6

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.

Chris Yang
  • 533
  • 2
    Related: https://physics.stackexchange.com/q/362305/2451 , https://physics.stackexchange.com/q/66429/2451 , https://physics.stackexchange.com/q/45498/2451 and links therein. – Qmechanic Feb 26 '22 at 07:54
  • Why would momentum be conserved in this system? That doesn't seem weird at all to me. – AfterShave Feb 27 '22 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. – Andrew Steane Feb 27 '22 at 12:08

1 Answers1

4

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).

Valter Moretti
  • 71,578