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QM, infinite square well

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Infinite square well of length L, from -L/2 to +L/2.
    Suddenly the box expands (symmetrically) to twice it's size, leaving the wave function undisturbed. Show that the probability of finding the particle in the ground state of the new box is [tex](8/3 \pi )^2[/tex].


    2. Relevant equations
    the ground state of the wave function is
    [tex] \psi = (2/L)^{(1/2)} cos(\pi x /L) [/tex]


    3. The attempt at a solution
    If the wave function remains undisturbed, shouldn't the probability of finding it in the interval (-L/2, L/2) remain the same, equal to 1, and everywhere else zero? So the probability of finding it in the interval -L,L would simply be 1/2. The question is just really confusing me.
     
  2. jcsd
  3. Sep 19, 2007 #2

    malawi_glenn

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    No, when the box expands, there is a new wave function for the gruond state. So you have to calculate how much your "old" wave function overlaps with the new one.

    And that you do by calc the following integral:
    [tex] \int_{-L}^{L} \psi _{new}^{*}\psi _{old} dx [/tex]


    Remember: the wave funtcion is the probabilty amplitude for finding particle at position x
     
    Last edited: Sep 19, 2007
  4. Sep 19, 2007 #3

    nrqed

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    well, the probability of finding the particle between -L and L is 1, of course! But it is zero between -L and -L/2, one between -L/2 and L/2, and zero between L/2 and L.

    However the question is not about the probability of finding the particle in some position! The question is about finding the particle in a certain quantum state! In which case you have to do the integral that malawi glenn gave.
     
    Last edited: Sep 19, 2007
  5. Sep 20, 2007 #4
    Doing the intergral
    [tex] \int_{-L}^{L} \psi _{new}^{*}\psi _{old} dx [/tex]
    with
    [tex] \psi_{old} = (2/L)^{(1/2)} cos(\pi x /L) [/tex]
    [tex] \psi_{new} = (1/L)^{(1/2)} cos(\pi x /2L) [/tex]
    gives the wrong answer, so I started to draw bit (always nice to draw a bit) and figured what I want is that area enclosed by with [tex]\psi_{old}[/tex] and [tex]\psi_{new}[/tex] in the interval (-L/2,L/2). So I did the same intergral, but over the inverval (-L/2, L/2), and then it works out to [tex](8/3 \pi )^2[/tex], which I suppose is correct, though I'm still not sure I really understand what happend here.

    Thanks for the help!
     
  6. Sep 20, 2007 #5

    malawi_glenn

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    You must have done wrong, First

    [tex] \int_{-L}^{L} \psi _{new}^{*}\psi _{old} dx [/tex]

    Is correct, but you must see that the old wave function is zero for x > L/2 and x < -L/2

    So of course you get the right answer if you do the integration only between -L/2 and L/2, it is equivalent since the first wave function is zero elsewere ;)

    i.e

    [tex] \psi_{old} = (2/L)^{(1/2)} cos(\pi x /L) [/tex] when -L/2< x < L/2, and zero elsewhere.

    And also:

    [tex] \psi_{new} = (1/L)^{(1/2)} cos(\pi x /2L) [/tex] when -L< x < L, and zero elsewhere. Thats why the integral I stated was between -L and L, same consideration must be done with the old one. And you solved this by drawing pictures, which is, as you said, always good.

    It is very important to state which domain gives which values for the wave function. I thought that you was aware of this, hence there was some misunderstanding.
     
    Last edited: Sep 20, 2007
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