quantum mechanics - Is the Wikipedia version of the Heisenberg equation of motion correct?

Back in 2011, this question asked about the Wikipedia version of the Heisenberg equation of motion for an operator $A$:

$$\begin{equation*} \frac{d}{dt} A(t) = \frac{i}{\hbar} \left[ H, A(t) \right] + \frac{\partial A}{\partial t} \end{equation*}$$

The accepted response asserted that "there is no mistake on the Wikipedia page", and indeed the formula is the same today as it was then.

But it's not correct, I think.

The first two occurrences of $A$ refer to $A_H$, "in the Heisenberg picture", where

$$\begin{equation*} A_H(t) = U^\dagger(t) A_S(t) U(t) \end{equation*}$$

where $U(t)$ is the unitary time-evolution operator and $A_S(t)$ is the operator in the Schrodinger picture, which may or may not be time-dependent.

So far so good. However, the last term is not

$$\begin{equation*} \frac{\partial A_H}{\partial t} \end{equation*}$$

but rather

$$\begin{equation*} \left(\frac{dA_S}{dt} \right)_H = U^\dagger(t) \frac{dA_S}{dt} U(t) \end{equation*}$$

In fact, if one wanted an equation of motion using exclusively $A_H$, I think one would write:

$$\begin{equation*} \frac{d}{dt} A_H(t) = \frac{i}{\hbar} \left[ H_H(t), A_H(t) \right] + U^\dagger(t) \frac{d}{dt} \left(U(t) A_H(t) U^\dagger(t) \right) U(t) \end{equation*}$$

But no one ever does, as far as I can see. Is it wrong, or too ugly for prime time, or???

Answer

You may see the idea on why the formula is written like that by a comparison with the classical phase space formula.

In classical phase space you have observables of the type $a(x,p,t)$, or more precisely $a(x(t),p(t),t)$; where $x(t)$, $p(t)$ and $t$ are considered independent variables. The equation of motion for $a$ is given by:

$$\frac{d}{dt}a(x(t),p(t),t)=\{h,a\}+\partial_t a\; ;$$ where $h$ is the classical hamiltonian, $\{\cdot,\cdot\}$ the Poisson bracket and the partial derivative in time is intended as deriving the functional form $a(\cdot,\cdot,\cdot)$ above w.r.t the third variable only (i.e. disregarding the fact that the first two argument may depend on the third). This is quite usual when dealing with function of many variables, and is a widely accepted convention (only the explicit dependence is taken into account when taking partial derivatives).

In quantizing, a formula with the same flavour is obtained replacing functions with operators (with suitable ordering, and in their Heisenberg version)---for me it will be replacing small letters with capital ones---and Poisson brackets with $\frac{i}{\hbar}[\cdot,\cdot]$. This is just a naïve procedure, and not an exact derivation of the formula; nevertheless you get:

$$\frac{d}{dt}A(X(t),P(t),t)=\frac{i}{\hbar}[H,A]+\partial_t A\; .$$ Again the meaning remains unchanged: the partial derivative is intended as the derivative of the functional form $A(\cdot,\cdot,\cdot)$ with respect to the third variable only.

Note (related to the OP comment in the other answer): As a disclaimer, I will not consider in the following any problem of domains of operators, convergence of series and so on; nevertheless also these problems are to be taken into account. The statements have to be intended as true on suitable domains and with suitably regular functions.

Let $A(X,P,t)$ be a ploynomial function (regardless of the ordering), whith $[X,P]\neq 0$, and $t$ commutes with everything. Let also $U(t)=U(X,P,t)$ be a unitary operator (in particular we will use $U^*(t)U(t)=U(t)U^*(t)=id$, where $id$ is the identity operator) such that $X(t):= U^*(t)XU(t)$, and $P(t)=U^*(t)PU(t)$. It is then clear that, using the property of $U$ above,

$$U^*(t)A(X,P,t)U(t)=A(X(t),P(t),t)\; .$$ This can be extended, roughly speaking, to any function that admits a series expansion.

Now given this equality, if you interpret the partial derivative as acting only explicitly on $t$, you see that again you obtain for polynomials/functions with series expansion

$$\partial_t A(X(t),P(t),t)=U^*(t)\partial_tA(X,P,t)U(t)\; ,$$ for the functional form of $\partial_t A(\cdot,\cdot,t)$ is always the same, and the "replacement" of $X,P$ with $X(t),P(t)$ (by means of $U$) can be done either before or after the derivative without changing the functional form (just changing $X$ in $X(t)$ and $P$ in $P(t)$).

As I said, this may not be true with more general functions, anyways this type of Heisenberg evolution formulas, written like that, are valid only in very special situations, if you want to be precise and consider domain problems and the proper definition of functions of unbounded operators.