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A little help with a particle on a ring and flux

  1. Dec 15, 2005 #1
    i uploaded my calculations, please check them out and point out the errors, it doesnt make any sense to me.
    i changed the file, now its readable (and if it needs more clearification please say so and i will rewrite it again.
    (see attachment)

    Attached Files:

    Last edited: Dec 16, 2005
  2. jcsd
  3. Dec 15, 2005 #2

    Physics Monkey

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    If I interpret you correctly, your system is a particle on a ring of circumference L with a uniform magnetic field applied, but what are you trying to do exactly? If you're just solving for the eigenstates, why not guess an angular momentum eigenstate? After all, the Hamiltonian commutes with the angular momentum so you can label physical states with angular momentum eigenvalues. It will simplify your life immensely.
  4. Dec 15, 2005 #3
    i'll try approaching it that way too, but just for practice, lets say im looking at the problem at just one dimension and make sure one end is connected to the other.

    you got me right, im trying to find the eigenstates and their energies, but i must be missing something... i got two equations and the solution i get doesnt satisfy them.

    i want to find what i did wrong in there, and after solving it this way i'll try other ways.
    Last edited: Dec 15, 2005
  5. Dec 16, 2005 #4
    ok, now its easier to follow my calcs.. pease spot the error...
  6. Dec 16, 2005 #5

    Physics Monkey

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    You need to be careful about dividing through by factors like [tex] 1 - e^{i K_+ L} [/tex], remember that you found that [tex] 1 - e^{i K_- L} [/tex] was zero. Just take a step back and think about your equations for second. Isn't it obvious that by choosing [tex] K_+ [/tex] correctly, you can make the A part of the equation trivial i.e. you can make A disappear from the equations? A similar reasoning applies to the B part. Hint: your equations are best thought of in terms of the factors [tex] A(1 - e^{i K_+ L})[/tex] and [tex] B(1-e^{i K_- L}) [/tex].
  7. Dec 16, 2005 #6
    so [tex] E_n=\frac{1}{2m}(\frac{2\pi n}{L}-\frac{e\Phi}{L})^2 [/tex] seems the only solution..
    Last edited: Dec 16, 2005
  8. Dec 16, 2005 #7

    Physics Monkey

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    But that's ok because you've forgotten that n can be negative or even zero.
  9. Dec 16, 2005 #8
    what bothers me here is that theres no way to flip the spin... if you send the right energy to the next level, the only thing the system can do is spin faster at the same direction, and will never change its direction...
  10. Dec 16, 2005 #9

    Physics Monkey

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    What exactly do you think is wrong with the solution [tex] \psi_m(x) = A \exp{\left( i \frac{2 \pi m}{L} x \right)} [/tex] and [tex] E_m = \frac{1}{2m} \left( \frac{2 \pi m \hbar}{L} - \frac{e \Phi}{L}\right)^2 [/tex]?

    Edit: Woops. Yes, that's what I mean.
    Last edited: Dec 16, 2005
  11. Dec 16, 2005 #10
    you mean [tex] \psi_m(x) = A \exp{\left( i \frac{2 \pi n}{L} x \right)} [/tex] and [tex] E_n=\frac{1}{2m}(\frac{2\pi n}{L}-\frac{e\Phi}{L})^2 [/tex] right?

    ok, looking at it again made me realize -there is- a difference between energy of up and down.

    and the certainty in the direcion is due to the fact the spin cant be sideways...
    thanks, i got it now :smile:
  12. Dec 16, 2005 #11
    ok, whats bothering me now is that: lets say that [tex] e\Phi=\pi [/tex]
    for negative state [tex] n_1 [/tex] and possitive n:
    [tex] \Delta E = \frac{\pi^2}{2mL^2}((2n_1 + 1)^2 - (2n - 1)^2) = \frac{\pi^2}{2mL^2}8n[/tex]
    and [tex] \Delta E = \frac{\pi^2}{2mL^2}((2(n+1) - 1)^2 - (2n - 1)^2) = \frac{\pi^2}{2mL^2}8n [/tex]
    [tex] \Delta E = \frac{\pi^2}{2mL^2}(8n) [/tex] will change the n state to both n+1 and -n...
    what will the system in n=2 do if i'll send a photon with [tex] \frac{\pi^2}{2mL^2}16 [/tex]?
    Last edited: Dec 16, 2005
  13. Dec 18, 2005 #12
    heh, seems like im answering my own questions most of the time =P
    i found my problem, negative n and possitive n are the same solutions, because this is not a traveling wave, but a stationary one.

    and i missed another level of energy when i though only on real A, if A is imaginary i get (2n+1)pi instead of 2npi,
    so the energy levels are: [tex] E_n=\frac{1}{2m}(\frac{\pi n}{L}-\frac{e\Phi}{L})^2 [/tex]
    but when you jump from one level to the next one you make all the nodes anti-nodes and vice versa.

    there, now i think i got it right (i hope i wont find another thing to ask about, since im starting to get on my nerves with all this questions i have to answer myself :tongue2: )
  14. Dec 18, 2005 #13

    Physics Monkey

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    Unfortunately, you seem to be answering your own questions incorrectly! The solutions are traveling waves, not standing waves, except in special circumstances. This is easy to see by virtue of the fact that each angular momentum eigenstate is an eigenstate of the Hamiltonian, and these all have non-degenerate energy eigenvalues unless [tex] \Phi [/tex] is specially chosen. You would need to have degeneracy between [tex] m [/tex] and [tex] - m [/tex] (i.e. [tex] \Phi = 0 [/tex]) to have a standing wave solution be possible, but clearly you don't.
  15. Dec 19, 2005 #14
    ok, back to the drawing board then...
    so what do you say about the degeneracy for n+1 and -n if [tex]e\Phi=\pi[/tex]?
    does it mean that the solution for the hamitonian eigenvalues is [tex]\Psi(x)=e^{i\frac{\pi x}{L}}(Ae^{ik_{(n+1)}x}+Be^{ik_{(-n)}x})[/tex]?

    by the way, am i atleast right about the new energy levels? [tex] E_n=\frac{1}{2m}(\frac{\pi n}{L}-\frac{e\Phi}{L})^2 [/tex]?
    Last edited: Dec 19, 2005
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