1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two Particle Schrodinger Equation in Polar Coordinates

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider two non-identical, non-interacting particles of mass M that are constrained to move on a circle of radius R. Write down the Schrodinger equation for this problem and find the eigenfunctions and energy levels of this system.

    2. Relevant equations

    (see below)

    3. The attempt at a solution

    I started by defining a coordinate system where the circle is centered at the origin of the x-y plane (z = 0). The particles don't interact and there is no external potential so V = 0. Also, using spherical coordinates, [tex] r = R, \ \theta = \frac{\pi}{2} [/tex] . I then wrote down the Schrodinger equation:

    [tex] [-\frac{\hbar^2}{2m_{1}}\nabla^{2}_{1} - \frac{\hbar^2}{2m_{2}}\nabla^{2}_{2} + V]\psi(r,\theta,\phi) = E\psi(r,\theta,\phi) [/tex]

    The Laplacian in spherical coordinates is:

    [tex] \frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}[\frac{1}{sin\theta}\frac{\partial}{\partial\theta}(sin\theta\frac{\partial}{\partial\theta}) + \frac{1}{sin^2\theta}\frac{\partial^2}{\partial \phi^2}] [/tex]

    Since r and theta are fixed, the Schrodinger equation reduces to:

    [tex] \frac{1}{2}(\frac{\hbar}{R})^2(m_1 + m_2)\frac{\partial^2\psi}{\partial\phi^2} + E\psi = 0 [/tex]

    letting [tex] a = \sqrt{\frac{2R^2E}{\hbar^2(m_1 + m_2)}} [/tex] the solution of the Schrodinger equation is:

    [tex] \psi = Ae^{a\phi} + Be^{-a\phi} [/tex]

    This is where I get stuck. I assume the next step is to apply boundary conditions which will then give the energy levels. How do you do this in this situation? I tried:

    [tex] \psi(0) = \psi(2\pi) [/tex]
    [tex] \frac{\partial\psi(0)}{\partial\phi} = \frac{\partial\psi(2\pi)}{\partial\phi} [/tex]

    but the results didn't make any sense. Any thoughts? Thank you in advance!
  2. jcsd
  3. Aug 5, 2011 #2
    How did you get the mass to the numerator?
    You are missing an [itex]i[/itex] in both terms. Also please check the dimension of [itex]a[/itex]. With that applying b.c. should give you the quantisation condition on E.
  4. Aug 5, 2011 #3
    Thank you for your help. Both careless mistakes, I need to be more careful. Now I am trying to determine the degeneracy of the first three energy levels. Here is my logic, I am not sure if it is correct.

    Ground state degeneracy = 2. Neither particle is excited, but since they are distinguishable you could swap their positions and get a different state with the same energy. So, state 1 = particle X in ground state at position 1 and particle Y in ground state at position 2; state 2 = particle X in ground state at position 2 and particle Y in ground state at position 1.

    First excited state degeneracy = 4. One particle is excited and the other one isn't, and again you can flip their positions to get another state.

    Second excited state. This one is more difficult. It seems to depend on which of the following situations has a larger energy, one particle in its second excited state and one particle in its ground state, or both particles in their first excited state. In the first situation the degeneracy would be 4, in the second situation the degeneracy would be 2.

    Here is one more thing that is confusing me. After I correct the mistakes from my first post I get the energy to be [tex] E = \frac{\hbar^2 n^2}{4MR^2} \ , \ n = 0,1,2,3,... [/tex] Clearly there is a unique value of energy for each value of [itex] n [/itex]. How could you determine the degeneracies just from this equation for the energy levels? Thanks again for any feedback.
  5. Aug 5, 2011 #4
    where [itex] m_1 = m_2 = M \ [/itex] (this is given in the original question) sorry for not mentioning that.
  6. Aug 10, 2011 #5
    fyi im just learning this too

    Was V=0 given in the problem?

    The particles are confined to a circle of radius R, I would have guessed there would be an infinite potential barrier at r=R.

    regardless, from what I have learned thus far,to calculate the eigenfunctions and energy levels, consider a general wave function that describes a system two particles

    psi( r1,ms1, r2,ms2) i think you only need polar rather than spherical coordinates

    where r1 is the radius of the first particle, ms1 is the internal angular momentum of the first particle and so forth

    Since the particles are non interacting, the ms term can be ignored
    leaving psi(r1,r2)=psi(r1)psi(r2). I believe you solve psi(r1)psi(r2) separately, but they should be the same wave function

    I would solve the energy levels for each particle separately, taking

    E(total)=E1n E2n

    Also, exchanging coordinates of non-interacting particles won't affect the probability distribution

    hence psi(r1,r2)=psi(r2,r1)
  7. Aug 10, 2011 #6
    V=0 was not explicitly given, but the particles do not interact and and no other external potential is given. I think you are confusing ON a circle with INSIDE a circle. If they were confined to move inside a circle, then yes, I think there would be an infinite potential at r=R. However this just has the particles moving ON the circle. It is like a particle moving in just the x-direction in Cartesian coordinates, there is no need to assume infinite potential barriers keeping it on the x-axis.

    I think you are right about solving for the energy levels separately. This would lead to:
    [itex] E^{(1)}_n = bn^2 [/itex] and [itex] E^{(2)}_{n'} = bn'^2 [/itex] where [itex] b = \frac{\hbar^2}{2MR^2} [/itex]. Therefore,
    [tex] E^{total}_{n,n'} = b(n^2 + n'^2) [/tex]
    Therefore, the ground state is [itex] n = 0 \ , \ n' = 0 [/itex] leading to [itex] E = 0 [/itex]
    And the first excited state is [itex] n = 1 \ , \ n' = 0 [/itex] OR [itex] n = 0 \ , \ n' = 1 [/itex] leading to [itex] E = b [/itex]
    And the second excited state is [itex] n = 1 \ , \ n' = 1 [/itex] leading to [itex] E = 2b [/itex]
    And the third excited state (which is not asked for in the problem, but I will include it anyway) is [itex] n = 2 \ , \ n' = 0 [/itex] or [itex] n = 0 \ , \ n' = 2 [/itex] leading to [itex] E = 4b [/itex]

    If [itex] g_n [/itex] is the degeneracy of the [itex] n^{th} [/itex] excited state then [itex] g_0 = 1, \ g_1 = 2, \ g_2 = 1 [/itex]. Is this correct?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Two Particle Schrodinger Equation in Polar Coordinates