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When solving the simple problem of a free particle in a box of volume $V = L^3$, we can impose either periodic boundary conditions $\psi(0) = \psi(L)$ and $\psi '(0)= \psi'(L)$ either strict boundary condition $\psi(0)=0$ and $\psi(L)=0$.

Now, when looking for the eigenstates of the Hamiltonian, one can easily solve the problem with the separation of variables method, and impose the boundary conditions.

For the case of periodic boundaries, one has that the eigenvalue in each direction $(x,y,z)$ of space must be $\lambda_k = 2\pi k/L$, while in the strict boundaries you find $\lambda_k = \pi k/L$.

Now, in every textbook I could find, they say that in the first case we have that $k\in \mathbb{Z}$ and in the second $k \in \mathbb{N}$.

For the case $k = 0$ I can understand the difference, but I do not understand why negatives $k$ in the case of boundary condition must be discarded.

Edit : (Sorry for bumping) I understand why "physically" the solutions with periodic boundary conditions differ from the one with strict boundary conditions, and you're answer is perfectly sensible in that way.

However, surely there must be a way of showing (mathematically, I mean) that in the case of strict conditions choosing $k \in \mathbb{Z}$ yields "duplicate" solutions? Could someone offer some insight concerning this proof?

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    $\begingroup$ For the hard-wall boundary condition, the negative $k$ gives the same wave function up to a phase factor $-1$. So physically, it' s the same state. $\endgroup$ – Praan Nov 25 '15 at 22:28
  • $\begingroup$ But where lies the difference with the periodic boundary condition ? Why can't I apply your argument to the solution of the periodic boundary condition ? $\endgroup$ – Frotaur Jan 19 '16 at 16:35
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    $\begingroup$ Notice that $e^{ikx}$ and $e^{-ikx}$ are linearly independent, whereas $\sin (kx)$ and $\sin (-kx) = -\sin (kx)$ are not. $\endgroup$ – higgsss Jan 19 '16 at 16:47
  • $\begingroup$ Oh thank you very much higsss ! I wrote my solution for the periodic boundary condition on the form $\psi(x) = A\exp(ikx) + B\exp(-ikx)$ and that's why I thought that for $k \rightarrow -k$ the new $\psi$ obtained was the same as the first one. But actually, since A and B are arbitrary, I indeed have twice as many solution as in the strict boundary case. You're remark made me tilt. $\endgroup$ – Frotaur Jan 19 '16 at 17:17
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By imposing periodic boundary conditions you end up with the 3D version of the particle on a ring (particle on a 3-torus?).

If we drop down to 1D to keep things simple, then the difference between the ring and the 1D box is that on a ring we have two waves propagating round the ring in opposite directions e.g. one clockwise and one anti-clockwise. The two opposite signs for $n$ refer to these two waves. By contrast in the box you have a standing wave for which (as Praan says in a comment) the two signs for $n$ are physically indistinguishable.

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  • $\begingroup$ There should also be the additional difference that on the circle the co-domain of the position operator doesn't lie within the domain of the momentum operator (and viceversa), making the uncertainty relations not satisfied (I don't remember the details on top of my head now, though). $\endgroup$ – gented Nov 26 '15 at 14:01
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For the strict boundary conditions, the -n solution is linearly dependent with the n solution: it's the same wavefunction multiplied by -1.

$\psi_{-n}(x)=\sin(-n\frac\pi Lx) = -\sin(n\frac\pi Lx)=-\psi_{n}(x)$

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