Heisenberg equation of motion

Question

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?

Answer 1

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.

Answer 2

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 $$ \frac{d}{dt}{O_H} = i[H,O_H]+(\partial_tO_s)_H. $$

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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.