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.


1 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 $$ \left(\hat{H}^{*}\psi^{*}\right)\Leftrightarrow \psi^{*}\hat{H} \ . $$

  • $\begingroup$ thank you for your answer. I don't understand some of the steps of the integration, though. In the integration by parts (from the second to third row), how did you get rid of $-\frac{\hbar^2}{2m}\nabla\psi^*\hat{A}\psi$? And after that (from the third to fourth row), why can you just take $\psi^*$ out of the gradient? $\endgroup$ Oct 20, 2023 at 13:25
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    $\begingroup$ @PedroNogueira That really is just integration by parts. The $\nabla^2\psi^*=\vec{\nabla}\cdot(\vec{\nabla}\psi^{*})$ is analogous to $g'$ in the $\int fg'=fg-\int f'g$ formula, so in this way, you can transfer $\vec{\nabla}$ to the other factor of the integrand (with a minus sign). The boundary term $(\vec{\nabla}\psi^{*})\hat{A}\psi$ vanishes in the infinite distance, so you don't have to deal with that. The step after that is just a second integration by parts (again with dropping the boundary term); I didn't take $\psi^{*}$ out of the gradient. $\endgroup$ Oct 20, 2023 at 14:30

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