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The finite square well and the infinite square well problem are well known, however is there a reason that there is almost no reference to the one sided infinite square well?

Consider a particle with mass $m$ moving in the one dimensional potential $V(x)= \infty$ for $x<0$ , $V(x) = -V_{0}$ for $0\le x \le L$; $V(x) = 0$ for $x>L$
i) Can you scale this problem so that all units drop out ?
ii) Can you find the boundstate eigenenergies and associated wavefunctions as a function of the parameter $\lambda$?
iii) Can you find the free eigenstates which are characterized by the eigenenergies $E\ge 0$?

I searched Griffiths Quantum Mechanics, but it didn't have any clue to how to solve this.

Can anybody tell me the proper formal name, so I can look it up, or tell me a reason why none such does exist?

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up vote 3 down vote accepted

The one-sided infinite square well eigenfunctions are all the odd numbered eigenfunctions of the finite square well twice as wide, by reflection symmetry. The odd-parity solutions obey the boundary conditions for the infinite square well, so this is exactly the same problem as the symmetric finite square well.

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I think I have heard this potential called the semi-infinite square well. The name makes sense, at least; if I were going to pick a name for it that's probably what I would choose.

In any case, whether the potential has a name or not should have no effect on your ability to solve the corresponding Schrödinger equation. :-P

share|cite|improve this answer that takes me back to the beginning of grad school. – dmckee Nov 28 '11 at 1:15

In ye olde nuclear physics (i.e. from the time of the liquid drop model and the semi-empirical mass formula) these were often called "hard core" potentials. The term is not exclusive to a rectangular well, however: the defining feature is the (effectively) infinite potential at low radius.

Very useful things, too.

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Surely a hard core potential means the potential is infinite near the center but finite on both sides? – Harry Johnston Nov 27 '11 at 23:48
@Harry Work the problem in spherical coordinates...there is no "other" side of the radial coordinate. – dmckee Nov 27 '11 at 23:50
But isn't the spherical problem fundamentally different to the 1D problem? – Harry Johnston Nov 28 '11 at 3:38
@Harry: Generally the radial coordinate separates from the angular coordinates and you really do solve it (the radial part) as a one dimensional problem. – dmckee Nov 28 '11 at 5:02

Refer this pdf for this problem (detailed solution of this problem)

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Nikhil, Welcome to Physics.SE. Can you include a summary of the links material in your answer? – Ali Oct 22 '13 at 16:00

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