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 ?

  • 1
    $\begingroup$ Your potential well is still single-dimensional, just infinitely deep. $\endgroup$ – Ruslan Feb 27 '14 at 17:34
  • $\begingroup$ There is no difference. Both families of functions span the same Hilbert space. $\endgroup$ – d_b Feb 27 '14 at 17:38

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 $\psi_b$.


Firstly, notice that if you take linear combinations of $\psi_f$ you get $\psi_b$ -- so the basis for both Hilbert spaces is the same.

Though I'm not sure how rigorous the following statement is, I'm tempted to say that: The wavefunctions do not sense any region outside the potential well, where we have infinite potential. So as far of the states of our theory are concerned, $V(x)=0$ -- i.e. even if you started with a finite-width infinite-depth potential well in 1-dimension, the problem reduces to that of a particle in a box. When acting on the basis states, $H$ and $p$ do commute in both cases.

Notice that the 'particle in a box' does not have translational symmetry just because $[H,p] = 0$. The boundary conditions choose special points and naively, those must break translational symmetry just as much as the (infinite-depth) finite-width well does.

If you hadn't broken the translational symmetry completely, you could imagine a bunch of boxes lined up along a line, or a bunch of infinite wells in succession (a periodic potential well somewhat like a crystal, where you still have some residual translational symmetry). Then you'd have had wavefunctions sitting inside each well. All those solutions would have been degenerate with corresponding solutions in your 'chosen' well. But since you've broken the translational symmetry and gotten rid of the other potential wells, you "lift" the degeneracy and the other states acquire a large (infinite) energy, and vanish.

This was somewhat of a braindump -- I'll come back later and try to make the response more coherent.

  • $\begingroup$ $\hat H$ and $\hat p$ can't commute if they don't have common eigenstates. So they do not commute for non-free particle. $\endgroup$ – Ruslan Feb 27 '14 at 17:51
  • $\begingroup$ What about 'particle in a box'? $\endgroup$ – Siva Feb 27 '14 at 19:39
  • $\begingroup$ For particle in a box you can't even consistently define $\hat p$, because its eigenstates must be $e^{ikx}$, and they contradict the boundary conditions, which demand that wavefunction vanishes at the boundaries. Another way to look at this is to use the whole $\mathbb{R}$ as the configuration space and then take a potential well with depth $U_{max}$ and take the limit $U_{max}\to\infty$ — you'll still get that $\hat H$ and $\hat p$ don't commute. $\endgroup$ – Ruslan Feb 27 '14 at 20:40
  • $\begingroup$ I agree that it's not obvious how to define a $p$-like operator for a compact interval. I thought that point of the question was to understand how one can do such a thing, and use that to solve the Schrodinger equation -- to understand how that is the same as solving for a particle in an infinite well. $\endgroup$ – Siva Feb 27 '14 at 20:51
  • $\begingroup$ Clearly, the wavefunctions you write in either case are eigenmodes of the $\frac{\partial^2}{\partial x^2}$ operator -- whether on the interval, or on the Real line. $\endgroup$ – Siva Feb 27 '14 at 20:53

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