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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 S equation being not invariant under charge conjugation, differently from KG equation?

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  • $\begingroup$ That Schrödinger equation there knows nothing about "charge". Why do you think $\psi\mapsto \psi^\ast$ is "charge conjugation"? $\endgroup$ – ACuriousMind May 30 '16 at 18:12
  • $\begingroup$ 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. $\endgroup$ – Luthien May 30 '16 at 18:18
  • $\begingroup$ 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. $\endgroup$ – Luthien May 30 '16 at 18:42
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Complex conjugation has nothing to do with charge conjugation. Charge conjugation flips quantum numbers, which don't appear at all in the standard Schrodinger equation.

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.

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  • $\begingroup$ 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? $\endgroup$ – Luthien May 30 '16 at 18:34
  • $\begingroup$ @Luthien Basically, yes. $\endgroup$ – tparker May 31 '16 at 3:39

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