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?


1 Answer 1


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.

  • $\begingroup$ Yes, I understand that. But how can I be sure that the three will commute? Is there a proof for it? $\endgroup$ Commented Mar 9, 2022 at 6:47
  • $\begingroup$ @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. $\endgroup$
    – Andrew
    Commented Mar 9, 2022 at 12:02
  • $\begingroup$ @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) $\endgroup$ Commented Apr 2, 2022 at 17:30

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