Let's say, initially the state is in first excited state of finite well potential and then I change the width & depth of the well, eventually to Dirac delta potential, then what happens to the state of the wave function?
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asked Apr 5, 2018 at 19:33
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$\begingroup$ Have you reviewed the unbinding of states in a finite potential as its depth or width varies? $\endgroup$– Cosmas ZachosCommented Apr 5, 2018 at 19:40
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$\begingroup$ Possible duplicate of Particle in infinite potential well which is doubled in size at $t_0$ $\endgroup$– sammy gerbilCommented Apr 7, 2018 at 15:16
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$\begingroup$ See also 1D Infinite Square Well: Box Suddenly Increases in Size. How treat this? If you are asking about slow changes see the Adiabatic Theorem. $\endgroup$– sammy gerbilCommented Apr 7, 2018 at 15:34
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1 Answer
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If the change is adiabatic, i.e. in the limit where the rate of change of the potential is slow, then your state will evolve as an instantaneous eigenstate of the potential.
However, if you keep squeezing the potential, then at some point the well will no longer support an excited state, i.e. it will be 'expelled' into the continuum. At that point, the potential's change can no longer be considered adiabatic, because the marker that sets what "slow" means in an adiabatic evolution is the energy gap between your state and the nearest eigenstate, which goes to zero in the continuum. Moreover, this 'expulsion' happens at a finite width for the well, so there's no need to wait until you get to a truly Dirac-delta potential.
(You could try to make the well deeper as you make it narrower in such a way that the first excited state will stay at a constant energy depth below the continuum. That won't work, though, because the ground-state energy will sink to $-\infty$, and it won't give you a well-defined operator as a limit.)
So: no, an isolated Dirac-delta potential cannot support more than one bound state.
answered Apr 19, 2018 at 9:06