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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 $\psi^*$?

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    $\begingroup$ The point is that $\bar{\psi}$ is not a solution of the Schrödinger equation $i\partial_t u=-\Delta u+V(x)u$ whenever $\psi$ is a solution. That is shown as you did, by taking the complex conjugate of the equation solved by $\psi$. Therefore, you cannot use $\bar{\psi}$ as a "variable symbol" in an equation that you already know it is not satisfied by it. It is essentially like using the same symbol for an extremum of integration and the variable of integration. $\endgroup$ – yuggib Apr 23 '16 at 9:52
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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.

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    $\begingroup$ This is correct. Maybe the confusion comes from believing that $\psi$ and $\psi^*$ are independent; they are not. If you solve for one, you determine the other. $\endgroup$ – anon01 Apr 23 '16 at 1:03
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$ψ$ 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.

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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.

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