# How to do time evolution of operators in the Heisenberg Picture while staying in the Heisenberg Picture

Consider the time evolution of an operator in the Heisenberg picture:

$$\tag{1}i\hbar \frac{d}{d t} \hat{A}_{H}(t) = \left([ \hat{A}_S(t), \hat H_S (t)] + i\hbar \frac{d}{d t} \hat{A}_S(t) \right)_H.$$

The question is how to evolve the Heisenberg operator without having to go back to the Schrödinger picture to find a new operator $\left( \frac{d}{d t} \hat{A}_S(t) \right)_{H}$ at every new time $t$ or pick a particular representation. It seems like the evolution of $\hat{A}_{H}$ requires we know $\left( \frac{d}{d t} \hat{A}_S(t) \right)_{H}$, so then we'd need an evolution equation for that, and hence need to know $\left( \frac{d^2}{d t^2} \hat{A}_S(t) \right)_{H}$ and so forth.

First, just to be clear I'd like to derive the above Heisenberg time evolution equation from the Schrödinger time evolution equation, partly just to make sure all the terminology is agreed upon and partly to show the manner in which I don't want to keep appealing back to the Schrödinger picture.

So, a Heisenberg operator $\hat A_H$ in the Heisenberg picture will be defined as a function of time, that for each time, $t,$ produces an operator $\hat A_H (t)$ on the Hilbert space $H$ that can take any abstract vector $|\psi_H\rangle$ in the Hilbert space and give a real number denoted $\langle\psi_H|\hat A_H (t)|\psi_H\rangle$ that just so happens to equal $\langle\psi_S(t)|\hat A_S (t)|\psi_S(t)\rangle$, where $\hat A_S$ is the corresponding operator in the Schrödinger picture, and $|\psi_S(t)\rangle$ is the abstract Hilbert space vector that is found from the abstract Hilbert space valued function of time that is the solution to the ODE $i\hbar \frac{d}{d t}|\psi_S(t)\rangle=\hat{H}_S|\psi_S(t)\rangle$ in the abstract Hilbert space subject to the initial conditions $|\psi_S(0)\rangle=|\psi_H\rangle$. So in particular, this means $\hat{A}_H(0)=\hat{A}_S(0)$, and note that I have not picked a representation, so these are just regular parameter-time derivatives of a parametrized curve of vectors in an abstract Hilbert space $H$. So both of these operators $\hat{A}_S(t)$ and $\hat{A}_H(t)$ are operators that take Hilbert space vectors and give Hilbert space vectors. And again I haven't picked a representation so $\hat{A}_S$ and $\hat{A}_H$ are functions of $\mathbb{R}$ that when they take a number $t$ they give the operators $\hat{A}_S(t)$ and $\hat{A}_H(t)$ respectively.

I'm aware that some people like to say that it isn't a function from real numbers to an operator on a Hilbert space and instead you get some kind of supposed parametrized family of operators with some parametrization by $(x,y,z,p_x,p_y,p_z, \dots ,t)$ (I have concerns about this idea for spin dependent operators, but that does not need to be addressed right now). Maybe I shouldn't be so harsh if it is just a hold over from quantization of classical systems, but I'm just trying to evolve a quantum system here, nothing fancy. That's nice for them if they allegedly want to take a partial derivative. However, that means if I want to evolve $\hat{A}_H(t)$ from one time to another, I'd have to find not just the operator corresponding to a later time, but a whole parametrized family at a later time, or maybe since they are parametrized already it was already there and there is nothing to solve. Either way, I don't see how that is going to come from the Heisenberg picture time evolution equation.

Which I think we are ready to derive. You can consider the quotient: $$Q(t,\Delta t)=\frac{\hat{A}_H(t+\Delta t)-\hat{A}_H(t)}{\Delta t}$$ and this is supposed to be (at the very least) an operator that takes any abstract vector $|\psi_H\rangle$ in the Hilbert space $H$ and gives a real number \begin{align} \langle\psi_H|Q(t,\Delta t)|\psi_H\rangle &= \langle\psi_H|\frac{\hat{A}_H(t+\Delta t)-\hat{A}_H(t)}{\Delta t}|\psi_H\rangle \\ & = \langle\psi_H|\frac{\hat{A}_H(t+\Delta t)}{\Delta t}|\psi_H\rangle-\langle\psi_H|\frac{\hat{A}_H(t)}{\Delta t}|\psi_H\rangle\\ & =\frac{\langle\psi_S(t+\Delta t)|\hat A_S (t+\Delta t)|\psi_S(t+\Delta t)\rangle-\langle\psi_S(t)|\hat A_S (t)|\psi_S(t)\rangle}{\Delta t}\end{align}

