So when the Hamiltonian is time-independent, we can define the Heisenberg state vectors by evolving the Schrödinger state vectors back in time:

$$ | \psi \rangle_H = \hat{U}^\dagger (t)|\psi(t) \rangle_S=e^{i\hat{H}t} |\psi(t)\rangle_S $$

and we define operators

$$ \hat{A}_H(t) = \hat{U}^\dagger (t) \hat{A}_S \hat{U}(t)$$

which gives us the Heisenberg equation: $$ \frac{d\hat{A}_H(t)}{dt} = -i[\hat{A}_H(t),\hat{H}]. $$

If, in the Schrödinger picture, we have a time-dependent Hamiltonian, the time evolution operator is given by

$$ \hat{U}(t) = T[e^{-i \int_0^t \hat{H}(t')dt'}] $$

If I define the Heisenberg operators in the same way with the time evolution operators and calculate $ dA_H(t)/dt $ I find

$$ \frac{d}{dt} \hat{A}_H(t)= \frac{d\hat{U}^\dagger(t)}{dt} \hat{A}_S\hat{U}(t) + \hat{U}^\dagger(t) \hat{A}_S \frac{d\hat{U}(t)}{dt} \\ = i \hat{U}^\dagger (t) \hat{H(t)} \hat{A}_S \hat{U}(t) - i\hat{U}^\dagger (t)\hat{A}_S\hat{H}(t)\hat{U}(t). $$

At this point, I am not sure how to proceed. I can't commute $\hat{H}(t)$ through $\hat{U}(t) $ because $[\hat{H}(t),\hat{H}(t')] \neq 0$. How do I show derive Heisenberg's equation for a time-dependent Hamiltonian?

  • $\begingroup$ Related 122687 . $\endgroup$ Commented May 16, 2017 at 13:27
  • $\begingroup$ Use $H_H\equiv U^\dagger H_S U\neq H_S$, as indicated in related question. $\endgroup$ Commented May 16, 2017 at 13:31
  • 1
    $\begingroup$ Related 293154. $\endgroup$ Commented May 16, 2017 at 16:33

4 Answers 4


You appear to be replicating all the steps that Dirac utilized with foresight to define his celebrated interaction picture. As indicated in the comment, the proper relationship is already in question 122687. The crucial point is that, as you implicitly noticed, the Heisenberg Hamiltonian is not the Schroedinger Hamiltonian, by contrast to the time-independent case.

That is, motion being a canonical transformation, the two hamiltonians are equivalent but, in general, not equal, $$ H_H=U^\dagger H_S U \neq H_S, $$ evident from the Dyson expansion.

Moreover, thinking of H as an observable, note $$ \frac{d H_H}{dt}= (\partial_t H_S)_H \neq 0, $$ not a constant of the motion.

Conversely, if you take your S-observable A without explicit time dependence, (think of x or p), your final equation is sound, and merely amounts to the customary convective term, $$ \frac{dA_H}{dt}= U^\dagger i[H_S,A_S]U = i[H_H,A_H], $$ as it should. Adding explicit time dependence yields the additional customary explicit term $(\partial_t A_S)_H$.

  • $\begingroup$ What does $(\partial_tH_S)_H$ mean? Also, why do you write a normal derivative with respect to time in front of the Heissenberg Hamiltonian but a partial derivative in front of the Schrödinger Hamiltonian? Does the Schrödinger Hamiltonian depend on anything more than just time? $\endgroup$ Commented Oct 22, 2021 at 1:12
  • $\begingroup$ It means what it says: differentiate the normally time independent $H_S$ w.r.t. time, now that it has a parametric dependence on it, and then perform the Heisenberg equivalence transform exemplified in the first formula here. One normally uses full derivatives to contrast the intrinsic versus the convective term, as illustrated. $\endgroup$ Commented Oct 22, 2021 at 1:28
  • $\begingroup$ Well, okay, thank you for the explanation, but I don't agree that that is what it "says". To me it's just an expression that has been given the label $H$, until I understand how to read it, but you answer me as if it is obvious how to interpret what you have written. Do you mean that $(\partial_tH_S)_H = U^\dagger\partial_tH_SU$? $\endgroup$ Commented Oct 22, 2021 at 1:35
  • $\begingroup$ Yes, of course; mainstream notation, unexceptional used. $\endgroup$ Commented Oct 22, 2021 at 2:45
  • $\begingroup$ Okay, thank you! $\endgroup$ Commented Oct 22, 2021 at 13:19

simply insert $UU^\dagger$ in between $H$ and $A_S$ in both term, here is the first term

$U^\dagger H A_sU=U^\dagger H U U^\dagger A_sU=H_HA_H$


From the Schrodinger Equation and rearranging under the assumption that the kets are time independent, we know $\frac{dU(t)}{dt} = -iH(t)U(t)$, or equivalently $\frac{dU^\dagger(t)}{dt} = iU^\dagger(t) H(t)$ (this is the crux of the problem here, so far as I can tell, and a piece that was incorrect in your original question)

Carrying out $\frac{dA_H(t)}{dt}$ as you did, we now have

$$ \frac{dA_H(t)}{dt} = i U^\dagger(t)H(t)A_SU(t) - i U^\dagger(t) A_S H(t)U(t) + \frac{\partial A_H(t)}{dt} \\ = U^\dagger(t)(i \left[H(t),A_S\right])U(t) + \frac{\partial A_H(t)}{dt}$$

When $U(t)$ does not commute with $H(t)$, I believe that this is as far as you can take things. If $U(t)$ does commute with $H(t)$, then we recover the Heisenberg equation and replace $A_S$ by $A_H(t)$, as we would expect.

  • $\begingroup$ Ideally you would like to transform to a frame where you have a time independent Hamiltonian, or do something like a transformation to the Interaction Picture where you can treat part of it as time independent. Unfortunately it may not always possible to find a totally time independent frame (I do not have a proof of this claim, someone please correct me if I am wrong). $\endgroup$
    – QtizedQ
    Commented May 16, 2017 at 2:54


$$ \frac{dU}{dt} = -iH(t) U(t) $$

I would write

$$ \frac{dU^{\dagger}}{dt} = \biggl(-iH(t) U(t)\biggr)^{\dagger} = iU^{\dagger}(t) H(t). $$

Notice that this is different from what you have.


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