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Having Trouble with QM Problem

  1. Aug 18, 2004 #1

    cepheid

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    I posted a small question about part of this problem earlier and got some tips that kept me going for a while. However, I'm still having trouble and I fear that my approach to the problem may be fundamentally wrong, since I'm not too confident with this stuff yet. So I thought I'd post the entire problem and my attempt at solving it so far:

    Given the following one-dimensional potential [tex] \inline{ V(x) = -g\delta(x) -g\delta(x-a)} [/tex] where [tex] \inline g [/tex] and [tex] \inline a [/tex] are positive constansts:

    a) Graph the potential

    This seemed easy enough. The difference between the two delta functions would be 0-0 everywhere except at x = 0, where the potential would equal [tex] \inline{ -\infty} [/tex], and at x = a, where it would also be [tex] \inline{ -\infty} [/tex]. So the graph of the potential would consist of two vertical lines starting at the x-axis at those two points and going down to negative infinity. Correct?

    b) Solve the time-independent Schrodinger equation for E < 0

    My approach was as follows. We derived the time independent equation in class. Assuming the wave function has the following form:

    [tex] \Psi(x,t) = \psi(x)T(t) [/tex]

    then: [tex] -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi = E\psi [/tex]

    I then divided the area being considered into three regions:

    i) x < 0

    [tex] E < 0 \therefore E = -|E| [/tex]

    So: [tex] \frac{d^2\psi}{dx^2} = -\frac{2m}{\hbar^2}(-|E|)\psi [/tex]

    [tex] \frac{d^2\psi}{dx^2} = \frac{2m|E|}{\hbar^2}\psi [/tex]

    Solving this differential equation:

    [tex] \psi(x) = Ae^{\kappa x} + Be^{-\kappa x} [/tex]

    where [tex] \inline{\kappa^2 = \frac{2m|E|}{\hbar^2}} [/tex]

    As far as I know, in the other regions, the differential equation ends up being the same, so we have:

    ii) 0 < x < a

    [tex] \psi(x) = Ce^{\kappa x} + De^{-\kappa x} [/tex]

    iii) x > a

    [tex] \psi(x) = Fe^{\kappa x} + Ge^{-\kappa x} [/tex]

    It also seemed to me that as [tex] \inline{ x \rightarrow -\infty, Be^{-\kappa x} \rightarrow \infty }[/tex]

    and as [tex] \inline{ x \rightarrow \infty, Fe^{\kappa x} \rightarrow \infty }[/tex]

    Neither of these situations is physically possible, so [tex] \inline{ B = F = 0 }[/tex]

    One continuity condition is [tex] \inline{ \psi(0^+) = \psi(0^-) }[/tex]

    [tex] Ae^0 = Ce^0 + De^0 \Rightarrow A = C + D [/tex]

    Also: [tex] \inline{ \psi(a^+) = \psi(a^-) }[/tex]

    [tex] Ce^{\kappa a} + De^{-\kappa a} = Ge^{-\kappa a} [/tex]

    [tex] Ce^{2\kappa a} + D = G [/tex]
    and so [tex] Ge^{-\kappa x} = Ce^{\kappa(2a-x)} + De^{-\kappa x} [/tex]

    To summarise, my solution is:

    [tex]\psi(x) = \left\{\begin{array}{cc}Ae^{\kappa x} & x < 0 \\Ce^{\kappa x} + De^{-\kappa x} & 0 < x < a\\Ce^{\kappa(2a-x)} + De^{-\kappa x} & x > a \end{array} [/tex]

    [tex] (A = C + D) [/tex]

    This may or may not satisfy my prof's requirements for part (b), but I decided to continue in the same way as the example problems in my notes did by integrating Schrodinger's equation:

    [tex]-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon}{\frac{d^2\psi}{dx^2}dx} -g\int_{-\epsilon}^{\epsilon}{\delta (x)\psi(x)dx} -g\int_{-\epsilon}^{\epsilon}{\delta(x-a)\psi(x)dx} = E\int_{-\epsilon}^{\epsilon}{\psi dx}[/tex]

    [tex] -\frac{\hbar^2}{2m}[\psi'(\epsilon) - \psi'(-\epsilon)] - g\psi(0) - g\psi(a) = 0 [/tex]

    Take the limit as [tex] \inline{ \epsilon \rightarrow 0} [/tex] :

    [tex] -\frac{\hbar^2}{2m}[\psi'(0^+) - \psi'(0^-)] - g\psi(0) - g\psi(a) = 0 [/tex]

    [tex] \frac{d\psi}{dx} = \left\{\begin{array}{cc}A\kappa e^{\kappa x} & x < 0 \\C\kappa e^{\kappa x} - D\kappa e^{-\kappa x} & 0 < x < a\end{array} [/tex]

    And so [tex] -\frac{\hbar^2}{2m}[(C\kappa - D\kappa) - (C\kappa + D\kappa)] - g\psi(0) - g\psi(a) = 0 [/tex]

    [tex] -\frac{\hbar^2}{2m}(-2D\kappa) = g(C+D) + g(Ce^{\kappa a} + De^{\kappa a}) [/tex]

    As it must be clear by now, I'm not exactly sure what I'm doing, and I'm at a loss how to proceed from here. If I had some inkling, then I wouldn't have bothered posting all of this. I'd really appreciate some help with this problem. The last two parts of the question are:

    c) Find the energy of the bound states (i.e. E < 0).
    d) Find normalized eigen-functions for the energies in part (c).
     
  2. jcsd
  3. Aug 18, 2004 #2

    cepheid

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    No replies...?

