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Double delta well separated by a delta barrier

  1. Mar 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider a one-dimensional time-independent Schrodinger equation for an electron in a double quantum well separated by an additional barrier. The potential is modelled by:

    V (x) = -γδ(x - a) -γδ(x + a) + βδ(x)​

    Find algebraic equations which determine the energies (or k-values) of electron bound states for γ > 0 and arbitrary real β (positive or negative). Describe the symmetry of their wave functions in terms of even and odd
    solutions. How many bound states do you expect for this system?

    2. Relevant equations

    You may nd it useful to work in units ħ = 1 and m = ½, and to introduce k defined as E = -k2, where E < 0 is the energy of a bound state, so that k is real.

    3. The attempt at a solution
    I've tried to solve this for arbitrary E, since I don't understand what happens at the barrier for E<0. So I split the problem into 4 parts: x<-a, -a<x<0, 0<x<a and x>a. This results in the wavefunctions:

    ψ1 = AeiE½x + Be-iE½x
    ψ2 = CeiE½x + De-iE½x
    ψ3 = FeiE½x + Ge-iE½x
    ψ4 = HeiE½x

    Using the conditions of continuity and differential continuity results in the following conditions for the coefficients:

    Ce-iE½a + DeiE½a = Ae-iE½a + BeiE½a
    -Ce-iE½a + DeiE½a = -(1+iγ/E½)Ae-iE½a + (1+iγ/E½)BeiE½a


    HeiE½a= Fe-iE½a + GeiE½a
    -He-iE½a = (1+iγ/E½)Fe-iE½a - (1+iγ/E½)GeiE½a
    And now I'm not sure what to do. I can start to solve for transmission and reflection coefficients, but as far as I am aware bound states don't have these? It feels like at this point I need to specifiy the sign for E, but for E<0 I don't know what happens at the barrier since barriers don't have bound states.
  2. jcsd
  3. Mar 18, 2016 #2


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    To simplify your life a little, you need to realize that if [itex]E < 0[/itex], then the corresponding wave function looks like:

    [itex]A e^{-kx} + B e^{+kx}[/itex]

    Where [itex]k = \sqrt{-E}[/itex] (in the units with [itex]\hbar = 1, m = \frac{1}{2}[/itex]. There is no "i" in the exponent. So when [itex]x \rightarrow -\infty[/itex], you have to have [itex]A = 0[/itex].

    The continuity and derivative conditions across a delta function potential have the same form, whether the energy is positive or negative.
  4. Mar 18, 2016 #3
    I know that but I was trying to solve it generically first because I don't understand what happens to an exponential at a barrier. I know that for a travelling wave, it would exponentially decrease throughout the barrier until it got to the other side with a lower amplitude. But if it's already exponential, and also below 0, and also width zero, then I have no idea.
  5. Mar 18, 2016 #4


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    Schrodinger's equation is:

    [itex] - \psi''(x) + V(x) \psi(x) = E \psi(x)[/itex]

    Suppose that you have a delta-function at [itex]x=a[/itex], so [itex]V(x) = g \delta(x-a)[/itex]. Then we can integrate both sides of this equation from [itex]x=a - \epsilon[/itex] to [itex]x=a + \epsilon[/itex].

    [itex]\int_{a-\epsilon}^{a+\epsilon} - \psi''(x) dx = - (\psi'(a+\epsilon) - \psi'(a-\epsilon))[/itex]

    [itex]\int_{a-\epsilon}^{a+\epsilon} g \delta(x-a) \psi(x) dx = g \psi(a)[/itex]

    [itex]\int_{a-\epsilon}^{a+\epsilon} E \psi(x) dx \approx E \epsilon \psi(a)[/itex]

    So when you take [itex]\epsilon[/itex] very small, the right side is approximately zero, so we just have:

    [itex]- (\psi'(a+\epsilon) - \psi'(a-\epsilon)) + g \psi(a) = 0[/itex]

