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

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?

• Technically the infinite well is an idealization – Aaron Stevens Dec 11 '18 at 4:13

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.
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 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.