I just got introduced to „more formal“ quantum mechanics and have a few questions about this Heisenberg equation:

I‘m interested in what assumptions we need to solve the system of differential equations (evolution of the spin operators):

$$ \frac{d}{dt}S_i(t) = \frac{1}{i\hbar}[S_i(t),\omega S_z(t)] $$

For $i \in \{x,y,z \}$ and $S_i(0)=\frac{\hbar}{2}\sigma_i$, and the $\sigma_i$ the Pauli matrices.

The solutions have been discussed on this site and are $S_z = 0, S_{j} = A_{j} e^{i \omega t}+B_{j} e^{-i \omega t}$, $A_{j}, B_{j} \in \mathbb{C}$ chosen appropriately.

I don’t think that the IVP, as is, will yield these solutions, in the derivations I checked, authors always make additional assumptions.

f.e. in this derivation: solution1, the author assumes that $[S_i,S_j]=i \hbar \epsilon_{ijk}S_k$ doesn’t only hold for $t=0$, but for all times, why would that be?

In this solution, discussed on this site: solution2, the OP assumes that $S_j(t)$ is of the form $\exp(-i \omega S_z t/\hbar)S_j(0) \exp(i \omega S_z t/\hbar)$ (and therefore actually already postulates the answer), why can we assume that?

If we only assumed $S_i(t)=U(t)^{\dagger}S_i(0)U(t)$, for $U \in \operatorname{U}(2)$, we would get:

$$ [S_i, \omega S_z]=\omega[U^{\dagger}S_i(0)U, U^{\dagger}S_z(0)U]= \omega U^{\dagger}[S_i(0),S_z(0)]U=\omega U^{\dagger} i \hbar \epsilon_{ijk}S_k(0) U = \omega i \hbar \epsilon_{ijk}S_k(t) $$

And so we’re back at the assumption of „solution1“, so maybe this is the way to go?

Maybe we could still weaken this assumption and demand only that $\forall t \in [0,\infty): S_i(t) \in \operatorname{U}(2)$, and then prove that this implies $\exists U \in \operatorname{U}(2): S_i(t)=U(t)^{\dagger}S_i(0)U(t)$, maybe thats not enough and we need to demand that $S_i$ stays self-adjoint?

I‘d be happy if someone could explain:)

  • $\begingroup$ maybe you can start by clarifying what $H$ is, and if it is time-dependent. Next, I’m not sure why you would think the commutation relations would not hold for all t (if these are equal-time CRs). $\endgroup$ Nov 5, 2022 at 3:07
  • $\begingroup$ but you have $S_z(t)$ yet $[S_i,\omega S_z]$ with no time-dependence. Are the latter same as $S_i(0)$? $\endgroup$ Nov 5, 2022 at 3:31
  • $\begingroup$ Yes, not considering that $S_z$ isn’t time dependent is exactly the error I made while thinking about all this, thank you! $\endgroup$
    – Henry T.
    Nov 5, 2022 at 3:33

1 Answer 1


The IVP should be sufficient to find the solution at any time. But with operator or matrix equations it does get confusing. A good approach is to solve the equation(s) first for regular functions and plug them in, seeing if they work for operators as well.

in this derivation: [solution1], the author assumes that $[S_i,S_j]=i \hbar \epsilon_{ijk}S_k$ doesn’t only hold for $t=0$, but for all times, why would that be?

I will show you here: We know that $U(t)U^\dagger (t) = U^\dagger(t) U(t) = 1$. And it is the case that we can write $S_i(t) = U^\dagger(t) S_i(0) U(t)$ - you can see this directly when trying to go from the schrödinger to the heisenberg picture, and also I will justify it in my answer to your second question below. So, assuming that at t=0 the commutation holds, we multiply by a bunch of $U^\dagger$ from the left, by $U$ from the right, and insert the identity in the middle in order to get the time evolved version of all three matrices:

$$S_i S_j - S_j S_i = i \hbar \epsilon_{ijk} S_k$$ $$\implies U^\dagger (t) S_i U(t)\,\, U^\dagger(t) S_j U(t) - U^\dagger (t) S_j U(t)\,\, U^\dagger(t) S_i U(t) = i \hbar \epsilon_{ijk} U^\dagger(t)S_k U(t)$$ $$\implies S_i(t) S_j(t) - S_j(t) S_i(t) = i \hbar \epsilon_{ijk} S_k(t)$$

In this solution, discussed on this site: [solution2][2], the OP assumes that $S_j(t)$ is of the form $\exp(-iHt/\hbar)S_j(0) \exp(iHt/\hbar)$ (and therefore actually already postulates the answer), why can we assume that?

This is always true of any operator satisfying the heisenberg equation of motion, as long as it doesn't explicitly depend on $t$:


$$\implies A(t) = e^{i t H/\hbar}A(0)e^{-i t H/\hbar}$$

You can check this by plugging it in. And this also means that $A(t)=U^\dagger (t)A(0)U(t)$, which is just the form of the equation above. It works with any observable operator, and any Hamiltonian. Pretty amazing that you can just write out the solution like that. But the caveat is that this only goes so far... there are much more useful ways to write the solution, as you saw in your post where we can write it explicitly in terms of pauli matrices for example. So some might not consider it actually "solved" if it is just written in this form.

Similarly, the solution in the Schrödinger picture for any wavefunction can be written

$$|\psi (t) \rangle = e^{-itH/\hbar}|\psi(0) \rangle$$

  • $\begingroup$ Thank you for your answer! Everything you say makes sense. So one would argue: $\partial_t S_z = \frac{1}{i \hbar} [S_z, \omega S_z]=0$ therefore $H$ doesn‘t depend on time and we can write the other to $S_i$ In the form $U^{\dagger}S_i U$, and then the argument in the solutions follows because of the things you said, right? $\endgroup$
    – Henry T.
    Nov 5, 2022 at 3:30
  • $\begingroup$ you have assumed $H$ does not depend on time yet the OP seems to have $S_z(t)$, which would suggest a time dependence somewhere… $\endgroup$ Nov 5, 2022 at 3:33
  • $\begingroup$ HenryT: You were probably given $H$, or else you cannot solve for the time dependence of $S_i$. Different operators $H$ lead to different solutions. If there is no $t$ in your $H$ (as is often the case), then the explicit exponential solution is correct. So then everything follows. @ZeroTheHero Writing $S_z(t)$ would be the case no matter what since we are in the Heisenberg picture. If $H$ depends on $S_z$ there is no issue, because this is not explicit time dependence. $\endgroup$ Nov 5, 2022 at 3:47

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