Now Schrödinger time evolution of vectors is unitary, so if the operator-norm of $\hat A_S (t+\Delta t)$ is uniformly bounded over an interval of $\Delta t$ containing zero and the Hilbert space vector was already finite norm then the limit as $\Delta t$ goes to zero $\lim_{\Delta t\rightarrow 0}\langle\psi_S(t)|Q(t,\Delta t)|\psi_S(t)\rangle$ exists just fine and equals

$$\left(\frac{d}{dt}\langle\psi_S(t)|\right)\hat A_S (t)|\psi_S(t)\rangle+ \langle\psi_S(t)|\left(\frac{d}{dt}\hat A_S (t)\right)|\psi_S(t)\rangle+ \langle\psi_S(t)|\hat A_S (t)\left(\frac{d}{dt}|\psi_S(t)\rangle\right).$$

And using Schrödinger time evolution this equals

$$\langle\psi_S(t)|\frac{\hat H_S(t)}{-i\hbar}\hat A_S (t)|\psi_S(t)\rangle+ \langle\psi_S(t)|\left(\frac{d}{dt}\hat A_S (t)\right)|\psi_S(t)\rangle+ \langle\psi_S(t)|\hat A_S (t)\frac{\hat H_S(t)}{i\hbar}|\psi_S(t)\rangle=$$

$$\langle\psi_S(t)|\frac{[\hat A_S (t),\hat H_S(t)]}{i\hbar}+\left(\frac{d}{dt}\hat A_S (t)\right)|\psi_S(t)\rangle.$$ From which we get

$$\frac{d}{dt}\hat A_H(t)=\left(\frac{[\hat A_S (t),\hat H_S(t)]}{i\hbar}+\frac{d}{dt}\hat A_S (t)\right)_H.$$

Which I'd like to write as an ODE in the space of operators on the abstract Hilbert space $H$.

Obviously, real life operators can always be written in a time independent manner by internalizing the external interactions. Same with the Hamiltonian, in which case everything is easy. Write $\hat U(t)=e^{-it\hat{H}/\hbar}$, which exists when the Hamiltonian is bounded, and the dynamics always start and stop on a subset of bounded energy so we might as well restrict to that subspace as our abstract Hilbert space $H$. Then the Heisenberg operator satisfies

$$\hat A_H(t)=\hat U(t)^{-1}A_S\hat U(t)$$ so has the following evolution equation ODE in the space of operators on the abstract Hilbert space:

$$i\hbar \frac{d}{dt}\hat A_H=[\hat A_H, \hat H_H].$$

And that's how straightforward and clear I wanted the ODE to be when we wanted to have externally driven operators and Hamiltonians. It's just that if you have to specify the external driving in the Schrödinger picture, it belies the advertised claim that the dynamics happen in the Heisenberg picture or that the time evolution equation is in the Heisenberg picture. And if the external driving happens directly in the Heisenberg picture, then there isn't really a time evolution equation, they just do the things you said they would by fiat. Seems like cheating either way.

If there is no answer, you could answer historically or soft-question if you can explain why or how this happened and why people say there is dynamical evolution in the Heisenberg picture when there are time dependent operators.

This is perhaps not what you were looking for, but one possibility would be to choose a fixed basis for all operators (say you have a finite Hilbert space), which we can denote $$\{\mathcal O_i\}$$. Then you can write your time-dependent Schrödinger-picture operator in terms of time-dependent coefficients and the time-independent basis operators $$\hat A_S(t) = \sum_i c_i(t)\hat{\mathcal O}_i.$$ Now you can go to the Heisenberg picture for all your basis operators $$\mathcal O_i\to \mathcal O_i(t)$$ and just plug them into your formula for $$\hat A$$ $$\hat A_H(t) = \sum_i c_i(t)\hat{\mathcal O}_i(t).$$ Now you said you don't want to choose a particular representation, by which you probably mean the above. Then the task is impossible though, as the time dependence of $$\hat A_S(t)$$ in the space of operators is completely arbitrary (you could arbitrarily set any of the $$\{c_i(t)\}$$ in any way you please), so any equation that does not take this into account must be wrong.