Substitute Schrödinger Equation by complex analysis, was the substitution correct?

Question

Known that

$$ \frac{\partial \Psi}{\partial t}=\frac{i\hbar }{2m}\frac{\partial^2\Psi}{\partial x^2}-\frac{i}{\hbar}V\Psi \tag 1 $$

By taking the complex conjugate of the equation, obtain

$$ \frac{\partial \Psi^*}{\partial t}=-\frac{i\hbar }{2m}\frac{\partial^2\Psi^*}{\partial x^2}+\frac{i}{\hbar}V\Psi^* \tag 2 $$

where the derivative of the complex conjugate equal to complex conjugate of the derivative.

However, if we substitute $\Psi$ by $\Psi^*$ directly, we obtain $$ \frac{\partial \Psi^*}{\partial t}=\frac{i\hbar }{2m}\frac{\partial^2\Psi^*}{\partial x^2}+\frac{i}{\hbar}V\Psi^* \tag 3 $$

Thus if we compare the last two equation, then we obtain the following function that $$ \partial_x^2\Psi=0 \tag 4 $$ and $$ \partial_t\Psi^*=\frac{i}{\hbar} V\Psi^*. \tag 5 $$

My question was that:

1. Was equation 3 correct? I think it does because if the function was continuous differentialble in $C$, then by complex analysis we could rewrite the function as $\Psi(x+iy)=u(x,y)+iv(x,y)$.

2. What happened in equation 4 and equation 5? was they correct? especially, why $\partial_x^2\Psi=0$ the particle was not accelerate?

(This was not a homework question, because usually people only consider eq. 2 which was in the textbook.)

Answer 1

The fact that some function $\Psi$ satisfies a differential equation does not imply that $\Psi^*$ satisfies the equation too, so everything after equation 2 is nonsensical.

It is, of course, possible that $\Psi^*$ also satisfies the differential equation, but that would be a special case. I'm not sure why you would think otherwise. For example, the equation $$y' + iy = 0$$ is solved by functions of the form $Ce^{-ix}$ but not by their complex conjugates.

Lastly, acceleration is not a particularly meaningful concept in quantum mechanics. Position is generally not a well-defined property of a state, much less the derivatives of position with respect to time.