In Griffiths, there is a worked-through derivation for the solutions to the wave function for a delta-square potential case and for the infinite square well case. The infinite square well solution takes on the form of sine. There is also a problem in Griffiths discussed here: Bound States in a Double Delta Function Potential, which shows that the solutions in this double delta function potential case take on the exponential form. They are visualized here:

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I know that the sum of exponentials can be equated to sine, so it could be such that the "inside" of these delta-potentials is the same as the infinite square well case. However, the "outside" region of the potentials is non-zero. My questions about this are

  1. Why can we not equate these two situations - the double delta function scenario and the infinite square well
  2. Why isn't it relevant where the particle starts? My intuition is that there should be a rather small probability of the particle being outside of the delta barriers if the particle starts between them, but the solutions are actually symmetric about them.
  3. Why is the probability not smaller on one side of the potential, and why do the solutions reach an extrema at the value of the potential? I would expect them to not have a very large chance of being there because of the infinite potential, and even if they were, I would expect the amount that tunnels through to be very low, so I would expect the probability to at least decrease substantially on one side.

1 Answer 1

  1. Why would you equate the two situations? They are completely different. In the case of the infinite square well, the particle is restricted to be in a finite region of space, because the potential is infinite outside that region. In the case of the attractive double-Dirac potential, the particle is allowed to be anywhere, because the potential is finite (in fact bounded above) everywhere. The particle is more likely to be found "at" the positions of the delta functions, because the potential is "smaller" there than everywhere else ($-\infty < 0 $, after all), but it still is allowed to be anywhere.

  2. It is relevant where the particle starts. But the solutions above are not the solutions to the problem of finding the time-dependence of the wave function of a quantum particle given some initial state. The two solutions above are the bound-state solutions of the time-independent Schrodinger equation (TISE). In other words, they are the stationary bound states of the system, or, alternatively, the bound states of well-defined energy.

    There are also scattering states of well-defined energy, states that will be oscillatory everywhere with cusps at the positions of the delta functions. These states are necessary for examining the time-dependence of the system given arbitrary initial states.

    That is, if instead you wanted the answer to the question you asked, i.e., what happens if the particle "starts" between the delta functions, then you need to specify an initial state and solve the time-dependent Schrodinger equation (TDSE) to get a time-dependent wave function. Let's suppose you start the system in a relatively well-localized Gaussian wave packet between the two delta functions. Then as time goes on, the wave packet will spread out, and when it "hits" the positions of the delta functions, there will be multiple reflections and transmissions that occur. Some part of the wave function will remain between the two positions, and some parts of the wave function will be transmitted across the two positions. Note that the bound-state solutions above don't really contribute to this calculation (except that they might play the part of resonances, which underlies the physics of the reflections that occur at the positions of the delta functions).

  3. Again, your misconceptions have to do with the fact that you are thinking about a time-dependent problem, and the solutions above are the stationary states, i.e., the ones that don't evolve in time other than trivially. (That is, recall that in going from the TDSE to the TISE, one assumes that the solutions have the form $e^{-iEt/\hbar}\psi(x)$, and the trivial exponential time-dependence has no physical consequences in the case of the isolated system.) The fact that the solutions have a certain symmetry comes from the fact that the system itself is symmetric under the parity operation.

    Finally, we can get some food physical intuition for why these stationary bound states are localized to the positions of the delta functions: this is where the potential energy is the smallest, and so if we're looking for states that don't evolve in time, we're sometimes looking for states at the bottom of a potential well, just like a classical harmonic oscillator is stationary when the particle is at the equilibrium position, i.e., at the bottom of the (quadratic) potential well. (Don't take this semi-classical picture too far, but it at least serves the purpose of getting use used to the structure of bound states for systems like this.)


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