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Equation for a bound energy state

  1. Jul 19, 2007 #1
    1. The problem statement, all variables and given/known data

    Let [tex]V(x) = -aV_0\delta(x)[/tex]
    Show that it admits a bound energy state of [tex] E = -ma^2V_0^2/2\hbar^2 [/tex]

    Hint 1: Solve Schrodinger's equation outside the potential E>0, and keep the solution that has the right behavior at infinity and is continuous at x = 0.

    2. Relevant equations



    3. The attempt at a solution

    So the first step would be to plug that potential into the time-independent version of the Schrodinger equation: [tex]\frac{d^2\psi}{dx^2} + 2m/\hbar^2( E - V)*\psi = 0 [/tex] which results in a rather ugly DE due to the term a*V_0*delta(x). Any suggestions on which method I should use to solve this DE?

    In regards to the hint, I am not sure how assuming that the potential is negative helps us solve the DE...

    Thanks and please just give me tips and not the entire solution.
     
    Last edited: Jul 20, 2007
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  3. Jul 19, 2007 #2

    nrqed

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    Are you sure that in the hint they did not say E<0?

    In any case, by "outside" the potential, they mean when x is not equal to zero. So there, you solve Schrodinger's equation with no potential at all!

    There are two linearly independent solutions on the left of the potential. Kepe the one that has the correct behavior as x goes to minus infinity. On the right, keep th esolution that has the right behavior as x goes to plus infinity. At that point you will have two functions depending on two arbitrary constants.

    Now impose that the wavefunction is continuous at x=0. That gies you a relation between the two constants.
    Next, impose the condition on the derivative of the wavefunction (the derivative of the wavefunction is not continuous at x=0 because of the infinite discontinuity in the potential. There is an equation giving the condition on the derivative in such a case that you must have in your textbook). Imposing that condition will give you a restriction on the energy and you will be done.

    Hope this helps.

    Patrick
     
  4. Jul 19, 2007 #3
    Yes that helps. And you're right --it should have been E < 0.

    I think I am miscalculating something though. I can solve the DE and get:

    psi = Ae^(sqrt(E)*x) when x is less than 0;
    = Be^(-sqrt(E)*x) when x is greater than 0;
    Also continuity implies A = B and normalized gives A = B = 1/sqrt(E).

    Now we can set

    [tex] \int_{-\epsilon}^{\epsilon}\left( \frac{d^2\psi}{dx^2} \right) [/tex]

    equal to the change of the slopes of our two functions at x = 0:

    1 - -1 = 2 and solve that equation.

    The real answer is E = -ma^2V_0^2/(2*hbar)
     
    Last edited: Jul 20, 2007
  5. Jul 19, 2007 #4

    Dick

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    Uh, e^(sqrt(E)*x) has something to do with the correct solution. But what happened to the hbar's, m's and 2's we've grown to know and love? You might notice they appear in the final solution. I don't think you have the V=0 solution right, yet.
     
  6. Jul 20, 2007 #5
    You're right Dick. The solution outside of the potential should be

    [tex] \psi(x) = A exp\left(ix\sqrt\frac{2mE}{\hbar}\right) for x< 0 [/tex] and
    [tex] \psi(x) = A exp\left(-xi\sqrt\frac{2mE}{\hbar}\right) for x> 0 [/tex].

    So we use the normalization requirement that \int_{-\infty}^{infty}\psi = 1 to get the value of A, correct?

    So, in integrating that, do we just use [tex]\int_{-\infty}^{0}A exp\left(xi\sqrt\frac{2mE}{\hbar}\right)dx + \int_{0}^{\infty}A exp\left(-xi\sqrt\frac{2mE}{\hbar}\right)dx= 1 [/tex]?

    How do we account for the potential at zero in this integral?
     
  7. Jul 20, 2007 #6

    Dick

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    Better, but those are the solutions for positive E and can't be normalized in the integral psi*conjugate(psi) sense. You want a bound state solution so E<0. What do the solutions look like then?
     
