I have been long itched by this issue of subtle difference between box-normalised free particle and infinite-dimensional potential well.
Choosing a one dimensional case, the Hamiltonian in two cases respectively are :
$$ \hat H_f = \frac{\hat p^2}{2m} $$
$$ \hat H_b = \frac{\hat p^2}{2m} + V(x) $$
where $V(x)$ is zero inside the box and infinity everywhere else.
Now the eigen-functions of both the hamiltonian satisfy similar boundary conditions(BC), that the wavefunction has to vanish at the end points of the box.
So,
$$ \psi_f(x) = L^{-1/2}e^{ikx} $$
$$ \psi_b(x) = \sqrt{\frac{2}{L}}sin(\kappa x) $$
where $k = \frac{2n\pi}{L} $ and $ \kappa = \frac{n\pi}{L} $
Now the difference in the functional form is what is reflected in the boundary conditions.
And I have tried to convince myself that the difference in the functional form is due to the fact that $\psi_f$ is eigenfunction of $\hat p$operator as well (since $ [\hat H_f,\hat p] = 0 $). But that it is not so for $\psi_b$ since its hamiltonian($H_b$) doesn't commute with momentum.
So my question is, is this all the difference it makes or is there more to it ?
Best Answer
First, note that $\hat H_f$ has degenerate spectrum: it has equal eigenvalues for states with $\left|k\right\rangle$ and for $\left|-k\right\rangle$. This in turn means that, in particular, the state $\frac{-i}{\sqrt{2}}\left(\left|k\right\rangle-\left|-k\right\rangle\right)$ is also an eigenvector of $\hat H_f$. But in position representation it will look like:
$$\left\langle x\right|\frac{-i}{\sqrt{2}}\left(\left|k\right\rangle-\left|-k\right\rangle\right)=\frac{-i}{\sqrt{2}}L^{-1/2}\left(e^{ikx}-e^{-ikx}\right)=\\ =\frac1{\sqrt{2L}}2\sin(kx)=\sqrt{\frac2L}\sin(kx),$$
Which is identical to your $\psi_b$.
In case of a potential well the translational symmetry is broken, and the Hamiltonian spectrum is no longer degenerate. Thus the only way to select a solution is $\psi_b$.