# Why do $\psi, \psi^*$ and the potential energy operator commute here?

This is the one-dimensional time-dependent Schrodinger Equation: $$i\hbar \frac{\partial \psi}{\partial t}= -\frac{\hbar^2}{2m}\frac{\partial^2 \psi }{\partial x^2} +\displaystyle {\hat {V}\psi}$$<br My textbook takes the complex conjugate of this equation(note that $$\psi^*$$ is the conjugate of $$\psi$$), $$-i\hbar \frac{\partial \psi ^*}{\partial t}= -\frac{\hbar^2}{2m}\frac{\partial^2 \psi ^* }{\partial x^2} +\displaystyle {\hat {V}\psi^*}$$
and multiplies it by $$\psi$$ $$-i\hbar\psi \frac{\partial \psi ^*}{\partial t}= -\frac{\hbar^2\psi}{2m}\frac{\partial^2 \psi ^* }{\partial x^2} +\displaystyle {\hat {V}|\psi|^2}$$
My Question: How can $$\psi \displaystyle {\hat {V}\psi^*}= \displaystyle {\hat {V}|\psi|^2}?$$Does it not depend on the nature of $$\displaystyle {\hat {V}}?$$
For example, if $$\displaystyle {\hat {V}}$$ was $$-i\hbar\frac{\partial}{\partial x}$$, given that $$\psi = e^{i(kx-\omega t)}$$($$k$$ and $$\omega$$ are constants),
$$\psi \displaystyle {\hat {V}\psi^*}=-e^{i(kx-\omega t)} i \hbar(ik e^{-i(kx-\omega t)})=\hbar k$$ whereas
$$\displaystyle {\hat {V}|\psi|^2} = \displaystyle {\hat {V}1}=0!$$
Or is $$\displaystyle {\hat {V}|\psi|^2}$$ just a notation?
Note: I know the $$V$$ I've chosen is the momentum operator; I'm only trying to show that the $$\psi, \psi^*$$ and an operator do not necessarily commute that way.
What's wrong with my understanding?

Conventionally, $$V$$ is used to denote the potential, and is a function only of position and not momentum. This follows the usage in classical mechanics, where the potential is a function of position only. You are right that if $$V$$ was a generic operator that could depend on momentum, then you would need to be more careful.
• @AmbicaGovind In the position basis, $\hat{x}$ is just an ordinary number. The same goes for any function of $x$, such as the potential. Since multiplication of ordinary numbers is commutative, you don't need to be careful about ordering. Commented Mar 9, 2022 at 12:02
• @AmbicaGovind multiplication of complex function and complex conjugate is squared of that. $$\psi\psi^*=|\psi|^2$$ that's identity in complex analysis. If you multiply both side by $\phi$ the identity doesn’t change (taken $\phi$ as just example) Commented Apr 2, 2022 at 17:30