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Motion on a ring. Quantum mechanics.

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Particle of mass [tex[m[/tex] is confined on the ring of constant radius [tex]r_0[/tex]. Solve Schroedinger equation for this problem.

    2. Relevant equations


    3. The attempt at a solution
    Problem is solved here
    http://www.physics.oregonstate.edu/~corinne/COURSES/ph426/notes3.pdf
    Why Schroedinger equation is written in the form
    [tex]\frac{d^2\Phi}{d\varphi}=-\frac{2I}{\hbar^2}[E-U(r_0)]\Phi(\varphi)[/tex]
    I am confused. First of all why moment of inertia [tex]I[/tex]? Second how potential [tex]U(r_0)[/tex] is defined? Is it [tex]\infty[/tex] everywhere accept at [tex]r_0[/tex]? Thanks for the answer.
     
    Last edited: Mar 14, 2015
  2. jcsd
  3. Mar 14, 2015 #2

    mfb

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    If you write ##\varphi## as function of x (or vice versa) and substitute it in the Schrödinger equation, you get a factor of ##mr_0^2## in the equation, which is ##I##.

    As your Schrödinger equation depends on the angle only, everything outside the ring "does not exist" in the world of this particle. If your potential depends on the radius only, it is constant for a particle at a specific radius.
     
  4. Mar 14, 2015 #3
    Thank you for the answer. How you mean write as function of [tex]x[/tex]? What is [tex]x[/tex] for you in this problem.

    I am not sure why particle moves around the ring? When I say particle is moving around the ring then I can right equation where wave function is only function of angle [tex]\varphi[/tex]. But first at all why particle is moving around the ring. Is it because of potential?
     
  5. Mar 14, 2015 #4

    mfb

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    x is a linear distance. The position variable you usually see in the Schrödinger equation.

    It does not have to move, but it has to be on the ring due to some unexplained and irrelevant external world. The usual reason would be a potential, yes.
     
  6. Mar 14, 2015 #5
    Thank you. If x is linear distance then Schroedinger equation takes a form
    [tex]\frac{d^2 \psi}{dx^2}=-\frac{2m}{\hbar^2}[E-U)\psi(x)[/tex]
    I am not sure how to go from this equation to
    [tex]\frac{d^2 \Phi}{d \varphi^2}=-\frac{2I}{\hbar^2}[E-U)\Phi(\varphi)[/tex]
     
    Last edited: Mar 14, 2015
  7. Mar 14, 2015 #6

    mfb

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    The first equation has a typo.
    How are linear distance (along the ring) and the angle phi related? The rest is using the chain rule on the left side.
     
  8. Mar 14, 2015 #7
    Thank you. ##x=r_0\varphi##
    So because ##r_0## is constant
    [tex]\frac{d^2\psi}{dx^2}=\frac{1}{r^2_0}\frac{d^2\psi}{d\varphi^2}[/tex]
     
  9. Mar 14, 2015 #8

    mfb

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    Right. And that allows to get the moment of inertia on the right side.
     
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