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Homework Help: 1-d Time Independent Schrodinger Equation Problem

  1. Feb 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m and energy E, where E >V1 >V2 travels to the right in a potential defined as

    V(x) = V1 for - b < x < 0

    V(x) = 0 for 0 < x < a

    V(x) = V2 for a < x < b

    (a) Write down the time-independent Schrodinger eq. and its general solution in each region. Use complex exponential notation.
    (b) Write down the boundary conditions which fix the undetermined constants in the solution of part (a).
    (c) Eliminate from those eqs. The normalization for the wave in region 3 and calculate the ratio of intensities of waves travelling to the left and to the right in region number 2.

    2. Relevant equations

    3. The attempt at a solution
    I really have no idea what I am doing. After writing the solutions as
    ψ(x) = A*exp(ikx) + B*exp(-ikx) with appropriate k for each of the 3 situations,
    I just cannot solve for the undetermined coefficients (and I've been doing this for the last 3 hours -_-). So yea, I'm completely lost.

    EDIT: Boundary Conditions are
    1)A$_{1}$e$^{-ik$_{1}$b}$ + B$_{1}$e$^{ik$_{1}$b}$ = 0
    2)A$_{1}$ + B$_{1}$ = A$_{2}$ + B$_{2}$
    3)k$_{1}$(A$_{1}$ - B$_{1}$) = k$_{2}$(A$_{2}$ - B$_{2}$)
    4)A$_{2}$e$^{ik$_{2}$b}$ + B$_{2}$e$^{-ik$_{2}$b}$ = A$_{3}$e$^{ik$_{3}$b}$ + A$_{3}$e$^{-ik$_{3}$b}$
    5)k$_{2}$(A$_{2}$e$^{ik$_{2}$b}$ - B$_{2}$e$^{-ik$_{2}$b}$) = k$_{3}$(A$_{3}$e$^{ik$_{3}$b}$ - B$_{3}$e$^{-ik$_{3}$b}$)
    6)A$_{3}$e$^{ik$_{3}$b}$ + B$_{3}$e$^{-ik$_{3}$b}$ = 0

    Look right to u guys?
    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 26, 2008 #2


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    the potential is not specified in all space... for the first line describing the potential you wrote down, is the "-b" supposed to be a negative infinity? otherwise, what is the for of the potential for x<-b
  4. Feb 26, 2008 #3
    It's not specified for all space, just a certain region, so b and -b is just any random number basically.
  5. Feb 26, 2008 #4


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    okay....... so then continue with your solution attempt. what boundary conditions are you going to use.
  6. Feb 26, 2008 #5
    Oh wow hold on that post just made me realize I left out 2 boundary conditions :-p (silly me)
  7. Feb 26, 2008 #6
    I can't read the correction you made to your original post but the problem only makes sense to me if there is an incoming particle (wave actually) from the left, a step in potential at x=0 and another one at x=a. Then there is a reflection at the first boundary, a middle zone with waves going both ways, and a transmitted beam at the second boundary.

    I think you already have expressions for the waves, which is to say their k numbers, based on the three potentials. So there are 5 undetermined coefficients, in general complex, for the wave amplitudes.

    However, as I see it, it gets a lot simpler when you consider what they are asking for: just the ratio between the left- and right- propagating waves in the middle zone. Let's take the outgoing wave in zone 3 to have unit amplitude; in fact, make the wave number something simple like 1 so it is just:

    exp(ikx - wt)

    (of course we won't worry to much about the wt).

    Then you just need to match up the waves in zone 2, let's give them a wave number like 3 or something:

    Aexp(i3x) + Bexp(-i3x)

    And I believe the boundary condition is that both the amplitudes and their derivatives have to match up at the transition. With the arbitrary numbers I put in above, I can solve pretty easily: I get A + B = 1 for the amplitudes, and 3A - 3B = 1 for the derivatives.

    So it seems I can solve the problem as stated without even worrying about what happened at the first boundary, with the incoming particle. Does this look right?

  8. Feb 28, 2008 #7
    how did u have only 5 undetermined coefficients, not 6?
  9. Feb 28, 2008 #8
    Because if the wave starts from the left, then in region 3 there is no left-moving (incoming) wave, only the outgoing one which I arbitrarily assigned a unit amplitude.
  10. Feb 28, 2008 #9
    I guess that would work, I just wasnt sure that there was not in fact some other potential past alpha, granted we are not told anything about it, so I guess you can take away the incoming wave.
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