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Wave function in Infinite/Finite Potential Wells

  1. Feb 14, 2014 #1
    1. The problem statement, all variables and given/known data
    What is the functional form of the wave function in the ground state in the
    five regions x<0, 0<x<a, a<x<b, b<x<L, and x>L?

    I've attached the picture of the potential well as well here: Screen Shot 2014-02-14 at 6.05.31 PM.png

    2. Relevant equations

    Schrodinger time independent equation

    3. The attempt at a solution

    I understand that for x < 0 and x > L, ψ = 0 because there's no possibility of the particle being there.

    And then between a and b: ψ = A*sin(kx) + B *cos(kx) where k = √(2mE)/(hbar)2

    It's between 0 < x < a and b < x < L that I'm having issues. Can anyone point me in the right direction?

    Thanks!
     
  2. jcsd
  3. Feb 14, 2014 #2
    So remember how you get k in the first place in the region where the potential is 0:

    [itex]\frac{-\hbar^2}{2m}\frac{∂^2\psi}{∂x^2}=E\psi[/itex]

    The solution comes from 'guessing' [itex]ψ = A\sin(kx) + B\cos(kx)[/itex], and plugging this into the differential equation and algebraically solving for k.

    Now consider the regions where [itex]V=U_0[/itex]


    [itex]\frac{-\hbar^2}{2m}\frac{∂^2\psi}{∂x^2}+U_0\psi=E\psi[/itex]

    [itex]\frac{-\hbar^2}{2m}\frac{∂^2\psi}{∂x^2}=(E-U_0)\psi[/itex]

    Notice how this is the same equation as above, the only difference being [itex]E-U_0[/itex] instead of [itex]E[/itex].

    At this point you just assume a similar solution for [itex]\psi[/itex] in the two regions with potential [itex]U_0[/itex] as you did with in the 0 potential region. Note however that the coefficients infront of the sines and cosines in each of the 3 regions are different in each region; to solve for each one continuity conditions, boundry values and normalizationmust be applied.
     
  4. Feb 14, 2014 #3

    TSny

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    You should consider if the the ground state wavefunction will be sinusoidal or exponential in regions 0<x<a and b<x<L.
     
  5. Feb 15, 2014 #4
    So I get that it will be the same as the function for a < x < b except that the constants in front of the sine/cosine will be different and k = √2m(E - Uo)/ (hbar)2?

    Is that right?
     
  6. Feb 15, 2014 #5

    vela

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    Maybe, maybe not. Did you consider what TSny said? How did you conclude the wave function would sinusoidal and not exponential in that region?
     
  7. Feb 15, 2014 #6
    The others have a point; the solutions you 'should' assume had the form

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


    For some z that may end up being real, imaginary or complex; this is how you get the original ('particle in a box') solution in the first place.
     
  8. Feb 16, 2014 #7
    Okay I think I see that. But how do we determine if the wavefunction will be sinusoidal or exponential?
     
  9. Feb 16, 2014 #8

    TSny

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    Recall the finite square well problem. What is the condition to get exponential behavior of the wavefunction outside the well?
     
  10. Feb 16, 2014 #9
    Oh I see. So if the particle is bound by the potential, we know V > E, so we "guess" the exponential solution. But if the particle is free and E > V, then we guess the sinusoidal solution? Is that the correct interpretation?
     
  11. Feb 16, 2014 #10

    TSny

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    Yes, although I wouldn't say it involves "guessing". If you study the form of the Schrodinger equation for the case where Vo > E, then you can see that the solution must be exponential. (But maybe you're using "guessing" more loosely.) Anyway, you're on the right track!
     
  12. Feb 17, 2014 #11
    Ah, yes, I'm using "guessing" in the the sense that as I'm learning differential equations, the book tells us to "guess" solutions to the second order differential equations. It's not really guessing, but I've gotten used to the term.

    Thanks so much!
     
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