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Particle in a BOX - what are allowed momenta?

  1. Sep 20, 2004 #1
    Particle in a BOX -- what are allowed momenta?

    Ok I am trying to come up with the first five eigenfunctions for the particle in a box of size 2L. Now, I gave the appropriate initial conditions and get as a solution phi(x) = bcos(kx) + asin(kx). I said that phi(-L) = phi(L) = 0 which game me bcos(kL) + asin(kL) = 0 and bcos(kL) - asin(kL) = 0 . Upon solving them, I get that for n being odd, k = nPi/2L and for even n, k = nPi/L. Thus for odd n, phi(x) = asin(nPix/2L) and for even n, phi(x) = bcos(nPix/L). Now, I can easily plot these for n = 1,2,3,4,5 but I am also asked WHAT ARE THE allowed values of momenta.

    Now, I managed to get the allowed values of energy since from the resulting differential equation, I defined E = (k^2)*(hbar^2)/2m. So clearly, once I have k for various n, I have E.

    Now my question: How do I relate energy to momenta?

    Also, the question says "DO NOT FORGET TO INCLUDE SYMMETRIC AND ANTISYMMETRIC EIGENFUNCTIONS in your plot" Do they mean even and odd or what?

    I am desparate for help with this nagging issue.

    student1938
     
  2. jcsd
  3. Sep 20, 2004 #2
    You would get two different values for the accepted energy. How come? Usually it works out to one expression and the only difference is what n you choose but there should be one expression for E.
     
  4. Sep 20, 2004 #3
    I know about that, but it depends on what the size of the box is. The only reason things worked out nicely to one expression was due to the fact that the box was centered at the origin and was of size L rather than what I have which is 2L.

    I think there was an expression relating energy to momentum. I just cannot recall that. I ' m sure someone must know it.

    And also the issue of symmetric and anti symmetric must be known by somebody.

    I am quite desparate guys.
     
  5. Sep 20, 2004 #4

    robphy

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    Recall de Broglie: [tex]p=\frac{h}{\lambda}=\frac{h}{2\pi} \frac{2\pi}{\lambda}=\hbar k [/tex].

    Yes, the symmetric wavefunction is the even function such that [itex]\psi(-x)=\psi(x) [/itex]. Similarly, the antisymmetric wavefunction is the odd function such that [itex]\psi(-x)=-\psi(x) [/itex].
     
  6. Sep 20, 2004 #5
    Was there even any need to come up with the expression for the allowed energy since it is clearly not related to the allowed momenta. It doesn 't hurt to show it but it was not necessary.
     
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