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Normalizing in Square Well

  1. Oct 28, 2005 #1
    This comes from problem set 7, problem 1 part a of the MIT coursework online. This problem seems straightforward to me and I believe I'm making a stupid math mistake of one kind or another, though its possible I'm (once again) missing something more important.
    The problem asks you to normalize a superposition of two states and show that the normalization is good for any time > 0.
    [tex]\Psi(x,t) = \sqrt{\frac{1}{a}}sin(\frac{\pi x}{a})e^{-i \omega_1 t} + \sqrt{\frac{1}{a}}sin(\frac{2\pi x}{a})e^{-i \omega_2 t}[/tex]
    [tex]1 = \int_l^{l+a} <\psi^*|\psi> \,dx = \int_l^{l+a} \Psi(x,t) = \sqrt{\frac{1}{a}}sin(\frac{\pi x}{a})e^{i \omega_1 t} + \sqrt{\frac{1}{a}}sin(\frac{2\pi x}{a})e^{i \omega_2 t} +[/tex]
    [tex] \sqrt{\frac{1}{a}}sin(\frac{\pi x}{a})e^{-i \omega_1 t} + \sqrt{\frac{1}{a}}sin(\frac{2\pi x}{a})e^{-i \omega_2 t}\,dx[/tex]
    Taking the complex square I get:
    [tex]= \int_l^{l+a} \frac{1}{a}sin^2(\frac{\pi x}{a}) + \frac{1}{a}sin^2(\frac{2 \pi x}{a}) + \left(e^{it(\omega_1 - \omega_2)} + e^{-it(\omega_1 - \omega_2)}\right)\left(\frac{1}{a}sin(\frac{\pi x}{a})sin(\frac{2\pi x}{a})\right)\,dx [/tex]
    And integrating the first two pieces (but not the third yet)
    [tex] =\left[\frac{x}{2a} - \frac{1}{4\pi}sin(\frac{2\pi x}{a})\right]_l^{l+a} + \left[\frac{x}{2a} - \frac{1}{4\pi}sin(\frac{4\pi x}{a})\right]_l^{l+a} [/tex]
    [tex]+ \int_l^{l+a} (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left(\frac{1}{a}sin(\frac{\pi x}{a})sin(\frac{2\pi x}{a})\right)\,dx[/tex]
    Now working out those first two results in a 1/2 for each, so there's my entire probability right there. It would seem to follow that the third will be zero, making the normalization independant to time. That's how I thought this would work out, but either I'm wrong about that logic or my math isn't working - or both.
    Looking at just that third part,
    [tex] \int_l^{l+a} (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left(\frac{1}{a}sin(\frac{\pi x}{a})sin(\frac{2\pi x}{a})\right)\,dx[/tex]
    [tex](e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\int_l^{l+a} \frac{2}{a} sin^2(\frac{\pi x}{a}) cos(\frac{\pi x}{a})\,dx[/tex]
    [tex]= (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left[\frac{1}{3\pi}sin^3(\frac{\pi x}{a})\right]_l^{l+a}[/tex]
    If I need to post any of the identities I'm using please let me know.
    [tex]= (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left[\frac{1}{3\pi}sin^3(\frac{\pi (l+a)}{a})\right] - \left[\frac{1}{3\pi}sin^3(\frac{\pi l}{a})\right][/tex]
    [tex]= (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left(\frac{1}{3\pi}\left[sin \frac{\pi l}{a}cos\pi + cos\frac{\pi l}{a}sin\pi\right]^3 - \left[\frac{1}{3\pi}sin^3(\frac{\pi l}{a})\right]\right)[/tex]
    [tex]= (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left[\frac{1}{3\pi}\left(-sin^3(\frac{\pi l}{a})\right) - \frac{1}{3\pi}sin^3(\frac{\pi l}{a})\right][/tex]
    Which is not zero by any means, nor would the sum be 1.
    Please let me know if you see any errors, and thanks for your time reading.
     
    Last edited: Oct 28, 2005
  2. jcsd
  3. Oct 28, 2005 #2

    Physics Monkey

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    Hi Locrian, if you put [tex] l = 0 [/tex] then you get zero like you should. So let me ask you, why should [tex] l = 0 [/tex]?

    Also, (and this a totally irrelevant point) but in the fourth line from the bottom, the result of the integration should be
    [tex]
    = (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left[\frac{\textbf{2}}{3\pi}sin^3(\frac{\pi x}{a})\right]_l^{l+a}
    [/tex]
    so that when you differentiate you cancel the three and the pi and get 2/a back.
     
    Last edited: Oct 28, 2005
  4. Oct 28, 2005 #3
    Thanks for the correction!

    As for [tex]l[/tex] being zero, it shouldn't have to be, should it? I should be able to make it any arbitrary value, so long as the other side of the box is at [tex]l + a[/tex]. Please hint me if I'm wrong about this.
     
  5. Oct 28, 2005 #4

    Physics Monkey

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    You're welcome!

    Here is a hint: You are free to choose your coordinates as you like, but the wavefunction does have to vanish at the edge of the box, so maybe the wavefunction you have written down already implicitly contains a choice for [tex] l [/tex]?
     
  6. Oct 28, 2005 #5
    Of course! Let me see if I understand:

    The choice of coordinates for the edges of the box must have been chosen before the wave functions of the superposed states is determined; I can't just assume any value for the edges of the box now because then the state function given doesn't make any sense.

    Am I to say then that, besides a (significant) error with a factor of two, I'm actually correct here, since plugging in 0 for [tex]l[/tex] will give me the answer to the problem?

    I appreciate your time physics monkey.
     
  7. Oct 28, 2005 #6

    Physics Monkey

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    Yes, you've got it. They have already assumed the box goes from [tex] x=0 [/tex] to [tex] x = a[/tex].

    You are quite welcome.
     
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