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If we have an infinite square well, I can follow the usual solution in Griffiths but I now want to impose periodic boundary conditions. I have

$\psi(x) = A\sin(kx) + B\cos(kx)$

with boundary conditions $\psi(x) = \psi(x+L)$

In the fixed boundary case, we had $\psi(0) = 0$ which meant $B=0$ and $\psi(L)=0$ which allows discrete values of $k$. I'm a little stuck with how to proceed with this in the periodic boundary condition case.

I think that $k = n\pi/L$ must still be true to satisfy the boundary condition (though I'm unable prove it). But now, I think that negative $n$ also matter and they're different to the positive $n$ case.

What is the general wavefunction solution in this case (for fixed boundaries, it was just $A\sin(kx)$)?

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It doesn't make sense to apply periodic boundary conditions to the infinite square well, since the solutions in each well will be independent from each other (it is not possible for a particle to tunnel between different wells).

Electrons will be strictly confined within the infinite square well of width $L$. When you consider the finite well, there is a tunnelling of electrons into adjacent wells. If the width of the well is $L$ and the period of the system is $D$ application of Bloch's theorem gives $|\psi(x+D)|^2 = |\psi(x)|^2$, but for an infinite well the wave function is strictly restricted to $0 \le x \le L$.

It is possible to consider a system made up of periodic $\delta$ function potentials (the so-called Dirac comb) and to make use of Bloch's theorem, but this makes use that though the height of the $\delta$ function is "infinite", it's "width" is zero. There is a limiting process that is going on in this case that doesn't occur in the infinite square potential of finite width (see D. J. Griffiths, Introduction to Quantum Mechanics).

If you look at the Kronig-Penney model, the dispersion relation is given by $$\cos ka = \cosh \beta b \cos \alpha(a - b) - \frac{\alpha^2 + \beta^2}{2 \alpha \beta} \sinh \beta a \sin \alpha(a-b)$$ (I'll leave you to look up the details on the link) there are no solutions when $V_0 \to \infty$ with $b$ (the width of the well) fixed (the Kronig Penney model considers the limit $V_0 \to \infty$ with $b \to 0$ such that $V_0 b =$ constant.

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  • $\begingroup$ Could you elaborate a little on why it doesn't make sense? I'm thinking of it just like we do for solid state where instead of the wavefunction decaying to zero at the edge of the sample, we instead imagine the sample is a ring. Can I not do the same thing here i.e. an infinite square well whose ends are connected instead of a having a wall? $\endgroup$ – user1936752 Jul 3 '16 at 5:52

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