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Potenitial well

  1. Jul 22, 2011 #1
    Hello again!

    Say I have a potential well, between 0 and a. I also know how the wave function looks like for (t=0):
    [itex] \psi(x,0)= \frac {2bx} {a} [/itex] for [itex] 0<x<\frac {a} {2} [/itex]
    [itex] \psi(x,0)= 2b(1- \frac {x} {a} ) [/itex] for [itex] \frac {a} {2} <x<a [/itex]

    Now, I wish to find the wave function of a general time, t.
    I know I need to find symmetry. Therefore I move the x axis:
    [itex] x\rightarrow x', x'\equiv x-\frac {a} {2} [/itex]

    So now I have the same well, only symmetrical.

    That means:

    [itex] \varphi_n =\sqrt {\frac {2} {a} } sin(\frac {\pi lx'} {a}) \rightarrow l=2n
    \varphi_n =\sqrt {\frac {2} {a} } cos(\frac {\pi lx'} {a}) \rightarrow l=2n+1 [/itex]

    For the question before the axis transition, I know I have to look for a sin- solution.
    But now, when I have this symmetry, I should look for a solution which vanishes at x=-a/2 instead of x=0

    Meaning, the cosine must stay. But in this problem's solution, I see that the sine stays...
    why is that so?
    Thank you!
  2. jcsd
  3. Jul 22, 2011 #2
    I assume you mean outside of [0, a], the potential is infinity (an infinite potential well).

    If that's the case, the wave function must vanishes at x=0 and x=a. [itex]\varphi_n = \sin \frac{\pi n x}{a}[/itex] do this, [itex]\varphi_m = \cos \frac{\pi m x}{a}[/itex] do not.
  4. Jul 22, 2011 #3
    Yes. of course. What I don't understand is why- if I change the axis so the problem becomes symmetric, do I still use sine instead of cosine.

    For this axis, the wave function has to be zero for a/2 and -a/2, instead of 0 and a...
  5. Jul 22, 2011 #4


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    You just did a mathematical trick to re-define your coordinates .. nothing about the problem has actually changed. If you transform back to your original coordinates, all of your wavefunctions will be sine waves. As to why you still have some sine solutions in the new coordinates, look closely at the conditions where the wavefunctions are sine solutions. Does that suggest anything to you?
  6. Jul 22, 2011 #5
    Thank you... but no... it doesn't suggest anything. I thought that there can't be a superposition of the two functions under such assumptions...
  7. Jul 22, 2011 #6


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    You only have sine functions for *even* values of n. Where are the nodes for such a sine wave? How does that match up with your boundary conditions? What if you tried to use cos functions for the even values of n?
  8. Jul 22, 2011 #7
    Why should it matter?

    once I know how the wave function looks like for t=0 I know it had collapsed into this form. Therefore- If I had sine function- it will not change...

    Or did I misunderstood you?
    Thank you so much for your enlightening answers...
  9. Jul 22, 2011 #8


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    You are missing the point of my answers ... for a given n (i.e. a given frequency), the only difference between a sine and a cosine is a phase shift of [itex]\frac{\pi}{2}[/itex]. So, whether or not a function is a sine or a cosine depends on your choice of origin, which you changed when you changed your coordinates. Now, the question you have to answer is what happens to the phase of the sine waves describing the eigenstates in your initial coordinate system when you shift to the new coordinate system. I suggest that you work it out on paper if it is not clear to you ...
  10. Jul 22, 2011 #9
    Oh... I did miss the point. Of course that depends on a phase only. I know that. I just can't get a clear answer about general things I don't understand.

    But writing the questions (and, of course, reading the answers) make everything more clear...
    Thank you so much!
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