Problem deducing Ehrenfest's theorem using Schrödinger's equation

Question

To derive Ehrenfest's theorem using Schrödinger's equation, we start from $$\frac{d\langle A\rangle}{dt}=\frac{d}{dt}\left(\int\psi^*\hat{A}\,\psi\,d^3r\right),$$ and then we get to $$\int\frac{d\psi^*}{dt}\hat{A}\,\psi\,d^3r+ \int\psi^*\frac{d\hat{A}}{dt}\,\psi\,d^3r+ \int\psi^*\hat{A}\,\frac{d\psi}{dt}\,d^3r,$$ where we can assume the middle term will be zero in most cases.

But my problem is with the next part. We can then replace the temporal derivatives by Schrödinger's equation and its complex conjugate, i.e. $$ i\hbar \frac{\partial}{\partial t} \psi= \hat{H} \psi = \left(-\frac{\hbar^2}{2m} \nabla^2 + V\right) \psi $$ $$ -i\hbar \frac{\partial}{\partial t} \psi^*= \hat{H} \psi^* = \left(-\frac{\hbar^2}{2m} \nabla^2 + V\right) \psi^*. $$ But all the derivations I've seen just change the order of the conjugate, as in $\hat{H} \psi^* = \psi^* \hat{H}$, and don't understand how that can possibly be the case, considering we stop having the Hamiltonian acting on $\psi^*$ when we do that.

Answer 1

This is because you do not have to take the complex conjugate, you have to take the adjoint: $$ \frac{\partial}{\partial t}|\psi\rangle=-\frac{i}{\hbar}\hat{H}|\psi\rangle \ , $$

$$ \frac{\partial}{\partial t}\langle\psi|=+\frac{i}{\hbar}\langle\psi|\hat{H} \ . $$ If the derivation really depended on the complex conjugation symmetry $\hat{H}=\hat{H}^{*}$, then the result would not hold whenever the Hamiltonian has factors of $i$ in it (like in the presence of a magnetic field).

But if you insist on doing the derivation in coordinate representation, you just need to integrate by parts twice, and use the vanishing of boundary terms: $$ \begin{aligned} \int\mathrm{d}^3x\left(\frac{\partial}{\partial t}\psi\right)^{*}\hat{A}\psi &= \frac{i}{\hbar} \int\mathrm{d}^3x\left(-\frac{\hbar^2}{2m}\nabla^2\psi ^{*}+V\psi^*\right)\hat{A}\psi \\ &= \frac{i}{\hbar} \int\mathrm{d}^3x\left[\left(-\frac{\hbar^2}{2m}\nabla^2\psi ^{*}\right)\hat{A}\psi+\psi^*V\hat{A}\psi\right] \\ &= \frac{i}{\hbar} \int\mathrm{d}^3x\left[\left(+\frac{\hbar^2}{2m}\vec{\nabla}\psi ^{*}\right)\cdot\vec{\nabla}\hat{A}\psi+\psi^*V\hat{A}\psi\right] \\ &= \frac{i}{\hbar} \int\mathrm{d}^3x\left[\psi ^{*}\left(-\frac{\hbar^2}{2m}\nabla^2\right)\hat{A}\psi+\psi^*V\hat{A}\psi\right] \\ &= \frac{i}{\hbar} \int\mathrm{d}^3x\psi ^{*}\hat{H}\hat{A}\psi \ . \end{aligned} $$ The derivation would of course also work with terms linear in momentum, e.g. a Zeeman term: $$ \left((\hat{L}_zB_z)^{*}\psi^{*}\right)\Leftrightarrow \psi^{*}\hat{L}_zB_z \ , $$ and in general $$ \left(\hat{H}^{*}\psi^{*}\right)\Leftrightarrow \psi^{*}\hat{H} \ . $$