# Complex Conjugate of Compound Wave Function

I came across a question which requires that we take the complex conjugate of $\psi_{1}+i\psi_{2}$ and the answer given was $(\psi_{1}-i\psi_{2})^{*}$. May I know why is it not $(\psi_{1}+i\psi_{2})^{*}$?

• That sounds incorrect, can you show exactly where you saw that, with more context? – knzhou Jul 19 '18 at 18:32
• I saw this on a problem set of the MIT course 8.04.1x, the question was to compute $\int_{-\infty}^{\infty} |\psi_{1}+i\psi_{2}|^{2}dx$ and the first step of the solution was $\int_{-\infty}^{\infty} |\psi_{1}+i\psi_{2}|^{2}dx$ = $\int_{-\infty}^{\infty} [\psi_{1}-i\psi_{2}]^{*}[\psi_{1}+i\psi_{2}]dx$. I'm not sure if there's an error in the solution. – Tianke Zhuang Jul 19 '18 at 18:40
• It is an error. The answer can be written either as $(\psi_1 + i \psi_2)^*$ or as $(\psi^*_1 - i \psi^*_2)$. – my2cts Jul 19 '18 at 18:55
• Yes, they made an error. I went to MIT and took this course. Unfortunately, the "EdX" problems are often put together by some overworked sleep-deprived undergrad that has just taken the course, paid about as much as a fast food worker. – knzhou Jul 19 '18 at 19:02
• Basically, you shouldn't trust something to be right just because it has the MIT logo on it. If it looks ridiculous, it is! – knzhou Jul 19 '18 at 19:03

For context, the question has you find $\int_{-\infty}^{\infty} |\tilde \Psi(x,t)|^2 dx$, where $\tilde \Psi(x,t) = \Psi_1(x,t) + i\Psi_2(x,t)$ or $\tilde \Psi(x,t) = \Psi_1(x,t) + \Psi_2(x,t)$, to show the effects of the phase difference of an arbitrary two normalizable states.
The definition of the absolute value acting on complex numbers is, $$|\tilde \Psi(x,t)|^2=\tilde \Psi^*(x,t)\tilde \Psi(x,t)$$ And since the solution doesn't follow this, they were incorrect.