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Consider the particle in a box problem in QM. The crux of the reason why QM is able to explain the physical phenomenon is not just the theory but also able to impose boundary conditions which eventually result in quantization. Now in the particle in a 1-d box problem, the wave function is assumed to be zero at the boundaries. It has been said that it is imposed, so that the wave function is continuous. Okay, but what about differentiability? In order for the wave function to satisfy Schrodinger equation,we also need differentiability right? Okay if we assume only left (from one side) derivative to exit, we could have as well assumed only left continuity (from one side). For continuity, we assume it should be from both sides, but for differentiability we need only one side? They also say the slope also must be continuous. I don't see any rationale behind these quantizations!

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The more I learn about QM, the more ugly it looks, and the mathematics involved is far from beautiful. –  Rajesh D Jul 9 '13 at 6:43
Have you learned the abstract Hilbert space formulation of QM, along the lines of Dirac's lectures? Or maybe the path integral formulation would be more to your liking. In any case the ugliness of the boundary conditions you are complaining about really comes from the highly idealised nature of the situations you are studying. Real systems have potentials with no infinities, so the wavefunctions are smooth, or at least $C^1$. –  Michael Brown Jul 9 '13 at 7:26
Michael Brown : yes to some extent, but I enjoyed it. I like the Hilbert space formalism and the measurement being defined as a Linear self adjoint operator, eigen values are the observables. –  Rajesh D Jul 9 '13 at 16:33
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1 Answer

Differentiability of the wave function is only required for finite changes in the potential. If the your potential is infinite (as it is outside the inifinitely deep potential well which you describe) the Hamiltonial is ill-defined anyways.

An other case where you can have an infinite potential is if you have a $\delta$-distribution as a potential, there again you will find that the wave-function must be continuous but not differentiable (the difference between left-side and right-side derivative is given by the strength of the $\delta$-potential in this case).

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If we assume Hamiltonian is ill defined, why can't we assume only left continuity then, I keep getting puzzled as to how these QM people keep making assumptions without any rationale! –  Rajesh D Jul 9 '13 at 6:41
On the intervals with $V = \infty$, you have $\Psi(x) \equiv 0$. Whether you take left-continuity and have $\Psi(-a) = 0, \Psi'(-a) = 0, \Psi'(-a + \epsilon) \neq 0$ if the potential drops from $\infty$ to $0$ at $-a$ or if you have $\Psi(-a) = 0, \Psi(-a-\epsilon) = 0, \Psi(-a) \neq 0$ makes no meaningful difference. As left-continuity is physically equivalent to right-continuity, you can simply work with discontinuity in general (for the derivative). –  Neuneck Jul 9 '13 at 6:47
I don't quite understand what you mean to say. Do you mean to say continuity from both sides of boundary is also not required and only one side is required? –  Rajesh D Jul 9 '13 at 6:53
My question is why do you deliberately make the wave function it go to zero at the boundaries? –  Rajesh D Jul 9 '13 at 6:57
The wave function must be continuous, as does any function that wishes to carry physical information. The only solution to $H \Psi = E \Psi$ for $H = \partial^2 + \infty$ is $\Psi = 0$, thus, where $V = \infty$ $\Psi = 0$. The wave function's continuity then demands $\Psi = 0$ at the boundary, but we have no information about $\Psi'$ at the boundaries, so we allow it to take any value. –  Neuneck Jul 9 '13 at 8:33
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