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I would like to know why there is a minus sign on the right-hand side of the Schrödinger's complex conjugate equation, whereas in the Schrödinger's equation there isn't. I know it is a simple question, but I don't know where this comes from. $$ -\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi = i \hbar \frac{\partial \psi}{\partial t} $$

$$ -\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* = -i \hbar \frac{\partial \psi^*}{\partial t} $$

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    $\begingroup$ Is "because $(i)^\ast=-i$" a good enough answer? Or what sort of answer are you expecting? $\endgroup$ – Emilio Pisanty Mar 10 '14 at 14:09
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It is the definition of complex number. Let's say

$z=x+iy\quad \Rightarrow z^*=x-iy$

$z=x-iy\quad \Rightarrow z^*=x+iy$

In simple words, you just have to change the sign of the Imaginary part. The thing is that $\psi(x)$ it's a imaginary number, so it's conjugate it's just $\psi^*(x)$. If you have the $\psi(x)$ function, then you can change $i\to -i$ or in the oposite way.

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I personally (maybe wrongly) see this feature as an early sign of the $CT$ symmetry where $C$ is the charge conjugate symmetry operation and $T$ is the time-reversal symmetry operation. Having no explicit charge in your equation, the charge conjugate symmetry operation would be simply taking the complex conjugate of the wave function while the $T$ operation would transform $t$ into $-t$. You can, as a matter of fact, notice that the minus sign you are bothered with disappears if you perform this $T$ transformation.

Hence the Schrodinger equation is invariant under $CT$ operation.

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Just because $\psi$ is a solution to the Schrodinger equation doesn't mean its complex conjugate is. After all, the Schrodinger equation is a pretty strong restriction, and not any random function is a valid wave function. So while it may be tempting to just substitute $\psi^*$ for $\psi$ in the Schrodinger equation, that's not something you're allowed to do.

But something you can do is take any arbitrary equation $A=B$ and take the complex conjugate of each side to conclude $A^*=B^*$. If you do that to the Schrodinger equation, you end up replacing all the $\psi$s with $\psi^*$s, leaving all the real bits alone (most of the constants and $V$)... and replacing $i$ with $-i$.

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