# Why is this Heisenberg EOM true even for time-dependent magnetic fields (spin dynamics)?

In what follows, I will use primes to denote Heisenberg picture operators (non-primed operators will be Schrödinger picture).

In his Chapter 12.1 on spin dynamics, Ballentine has us first consider a spin in a uniform magnetic field in arbitrary direction. Crucially, $$\mathbf{B}$$ is taken to be time-independent. Because of this, we have $$H' = H = - \boldsymbol{\mu} \cdot \mathbf{B} = -\gamma\, \mathbf{S} \cdot \mathbf{B}$$, from which we find the EOM for the $$\mathbf{S}'$$ observable as $$\frac{d \mathbf{S}'}{dt} = \frac{i}{\hbar}\left[H', \mathbf{S}'\right] = \mathbf{S}' \times \gamma\,\mathbf{B}.$$ But (and here is where I am lost), Ballentine further claims that this EOM is valid even if $$\mathbf{B}$$ depends on time! Why on earth should this be? In this case, it is not true that $$H'= H$$ (at least a priori). Rather, we have from the Heisenberg picture definition that $$H'(t) = U^\dagger(t) H(t) U(t),$$ where $$U$$ is some ugly time-ordered exponential. Why should we have $$H' = H$$ nevertheless (clearly this is sufficient to yield Ballentine's claim and I'm guessing it's necessary too)?

• Have you tried to re-derive the expression for the Heisenberg picture EOM? See e.g. Wikipedia. Commented Sep 9, 2023 at 21:57
• @TobiasFünke I don't even know how to begin to use the BCH formula once I have $U$ as a time-ordered exponential. That part is way above my pay grade (I haven't even seen it in Ballentine to be honest -- it's just something I know of somewhat vaguely from elsewhere).
– EE18
Commented Sep 9, 2023 at 22:00
• Why would you need the BCH? $U$ obeys the Schrödinger equation (the time-ordered exponential is the solution, so this holds by construction). Then simply compute the time derivative of $A_H(t)$, where $A_H$ is the Heisenberg operator corresponding to some $A(t)$ in the Schrödinger picture, and see where it gets you... This is probably the easiest thing to do (and again, it is basically done in Wikipedia) and then check what possible implicit assumptions Ballentine could meant... Commented Sep 9, 2023 at 22:59
• I don't think I follow what you mean. I looked at Wikipedia but all it does is prove $\frac{d \mathbf{S}'}{dt} = \frac{i}{\hbar}\left[H', \mathbf{S}'\right]$, but I'm afraid I still don't see how to evaluate the commutator on the RHS. The commutator would be $U^\dagger H UU^\dagger \mathbf{S}U - U^\dagger \mathbf{S}UU^\dagger H U = U^\dagger[H,\mathbf{S}]U$, but it's not clear to me why that equals $[H,U^\dagger\mathbf{S}U]$ which is what would be necessary to use the manipulation I gave in the OP. @TobiasFünke
– EE18
Commented Sep 10, 2023 at 0:11
• I suppose $[H,U^\dagger\mathbf{S}U] = HU^\dagger\mathbf{S}U - U^\dagger\mathbf{S}UH = U^\dagger H\mathbf{S}U + [H,U^\dagger]\mathbf{S}U - U^\dagger\mathbf{S}HU - U^\dagger\mathbf{S}[U,H] = U^\dagger[H,\mathbf{S}]U + [H,U^\dagger]\mathbf{S}U - U^\dagger\mathbf{S}[U,H]$. Is $[H,U^\dagger]\mathbf{S}U - U^\dagger\mathbf{S}[U,H] = 0?$ Well, $([H,U^\dagger]\mathbf{S}U)^\dagger = U^\dagger \mathbf{S} [H,U^\dagger]^\dagger = U^\dagger\mathbf{S}[U,H]$ so I guess so? Is this the manipulation you had in mind @TobiasFünke
– EE18
Commented Sep 10, 2023 at 0:17

