The time evolution of a quantum mechanical operator $A$ (without explicit time dependence) is given by the Heisenberg equation

$$ \frac{d}{dt}A = \frac{i}{\hbar} \left[H,A\right] \tag{1}$$

where $H$ is the system's Hamiltonian. The time evolution of the corresponding expectation value is given by the Ehrenfest theorem

$$ \frac{d}{dt}\left\langle A\right\rangle = \frac{i}{\hbar} \left\langle \left[H,A\right]\right\rangle \tag{2} $$

However, as I have noticed, these can yield differential equations of different forms if $\left[H,A\right]$ contains expressions that do not "commute" with taking the expectation value. For example, let

$$\left[H,A\right]=\frac{dA}{df} \tag{3}$$

for some quantum number $f$. The states used for taking the expectation values in (2) clearly depend on $a$. From (1) and (2) the different time evolutions calculated using (3) are

$$ \frac{d}{dt}A = \frac{i}{\hbar}\frac{dA}{df} \tag{4}$$ and $$ \frac{d}{dt}\left\langle A\right\rangle = \frac{i}{\hbar}\left\langle \frac{dA}{df}\right\rangle \tag{5}$$

Note that clearly (5) is not the same as

$$ \frac{d}{dt}\left\langle A\right\rangle = \frac{i}{\hbar}\frac{d\left\langle A\right\rangle}{df}\tag{6}$$

which I would have expected, since then the equations looks equal to (4).

Did I make some mistake? Is this correct? What is the difference between (4)-(6)? Clearly, (4) and (6) give rise to the same solution (one for the operator and one for the expectation value) since the differential equations are the same, while this may not be the case for (5). In fact, from (5) alone, I don't see a clear way on how to construct a solution since $\left\langle A\right\rangle$ does not occur on the right hand side.

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    $\begingroup$ where does (3) come from? $\endgroup$
    – Phoenix87
    Jul 9, 2020 at 14:27
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    $\begingroup$ Isn't it two physically completely different things, if I observe how that state in which a system is evolves by time, or if I investigate the time evolution of the operator itself? $\endgroup$ Jul 9, 2020 at 14:27
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    $\begingroup$ You've skipped an $i/\hbar$ in (3) needed for (4). $\endgroup$ Jul 9, 2020 at 14:47
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    $\begingroup$ @Phoenix87 That's just a "pathological example". $\endgroup$ Jul 9, 2020 at 14:51
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    $\begingroup$ It's then clear from (3) that A is solely a function of $t-i\hbar f$, which is the time-shifted Heisenberg picture operator when f is imaginary. $\endgroup$ Jul 9, 2020 at 15:03

2 Answers 2


The "example" eq. (3) here is kind of a red herring. Let's just take the evolution equation eq. (1) and Ehrenfest's theorem eq. (2). What you're asking about is that taking the expectation value of eq. (1) and using eq. (2) results in $$ \frac{\mathrm{d} \langle A\rangle}{\mathrm{d}t} = \left \langle \frac{\mathrm{d}A}{\mathrm{d}t}\right\rangle,$$ which you seem to think isn't true, but the fact that taking the expectation value commutes with the time derivative is exactly the content of Ehrenfest's theorem.

Your example is just re-parametrizing the time evolution of the Heisenberg operator by $f = \frac{\mathrm{i}}{\hbar} t$, so this doesn't change anything. That you are really just reparametrizing you can see by plugging eq. (3) into eq. (1), yielding $$ \frac{\mathrm{d}A}{\mathrm{d}t} = \frac{\mathrm{i}}{\hbar}\frac{\mathrm{d}A}{\mathrm{d}f},$$ meaning $\frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\mathrm{i}}{\hbar}$ by the chain rule.

  • $\begingroup$ Does that always work? What if, for example $f$ is the crystal momentum? (I have seen this in one lecture) Or what if $f$ is the spatial coordinate in a 1D system? $\endgroup$ Jul 9, 2020 at 15:40
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    $\begingroup$ @HerpDerpington It follows directly from the second equation in my post, but note that it only holds in the context of the specific observable $A$. I'm not claiming that $f=\frac{\mathrm{i}}{\hbar}t$ holds in some general sense (which would be unclear to begin with), just that for this specific $A$ you get the expression for $A(f)$ by plugging $t = -\mathrm{i}\hbar f$ into $A(t)$. $\endgroup$
    – ACuriousMind
    Jul 9, 2020 at 15:45
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    $\begingroup$ "which you seem to think isn't true" that was a little bit misleading, what i actually wanted to say is that $\left\langle dA/df \right\rangle \neq d\left\langle A \right\rangle /df$. You cannot swap this in the general case. (Derivatives in $f$, not in $t$) $\endgroup$ Jul 9, 2020 at 17:06
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    $\begingroup$ See also my comment to @CosmasZachos under the question. $\endgroup$ Jul 9, 2020 at 17:20

Note that $[H,A]$ is always the time derivative of $A$, modulo constants. If you want something physically different from a time derivative, then you should change the "generator" (of translations in time, i.e. time evolution) $H$. For example, the momentum operator $p$ generates translations, so $[p,A]$ gives you the spatial derivative of $A$ along the direction of $p$. That is, modulo constants, you always have $[p, A]=\nabla A$.

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    $\begingroup$ How does this answer the question on expectations? $\endgroup$ Jul 10, 2020 at 10:48

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