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Finding the energy eigenfunctions of infinite well with potential step

  1. Oct 3, 2014 #1
    1. The problem statement, all variables and given/known data
    The potential for a particle mass m moving in one dimension is:
    V(x) = infinity for x < 0
    = 0 for 0< x <L
    = V for L< x <2L
    = infinity for x > 2L

    Assume the energy of the particle is in the range 0 < E < V

    Find the energy eigenfunctions and the equation that determines the energy eigenvalues. Don't worry about normalizing the eigenfunctions.

    2. Relevant equations


    3. The attempt at a solution
    I believe I've found the equation for the energy eigenvalues....
    in the region 0 < x < L the wave equation has the form:
    [itex] \psi_1(x) = e^{ikx} + Be^{-ikx} [/itex] where [itex] k = \sqrt{2mE}/\hbar[/itex]
    for L < x < 2L, the wave equation is:
    [itex] \psi_2(x) = Ce^{-qx} + De^{qx} [/itex] where [itex] q = \sqrt{2m(E-V)}/\hbar[/itex]

    applying the boundary condition [itex] \psi_1(x=0) = 0 [/itex],
    I find B = -1 and the wave equation can be rewritten and simplified as [itex] \psi_1(x) = A' sin(kx) [/itex]

    applying the boundary condition [itex] \psi_2(x=2L) = 0 [/itex],
    I find [itex]D = -C e^{-4qL} [/itex] and the wave equation can be simplified to [itex] \psi_2(x) = C' sinh[q(x-2L)] [/itex]

    Now, equating these two equations at x = 0 and their derivatives at x=L, and then dividing the two results gave me a solution i can use to solve for E, the energy eigenstates: [itex] q tan(kL) = -k tanh(qL) [/itex]

    But now, I'm not sure how to go about finding the energy eigenfunctions... For the plain infinite well, it was easy because you just had one equation, sin(something) = 0 , so something = n*pi. Then you just rewrote the equation with n in it...
    I'm thinking this involves another application of the boundary conditions, but I don't see how quantized energy functions will arise from equating my solutions at x = L...
     
  2. jcsd
  3. Oct 4, 2014 #2

    vela

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    ##\psi_1## and ##\psi_2## are not wave equations; they're solutions to the wave equation.

    Since E<V and you want q to be real, it should be ##q = \sqrt{\frac{2m(V-E)}{\hbar^2}}.##

    You've found them for the most part. You have
    $$\psi(x)=\begin{cases}
    A' \sin kx & 0 \le x \le L \\
    C' \sinh q(x-2L) & L < x \le 2L
    \end{cases},$$ where ##k## and ##q## satisfy the relationship you found above. You also have ##k^2+q^2 = k_0^2## where ##k_0^2 = \frac{2mV}{\hbar^2}.## You can't solve for ##k## and ##q## exactly. You should write C' in terms of A'.
     
  4. Oct 4, 2014 #3
    thanks vela - Just to make sure I understand what is going on in this problem: so to find the different energy eigenfunction equations for the different energy states, I solve my energy eigenvalue equation for E (my book uses a plotting and intersection method for the infinite square well), the number of intersections = my number of different every states? So the energy eigenfunction equations would all be the same except substituting in the different E values for each one? And these energy eigenfunction equations take two different forms over the two regions - before and in the potential well?

    Other than that, once I clean up my two equations writing C' in terms of A' and relating k and q, that should just be the answer.
     
    Last edited: Oct 4, 2014
  5. Oct 4, 2014 #4

    vela

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    That's correct.
     
  6. Oct 4, 2014 #5
    great- thanks
     
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