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Harmonic oscillator inside square well again

  1. Apr 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider the SHO inside a square well, which looks like a soda can cut in half inside of a box.

    [tex]

    \[
    V(x,y) = \left\{ \begin{array}{l}
    \frac{1}{2}kx^2 ,{\rm{ }}y{\rm{ }} < {\rm{ }}|a| \\
    \infty ,{\rm{ }}y{\rm{ }} \ge {\rm{ }}|a|{\rm{ }} \\
    \end{array} \right\}
    \]

    [/tex]

    a. Find the energy levels in this potential.
    b. Write the wave function for the ground state.
    c. Find a relation between the constants of the problem that makes the first excited state degenerate.

    2. Relevant equations

    Since the potential is time-invariant (which means the Hamiltonian is also time-invariant, right..?) then we should use the time-independent schrodinger eqn:

    [tex]
    \[
    - \frac{{\hbar ^2 }}{{2m}}\nabla ^2 \psi + V\psi = E\psi
    \]
    [/tex]

    3. The attempt at a solution

    So (in a previous post) I indicated I would try to solve this by finding a separable solution, of the form [tex] \[\psi (x,y) = \psi _x (x) \cdot \psi _y (y)\] [/tex]. So that gives us

    [tex]
    \[
    - \frac{{\hbar ^2 }}{{2m}}\left( {\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) + V\left( {\psi _x \psi _y } \right) = E(\psi_x \psi_y)
    \]
    [/tex]

    According to my text (griffiths), we should be able to make this into two equations of x and y only, which each have to be a constant. But I'm having trouble with that, since I'm not sure how to separate the potential such that [tex] \[V(x,y) = V_x + V_y \] [/tex]. Assuming I can even do that, I'm not sure how that would help me. I'd get something like this:

    [tex]
    \[
    \left( {\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) - \frac{{2m}}{{\hbar ^2 }}\left( {V_x + V_y } \right)\left( {\psi _x \psi _y } \right) = - l^2 \left( {\psi _x \psi _y } \right),{\rm{ }}l \equiv \frac{{\sqrt {2mE} }}{\hbar }
    \]
    [/tex]

    Divide by [tex] \psi _x \psi _y [/tex] to get

    [tex]
    \[
    \frac{1}{{\psi _x \psi _y }}\left( {\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) - \frac{{2m}}{{\hbar ^2 }}\left( {V_x + V_y } \right) = - l^2
    \]
    [/tex]

    ...and so I'm stuck here.. I'm not sure how to write this in order to make this more conducive to separation, so I could use some hints here.

    Once I get a differential equation for just x and y though, I can solve that for [tex] \psi [/tex] and get a general solution, upon which I can impose the boundary conditions of the problem to get the energies. (..right?) The thing that concerns me, though, is that there are only boundary conditions for the y-direction; there are no constraints on the x-direction. Based on that, wouldn't I get two energies for the x-direction and y-direction? How do I resolve the two into one expression for the energy? Do I just sum them?

    Once I have the expression for the energy, it should be trivial to just tack the standard time dependence [tex] \[\left( {\varphi = e^{\frac{{ - iE_n }}{\hbar }t}} \right)\][/tex] onto the general solution and plug in the corresponding ground state energy. Or is that not the right way to go about that?

    Regarding part c, I have no idea where to even start. I know that degeneracy means a measurement of the total energy of two energy eigenstates returns the same eigenvalue, but I'm not sure what the problem means when it says that the first excited state itself is degenerate. Usually I think of degenerate (bound) states as two separate states that somehow have the same energy.. I guess what I'm saying is I would have expected the problem to say something like 'fiddle with the constants such that the first _and third_ excited states are degenerate'. Any ideas here?

    Thanks very much for any help at all.
     
    Last edited: Apr 12, 2008
  2. jcsd
  3. Apr 12, 2008 #2

    nrqed

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    for y greater than |a| the wavefunction is obviously zero. So consider only y < |a|.

    Then your potential is simply 1/2 x^2 and the separation will be easy.
     
  4. Apr 12, 2008 #3

    siddharth

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    That doesn't look right. Since the wavefunction is written as the product of [tex]\psi_x[/tex] and [tex]\psi_y[/tex], it should be

    [tex]
    - \frac{{\hbar ^2 }}{{2m}}\left( \psi_y{\frac{{\partial ^2 \psi _x }}{{\partial x^2 }} + \psi_x\frac{{\partial ^2 \psi _y }}{{\partial y^2 }}} \right) + V\left( {\psi _x \psi _y } \right) = E(\psi_x \psi_y)
    [/tex]
     
  5. Apr 12, 2008 #4
    Okay yeah you're right. I didn't write it correctly. So now we have

    [tex]
    \[
    \left( {\frac{1}{{\psi _x }}\frac{{d^2 \psi _x }}{{dx^2 }} + \frac{1}{{\psi _y }}\frac{{d^2 \psi _y }}{{dy^2 }}} \right) + \frac{{2m}}{{\hbar ^2 }}\left( {\frac{1}{2}kx^2 } \right) - l^2 = 0
    \]

    [/tex]

    This does make it easier to separate. I'll try this and post more questions later.
     
  6. Apr 12, 2008 #5

    nrqed

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    Of course it's easier to separate! Move everything that depends on x on one side and on y on the other side. Since x and y are independent variables, both sides must be equal to a common constant
     
  7. Apr 12, 2008 #6
    Okay, so now I have

    [tex]

    \[
    \begin{array}{l}
    \frac{{d^2 \psi _x }}{{dx^2 }} = \psi _x \left( {\frac{m}{{\hbar ^2 }}kx^2 - l^2 + n} \right),{\rm{ }}l \equiv \frac{{\sqrt {2mE} }}{\hbar } \\
    \frac{{d^2 \psi _y }}{{dy^2 }} = n\psi _y \\
    \end{array}
    \]

    [/tex]

    Does this look reasonable? What is the significance of the separation constant? How do I even solve that first equation?
     
  8. Apr 13, 2008 #7
    Does the way I've proposed to solve the rest of the problem seem reasonable?
     
  9. Apr 13, 2008 #8
    I have no idea how to solve that first diff. eq. Am I even going about it in the right way?
     
  10. Apr 13, 2008 #9
    everyone hates me
     
  11. Apr 14, 2008 #10
    i like mangoes
     
  12. Apr 14, 2008 #11
    alright, so my friend told me that i can resolve the potential into an x and a y component, giving me two one dimensional potentials that i have already solved (aka inf square well in the y direction, SHO in the x direction). what I don't understand is why I didn't get the differential equations (i.e., why my separated equations have that stupid separation constant). shouldn't i just get the equations for the one dimensional cases?
     
  13. Apr 15, 2008 #12

    siddharth

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    i think you made a mistake. If you group the terms, you get

    [tex]\frac{-\hbar ^2}{2m}\left(\frac{\psi_x^{''}}{\psi_x}\right) + 1/2kx^2 = E + \frac{\hbar ^2}{2m} \left(\frac{\psi_y^{''}}{\psi_y}\right)[/tex]

    The LHS is only a function of x, the RHS only a function of y, and they are equal. This must imply that both is equal to some constant. (Say [tex]\lambda[/tex])

    So, you have
    [tex]\frac{-\hbar ^2}{2m}\left(\frac{\psi_x^{''}}{\psi_x}\right) + 1/2kx^2 = \lambda[/tex]

    [tex]E+\frac{\hbar ^2}{2m} \left(\frac{\psi_y^{''}}{\psi_y}\right)=\lambda[/tex]

    Can you take it from here?
     
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