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Double Well Potential

  1. Sep 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the Hamiltonian ##H=-\frac{d^2}{dx^2}+V(x)##

    ##V(x) =\begin{cases} \infty & x < 0 ,x>\pi \\V_0 & \frac{\pi}{2}-\frac{a}{2} \geq x \leq \frac{\pi}{2}+\frac{a}{2} \\ 0 &elsewhere\end{cases}##

    Use the boundary conditions at x=0 and x=##\pi## to set two of the constants and then use the boundary conditions of continuity of the wave functiond and the first derivative at ##x=\frac{\pi}{2}-\frac{a}{2}## and when ##x=\frac{\pi}{2}+\frac{a}{2}## as well as normalization to solve for the lowest 4 eigenstates.



    2. Relevant equations



    3. The attempt at a solution

    So when Schrodinger's is zero: ##\frac{d^2\psi(x)}{dx^2}=E\psi(x)##

    So I found that ##\psi_1(x)=Asin(\sqrt{E}x)+Bcos(\sqrt{E}x)##

    Then when ##V(x)=V_0##:##\frac{d^2\psi(x)}{dx^2}+V_0\psi(x)=E\psi(x)##

    Then if E>V: ##\psi_2(x)=Csin(\sqrt{E-V_0}x)+Dcos(\sqrt{E-V_0}x)##

    Or if V>E: ##\psi_3(x)=Le^{\sqrt{V_0-E}x}+Fe^{-\sqrt{V_0-E}x}##

    I know that if x=0 then A must be zero. I'm not sure what to do with pi though. Another student showed me that they had an additional equation for ##\psi## but I'm not sure where it came from:

    ##\psi(x)_4=Gsin(\sqrt{V_0-E}(x-\pi))+Hcos(\sqrt{V_0-E}(x-\pi))##

    Before I can even give the second part of the question a shot I think I need to reduce the number of coefficients first. If the above equation is accurate (though I have no clue where it came from at the moment) then G would need to be 0 as well. That's 6 coefficients left.

    From here I believe that ##\psi_1=\psi_2=\psi_3## (I can't guess about the fourth). I'm just not sure what my next steps should be to get rid of so many variables.
     
  2. jcsd
  3. Sep 22, 2014 #2

    Orodruin

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    You should not have three functions that are equal to each other, you should have three different functions, each valid in a particular range of ##x## for the solution. You essentially have the functional form for the solutions in the different parts already and you need to determine the constants. The big problem here is finding out the boundary conditions and adapting the constants to them. What boundary conditions will be valid at ##x = 0## and ##x = \pi##? What boundary conditions must be fulfilled at the boundary between the ##V = 0## and ##V = V_0## regions?
     
  4. Sep 22, 2014 #3
    I guess I'm stuck at this then.

    Well, I know that A must be zero so the function that describes the range where ## 0\leq x <
    \frac{\pi}{2}-\frac{a}{2} ## and ##\frac{\pi}{2}+\frac{a}{2} <x\leq \pi## is##\psi_1(x)=Bcos(\sqrt{E}x)##

    For ##\psi_2## and ##\psi_3## I can't apply the same constraints because it is outside their range. However, I thought it would be true that they should be equal to each other since they describe the function in the same area of the well. And, I assumed that ##\psi_1## would be continuous at the barrier with 2 and 3.

    I imagine that I should have something like B=C+D or L+F=B depending on the relationship of E and V.
     
  5. Sep 22, 2014 #4

    Orodruin

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    The boundary condition at ##x = 0## (and ##x=L##) should be that the wave function is zero. This is essentially because the wave function has to be zero when the potential is infinite and it is continuous. Also note that you will have different constants in front of the solution for ##\psi_1## in the disjoint regions you described in the last post.

    In the middle region you will have either ##\psi_2## or ##\psi_3## depending on the energy. The boundary conditions (apart from continuity) can be inferred from integrating the Schrödinger equation in a small (infinitesimal) region around the boundary.
     
  6. Sep 22, 2014 #5
    I don't understand quite what mean by ##x=L##. L isn't a boundary but a coefficient.

    Do I need to treat this as two different double wells since I don't know the energy?

    I think I understand the equation the other student had. It is describing the function on the right side of the well. And the trig argument is ##x-\pi## so that it is equal to the coefficient at the "wall."
    So ##\psi_1## should be ##\psi_{1a}(x)=Bcos(\sqrt{E}x)## on one side and ##\psi_{1b}(x)=Acos(\sqrt{E}(x-\pi)## on the other.

    Then, I should take the derivative of 2 and 3 and set them equal to what other derivative? I guess I can't say that 1a or 1b are equal to 3 or 4 or else they must equal each other. I want to believe that 1a=2 and 1b=3 but I can't physically reconcile that.
     
  7. Sep 22, 2014 #6

    Orodruin

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    Sorry, I have been lecturing on an oscillating string with length L all evening so it was a backbone reaction. By L I mean pi.
     
  8. Sep 22, 2014 #7
    It's no problem. So, I guess I have the 4 equations that I should need, I'm just not sure about their relationships at the boundary conditions.
     
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