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Quantum harmonic oscillator problem

  1. Jan 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Is there any way to find [tex]<\varphi_{n}(x)|x|\varphi_{m}(x)|>[/tex] (where phi_n(x) , phi_m(x) are eigenfunction of harmonic oscillator) without doing integral ?


    2. Relevant equations

    perhaps orthonormality of hermite polynomials ?

    [tex]\int^{+\infty}_{-\infty}H_{n}(x)H_{m}(x)e^{-x^{2}}dx=\delta_{nm}(Pi)^{1/2}2^{n}n![/tex]



    3. The attempt at a solution

    Actually i need to find <x> from known [tex]\psi(x,t)[/tex].
    [tex]<x>=<\psi(x,t)|x|\psi(x,t)|>[/tex]

    This gives me a lot of [tex]<\varphi_{n}(x)|x|\varphi_{m}(x)|>[/tex] terms, some of them cancel out (for m=n) ,
    but at the end I'm left with three terms thath I have to calculate ([tex]<\varphi_{1}(x)|x|\varphi_{2}(x)>,<\varphi_{2}(x)|x|\varphi_{3}(x)>, <\varphi_{3}(x)|x|\varphi_{1}(x)>[/tex]
    and that also looks like a lot of work.

    final result is <x>=0

    (This is problem from Schaums QM (supplementary prob.))
     
  2. jcsd
  3. Jan 9, 2008 #2

    malawi_glenn

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  4. Jan 9, 2008 #3
    There is a recurrence relation, which is satisfied by the Hermite polynomials

    [tex]H_{n+1}(x)=2\,x\,H_{n}(x)-2\,n\,H_{n-1}(x)\Rightarrow x\,H_n=\frac{1}{2}\left(H_{n+1}(x)+2\,n\,H_{n-1}(x)\right)[/tex]

    Thus you do not have to do the integrals, i.e.

    [tex]\int^{+\infty}_{-\infty}H_{1}(x)\,x\,H_{2}(x)e^{-x^{2}}dx=\int^{+\infty}_{-\infty}H_{1}(x)\left(\frac{H_{3}(x)+4\,H_{1}(x)}{2}\right)e^{-x^{2}}dx=2\,\int^{+\infty}_{-\infty}H_{1}(x)\,H_{1}(x)e^{-x^{2}}dx=4\,\sqrt{\pi}[/tex]

    Similar expressions hold for the other integrals.
     
  5. Jan 9, 2008 #4
  6. Jan 9, 2008 #5

    malawi_glenn

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    ok, it is very powerful and it is used very often in QM (more than just about harm osc). Send me a private message and I'll teach you if you want ;)
     
  7. Jan 9, 2008 #6
    Thank's Rainbow Child, that is really useful.
    Using this recurrence relation for hermite polynomials and [tex]<\varphi_{n}(x)|x|\varphi_{m}(x)>=<\varphi_{m}(x)|x|\varphi_{n}(x)>* [/tex]
    I was able to reduce <x> to the form:
    [tex]\frac{2}{7}(4\sqrt{\pi}A_{1}A_{2}e^{i\omega t}+4\sqrt{\pi}A_{1}A_{2}e^{-i\omega t}+24\sqrt{\pi}A_{2}A_{3}e^{-i\omega t}+24\sqrt{\pi}A_{2}A_{3}e^{i\omega t})[/tex]
    Where An's are normalization constants, but I don't see how this leads to <x>=0 ?
    Even if I expand, just Sin(wt) terms will cancel out, but not Cos(wt).
    (It is problem 5.15 from Schaum's QM)
     
    Last edited: Jan 10, 2008
  8. Jan 9, 2008 #7

    malawi_glenn

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    \pi and \cdot

    if you want to learn real QM, try ladder operators (A)
     
  9. Jan 9, 2008 #8
    If I don't solve the problem this way, I'll send you a message again.
     
  10. Jan 10, 2008 #9
    Your answer is correct (up to constants, I didn't carried out the whole calculations). The only case where [itex]<x>=0[/itex] would be if we had eigenfunctions of harmonic oscillator of the form

    [tex]\varphi_{n}(x),\varphi_{m}(x)[/tex]

    with [tex]n\neq m\pm 1[/tex]

