# Schrodinger equation in terms of complex conjugate

## Main Question or Discussion Point

I know there's a similar post, but i didn't understand it. Why the derivative respect to t in terms of the complex conjugate of ψ is:

instead of being the original S.E in terms of ψ*

or the equation in terms of ψ with the signs swapped

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Paul Colby
Gold Member
One is simply applying conjugation to the equation as a whole. This yields an equation which is valid provided the original is valid. Some c-number conjugation rules that may be helpful are,

1. $(a b)^\ast = a^\ast b^\ast$
2. $(a + b)^\ast = a^\ast + b^\ast$
3. $(i)^\ast = -i$
4. $(\frac{\partial \psi}{\partial t})^\ast = \frac{\partial \bar\psi}{\partial t}$

I got it!. I always forget to use the product rule, Thanks