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Infinite potential well/eigenfunction problem!

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a particle of mass, m inside the potential:
    V(x) = 0 for 0<x<L and infinite Otherwise

    (i) Write down the normalised ground state and first excited state energy eigenfunctions?

    (ii) Use the ground state wave function to calculate the expectation value of the momentum operator


    3. The attempt at a solution

    I did what I normally do to waves(!) ang got the ground state to be:
    root(2/L)*sin((pi*x)/L)

    and the first excited state to be:
    root(2/L)*sin((2*pi*x)/L)

    First of all are these right for part (i)?

    Then I have no idea how to do the second part so if anyone can point me in the right direction, I'll be very grateful!!:smile:
     
  2. jcsd
  3. Nov 3, 2009 #2
    You know the wave function, and thus everything there is to ever know about the system, including the momentum.

    Look for an operator [tex]\hat p[/tex] that yield the momentum when it operates on the wavefunction. Then use:

    [tex]\langle p\rangle=\int\psi_0^{*}(x)\hat p \psi_0(x)\, dx[/tex]
     
  4. Nov 3, 2009 #3
    The momentum operator is proportional to the derivative d/dx. Calculate the average of this operator. It should be zero since on average the particle goes nowhere.
     
  5. Nov 3, 2009 #4
    That all makes sense to me and I think I could do it if I knew what the 'ground state' wave function was!!

    Do I just use psi = root(2/L)*sin((pi*x)/L)
    or do i need to use e^i..... as the expectation value includes a complex conjugate?
     
  6. Nov 3, 2009 #5
    Whatever. The grounsd state (and all excited states too) wave functions can be choosen real. The complex factor is possible, like exp(iφ) but it disappears in the complex conjugate product.

    The stationary state wave function has the following general form:

    ψn(x)=√(2/L)*sin(πnx/L), with n=1, 2, 3, ...

    The ground state is n=1.
     
  7. Nov 3, 2009 #6
    Ok here goes....

    [tex]<p> = -i\hbar[/tex] [tex]\int^{\infty}_{-\infty}\sqrt{\frac{2}{L}}sin(\frac{x\pi}{L})[/tex][tex]\frac{\partial}{\partial x}\sqrt{\frac{2}{L}}sin(\frac{x\pi}{L}) dx[/tex]

    [tex]=-i\hbar[/tex] [tex]\int^{\infty}_{-\infty}\sqrt{\frac{2}{\pi}}sin(\frac{x\pi}{L})cos(\frac{x\pi}{L})dx[/tex]

    [tex]=-i\hbar\left[\frac{L}{\pi}sin^{2}(\frac{x\pi}{L})-\frac{2x\pi-Lsin(\frac{2x\pi}{L})}{4\pi}\right]^{\infty}_{-\infty}[/tex]

    Any help from here - I'm not sure what to do with the infinity's or am i supposed to use 0 and L as the limits?!

    Sorry to be a pain - this is one subject that I'm really struggling to get my head around!!
     
  8. Nov 3, 2009 #7

    gabbagabbahey

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    Isn't your wavefunctiuon zero except for [itex]0<x<L[/itex]?:wink:
     
  9. Nov 3, 2009 #8
    Yes, use 0 and L as limits. Anyway, the wave function "inside" the potential well is zero: it follows from the preponderant equation term U(x)ψ(x) = 0. The higher U, the smaller ψ.
     
  10. Nov 3, 2009 #9
    silly me!!! So the answer is: [tex]\frac{-iL\hbar}{2}[/tex]??

    Now I have to find the expectation value of the momemtum operator squared so I may be back!!!!

    Thanks for all your help so far! :smile:
     
  11. Nov 3, 2009 #10
    I promise this is the end!!!!

    I now have a value of:

    [tex]\frac{-\hbar \sqrt{1/L}\pi^{2}}{L^{\frac{5}{2}}}[/tex]

    for the expectation value of the momentum operator squared. Could this be right? It seems very long winded to me!!


    The last part of my question says that if the wave function is given by:

    [tex]\phi(x)=\frac{1}{\sqrt{10}}\phi_{1}(x)+\frac{3}{\sqrt{10}}\phi_{2}(x)[/tex]

    what is the probability of measuring the ground state energy?


    Now this seems to simple after everything else!!
    Is it just [tex](\frac{1}{\sqrt{10}})^{2}[/tex]
    = 0.1?
     
  12. Nov 3, 2009 #11

    gabbagabbahey

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    No, try again; you have made a simple error somewhere in your calculation.

    [tex]\langle p\rangle=\int_{-\infty}^{\infty}\psi_0^*(x)\left(-i\hbar\frac{d}{dx}\right)\psi_0(x)dx=-i\hbar\left(\frac{2}{L}\right)\int_0^L \sin\left(\frac{\pi x}{L}\right) \frac{d}{dx} \sin\left(\frac{\pi x}{L}\right) dx[/tex]
     
  13. Nov 3, 2009 #12
    I am afraid, not. You have to get zero. Your integrand is just a product of sine and cosine, and they are "orthogonal".
     
  14. Nov 3, 2009 #13
    I do not know. Simplify it. You have to obtain something of the right dimension. h_bar should be squared at least. I vote for (pi*h_bar/L)2.
     
  15. Nov 3, 2009 #14
    You can simply find <x> instead. you should find that to be zero. then the time derivative of zero is also zero. so <p> should also be zero
     
  16. Nov 3, 2009 #15
    Ok, thanks for that (it was just a case of multiplying instead of dividing by a fraction!!)

    Now I have: [tex]<p> = 0[/tex]

    and [tex]<p^{2}> = -\frac{\pi^{2}\hbar^{2}}{L^{2}}[/tex]



    Please let this be right before I go insane!!!
     
  17. Nov 3, 2009 #16
    Now I'll try the last part again!!

     
  18. Nov 3, 2009 #17
    It should be positive: one "mines" comes from (i)2 of momentum operator, the other comes from the cosine derivative, so the total sign is plus.
     
  19. Nov 3, 2009 #18
    Thank you everyone for your help. I don't know how I'd have got through that without people pointing out my silly little (and bigger!) mistakes.
     
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