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Double delta function potential: two bound states vs one ?

  1. Oct 21, 2013 #1
    In the double delta function potential well, where one delta function ( -αδ(x) ) is at -a and one at +a, if the energy is less than zero, there can be either one or two bound states, depending on the magnitude of α....if α is large enough, there can be two bound states, but if α is small, there is only one bound state.

    My question is, is there any physical intuition explaining why this is the case? Before solving this problem, I thought that since there are two delta wells, there will be two bound states, but this is not always the case.

    For example, if I have two electrons, and my α is large enough to admit two bound states, I assume that I can place each electron into a well.
    If however, α is small such that there can be only one bound state, then does this mean that I can only put one electron into a well, and the other well must be empty?

    I appreciate any insight!
  2. jcsd
  3. Oct 21, 2013 #2
    No, the single state overlaps both wells. They are both full.
  4. Oct 22, 2013 #3


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    In any even potential in one dimension, each energy eigenfunction is either even or odd. The ground state is always even, and the first excited state is always odd.

    If the two delta's are far apart, then (to a good approximation) the ground state is just the even combination of the two single-well bound states, and the first excited state is the odd combination.

    As we bring the delta's closer together, the odd combination has to switch more and more quickly from positive to negative between the two wells. This results in greater kinetic energy (from the large derivative of the wave function between the wells), which eventually becomes big enough to make the total energy positive. At that point, only the even combination (which does not have to change sign between the wells) continues to exist as a negative-energy bound state.
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