Physical significance of no self-adjoint momentum operator on half line?

Question

I am watching a quantum mechanics lecture by professor Schuller. He mentioned that there does not exist any self-adjoint momentum operator defined on the half line. What is the physical significance of this mathematical fact? Is it impossible to create a single impenetrable barrier for a quantum particle? This makes little sense to me because I know you can create an infinite well which consists of two barriers. Is it that a particle constrained like this has no notion of momentum?

Answer 1

1. Minutes before in the lecture Prof. Schuller is calculating the deficiency indices $d_{\pm}$. The kernels of $P^{\ast}\pm i$ are given by exponentially increasing/decreasing functions, respectively. On a half line $\mathbb{R}_{+}$ precisely one of these exponential functions are not square integrable, so that the two deficiency indices $d_{\pm}$ becomes different, and no self-adjoint extension exists.

2. On the other hand, a physics laboratory/experiment has in practice finite extent, so a more realistic mathematical model is given by a compact interval $I=[a,b]$, where the two deficiency indices $d_{\pm}$ match, and the self-adjoint extension exists.

Related: What's the deal with momentum in the infinite square well?

Answer 2

From a Mathematician's perspective, you can define a "momentum" operator on the half line, but you need to allow 2d vector functions instead of scalar functions. The problem is that differentiation on $[0,\infty)$ has only one endpoint condition at $x=0$, and any setting (or lack of setting) results in a non-self-adjoint operator. By introducing a 2-d momentum variable, there are then two conditions and a one-parameter family of conditions that are equivalent to connecting the two complex components at $0$ by a multiplier $e^{i\theta}$. Whether or not such a thing is Physical is something I cannot judge.