    Look, I hate to resort to bumping my own thread...but I'm really in a bind here, and as you can see, I'm not asking for a freebie here. I've put a lot of work into this so far and have been thinking about it non-stop. Nevertheless, I'm just not clear enough on this material for that to be fruitful, so I'd really appreciate any assistance that you could offer.

    Thanks
     
  4. Aug 18, 2004 #3

    Tom Mattson

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    Right, but speaking for myself I actually have to do the problem to answer your question.

    Sit tight, I'll get back to you later today.
     
  5. Aug 18, 2004 #4

    Dr Transport

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    also remeber to take into account the continuity of the wave function at both x = 0 and x = a, this will help you along......

    dt
     
  6. Aug 18, 2004 #5

    cepheid

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    Hmm...

    Interesting Dr Transport. That's what I thought I was doing here^^. But maybe you were referring to something else that I'm missing??

    btw Tom...I apologise for my impatience before.
     
  7. Aug 21, 2004 #6

    cepheid

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    Sorry, but I've still no idea how to proceed with this. Have I gone wrong somewhere? I've posted everything I did so far. Help is much appreciated, thanks.
     
  8. Aug 21, 2004 #7

    krab

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    You've integrated incorrectly in:

    [tex]-\frac{\hbar^2}{2m}\int_{-\epsilon}^{\epsilon}{\frac{d^2\psi}{dx^2}dx} -g\int_{-\epsilon}^{\epsilon}{\delta (x)\psi(x)dx} -g\int_{-\epsilon}^{\epsilon}{\delta(x-a)\psi(x)dx} = E\int_{-\epsilon}^{\epsilon}{\psi dx}[/tex]

    The integral over the second delta function is 0, since the interval [itex]-\epsilon<x<\epsilon[/itex] does not contain x=a.
     
    Last edited: Aug 21, 2004
  9. Aug 21, 2004 #8

    cepheid

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    Good point!

    That simplifies matters considerably. Thanks!
     
  10. Aug 21, 2004 #9

    cepheid

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    Or not...

    So instead I get:

    [tex] -\frac{\hbar^2}{2m}[(C\kappa - D\kappa) - (C\kappa + D\kappa)] - g\psi(0) = 0 [/tex]

    Which gives me:

    [tex] -\frac{\hbar^2}{2m}(-2D\kappa) = g(C+D) [/tex]

    [tex] \frac{\hbar^2 D\kappa}{m} = gA [/tex]

    [tex] \kappa = \frac{mgA}{\hbar^2 d} [/tex]

    And since [tex] \kappa^2 = \frac{2m|E|}{\hbar^2} [/tex]

    We get: [tex] |E| = \frac{m^2 g^2 A^2 \hbar^2}{2m\hbar^4 D^2} [/tex]

    Hopefully that answers part c. I don't know. It bothers me that A and D are still there.

    Now for the normalization: [tex] \int{\psi^*\psi}dx = 1 [/tex]

    I get: [tex] A^2\int_{-\infty}^{0} {e^{2 \kappa x}}dx + \int_{0}^{a} {(Ce^{\kappa x} + De^{-\kappa x})^2}dx + \int_{a}^{\infty} {(Ce^{\kappa(2a-x)} + De^{-\kappa x} )^2dx =1 [/tex]

    :eek: Any suggestions?!!?!??!
     
    Last edited: Aug 21, 2004
  11. Aug 21, 2004 #10

    cepheid

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    I've no idea how to evaluate/simplify that^
     
  12. Aug 22, 2004 #11

    cepheid

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    ^In previous examples it involved combining the integrals somehow, but that doesn't seem possible here
     
  13. Aug 23, 2004 #12

    cepheid

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    Well anyways, I gave up on it, so I guess that's that. Thanks to all those who did reply, although I have to say that for a thread that's been going on for more than ten days, the response was not what I had hoped, especially since I took great pains to show what work I had done.

    I'm not saying that I don't still want help learning how to solve the problem, I'm just saying that based on the past few days, I'm confident that that help will not be forthcoming.
     
  14. Aug 23, 2004 #13

    ehild

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    Cepheid,

    this is a weird problem, you'd better ask your teacher to explain it if you have given such a problem for beginner students. You did everything well, but you do not have enough skills yet. The normal way of solving a Schroedinger equation is as you did. You assume a solution function, you get two parameters, [tex] \pm \kappa[/tex]. You combined the wavefunction from these functions using constants, A, B, C, G. Now you need to eliminate all the constants but one. For these, you use the continuity of the wavefunction and the conditions for its derivative. For this, you used one at x=0, but missed the other, at x=a. If you write the integral from [tex](a-\epsilon ) [/tex] to [tex] (a + \epsilon ) [/tex] you get an other relationship between C , D and G. Now you can express C and D with both A and G. Eliminate C and D, and you get two homogeneous equation for A and G. This system of equations has nontrivial solutions only (that is not A=0 and G=0) if its determinant is zero. From that condition you get an equation for [tex] \kappa [/tex]. This equation is transcendental. Something like [tex] m \kappa = 1 \pm \exp (- \kappa a) [/tex], m being a constant that contains m, g and [tex]\hbar [/tex]. The next step is to express G with A , and then C and D with A using the value of [tex] \kappa [/tex] you got to built up the wavefunction. As it contains only one free parameter now, you can calculate it (in principle) from the normality condition.
    This is a lot of work, and I am not sure if there is no other solution. Working with delta potentials is really difficult.

    ehild
     
  15. Aug 23, 2004 #14

    Dr Transport

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    The one thing I have not seen in any other posts to this is the following: continuity of the wave function at [tex] x = a/2 [/tex]. There will then be a boundary condition relating C and D. This problem has intrigued me a little, so I'll try to sit down and work on it some.

    dt
     
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