- #1
Tony Hau
- 101
- 30
- TL;DR Summary
- Given a wavefunction ##\Psi##, why is it that ##\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}##?
It is a rather simple question:
In my textbook it writes something like: $$\frac {\partial \Psi} {\partial t}= \frac{i\hbar}{2m}\frac {\partial^2 \Psi} {\partial x^2}- \frac{i}{\hbar}V\Psi$$
$$\frac {\partial \Psi^*} {\partial t}= -\frac{i\hbar}{2m}\frac {\partial^2 \Psi^*} {\partial x^2}+\frac{i}{\hbar}V\Psi^* $$
I don't understand why, because I have never learned the topic of complex function. My educated guess is as follows:
Assume that ## \Psi^{*}(x,t) = |Z|e^{i\omega t}##, and ##\Psi(x,t) = |Z|e^{-i\omega t}##. Obviously ##\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}##, which then can be generalized into any case. Am I right?
In my textbook it writes something like: $$\frac {\partial \Psi} {\partial t}= \frac{i\hbar}{2m}\frac {\partial^2 \Psi} {\partial x^2}- \frac{i}{\hbar}V\Psi$$
$$\frac {\partial \Psi^*} {\partial t}= -\frac{i\hbar}{2m}\frac {\partial^2 \Psi^*} {\partial x^2}+\frac{i}{\hbar}V\Psi^* $$
I don't understand why, because I have never learned the topic of complex function. My educated guess is as follows:
Assume that ## \Psi^{*}(x,t) = |Z|e^{i\omega t}##, and ##\Psi(x,t) = |Z|e^{-i\omega t}##. Obviously ##\frac {\partial \Psi^*} {\partial t} = - \frac {\partial \Psi} {\partial t}##, which then can be generalized into any case. Am I right?