    So that's the derivative condition. You have to apply that for all three delta-functions (which have different values for [itex]a[/itex] and [itex]g[/itex]). So that gives you three equations. You also have to apply continuity at all three delta-functions. That gives you three more equations. So you have 6 equations and 6 unknowns (the coefficients for [itex]B, C, D, F, G, H[/itex]). If you solve them, that'll end up giving you an equation for [itex]k[/itex].
  6. Mar 18, 2016 #5
    So doing that, I get
    [itex] C=D\frac{2k}{\gamma} e^{2ka} (\frac{2k}{\gamma}-1) [/itex] from the first well
    [itex] G=F\frac{2k}{\gamma} e^{2ka} (\frac{2k}{\gamma}-1) [/itex] from the second well
    And then I use these with C+D=F+G from the barrier to get:
    D=F and C=G

    So that I have:
    [itex] Be^{kx}[/itex] for x<-a
    [itex] Ce^{kx}+De^{-kx}[/itex] for -a<x<0
    [itex] De^{kx}+Ce^{-kx}[/itex] for 0<x<a
    [itex] He^{-kx} [/itex] for x>a

    Am I on the right track here?

    Edit: I can also get B=H by using the D=F relation with the alternate solution from the wells:
    [itex] D=B\frac{\gamma}{2k} e^{-2ka} [/itex] from the first well
    [itex] F=H\frac{\gamma}{2k} e^{-2ka} [/itex] from the second well
    Last edited: Mar 18, 2016
  7. Mar 18, 2016 #6


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    I think that should be
    [itex] C=De^{-2ka} (\frac{2k}{\gamma}-1) [/itex]
  8. Mar 18, 2016 #7
    Yes. Yes it should.

    If I use the differential continuity at the barrier I come up with a different relation between C and D:
    [itex] D=C\frac{2+\beta/k}{2-\beta/k} [/itex]

    These seems at odds with the relation from the first well:
    [itex] C=D(\frac{2k}{\gamma}-1)e^{2ka} [/itex]

    Mathmatically they both seem correct but they don't agree with each other. Is this maybe because they're only valid at their respective potentials or are they just wrong?
  9. Mar 18, 2016 #8


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    Let's recap: You have the four regions:
    1. [itex]B e^{+kx}[/itex]
    2. [itex]C e^{kx} + D e^{-kx}[/itex]
    3. [itex]F e^{kx} + G e^{-kx}[/itex]
    4. [itex]H e^{-kx}[/itex]
    From the boundary between regions 1&2, you get:
    [itex]B e^{-ka} = C e^{-ka} + D e^{+ka}[/itex]
    [itex]k (C e^{-ka} - D e^{+ka} - B e^{-ka}) = -\gamma B e^{-ka}[/itex]

    From the boundary between regions 2&3, you get:
    [itex]C + D = F + G[/itex]
    [itex]k (F - G - C + D) = \beta (F+G)[/itex]

    From the boundary between regions 3&4, you get:
    [itex]F e^{ka} + G e^{-ka} = H e^{-ka}[/itex]
    [itex]k (F e^{ka} - G e^{-ka} + H e^{-ka}) = -\gamma H e^{-ka}[/itex]

    I think you made a mistake in solving these. However, here is a trick to simplify things: Try two guesses about the parity:

    First guess: [itex]\psi(-x) = \psi(x)[/itex] This implies: [itex]B = H, C=G, D=F[/itex]
    Second guess: [itex]\psi(-x) = -\psi(x)[/itex] This implies: [itex]B = -H, C=-G, D=-F[/itex]
  10. Mar 18, 2016 #9
    Ok, going through the calculations again, I've arrived at 3 equations for k, 1 for each boundary:

    For even symmetry

    Boundary 1: [itex] k=-k\frac{De^{ka}-Ce^{-ka}}{De^{ka}+Ce^{-ka}} +\gamma[/itex]
    Boundary 2: [itex] k=\frac{\beta}{2} \frac{D+C}{D-C} [/itex]
    Boundary 3: [itex] k=-k\frac{De^{ka}-Ce^{-ka}}{De^{ka}+Ce^{-ka}} +\gamma[/itex]

    And for odd symmetry:

