Question about the linearity of wave functions

Question

For piece-wise constant potential, the potential energy is constant so the time dependent wave function can take the form $\psi(x,t)=C_1e^{i(kx- \omega t)}+C_2e^{i(-kx-\omega t)}$ where $k=\frac{p}{h}=\frac{\sqrt{2mE}}{h}$ and $E$ is just the difference between the energy of the particle and the constant potential.

When testing $\psi(x,t)$ as an eigenfunction of the momentum operator, the result is that $-i\hbar \frac{\partial \psi(x,t)}{\partial t}=\hbar k(C_1e^{i(kx- \omega t)}-C_2e^{i(-kx-\omega t)}) \not= p\psi(x,t)$

But when testing the two components of the wave function separately, it is found that they are eigenfunctions.

$-i\hbar \frac{\partial \psi_1(x,t)}{\partial t}=-i\hbar \frac{\partial}{\partial t} (C_1e^{i(kx-\omega t)})=-i \hbar (ik\psi_1)=\hbar k\psi_1=p_x\psi_1$

$-i\hbar \frac{\partial \psi_2(x,t)}{\partial t}=-i\hbar \frac{\partial}{\partial t} (C_2e^{i(-kx-\omega t)})=-i \hbar (-ik\psi_2)=-\hbar k\psi_2=-p_x\psi_2$

$\psi_1$ describes the particle moving in the $+x$ direction and $\psi_2$ describes the particle moving in the $-x$ direction.

So if two solutions to the S.E. decribes definite states of momentum, the sum $\psi(x,t)$ does not necessarily have to describe a definite state of momentum?

The functions $\psi(x,t),\psi_1(x,t), \psi_2(x,t)$ are eigenfunctions of the energy operator $i\hbar \frac{\partial}{\partial t}$, and the commutator $[H,P]=i\hbar \frac{\partial V(x)}{\partial x} $ which means that definite states of energy and momentum can be found simultaneously as long as we have a constant potential. (However, $H$ can only be used in the time-independent case right? So I'm not sure if this applies.)

It's evident from the math above that $\psi(x,t)$ is not an eigenfunction of the momentum operator, but is there a physically reason for why it shouldn't be?

Answer 1

Correct. The state $\psi$ you've described is a superposition of two momentum eigenstates, with momentum $p$ and $-p$. So $\psi$ does not itself have a definite momentum - if you measured the momentum of a particle in this state you'd get either $p$ or $-p$ with a probability distribution that depends on the $C_1, C_2$.

The energy of a free particle depends only on its speed, not its direction of movement, so the energy basis is degenerate. Here your state has a definite value of energy because in all cases, the particle is moving with the same speed. However, momentum is a vector quantity and does depend on the direction of movement.

In general, a linear combination of two eigenstates of a given operator is not going to be another eigenstate of that operator - unless the two eigenstates also have the same eigenvalue. For example, here $\psi_1$ and $\psi_2$ are both eigenstates of $H$ with the same eigenvalue (energy) $E$, so $\psi$ is still an eigenvalue of $H$. But since $\psi_1$ and $\psi_2$ are eigenstates of $P$ with different eigenvalues (momenta) $p$ and $-p$, then $\psi$ is not an eigenstate of $P$.

(By the way, you wrote: "$H$ can only be used in the time-independent case right?" It requires a time-independent potential. It's fine to use it with time-dependent wavefunctions.)