    Only in this case the integrals

    [tex]\int_{-\infy}^{+\infty}H_n(x)\,x\,H_m(x)\,e^{-x^2}\,d\,x=\int_{-\infty}^{+\infty}H_n(x)\frac{1}{2}\left(H_{m+1}(x)+2\,m\,H_{m-1}(x)\right)\,d\,x=\frac{1}{2}\int_{-\infty}^{+\infty}H_n(x)H_{m+1}(x)\,d\,x+m\,\int_{-\infty}^{+\infty}H_n(x)\,H_{m-1}(x)\,d\,x[/tex]

    would vanish. Thus there must be a typo error in the book! :smile:
     
  11. Jan 13, 2008 #10
    I didn't get the constants right up there.
    I did the thing again and find

    [tex]<x>=\frac{2(2+\sqrt{3})}{7}Cos(\omega t)\sqrt{\frac{\hbar}{m\omega}}[/tex]


    At last I've tried also ladder operators (thanks malawi_glenn).
    [tex]x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{+})[/tex]

    then using
    [tex]a|k>=\sqrt{k}|k-1>[/tex]
    [tex]a^{+}|k>=\sqrt{k+1}|k+1>[/tex]

    arrive at
    [tex]<n|x|k>=\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{k}<n|k-1>+\sqrt{k+1}<n|k+1>)[/tex]
    where n and k are eigenfunctions of H.O.
    From here it's easy, I just use the dot product.
    There is a solved problem in Schaums QM for this.

    I got the same result for <x> using ladder op. ,that's nice :smile:.
    Ok, I'm gonna mark this thread as solved even though the solution in the book is <x>=0.
     
  12. Jan 13, 2008 #11

    malawi_glenn

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    how about this: ?

    [tex]a|n>=\sqrt{n}|n-1>[/tex]
    [tex]a^{\agger}|n>=\sqrt{n+1}|n+1>[/tex]

    [tex]<n|x|n> =\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n}<n|n-1>+\sqrt{n+1}<n|n+1>) = 0[/tex]
     
    Last edited: Jan 13, 2008
  13. Jan 13, 2008 #12
    Ok, but I have <1|x|2> and <2|x|3>, so I can't have zero, right ?
     
  14. Jan 13, 2008 #13

    malawi_glenn

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    no of course not, but the case <n|x|n> = <x> = 0, I thought you said that you did not understand why you did not get it.

    The thing you have: <n|x|k> is correct, just plug in your values for n and k and you will get the correct answer.
     
  15. Jan 13, 2008 #14
    Ok, no problem .
    As Rainbow Child said "Thus there must be a typo error in the book!" :smile:
     
  16. Jan 13, 2008 #15

    malawi_glenn

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    ok, but <x> always refer to be taken between same two states.

    So your tasks was to calculate:

    <x> and <1|x|2> and <2|x|3> ?

    What does your book give for answers?
     
  17. Jan 14, 2008 #16
    Here is the text from the book:
    The wave function of a harmonic oscillator at time t=0 is
    [tex]\psi(x,0)=\sqrt{2}A\phi_{1}+\frac{1}{\sqrt{2}}A\phi_{2}+A\phi_{3}[/tex]
    Where phi_n is stationary eigenfunction of the harmonic oscillator for the nth state,
    and A is normalization constant.
    a) compute A.
    b) find psi(x,t) for all values of t.
    c) calculate the average values of <E> at times t=0, t=Pi/w, t=2Pi/w.
    d) find expectation values <x> and <p> for t>=0.


    a) done that , got result as in the book
    b) done that, got result as in the book
    c) skipped that
    d) calculated just <x>, <x>=2(2+sqrt(3))/7 * sqrt(h/(2Pi*mw))*Cos(wt)



    Solutions from the book:
    a) [tex]A=\sqrt{2/7}[/tex]

    b) [tex]\psi(x,t)=\sqrt{2/7}(\sqrt{2}\phi_{1}e^{\frac{-3i \omega t}{2}}+\frac{1}{\sqrt{2}}\phi_{2}e^{\frac{-5i \omega t}{2}}+\phi_{3}e^{\frac{-7i \omega t}{2}})[/tex]

    d) <x>=0 , <p>=0
     
  18. Jan 14, 2008 #17

    malawi_glenn

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    Great, always post the original question in the first post so no confusion aries :)

    You now also know how to write p as a combination of a and a(dagger).

    shall see if can get the right values for <x>
     
  19. Jan 14, 2008 #18
    Yes, I should have done that :redface:
     
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