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Quantum Mechanics Positional Operator

  1. Oct 11, 2007 #1
    [tex]\Psi_{n}(x)[/tex] = [tex]\sqrt{2/L}[/tex]sin([tex]\pi[/tex]nx/L), 0 [tex]\leq[/tex]x [tex]\leq[/tex] L

    [tex]\Psi_{n}(x)[/tex] = 0, x<0, x>L

    [tex]\pi[/tex] is meant to be just normal, not superscript, sorry

    n is an integer

    show that <[tex]\hat{x}[/tex]> = L/2

    <[tex]\hat{x}[/tex]> is the expectation value of the positional operator [tex]\hat{x}[/tex] right?

    [tex]\hat{x}[/tex] [tex]\Psi[/tex]= x [tex]\Psi[/tex] i think so....

    [tex]\int[/tex]2x/L sin [tex]^{2}[/tex]([tex]\pi[/tex]2nx/L) dx

    which gets me L/2 - L[tex]^{2}[/tex]/4n[tex]^{2}[/tex] [tex]\pi[/tex][tex]^{2}[/tex]

    but it is supposed to be just L/2 sigh.....

    any help will be much appreciated :confused:
  2. jcsd
  3. Oct 11, 2007 #2


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    Hint: Put everything inside a single tex quote! As in

    [tex]\int_0^L ~\frac{2x}{L}~ \sin^{2}(\pi 2 n x / L) dx [/tex]

    This seems right (you used the limits to be 0 and L, right?)
    Can you show your steps? You must have made a mistake in the integration.
  4. Oct 11, 2007 #3
    ok here goes, i was wary of doing that because, as you can see, that was my first post and it took forever!

    sorry that last one should have read:

    [tex]\int_0^L ~\frac{2x}{L}~ \sin^{2}(\pi n x / L) dx [/tex]

    [tex]\int_0^L ~\frac{x}{L}~- ~\frac{x}{L}~ \cos(2 \pi n x / L) dx [/tex]

    integrating by parts:

    [tex]~\frac{x^{2}}{2L}-(~\frac{x}{L}~~\frac{L}{2n \pi }~\sin(~\frac{2 n \pi x}{L}~)- \int_0^L ~\frac{L}{2 n \pi L}~\sin~\frac{2n \pi x}{L}~dx )[/tex]

    [tex]\left[ ~\frac{x^{2}}{2L}-~\frac{x}{2n \pi}~\sin(~\frac{2 n \pi x}{L}~)- ~\frac{L}{4 n^{2} \pi^{2}}~\cos~\frac{2n \pi x}{L}~ \right][/tex] (between L and 0)

    = [tex]~\frac{L}{2}~ - ~\frac{L}{4 n^{2} \pi^{2}}~[/tex]

    but it should just be [tex]~\frac{L}{2}~[/tex] ?????
    Last edited: Oct 11, 2007
  5. Oct 11, 2007 #4
    you have basically worked out the answer, it's just that you forgot that the when zero is put in as a limit in the second part (the cosine part), it gives 1 aswell. i.e. [itex] cos(2 \pi n) = cos(0) [/itex], so the second part that you got disappears, leaving you with just the L/2 which is what you want.

    [tex] <x> = ~\frac{L}{2}~ - ~\frac{L}{(2 \pi n)^2}~ + ~\frac{L}{(2 \pi n)^2}~ [/tex]
    Last edited: Oct 11, 2007
  6. Oct 11, 2007 #5
    AHHHHHH ive done that so many times, still i haven't learnt, thank you very much, both of you!

    i read your blog, a few days ago id never done any quantum, they threw us in by asking for the wave function of the 'particle in a box' . It was a steep learning curve that lasted all night and several litres of coffee....got there in the end!
  7. Oct 11, 2007 #6
    That's ok. Only too glad to help. Steep learning curves are the best in the long run, take my word for it :P
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