Now, for <p>, can I just have:
    [tex]<p>=m\frac{d<x>}{dt}=m\frac{d}{dt}\frac{a}{4}=0[/tex]
    Is this alright or do I have to plug the operator in and integrate?
     
  2. jcsd
  3. Sep 20, 2009 #2
    I think I spot your mistake: from the line in <x^2>=.... in the integral you pass from
    [itex]\int_{\frac{-a}{4}}^{\frac{3a}{4}}2x(\frac{x}{2})dx[/itex] in the firs line to [itex]\int_{\frac{-a}{4}}^{\frac{3a}{4}}x^3dx[/itex] in the second one
     
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