Regarding derivation of Probability Current The question for the full derivation of Probability Conservation -> Probability Current was already asked here: 
Probability current.  I apologize for not retyping it out, but it's already beautifully done in latex which I have yet to become proficient at.
I can follow most of the derivation. However, I'm still confused as to why one can take the complex conjugate of the actual Schrodinger Equation in the second step. What allows one to do this? It's not as if it's a wave-function, which can have a complex conjugate because it describes a state. I guess my question refers to how one can take the conjugate of a (momentum) operator, because that appears to be what's happening.
And if one is allowed to take complex conjugates of operators, why isn't this prevalent anytime we do use Schrodinger or other Equations??  Shouldn't they also be part of the full solution?
Thanks in Advance!
 A: Let's start from Dirac formalism:
when you have a ket $ | \psi \rangle$ and you apply an operator $P$, you obtain a new ket $P | \psi \rangle$. In Dirac representation the complex conjugate of a function $(\psi ^*)$ correspond to the bra $\langle \psi| $. Otherwise the bra of $P | \psi \rangle$ is $\langle \psi | P^+$ where $P^+$ is the adjoint operator of $P$. 
Fortunately in our case we use the momentum operator that's a self-adjoint one.
This fact and the reality of the potential ($V^* = V$) allow you to take the complex conjugate of the Schroedinger equation and to obtain the result that you see in the Probability current proof.
A: If you look at ChrisR's answer in Probability current, you can see that the complex conjugation is applied to both sides of the equation.
Since the terms on both sides are complex numbers, this is easy to do. It is not necessary to think deeply about what complex conjugation of some operator should be. Just write all the terms down and apply the conjugation to each term. 
