# Is a state with definite momentum possible in presence of an infinite square-well potential? [duplicate]

The momentum operator is usually defined as

$$\hat p = -i \hbar \frac d {dx}$$

The eigenfunctions of this operator are plane waves

$$\tag{1} \label{1} \psi_p(x) = A \exp\left(\frac{i p x}{\hbar}\right)$$

and they are not square-integrable, i.e.

$$\int_{\mathbb R} | \psi_p(x)|^2 dx = \infty$$

therefore, it is usually said that a free particle with a well-defined momentum is a physical impossibility.

But we have considered a case in which the wavefunction $\psi$ can extend from $-\infty$ to $\infty$. In general, the wavefunction $\psi$ should satisfy the following conditions:

$$\tag{2} \label{2} \psi \in C^0 (\mathbb R)$$ $$\tag{3} \label{3} \int_\mathbb{R} | \psi(x)|^2 dx = 1$$

In the case of an infinite square-well potential

$$V(x) = 0 \ \ \ \text{if} \ x \in [0,L]$$ $$V(x) = \infty \ \ \ \text{if} \ x \notin [0,L]$$

we usually require that

$$\tag{4} \label{4} \psi(x) = 0 \ \ \ \text{if} \ x \notin [0,L]$$

so that \ref{2} implies

$$\tag{5} \label{5} \psi(L)=\psi(0)=0$$

But by applying \ref{5} to \ref{1}, we obtain

$$\tag{6} \label{6} e^{ipL/\hbar}=1=0$$

which is clearly absurd and cannot be valid.

However, in this article the authors introduce a self-adjoint extension of the operator $\hat p$ (which they call $P_\theta$) by replacing \ref{5} with

$$\tag{7}\label{7} \psi(L)=e^{i \theta} \psi(0)$$

from which they obtain the "momentum eigenfunctions"

$$\tag{8} \label{8} \phi_n(x,\theta) = \frac 1 {\sqrt{L}} \exp \left(2 i \pi \nu \frac x L \right), \ \ \ \nu = n + \frac{\theta}{2 \pi}, \ \ \ n \in \mathbb Z$$

They conclude with the remark

For a particle in a box, it is often argued that the “physical” wave function should continuously vanish on the walls $x = 0$ and $x = L$, ensuring that the presence probability vanishes continuously for $x ≤0$ and for $x ≥ L$. One should realize that the continuity of the measurable quantity

$$Pr(0 ≤ x ≤ u) = \int_0^u |\phi(x)|^2 dx, \ \ \ u \in[0,L]$$ is ensured as soon as the integral $\int_0^L |\phi(x)|^2 dx$ does converge and does not require any continuity property of $\phi(x)$. Specializing this remark to the eigenfunctions of $P_\theta$ we observe that $|\phi_n(x,\theta)|^2$ does not vanish continuously at $x=0$ but nevertheless the physical quantity $$Pr(0 ≤ x ≤ u) = \frac{u}{L}$$ vanishes continuously, as it should, for $u \to 0$.

They also say that the Heisenberg inequlity is not valid for the "momentum eigenfunctions" of Eq. \ref{8}, and indeed we would obtain

$$\sigma_p = 0$$ $$\sigma_x = \frac{L}{\sqrt {12}}$$

From a mathematical point of view, their reasoning is more or less clear to me: if we replace \ref{5} with \ref{7} and restrict ourselves to the interval $[0,L]$, it is possible to extend the momentum operator, making it self-adjoint, and to introduce eigenfunctions for this extended version. But is this reasoning physically meaningful?

In particular, what about the value of the wavefunction outside the interval $[0,L]$? Also, what about the Heisenberg inequality? Is the fact that it is not valid acceptable from a physical point of view?

Update

I don't understand if the authors are implying that the wave function is set to $0$ for $x \notin [0,L]$; if this was the case, shouldn't we have the problem that $Pr(0\leq x \leq u)$ is discontinuous at $u=L$?