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Find wavefunction of a particle

  1. Nov 4, 2005 #1
    I am not sure how to solve this question. I can only say that, the wavelength is an integral number multiple of the circumference. Then?

    A particle is confined to move in a circular tube with radius R. Work out the wavefunction of this particle.
     
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  3. Nov 4, 2005 #2

    lightgrav

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    Are they talking about a torus-shaped tube (donut)? what's the other dimension?
    or a circular cross-section cylinder (what length?)
     
  4. Nov 5, 2005 #3
    More like a string with two ends connected to each other to form a circle. Like a big letter "O" .
     
  5. Nov 5, 2005 #4

    siddharth

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    You mean a ring of radius 'r'.
    Have you learnt the expression for the Hamiltonian in cylindrical co-ordinates?
     
  6. Nov 6, 2005 #5

    lightgrav

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    The only thing you can say is that there are an integer number of wavelengths.
    You can write it as Aexp(imΦ) multiplied by a delta-function (r-R) .
    but then you would be specifying a Φ=0 orientation,
    so you should include an arbitrary phase angle in the exponent.

    But this is really the same thing.
     
  7. Nov 6, 2005 #6

    siddharth

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    To find the wavefunction, you have to solve the time independent Schrodinger equation [tex] H \Psi = E \Psi [/tex]. Because of the symmetry for a particle confined to move in a ring, working in the cylindrical polar co-ordinates will save you a lot of time.
    So, if you write down the Hamiltonian Operator in the cylindrical polar co-ordinates(remember that the radius is constant), you will be able to work out the wavefunction of the particle.
     
  8. Nov 7, 2005 #7

    dextercioby

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    Separate variables in the time-indep SE and then, using boundary conditions (the wavefunction must be 0 on the boundar) find the solution. Hope you've seen and worked with Bessel functions before.

    Daniel.
     
  9. Nov 7, 2005 #8

    siddharth

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    dextercioby, I think it's possible to solve this question without any knowledge of Bessel fuctions.
    Also, won't we be using the periodic boundary condition [tex] \Psi(\theta) = \Psi(\theta + 2\pi) [/tex]? How do we use the fact that the wavefunction must be 0 on the boundary?
     
  10. Nov 9, 2005 #9
    There is no as such some fixed wavefunction of the partyicle in a ring. Generally it is some exponential function , like for an electron in Bohr radius it is [itex]e^(-r)[/itex] . For deciphering the wavefunction , you also need to satisfy the boundary conditions , like how potential varies , it can assumed to be zero on the path particle moves , infinite potential generally implies the particle is forbidded to pass the barrier.

    BJ
     
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