# Schroedinger and Klein-Gordon equation and their complex conjugate

Let's consider the Schroedinger equation: $$\begin{equation} i\hbar\frac{\partial}{\partial t}\psi=-\frac{\hbar}{2m}\nabla^2\psi \end{equation}$$ If I have a wavefunction $$\psi$$ as a solution, then its complex conjugate $$\psi^*$$ is not a solution. If I'm not mistaken this means that the Schroedinger equation is NOT invariant under charge conjugation C, am I right? And what does $$\psi^*$$ represent, if $$\psi$$ is the wave function describing my particle?

When I move to Klein-Gordon equation my book says that $$\psi$$ and $$\psi^*$$ are both solutions of KG equation, where $$\psi$$ contains the operators for destruction of particle and creation of antiparticle and $$\psi^*$$ for creation of particle and destruction of antiparticle. Does this mean that KG equation is invariant if I "complex-conjugate" it? And if this is the case what is the physical meaning of Schroedinger equation being not invariant under charge conjugation, differently from KG equation?

• That Schrödinger equation there knows nothing about "charge". Why do you think $\psi\mapsto \psi^\ast$ is "charge conjugation"? May 30, 2016 at 18:12
• Found a question here on stackexchange (physics.stackexchange.com/q/102838) where it was said "Having no explicit charge in your equation, the charge conjugate symmetry operation would be simply taking the complex conjugate of the wave function". But I'm new in symmetries concepts and so on. May 30, 2016 at 18:18
• Ok I thought about it and you're right, it's totally wrong in this context, i was thinking about "complex conjugation" when I wrote this question. May 30, 2016 at 18:42

The actual symmetry related to complex conjugation is time reversal. However, to actually perform time reversal, you must also replace $i$ with $-i$, so the time reversed Schrodinger equation is $$-i\hbar \frac{\partial}{\partial t} \psi^* = - \frac{\hbar}{2m} \nabla^2 \psi.$$ This is equivalent to the original equation, so the Schrodinger equation is time symmetric. More generally, this is because the Hamiltonian is Hermitian.
• Does this mean that $\psi^*$ is my solution $\psi$ that has been time-reversed? I mean, if $\psi$ is a solution in my "original" world, is $\psi^*$ a solution in a "time-reversed" world? May 30, 2016 at 18:34