Confusion of Schrödinger equation and complex conjugates I have a similar question that was asked in the following link:
(Schrödinger's Equation and its complex conjugate). But I find both the question and answers not specific enough. 
So let me rephrase the question. The Schrödinger equation for $\psi$ is given by
$$-\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi =  i \hbar \frac{\partial \psi}{\partial t}$$
So it is clear when one takes the complex conjugation of the above equation, it becomes $$-\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* =  -i \hbar \frac{\partial \psi^*}{\partial t}$$
Therefore the Schrödinger equation for $\psi^*$ has the minus sign in front of the time derivative term.
However, when one treats $\psi$ in the first equation as a placeholder, or a dummy variable, and replace it with $\psi^*$, the equation becomes $$-\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* =  i \hbar \frac{\partial \psi^*}{\partial t}$$ which cannot be right. 
My question then is why one cannot treat $\psi$ in the first equation as a placeholder? Where is the logical pitfall in replacing $\psi$ with $\psi^*$?
 A: You can always change the symbol that stands for a dummy variable, but you can't change its interpretation. Your mistake is tantamount to starting from the equation
$$x + 1 = 2$$
which has solution $x = 1$, then declaring $x$ is a dummy variable and replacing it with $-x$, for
$$-x + 1 = 2$$
which has solution $x = -1$. This is totally valid, but these two $x$'s don't mean the same thing.
A: $ψ$ and $ψ^*$ are not the same but are mutually connected.Taking a complex conjugate means finding the reflection of a point in the Argand plane about the real axis. So you cannot simply put $ψ$ as a substitute for $ψ^*$ and vice versa. So, if you need to replace  $ψ$ by $ψ^*$, you need to put a negative sign for where you see the operator $i$ as an image and it's mirror image are not same. By simply replacing $ψ$ by $ψ^*$ as you did in the above question, you are just treating $ψ$ and $ψ^*$ as same. That's not right.
A: I came into this question for couple of times which confused me often, so for the sake of my momorization I tried to write it out here.
Say the equation (1) (2) (3), from mathematics, we knew that complex conjugate of derivative = derivative of a complex conjugate for a differentialable function (https://math.stackexchange.com/questions/878258/showing-that-derivative-of-conjugate-is-conjugate-of-derivative-using-chain-rul ). However, this is exactly why the eq 3 was not always true.
Set $f(x)=u(x)+iv(x)$ and $V(x)=0$ in eq 1, 
there was 
$-\partial_x^2u(x)- i \partial_x^2v(x)=i\partial_t f(x)=i\partial_tu(x)-\partial_tv(x)$. for eq 1.
Notice $- i \partial_x^2v(x)=i\partial_tu(x)$ and $-\partial_x^2u(x)=-\partial_tv(x)$.
However, by expanding eq 3 there was $-\partial_x^2u(x)+ i \partial_x^2v(x)=i\partial_tu(x)+\partial_tv(x)$
Notice $i \partial_x^2v(x)=i\partial_tu(x)$ and $-\partial_x^2u(x)=\partial_tv(x)$.
However, this is not the case if one take the entire complex conjugate of the eq 1. 
Thus eq 3 is not always true.
