Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a wavefunction $\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$. $A$ and $B$ are complex constants.

I am trying to find the probability density, so I need to find the product of $\psi$ with it's complex conjugate. The problem is, im not sure what is it's complex conjugate, I know the complex conjugate of $5+4i$ is $5-4i$, but what would be the complex conjugate of $\psi$? Is it just $-Ae^{i(kx-\omega t)}-Be^{-i(kx+\omega t)}$?

share|cite|improve this question

The complex conjugation factors through sums and products. So you can take the complex conjugate of the factor with A and B separately. The constant A and B form know problem, this goes according to the usual rules. This leaves something of the form $e^{(a+bi)}$. Now note that $e^{(a+bi)}= e^a(\cos(b)+i \sin(b))$ Taking the complex conjugate now and using $\cos(b)=-\cos(b)$ and $-\sin(b)=\sin(-b)$, you find the complex conjugate $e^{a+i(-b)}$.

This means: $\bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)}$

Note the use of the minus sign to compactly write the complex conjugate of $e^{a+ib}$. In computation this is what write, but you might want to keep the explanation of the $\cos$ and $\sin$ in the back of your head.

share|cite|improve this answer
It's easier to note once and for all that $\overline{e^z}=e^{\overline z}$, because the exponential function can be defined uniquely without committing to a choice between $i$ and $-i$ (e.g., by power series). – Henning Makholm Jan 1 '12 at 18:42
Thanks, so I want to find $\psi\bar{\psi}$. Do I just multiply it out like this? $(Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)})(\bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)})$, and so i get: $A\bar A+B\bar B+A\bar Be^{i(kx-\omega t)+i(kx+\omega t)}+B\bar Ae^{-i(kx+\omega t)-i(kx-\omega t)}$? Is that correct? – Thomas Jan 1 '12 at 18:46
Yes, if needed you can further simply to: $A\bar A+B\bar B+A\bar Be^{2ikx}+B\bar Ae^{-2ikx}$ – MrOperator Jan 1 '12 at 18:52

The complex conjugation will map $A \to \bar{A}$ and $B \to \bar{B}$.

If, say $A= 5 + 4 i$, then $\bar{A} = 5 - 4 i$, as you noted. So $$ \bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.