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Semi-infinite potential well

  1. Nov 27, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Let V(x) = +∞ for x ≤ 0, -V1 for 0 < x < b and 0 for x > b. V1 and b are positive. The solutions in each of the physical regions are ##\psi_1 = P \exp(ik_1 x) + Q \exp (-ik_1 x)## and ##\psi_2 = R \sin (k_2x + \gamma)##.

    Show that ##\lim_{V_1 \rightarrow 0} \gamma = 0##

    3. The attempt at a solution
    I am trying first to understand what ##\gamma## actually represents. I know to obtain the solution in region 2, it is a phase shift constant and so that solution is equivalent to a LC of sin and cos solutions. I also derived a relation between ##k_1, k_2 ##and ##\gamma## using the continuity conditions, but I am not sure how (or if) to apply it here.
    Any hints in the right direction would be great.
    Thanks.
     
  2. jcsd
  3. Nov 27, 2013 #2

    BruceW

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    yeah, it sounds like you are on the right track. exactly what is the relation have you found so far? remember there is more than one continuity condition that needs to be satisfied.
     
  4. Nov 27, 2013 #3

    CAF123

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    Hi BruceW,
    The continuity condition I obtained by imposing the continuity of the first derivative of the wave function and the fact it must be single valued, was ##k_1 \tan (k_2 b + \gamma) = k_2 \tan (k_1 b)## The other condition I got was that simply P = -Q. But what should I do with these?

    Thanks.
     
  5. Nov 27, 2013 #4

    BruceW

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    right. that's good so far. You have used both of the continuity equations. Now, when ##V_1## tends to zero, what does ##k_1## tend to? hint: it's almost intuitive really, but you can use the Hamiltonian to find the exact relationship.
     
  6. Nov 27, 2013 #5

    CAF123

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    It tends to ##k_2##. I agree that this is intuitive, but I think to prove it, we should take the limit as V1 tends to zero in the Hamiltonian expression for the second region. So $$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \lim_{V_1 \rightarrow 0} \psi_{1} - \lim_{V_1 \rightarrow 0} V_1 = E \lim_{V_1 \rightarrow 0} \psi_{1}$$ The second term on the LHS goes to zero since the potential is continuous in this region. The wave function in the second region will tend to the wave function in the third region. This means we end up with precisely the form of the Hamiltonian that would have previously described region 3.

    The continuity condition then becomes as ##V_1## tends to zero: $$k_2 \tan (k_2 b + \gamma) = k_2 \tan(k_2 b)$$ For this to hold for all ##k_2##, ##\gamma \rightarrow 0## too. Is this okay?
     
  7. Nov 27, 2013 #6

    BruceW

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    yep. looks good to me. although, technically, since tan is a periodic function, you can have gamma tend to other values, but it doesn't change the physics of the system, right? so maybe you should say a few words about that.
     
  8. Nov 28, 2013 #7

    CAF123

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    Ok thanks. What do these quantities ##\gamma, k_1, k_2## actually represent? I think ##\gamma## is a phase shift and ##k_1, k_2## are a measure of the momentum of the particles in each region.
     
  9. Nov 28, 2013 #8

    BruceW

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    yep. ##\hbar k_1,\hbar k_2## are definitely the absolute value of the momentum in each region. And just by looking at how ##\gamma## enters into the wavefunction, its role is to give a phase shift, as you say.
     
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