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Find the wave function of a particle bound in a semi-infinite square well

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the semi-infinite square well given by V(x) = -V0 < 0 for 0≤ xa and V(x) = 0 for x > a. There is an infinite barrier at x = 0 (hence the name "semi-infinite"). A particle with mass m is in a bound state in this potential with energy E ≤ 0. Solve the Schrodinger equation to derive ψ(x) for x ≥ 0. Use the appropriate boundary conditions and normalize the wave function so that the final answer does not contain any arbitrary constants.


    2. Relevant equations
    [-h_bar2/2m]ψ'' + V(x)ψ = Eψ


    3. The attempt at a solution
    • Schrodinger Equation for 0 ≤ xa and x > a:
      [-h_bar2/2m]ψ'' - V0ψ = Eψ, 0 ≤ xa
      [-h_bar2/2m]ψ'' = Eψ, xa
    • Rewrite Schrodinger equations:
      ψ'' + 2m(E+V0)/h_bar2 = 0, 0 ≤ xa
      ψ'' + 2mE/h_bar2 = 0, xa
    • Solve Schrodinger equations:
      ψ1 = A1eik1x + B1e-ik1x, k1 = sqrt[2m(E+V0)]/h_bar, 0 ≤ xa
      ψ2 = A2eik2x + B2e-ik2x, k2 = sqrt[2mE]/h_bar, xa
    • k2 is negative, and the wave function must not blow up at x = ∞, so A2 = 0:
      ψ1 = A1eik1x + B1e-ik1x, k1 = sqrt[2m(E+V0)]/h_bar, 0 ≤ xa
      ψ2 = B2e-ik2x, k2 = sqrt[2mE]/h_bar, xa
    • Apply boundary conditions:
      ψ1(0) = 0
      ψ1(a) = ψ2(a)
      ψ'1(a) = ψ'2(a)

      1st condition: A1 + B1 = 0
      2nd condition: A1eik1a + B1e-ik1a = B2e-ik2a
      3rd condition: ik1A1eik1a - ik1B1e-ik1a = -ik2B2e-ik2a

    Now I have 3 equations for 3 unknowns, A1, B1, and B2. But I have been trying to solve this algebraically for quite awhile, and I just can't get it to work. When I solve A1 and B1 in terms of B2 and try to plug them into the third condition, I just get B2 cancelling on both sides. Maybe I'm being really dumb about basic math but I would really appreciate if someone could help with this. Thanks!
     
  2. jcsd
  3. Jul 3, 2012 #2

    vela

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    Actually, the way you defined k2, it's imaginary because E ≤ 0. It's usually better to define ##k_2 = \sqrt{-2mE}/\hbar##, so you can avoid those error-inducing factors of i.
    Once you get it down to everything in terms of one constant, you're finished with this part. To determine the constant, you require the wave function to be normalized.

    What conditions did you get on k1 and k2? A non-trivial solution exists for only certain values.
     
  4. Jul 3, 2012 #3
    Thank you so much for your response! I can see why you would define k2 that way, I guess I kept it imaginary so that ψ1 and ψ2 would have a similar format. The way I defined them should still theoretically work right? Or do I have to change it?

    So I should solve in terms of one constant, but then how do I normalize when there are two wave functions? Do I integrate ψ1 from 0 to a and then add that to the integral of ψ2 from a to ∞, and set that equal to 1?
     
  5. Jul 3, 2012 #4

    vela

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    It'll work out either way.

    Yes, that's what you would do.
     
  6. Jul 4, 2012 #5
    Okay, when I solve that system of equations, I get:

    A1 = B2e-ik2a / (eik1a - e-ik1a)
    B1 = -B2e-ik2a / (eik1a - e-ik1a)

    Plugging these in to normalize:
    Let j = e-ik2a / (eik1a - e-ik1a)

    jB22∫eik1x - e-ik1xdx + B2∫e-ik2x = 1

    Then I get
    jB2/ik1 [eik1a-eik2(0)] - jB2/-ik1 [e-ik1(a) - e-ik1(0)] + B2/-ik2 (e-ik2(∞) - e-ik2(0)) = 1

    What is e-ik2(∞)? Is that still 0? Even if it is, my answer looks unbelievably ugly.
     
  7. Jul 4, 2012 #6

    vela

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    You can simplify that a bit using
    $$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ In fact, if you step back a bit, your first boundary condition tells you A1=-B1 so that you can write ##\psi_1(x) = A \sin k_1x##.
    I just noticed a mistake in your original post. The way you defined k2, it's purely imaginary and equal to
    $$k_2 = i\left(\frac{\sqrt{-2mE}}{\hbar}\right)$$ where the quantity in the parentheses real and positive. Consequently, ik2 is purely real and negative. This implies that the A2 term vanishes as x→∞ but the B2 term blows up.
     
  8. Jul 4, 2012 #7
    Okay, if we are going to include trig functions anyway, there is a simpler way that I saw to set this up:

    ψ1 = A1sin(k1x) + B1cos(k1x)
    Since ψ1(0) = 0, B1 = 0

    So:

    ψ1 = A1sin(k1x)
    ψ2 = A2eik2x

    Since ψ1(a) = ψ2(a):
    A2 = A1sin(k1a) / eik2a

    Now normalize to find constants:
    ∫A1sin(k1x)dx + ∫A2eik2xdx

    Solving this, I get C = [1-cos(k1a)-sin(k1a)/eik2a]-1

    Which doesn't really look simple enough to be right...
     
  9. Jul 4, 2012 #8

    vela

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    Remember you have to integrate the square of the modulus to normalize the wave function.
     
  10. Jul 4, 2012 #9
    Ah! You're right. So what I got was:

    A12∫sin2(k1x)dx + A12sin2(k1a)/e2ik2a∫eik2xdx = 1

    That works out to:

    1 = A12[a/2 - (1/4)sin(2k1a) - sin2(k1a)/(ik2eik2a)]

    Is that the right way to get A1?
     
  11. Jul 5, 2012 #10

    vela

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    Right idea. I really suggest you get rid of the unnecessary i's.
     
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