In general, if you have a time dependent Hamiltonian $$H$$, then this gives an evolution operator $$U$$ obeying: $$i\frac{dU}{dt} = HU \tag{1}$$ which is solved by time ordered exponential: $$U(t_2,t_1)=\mathcal T\exp\left(-i\int_{t_1}^{t_2}H(s)ds\right)$$ From this, you can construct the Schrödinger picture and the Heisenberg picture. The former is by evolving the kets: $$|\psi\rangle_S=U|\psi\rangle$$ which satisfy Schrödinger’s equation: $$i\frac{d}{dt}|\psi\rangle_S=H|\psi\rangle_S$$ The latter is obtained by evolving the operators: $$O_H=U^{-1}OU$$ You’ll need: $$i\frac{dU^{-1}}{dt} = -U^{-1}H \tag{2}$$ By using the product rule of the time derivative, you obtain Heisenberg’s equations of motion: \begin{align} i\frac{dO_H}{dt} &= i\frac{dU^{-1}}{dt}O_HU+U^{-1}O_H i\frac{dU}{dt}\\ &= -U^{-1}H O_HU_H+ U^{-1}O_H HU \\ &= [O_H,H_H] \end{align} Note that if $$O$$ has an explicit time dependence, you’ll need to add the extra term: $$i\frac{dO_H}{dt} = [O_H,H_H] + i(\partial_tO)_H$$

In the case of a time independent Hamiltonian, $$U,H$$ commute, so $$H_H=H$$. However, this is not the case in general. Physically, a time dependent system does not necessarily conserve energy, which is consistent with Noether’s theorem.

Since the Heisenberg picture is obtained by conjugating operators, it preserves algebraic expressions. For example, in your case $$B$$ is a scalar so: \begin{align} H_H &= -U^{-1}\gamma B\cdot SU \\ &= -\gamma B \cdot U^{-1}SU \\ &= -\gamma BS_H \end{align} As corollary, the commutation relations are also preserved, so you still have: $$S_H\times S_H=iS_H$$ Thus, from the general equations of motion, the expression of the Hamiltonian and the conjugation relations, you recover the announced equation of motion: \begin{align} i\dot S_H^k&= [S_H^k,H_H] \\ &= -\gamma B_i[S_H^k,S_H^i]\\ &= -\gamma B_i i\epsilon_{ijk}S_H^j \\ \dot S_H &= S_H\times \gamma B \end{align} Btw, you recover the classical EoM by taking the expected value.

Note that the equation of motion of the Hamiltonian is an example where you need to include the additional explicit time dependence term in the equation of motion: $$\dot H_H= -\gamma \dot B\cdot S_H$$ This is consistent with the $$S$$ equation of motion since $$\dot S_H\cdot B=0$$.

In general, the Hamiltonian’s expression is obtained just by replacing the operators by their Heisenberg picture versions. This is because the explicit time dependencies are chosen to be scalars that commute with the conjugation via the time evolution.

Hope this helps.

• Thank you for this very helpful answer. I wanted to focus on this: "In general, the Hamiltonian’s expression is obtained just by replacing the operators by their Heisenberg picture versions. This is because the explicit time dependences are chosen to be scalars that commute with the conjugation via the time evolution." I am still trying to make sense of this as it seems to be the key. Can you elaborate if you get the chance, and/or show exactly what you mean in this specific case? Does what you said here get justified by my ugly commutator computations in some comments above?
– EE18
Commented Sep 13, 2023 at 13:14
• I've corrected an equation regarding time ordering (also, technically, I think this requires $t_2>t_1$, else we'd have to work with the anti-time ordering, no?)...Anyway, your answer is what I meant in the comments under the question. +1 Commented Sep 13, 2023 at 13:31
• @TobiasFünke Thanks, I forgot the most important part! Well the time ordering automatically sorts the order of the integral, so I guess it depends on your conventions.
– LPZ
Commented Sep 13, 2023 at 17:56
• What Tobias Fünke meant (I think) is that the components of $B$ are not operators on the Hilbert space, they are mere scalars. By linearity of the operator multiplication, you can take it out of the conjugation.
– LPZ
Commented Sep 13, 2023 at 18:32
• OMG, I can't believe I missed that. I was getting caught up in $U^{-1}S_iU$ and how I work that out, but your points are that $U^{-1}S_iU$ is $S_i'$ and that therefore $\frac{d \mathbf{S}'}{dt} = \frac{i}{\hbar}\left[H', \mathbf{S}'\right] = -\frac{i}{\hbar}\gamma B_i\left[S_i', \mathbf{S}'\right]$ which is of the same algebraic form as before given that the $S_i'$ obey the same commutation relations as the $S_i$, hence we get the same result...apologies to both of you for my slowness here @TobiasFünke
– EE18
Commented Sep 13, 2023 at 18:38