  8. Jul 20, 2007 #7
    You're right again. So, it would be

    [tex] \psi = A*exp\left(x\sqrt{-2mE/\hbar^2} \right) \text{ for } x < 0 [/tex] and

    [tex] \psi = A*exp\left(-x\sqrt{-2mE/\hbar^2} \right) \text{ for } x > 0 [/tex].
    Firstly, why can we not normalize "in the integral psi*conjugate(psi) sense"? Secondly, how do we find A then?
     
    Last edited: Jul 20, 2007
  9. Jul 20, 2007 #8

    Dick

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    For the E>0 case, psi*conjugate(psi)=A^2. Integrating a constant over an unbounded interval is infinity. Unbound wave functions can't be normalized that way. (They are normalized by flux, but that's a totally different problem). For the E<0 case you could now normalize, but you don't have to to determine E. The Schrodinger equation is now schematically psi''(x)=K*delta(x). Integrate it in a region around x=0 and you will get (psi'(0+)-psi'(0-))=K*psi(0). The K determines the jump in the derivative of psi at x=0. But if you look at that equation the normalization constant A occurs on both side of the equation. So it will cancel out.
     
  10. Jul 20, 2007 #9
    So [tex]\frac{d^2\psi}{dx^2} = (\psi'(0+)-\psi'(0-)) = 2A\sqrt{-2mE/\hbar^2}[/tex], right?

    And if we integrate the second derivative in a region around zero, we get[tex] \int_{-\epsilon}^{\epsilon}\left( \frac{d^2\psi}{dx^2}dx \right) = \int_{-\epsilon}^{\epsilon}2m/\hbar^2( E + aV_0\delta(x))\psi dx [/tex], right?

    So, [tex] \int_{-\epsilon}^{\epsilon}\delta(x)dx = 1 [/tex], right?
    Now I am unsure what to do with those results...
     
    Last edited: Jul 20, 2007
  11. Jul 20, 2007 #10

    Dick

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    Ouch. I was way too 'schematic'. The first equation is nonsense. The second equation is pretty good. As epsilon goes to zero, the integral of the second derivative yeilds the difference of the two one-sided derivatives. The integral of the delta function is 1, so the result for that term is all of the constants times psi(0). The integral of the E term goes to zero, since the wave function is finite and the size of the interval goes to zero. This should give you an algebraic equation to solve for E. (The E coming from the derivatives of psi). Oh, and be careful of the signs.
     
  12. Jul 20, 2007 #11

    Dick

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    Your first line is better than I thought. Just ignoring the second derivative of psi, you've correctly computed the difference of the one-sided derivatives. Just equate that to what you get from the delta function. You are almost there.
     
    Last edited: Jul 20, 2007
  13. Jul 20, 2007 #12
    So, [tex] \int_{-\epsilon}^{\epsilon}2m/\hbar^2( E + aV_0\delta(x))\psi dx = A*a*V_0 [/tex].

    And the other expression for
    [tex] \int_{-\epsilon}^{\epsilon}\left( \frac{d^2\psi}{dx^2}dx \right) [/tex]

    is

    [tex] (\psi'(0+)-\psi'(0-)) = 2A\sqrt{-2mE/\hbar^2} [/tex]

    If I set those two expression equal, I get something close to, but not exactly the answer. However, if I set

    [tex] AaV_0 = 2A\left( \sqrt{-2mE/\hbar^2}\right)^{-1} [/tex]

    I get exactly the answer. Now why would that be inverted...
     
  14. Jul 20, 2007 #13

    Dick

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    Why is there no 2m/hbar^2 on the right hand side of your first line?
     
  15. Jul 20, 2007 #14
    There we go. Thanks.
     
  16. Jul 20, 2007 #15

    Dick

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    And your improvised inversion to get the right answer doesn't work either. If you solve that then a*V0 will turn up in the denominator. And I think you are still improvising a correct sign. We can pass on that for now but when you finally get to the answer go back and make sure you the right answer without fudging anything.
     
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