# Is the momentum in the infinite square well observable or not?

I've read in posts such as this and this that the momentum operator is not self-adjoint in the infinite square well because the geometric space is a bounded region of $$\mathbb R$$, for example $$[0,a]$$ for a well of width $$a$$. As such, it leads to weird stuff happening like momentum not being conserved.

What I don't understand is why the domain of the wave functions cannot be extended to $$\mathbb R$$ and have $$\psi$$ simply equal $$0$$ outside the well. That way, instead of integrating from $$0$$ to $$a$$, we can integrate from $$-\infty$$ to $$+\infty$$. Then $$\langle \psi | \hat p \psi \rangle = \frac{\hbar}{i}\psi^*\psi \bigg\rvert_{-\infty}^{+\infty} + \int_{-\infty}^{+\infty} \left(\frac{\hbar}{i} \frac{\mathrm d \psi}{\mathrm dx} \right)^* \psi \; \mathrm dx = \langle \hat p \psi | \psi\rangle,$$ and $$\hat p$$ would still be self-adjoint. Also, does the boundary condition for the stationary states $$\psi(0) = \psi(a) = 0$$ not predicate on the assumption that $$\psi = 0$$ outside the well? If $$\psi$$ was not defined outside the well, $$\psi$$ does not have to be continuous at the walls of the well, so $$\psi(0)$$ and $$\psi(a)$$ could equal any value.

• Commented Feb 26, 2022 at 7:54
• Why would momentum be conserved in this system? That doesn't seem weird at all to me. Commented Feb 27, 2022 at 10:54
• For anyone learning quantum mechanics and puzzled by this issue: don't worry, in a finite well the ordinary momentum operator is perfectly ok and you can always apply the ideas of standard quantum theory to a finite but very deep well. Commented Feb 27, 2022 at 12:08

## 1 Answer

It is not necessary to extend the wavefunctions to the whole real line. As far as I can understand, you are defining an operator $$\hat{p}$$ in $$L^2([0,a], dx)$$ with the domain $$D(\hat{p}) := \{ \psi\in C^1([0,a])\:|\: \psi(a) = \psi(0)=0\}$$ and acting in that way $$(\hat{p}\psi)(x):= -i\hbar \psi'(x)\:.$$ As you notice, that operator is Hermitian. However, in QM observables need to be selfadjoint operators, which is a much stronger requirement. Selfadjoint operators admit a spectral decomposition, simply hermitian ones do not.

From a mathematical perspective, posts you found yourself and other posts indicated by Qmechanic focus on related issues. In particular, if candidate momentum operators defined as above or in a similar way (with Dirichlet boundary conditions) admit a unique selfadjoint extension. The answer is negative. There is no good momentum observable in an infinite square well (i.e., with vanishing boundary conditions).