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Bizarre results of 1-dimensional delta-function potential well

  1. Jul 12, 2010 #1
    I was curious what a delta-function potential barrier would do to a free (E>0) 1-dimensional particle, so I figured it out. I used V = b δ(x) for the potential energy. I got the expected result that full reflection will occur as |b|→∞ or as E→0. These results make sense intuitively when b>0. My intuition is mystified because I get the same result when b<0. Why should a potential well cause complete reflection? I was hoping someone could help my intuition grasp this result. Or maybe I did something wrong in my work, so I am showing my work below...

    First of all, I give the particle mass, which I call m, and define
    k = √(2 m E / ħ²)
    Since the wave function must be continuous at x=0, I get
    ψ1 = A (ei k x+B e-i k x)
    ψ2 = A (1+B) ei k x
    where ψ1 is the incoming and reflected particle on the left, and ψ2 is the transmitted particle on the right. A is the normalization constant, but, since the particle is free, there is no need to solve for it. I then worked out that the fraction reflected is
    R = |B
    Using the rule about how the slopes of wave functions change across delta-function potentials, I solved for B, and I got
    R = 1/(1+W) where W ≡ 2 ħ² E / (m b²)

    Thanks in advance for any insight!
     
  2. jcsd
  3. Jul 12, 2010 #2
    The result that you have got is correct. Try to imagine the scenario in terms of wave mechanics. Whenever a wave is incident on a boundary, no matter how strange, it has to split into two components: a reflected wave and a transmitted wave. It is from this principle that you see that even while having a potential well, rather than a barrier, there is still a reflected component. As quantum mechanics also has a wave interpretation (even though not exactly the wave behavior that one would expect, as in classical mechanics), the same result holds for the case of quantum mechanics.
     
  4. Jul 13, 2010 #3
    Excellent! You have definitely helped me remove a bunch of wrong ideas floating around in my head, but I am still confused about something.

    After you mentioned that ANY boundary creates reflections, I looked at the situation of a free particle on the left where V=0 encountering a potential drop on the right. I got that, as the potential drop becomes -∞, R→1 and the wave function on the right becomes 0. Since real particles are sums of these completely-reflected eigenfunction, my results means that real particles should also be completely reflected. But real particles are NOT completely reflected by potential drops!
     
  5. Jul 13, 2010 #4
    If I am right, then you are taking b to approach infinity to show that R approaches 1 in that case. But, to make a Dirac delta potential, this is not how you do it. The total area (represented by b) in a Dirac delta function is finite; it is the magnitude of the function that is infinite. But when you take b to approach infinity, you make the area of the function approach infinity, and no longer get the results for a Dirac delta potential. What you then have (when taking b to infinity) is a derivative of Dirac delta function, which is even more ill-defined than the original function, and forget about the same relation (R and T) holding for that case, I think that even the wavefunction becomes discontinuous at the boundary for such a potential!
     
  6. Jul 13, 2010 #5

    DrDu

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    I don't know what you mean by a "real" particle. However, the situation is analogous to light ("photons") falling upon a medium of high index of refraction. The higher the index of refraction the higher the percentage of light reflected.
    To compare with classical mechanics, take in mind that the classical approximation holds when the distance, over which the potential changes is large compared to the de Broglie wavelength of the particle. For a sharp step or a delta function this is never the case.
    There are other potentials which are exactly solvable and where the influence of this condition can be studied, namely the Eckart potential (Google is your friend).
     
  7. Jul 13, 2010 #6

    DrDu

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    No Jivesh, you don't get the derivative of a delta function when b goes to infinity.
    Considering the limit of b to infinity for the Reflection is certainly ok.
     
  8. Jul 13, 2010 #7
    But when we take b to approach infinity, then aren't we saying that the area under the Dirac delta curve is infinite? If I integrate a derivative of delta function, then I get back the same delta function and with a suitable limits, I can get the required infinite area. Is there any other function that can have the same definition as the delta function for b approaching infinity?
     
  9. Jul 13, 2010 #8

    DrDu

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    bknock already defined the delta function and b is only a prefactor. There is nothing wrong in considering the limit that b becomes arbitrarily large.

    The derivative of a delta function, in the sense you use it, would not lead to a well defined Hamitonian. That integrating over the derivative of a delta function does yield a delta function has nothing to do with the problem at hand. After all, there is no integration present in the hamiltonian, is it?
     
  10. Jul 13, 2010 #9
    DrDu, you have completely fixed my confusion! Thank you. And thank you Jivesh :)
     
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