    Boundary 1: [itex] k=-k\frac{De^{ka}-Ce^{-ka}}{De^{ka}+Ce^{-ka}} +\gamma[/itex]
    Boundary 2: k is kind of undefined here because [itex] F-G-C+D =0[/itex], I assume because the wavefunction is zero here?
    Boundary 3: [itex] k=-k\frac{De^{ka}-Ce^{-ka}}{De^{ka}+Ce^{-ka}} -\gamma[/itex]

    I'm a bit confused about this energy at the barrier, it doesn't depend on distance or anything.
  11. Mar 18, 2016 #10
    I understand that if this were a non-delta well, I could compare [itex] k=\sqrt{-E} [/itex] at the V=0 portion to [itex] \kappa =\sqrt{E+V_0} [/itex] inside the well by forming an equation of k as a function of [itex]\kappa[/itex] and finding the intersect of the two functions. But here that isn't possible here because there is no "inside the well".
  12. Mar 19, 2016 #11


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    I haven't checked whether those are right, but to get an eigenvalue for [itex]k[/itex], all three (or the first two, since the third is a repeat) equations must be true. So you have to solve the first equation for [itex]C[/itex] in terms of [itex]D[/itex] (which you already did, earlier). Plug that into the second equation. [itex]D[/itex] should then cancel out, giving an equation involving [itex]k[/itex] alone.
  13. Mar 19, 2016 #12
    Well, I get [itex] C=De^{2ka}(\frac{2k}{\gamma}-1) [/itex] as before
    Which I put into [itex] k=\frac{\beta}{2}\frac{D+C}{D-C} [/itex]
    Which comes out with [itex]k=\frac{\beta}{2}\frac{1+e^{2ka}(\frac{2k}{\gamma}-1)}{1-e^{2ka}(\frac{2k}{\gamma}-1)}[/itex]

    I'm having a hard time solving this for k, almost every term has a k in it. I've been checking the maths and it all seems to work, I'll keep checking though.
  14. Mar 19, 2016 #13


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    That's as good as you can get. It's a transcendental equation involving [itex]k[/itex]. To solve for [itex]k[/itex], you can either do it numerically, or you can try expanding [itex]k[/itex] in a power series (in powers of [itex]\gamma[/itex], maybe).
  15. Mar 19, 2016 #14
    The question wants algebraic solutions for E though. Although I'm not sure if this forbids a power series.
    Last edited: Mar 19, 2016
  16. Mar 19, 2016 #15


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    Well, I don't think it has a closed-form solution. If [itex]\beta=0[/itex], then I think the equation for [itex]k[/itex] turns out to be:

    [itex]k = \frac{\gamma}{2}(1 + e^{-2ka})[/itex]

    There is no closed-form solution to that equation. What you can do is to graph both sides as a function of [itex]k[/itex], and see for what value of [itex]k[/itex] the two curves intersect.
  17. Mar 19, 2016 #16
    I've been trying to solve it graphically but then I need values for [itex]\beta,\gamma[/itex] and a and the functions change drastically depending on the values of these.

    I'm also unsure about this solution since it's currently:

    [itex]Be^{kx}[/itex] for x<-a
    [itex]De^{2ka}(\frac{2k}{\gamma}-1)e^{kx}+De^{-kx}[/itex] for -a<x<0
    [itex]De^{kx} + De^{2ka}(\frac{2k}{\gamma}-1)e^{-kx}[/itex] for 0<x<a
    [itex]Be^{-kx}[/itex] for x>a

    So going through the barrier has no effect on the wavefunction and going through the second well somehow rids the wavefunction of the [itex](\frac{2k}{\gamma}-1)e^{2ka}[/itex] term. I'm just worried that the question calls for some kind of equation for E and I'm unable to provide one.
  18. Mar 24, 2016 #17
    So just in case people find this and want an answer. Yes, this is right and by "algebraic equations" the question didn't mean analytic so a relation that can be solved graphically is fine. Also, if you find that the odd solution ignores the central delta function completely, then you're headed in the right direction. Quantum mechanics is fun like that :).
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