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

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.