In the Heisenberg picture (using natural dimensions): $$ O_H = e^{iHt}O_se^{-iHt}. \tag{1} $$ If the Hamiltonian is independent of time then we can take a partial derivative of both sides with respect to time: $$ \partial_t{O_H} = iHe^{iHt}O_se^{-iHt}+e^{iHt}\partial_tO_se^{-iHt}-e^{iHt}O_siHe^{-iHt}. \tag{2} $$ Therefore, $$ \partial_t{O_H} = i[H,O_H]+(\partial_tO_s)_H \, , \tag{3} $$ but this is not equivalent to what many textbooks list as the Heisenberg equation of motion. Instead they state that $$ \frac{d}{dt}{O_H} = i[H,O_H]+(\partial_tO_s)_H. \tag{4} $$ Why, in general, is this true and not the former statement? Am I just being pedantic with my use of partial and total derivatives?

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    $\begingroup$ Why you applied partial derivative? In Heisenberg formalism, the state kets are fixed in time and operators do vary in time. So you can take the total time derivative of the operator on the LHS. $\endgroup$
    – UKH
    Dec 29, 2016 at 5:28
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    $\begingroup$ Sorry I can't understand your logic there. Here the $O_s$ is allowed to vary with time and so does $O_H$, but it is very clear that on the LHS there is a total time derivative of $O_H$, and there is a partial time derivative appearing on the RHS. Why aren't they both partial derivatives in time? $\endgroup$
    – I.E.P.
    Dec 29, 2016 at 7:48
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    $\begingroup$ @I.E.P. In Eq. (2), On the left hand side, why isn't it $\frac{d\, O_H}{dt}$? $\endgroup$
    – Wein Eld
    Dec 29, 2016 at 7:54
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    $\begingroup$ @I.E.P., On the left hand side, you shall use $\frac{d\, O_H}{dt}$, and the total derivative can be expressed as the sum of partial derivatives. $\endgroup$
    – Wein Eld
    Dec 29, 2016 at 8:00
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    $\begingroup$ @I.E.P. I think here, what you are missing is the mathematical difference of total derivative and partial derivative. On the left $O_H$ as function of $t$, hence the total derivative, on the right, $O_H$ as a composed function via the relation (1), hence the partial derivative for every component function. $\endgroup$
    – Wein Eld
    Dec 29, 2016 at 9:13

2 Answers 2


With some definitions to make time dependences explicit, your equation (4) can be made sense of. Let's take the following:

Let $O_s$ be an operator depending on time and other parameters $O_s:\mathbb{R}\times S\rightarrow \mathrm{Op}$, where $S$ is the space of the other parameters and $\mathrm{Op}$ is the space of operators on the Hilbert space. Let $\phi:\mathbb{R}\times\mathrm{Op}\rightarrow\mathrm{Op}$ denote time evolution of operators in the Heisenberg picture, given by $\phi_t(O)=e^{iHt}Oe^{-iHt}$.

Note that $(\partial_t \phi)_t(O)=i[H,\phi_t(O)]$ and $\partial_O\phi=\phi$ (because $\phi$ is linear in $O$). Now, given a parameter $p\in S$ we can define the function of time: $O_H:\mathbb{R}\rightarrow \mathrm{Op}$ with $O_H(t)=\phi_t(O_s(t,p))$. Our function $O_H$ is a one-parameter one, so it only makes sense to take its total derivative: \begin{align} \frac{dO_H}{dt}(t)=&(\partial_t\phi)_t(O_s(t,p))+(\partial_O\phi)_t\left[(\partial_tO_s)(t,p)\right]\\ =& i[H,\phi_t(O_s(t,p))]+\phi_t\left[(\partial_tO_s)(t,p)\right]\\=& i[H,O_H(t)]+e^{iHt}(\partial_tO_s)(t,p)e^{-iHt}, \end{align}

where in the first step I have applied the chain rule and in the others, the equalities we already had.


No, you are not "just" being pedantic with your misuse of partial derivatives: your Eqns (2) and (3) are flat wrong. You simply did not apply the definitions right, as @WeinEld has been pointing out. (You might have spared yourself grief if you illustrated your very question for a simple system, such as the SHO.)

$$ O_H \equiv e^{iHt}O_se^{-iHt} , $$ so for $$ O_S=f(x,p;t) \qquad \Longrightarrow \qquad O_H=f(x(t),p(t);t), $$ where $x(t)= e^{iHt}xe^{-iHt} $ and likewise for p.

The time-derivative of $O_H$ consists of the partial derivative w.r.t. t after the semicolon, plus the convective derivative due to the flow of x and p in the Heisenberg picture, $$ \frac{\partial O_H}{\partial x(t)} \dot{x} + \frac{\partial O_H}{\partial p(t)} \dot{p} = i[H,O_H] = e^{iHt}(i[H,O_S])e^{-iHt}. $$ (Prove this! Unless you did, the discussion is all vapor.)

The partial derivative is $$ \frac{\partial O_H}{\partial t}=e^{iHt} \frac{\partial O_S}{\partial t}e^{-iHt}=\left (\frac{\partial O_S}{\partial t}\right ) _H. $$ (Some express this as $ \frac{\partial O_H}{\partial t}$, trusting the reader would properly understand the evident differentiation of only the argument after the semicolon, but this very question may make them think twice. Now, to be sure, since $O_S$ has vanishing convective derivative, $dO_S/dt=\partial O_S/\partial t$, as raised in a comment, so this is a non-issue.)

In any case, putting the two pieces together nets the conventional $$ \frac{d}{dt}{O_H} = i[H,O_H]+(\partial_tO_s)_H. $$

Monitor the evident behavior of a simple observable such as $O_S=tx$ in the SHO, $H=(p^2+x^2)/2 $, the celebrated rigid classical-like rotation in phase space, $x(t)=x\cos t +p \sin t$, $p(t)=p\cos t - x \sin t$; thus $O_H=tx(t)$. Hence $dO_H/dt= t p(t)+x(t)$: now appreciate efficiencies and differences of the respective pictures. (Such as $$dO_H/dt=\exp(itH) (it[p^2/2,x] + x)\exp(-itH)=e^{it~[(x^2+p^2)/2,}~ (tp + x)~,$$ with the physicists' customary avoidance of the mathematician's ad map notation.)

You might find your bearings by thinking of the S picture as the Eulerian frame, and the H picture as the Lagrangian, comoving frame.


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