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Infinite square well

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data
    I am calculating [tex]p_{nm} = <\psi_n|p|\psi_m>[/tex] where [tex]\psi_n = \sqrt{2/a}\sin(n\pi x/2)[/tex]. This is for the infinite square well from 0 to a.

    I think I am messing up: I get that p_{nm} = 0 when n,m are both odd, even and something nonzero otherwise.

    If I am not messing up I have some follow-up questions, but does anyone else get different results?

    EDIT: the wavefunction should be [tex]\psi_n = \sqrt{2/a}\sin(n\pi x/a)[/tex]

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Dec 13, 2007
  2. jcsd
  3. Dec 12, 2007 #2
    I believe that is correct. Perhaps it is easier to see why this occurs if you displace the whole well to -a/2 to a/2 - the situation is physically the same but you can immediately see a difference between the odd and even states.
     
  4. Dec 12, 2007 #3
    H is the Hamiltonian.
    Now, I am trying to show that p^2/(2M) = H which in matrix representation is:

    [tex]1/(2M)(p^2)_{nm} = H_{nm} [/tex]

    The Hamiltonian is diagonal. So how would I write out (p^2)_{nm} with delta's in such a way that I can show that the above expression is true. The rhs is equal to

    prefactor times [tex] \delta_{nm}[/tex]

    so the lhs must be equal to that also somehow.
     
  5. Dec 12, 2007 #4
    Ah I didn't notice before but your wavefunction is slightly wrong (can you see how? - it might just be a typo)

    The lhs is diagonal - check it explicitly! (It's not hard!) You're guaranteed it is equal to the right hand side in side the well (that is, in fact, how you would derive H - from p^2/(2m)).
     
  6. Dec 12, 2007 #5
    Just from memory, shouldn't <p> always be zero in the infinite square well, regardless of your choice of n,m.

    Classically this obviously makes sense, since setting the particle in motion in a 1 dimensional box, you find that it'll be going some velocity one way, just as often as that velocity in the opposite direction, which obviously averages over time to be zero.

    Quantum mechanically, I believe this still holds very true.
     
  7. Dec 13, 2007 #6
    [tex] <\psi_n| p |\psi_n >[/tex] is always zero but [tex]<\psi_n| p |\psi_m >[/tex] is not always zero. Indeed n= m is just a special case of n and m having the same parity.
     
  8. Dec 13, 2007 #7
    Gotcha.. it's been awhile since I've touched on specifics of the square well.
     
  9. Dec 13, 2007 #8
    Maybe I am just missing something obvious, but I cannot even figure out how to write LHS down.

    I get that p_{nm} = i hbar 2n/(n^2-m^2) when n and m have opposite parity. I just don't see how to square p_{nm}.
     
  10. Dec 13, 2007 #9
    anyone?
     
  11. Dec 13, 2007 #10
    There are two ways I can see you getting to it (I'm personally not convinced by your p_{nm} - I got something that looks similar with a couple of extra factors, but I just did a rough so I can't be sure)

    The way to do square p_{nm} directly would be to perform the summation
    [tex]p^2_{nm}\sum_l p_{nl} p_{lm} [/tex]
    Remembering that only terms where both n and l AND l and m have different parity contribute. It's not obvious (to me) how the different terms contribute, you might be able to evaluate it, but you're gonna need some tricks with infinite series (try for p^2_{11}).

    Another way would be to directly calculate the elements in the Schrodinger formulation (which is what I was thinking about when I said it wasn't hard)
    [tex]p^2_{nm}=-\hbar^2<n|\frac{d^2}{{dx}^2}|m>[/tex]

    You can compare these to find the answers to some infinite sums!
     
  12. Dec 13, 2007 #11
    With these types of values you can't just multiply them together to get p^2, as fantis first suggested. To find p^2, you must do it through the second method he has listed. The reason the first one doesn't work is because you're working with operators.

    However, I'm not quite sure what you're trying to show for that?? If I'm reading it correctly, then just write down your H operator, show that this is the same as the p^2/2m operator, and thus their outcomes are the same-- literally one line of work.

    Secondly, if you want some intuition into the type of matrix that is setup by the operator, first notice that the matrix is symmetric about the diagonal, that is p_nm is the same as p_mn. Second (for the p^2/2m matrix), notice that your operator is d^2/dx^2, which means that you're differentiating twice the sin fcn, which returns the negative sin fcn. Since the functions are orthonormal, all the entries on the off diagonal (i.e. m not equal to n), will be 0.
     
  13. Dec 13, 2007 #12
    Are the missing factors -2 and 1/a ?
     
  14. Dec 14, 2007 #13
    OK. I figured out p^2_mn.

    I also need to calculate x_{nm}. I get that is equal to 0 when n and m have the same parity and

    [tex] 2/a \left( -\frac{2}{(\pi/a (n-m))^2} + \frac{2}{\pi/a (n+m))^2} \right)[/tex]

    when they have the same parity but are not equal and a/2, when n =m.

    EDIT: the above sentence should be "when they have the opposite parity but are not equal and a/2, when n =m. "

    Is that right? I am trying to calculate the commutator [x,p] using matrices.
     
    Last edited: Dec 14, 2007
  15. Dec 14, 2007 #14
    I'd just like to justify why the first method (matrix multiplication) I suggested DOES work:
    The set of all energy eigenstates |n> corresponds to a complete set, so
    [tex]\sum_n |n><n|=I[/tex] where the sum is taken over all eigenstates and I is the identity operator.

    Consequently
    [tex]p^2_{mn}= <m|p^2|n>=<m|p p|n>=\sum_l <m|p|l><l|p|n> = \sum_l p_{ml} p_{ln} [/tex]
    (That's why it's matrix mechanics - the operators multiply like matrices)

    Coto - can you give me a reason you can't do this? Since any state can be written as a superposition of sharp energy states, and the energy states are all orthogonal they make a complete set.

    Ehrenfest - that looks about right to me - though you said same parity twice, I'm assuming you just made a typo and know which one corresponds to which parity.

    Now you should, in principle, be able calculate [x,p]=xp-px directly from matrix multiplication (or equivalently in the Shrodinger representation)
     
    Last edited: Dec 14, 2007
  16. Dec 14, 2007 #15
    Hi Ehrenfest,

    Maybe I'm a little slow here, but can you show me how you arrived at [tex]p_{nm}[/tex]?
     
  17. Dec 14, 2007 #16
    OK. I was able to calculate xp and px, and that works out well. I think fantispug is right that the first method will work, at least theoretically. I thought about it for awhile and I really could not figure out how actually carry it out though. Maybe Coto just meant that it is extremely hard.
     
  18. Dec 14, 2007 #17
    [tex]p_{nm} = <\psi_n|p|\psi_m>[/tex]

    Evaluate the inner product using the usual integration over the domain of the wavefunction using the explicit forms \psi_n and \psi_m which are in a previous post of mine in this thread. Use the operator representation of momentum.
     
  19. Dec 14, 2007 #18
    Ok nvm. I guess I see what you're saying fantis, however, I'm still going to have to go with it's not a very efficient way to do this problem.
     
    Last edited: Dec 14, 2007
  20. Dec 14, 2007 #19
    Deleted.. I'm a dummy.
     
    Last edited: Dec 14, 2007
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