Correct way to take complex conjugate of the Schrodinger equation

Question

There are one or two questions on this site regarding the complex conjugate of Schrodinger equation but they do not clear my doubt.

Question : The Schrodinger equation is $$iħ \frac{∂Ψ}{∂t} = HΨ $$ How to take the complex conjugate of it, I think there are two ways to do it.

1. $$\left( iħ \frac{∂Ψ}{∂t}\right)^* = \left(\frac{-ħ^2}{2m}\frac{\partial^2Ψ}{\partial x^2}+VΨ \right)^*$$ $$\implies \frac{∂Ψ^*}{∂t} = \frac{1}{-iħ}\left(\frac{-ħ^2}{2m}\frac{\partial^2Ψ^*}{\partial x^2}+VΨ^* \right) \tag{1}\label{e1}$$

2. $$\left( iħ \frac{∂Ψ}{∂t}\right)^* = \left(H Ψ \right)^*$$ $$\implies \frac{∂Ψ^*}{∂t} = \frac{1}{-iħ} Ψ^* H^* = \frac{1}{-iħ} Ψ^* H $$ $$\implies \frac{∂Ψ^*}{∂t} = \frac{Ψ^*}{-iħ}\left(\frac{-ħ^2}{2m}\frac{\partial^2}{\partial x^2}+V \right) \tag{2}\label{e2}$$ Which one of the above two is correct? It seems to me that first one is correct as H is nothing but a double derivative with respect to x(in 1D). However when I was trying to prove Ehrenfest Theorem(see description), the second one worked and gave me correct result, but I was unable to prove it with first one. So is the $\eqref{e2}$ correct and $\eqref{e1}$ wrong?

Description of question : I was trying to prove Ehrenfest Theorem, $$\frac{d\langle p\rangle}{dt} = \langle \frac{-\partial V}{\partial x}\rangle$$ for that I did, $$\frac{d\langle p\rangle}{dt} = \frac{ħ}{i}\frac{d}{dt}\int\left( Ψ^*\frac{\partial Ψ }{\partial x}\right)dx = \frac{ħ}{i}\int\frac{\partial}{\partial t}\left( Ψ^*\frac{\partial Ψ }{\partial x}\right)dx = \frac{ħ}{i}\int\left( \frac{\partial Ψ^*}{\partial t}\frac{\partial Ψ }{\partial x} + Ψ^*\frac{\partial^2 Ψ}{\partial x \partial t}\right)dx$$ In the next step I substituted $\frac{∂Ψ}{∂t} $ and $\frac{∂Ψ^*}{∂t} $ from schrodinger equation. Now I have two options, to proceed according to $\eqref{e1}$, or $\eqref{e2}$.

According to $\eqref{e1}$, I am getting, $$ \int\left( \left(\frac{-ħ^2}{2m}\frac{\partial^2Ψ^*}{\partial x^2}+VΨ^* \right)\frac{\partial Ψ }{\partial x} - Ψ^*\frac{\partial }{\partial x }\left( \frac{-ħ^2}{2m}\frac{\partial^2Ψ}{\partial x^2}+VΨ \right)\right)dx \tag{3}\label{e3}$$ After that I am unable to prove the theorem, I don't know if I am lacking in the math part or if the use of $\eqref{e1}$ is incorrect.

Going according to $\eqref{e2}$, I am getting, $$ \int\left( Ψ^*\left(\frac{-ħ^2}{2m}\frac{\partial^2}{\partial x^2}+V \right)\frac{\partial Ψ }{\partial x} - Ψ^*\frac{\partial }{\partial x }\left( \frac{-ħ^2}{2m}\frac{\partial^2Ψ}{\partial x^2}+VΨ \right)\right)dx \tag{4}\label{e4}$$ This easily reduces to $\langle \frac{-\partial V}{\partial x}\rangle$, proving the theorem.

Hence it appears $\eqref{e2}$ is correct, but why is that so? Why is $\eqref{e1}$ incorrect? I got to know about $\eqref{e2}$ from here

Edit - This question was closed because it looked like homework question or check my work question, however my question was which of $\eqref{e1}$ or $\eqref{e2}$ is the correct way to take complex conjugate. It is a Physics/Math conceptual question that what is the correct way and why so. According to the comment under the question, I got the result from $\eqref{e1}$ as well, but this raises even more questions are $\eqref{e1}$ and $\eqref{e2}$ equivalent, because they do not look so, in one $\frac{\partial ^2}{\partial x^2}$ is operating on $Ψ^*$ and in other $Ψ^*$ is merely multiplied to $\frac{\partial ^2}{\partial x^2}$ which will operate on something else.

Answer 1

The first approach is correct.

$H$ is an operator that acts on the wave function, so it doesn't commute with it. It seems like in the second approach you (incorrectly) mixed up the hermitian conjugate (which also involves transposing the equation) with the complex conjugate.