Is the momentum wave function's square amplitude always time-invariant for a free particle?

Question

I have noticed whenever working with free particles that the square amplitude of the momentum wave function $|\Phi(p)|^2$ ends up being time invariant, so I followed this chain of logic supporting the conclusion that this is always the case: $$\frac{\partial}{\partial t}\Phi(p,t) = -i\frac{p^2}{2m\hbar}\Phi(p,t)$$ $$\frac{\partial}{\partial t}\Phi^*(p,t) = i\frac{p^2}{2m\hbar}\Phi^*(p,t)$$ $$ \frac{\partial}{\partial t}|\Phi(p)|^2=\Phi^*(p,t)\frac{\partial}{\partial t}\Phi(p,t)+\Phi(p,t)\frac{\partial}{\partial t}\Phi^*(p,t) $$ $$=[-i\frac{p^2}{2m\hbar}+i\frac{p^2}{2m\hbar}]\Phi^*(p,t)\Phi(p,t)$$ $$=0$$ Where I adapted the momentum form of the Schrodinger equation from this post.

Have I done this correctly? Are there more general approaches to demonstrating invariance?

As a side-note, I find this result fascinating: according to the classical analogy, I would expect that the expectation value of the momentum probability would be time invariant for a free particle e.g. $\frac{\partial}{\partial t}\left\langle \Phi | p |\Phi \right\rangle$. However a much more strict case appears true in general: in the absence of any forces, the observable momenta of a particle never change, though the observable positions do.

Answer 1

Yes, the momentum-space probability distribution is time-independent for a free particle. This is because $\hat{p}$ commutes with the Hamiltonian $\hat{p}^2/2m$ of a free particle, and so the eigenstates of the Hamiltonian can be chosen to also be eigenstates of $\hat{p}$. In fact, these states are exactly the momentum eigenstates $\lvert p \rangle$.

This is general: If an operator (corresponding to a physical observable) $\hat{A}$ commutes with $\hat{H}$, then we can choose the eigenstates of $\hat{H}$ to also be eigenstates of $\hat{A}$, say, $\lvert E, a\rangle$, where \begin{align} \hat{H}\lvert E_n, a_n\rangle &= E_n \lvert E_n, a_n\rangle\,,\\ \hat{A}\lvert E_n, a_n\rangle &= a_n\lvert E_n, a_n\rangle\,, \end{align} where for convenience we are assuming no degeneracy; the generalization is not difficult. Then, the time-evolution of some initial state $\lvert \Psi (0)\rangle$ is $$ \lvert \Psi (t)\rangle = \sum_{n}e^{-iE_nt/\hbar}\lvert E_n, a_n\rangle\langle E_n, a_n \rvert \Psi(0) \rangle\,, $$ from which you can directly see that the probabilities $$ \Pr(A=a_n) = \lvert e^{-iE_nt/\hbar}\langle E_n, a_n \rvert \Psi(0) \rangle\rvert^2 =\lvert \langle E_n, a_n \rvert \Psi(0) \rangle\rvert^2 $$ for measurements of $\hat{A}$ are time-independent, because they are the same as the corresponding probabilities for measurements of the energy.

(Note that the translation between this general case and the momentum-space case is that $$ \lvert \langle E_n, a_n \rvert \Psi(0) \rangle\rvert^2 \to \lvert \langle p | \Psi(0) \rangle \rvert^2 = \lvert \tilde{\Psi}(p,0) \rvert^2\,. $$

Answer 2

Yes. An eigenstate of any operator that commutes with the Hamiltonian will have a time-invariant square amplitude, and the momentum operator commutes with the Hamiltonian for a free particle.