UPDATE: See update at bottom of post
Short Version:
$\sigma^-_H$ in the Heisenberg picture. $U = e^{+i\omega (|e\rangle \langle e)_H}$. Prove that
$$ \frac{d \sigma^-_H}{dt} e^{+i\omega t} = U^{\dagger} \frac{d\sigma^-_H}{dt} U \tag{*} $$
Long Version:
Suppose we have the operator $\sigma^-_H = \left(|e\rangle \langle e\rangle\right)_H$, the Heisenberg lowering Pauli operator. It is in the Heisenberg picture so it has time dependence (For now I don't specify the Hamiltonian as I don't think that should matter for the problem).
Consider the operator
$$ \sigma^-_A = \sigma^-_H e^{+i\omega t} $$
We can calculate, using the product rule that
$$ \tag{1} \frac{d \sigma^-_A}{dt} = \left(\frac{d\sigma^-_H}{dt} + i\omega \sigma^-_H \right)e^{+i\omega t} $$
Now consider the unitary transformation
$$ U = e^{+i\omega (|e\rangle \langle e| )_H t} $$
and the new operator
$$ \sigma^-_B = U^{\dagger} \sigma^-_H U $$
We have that $U^{\dagger} \sigma^-_H = \sigma^-_H$, $\sigma^-_H U = \sigma^-_H e^{+i\omega t}$ and $\frac{d U}{dt} = i\omega (|e\rangle \langle e|)_H U$
We can see that
$$ \sigma^-_B = \sigma^-_H e^{+i\omega t} = \sigma^-_A $$
This means we should have $\frac{d \sigma^-_B}{dt} = \frac{d\sigma^-_A}{dt}$ I now calculate $\frac{d \sigma^-_B}{dt}$ using the product rule on the definition of $\sigma^-_B$.
\begin{align} \frac{d \sigma^-_B}{dt} &= \frac{d U^{\dagger}}{dt} \sigma^-_H U + U^{\dagger} \sigma^-_H \frac{dU}{dt} + U^{\dagger} \frac{d\sigma^-_H}{dt} U\\ &= -i\omega U^{\dagger}(|e\rangle \langle e|)_H \sigma^-_H U + i\omega U^{\dagger} \sigma^-_H (|e\rangle \langle e|)_H U + U^{\dagger} \frac{d \sigma^-_H}{dt} U\\ &= i \omega \sigma^-_H e^{+i\omega t} + U^{\dagger} \frac{d \sigma^-_H}{dt} U \tag{2} \end{align}
Noting that $(|e\rangle \langle e|)_H \sigma^-_H = 0$.
Comparison of $(1)$ and $(2)$ indicates that if $\frac{d \sigma^-_B}{dt} = \frac{d\sigma^-_A}{dt}$ then we should have
$$ \frac{d \sigma^-_H}{dt} e^{+i\omega t} = U^{\dagger} \frac{d\sigma^-_H}{dt} U \tag{*} $$
I think I should be able to prove this formula only using the properties of $U$ and $\sigma^-_H$ indicated above but I am not able to.
My Best Attempt:
My best attempt is to invoke the time dependence of $\sigma^-_H$ under the Heisenberg picture Hamiltonian, $H_H$ to expand out $\frac{d\sigma^-_H}{dt}$. I don't think $H_H$ should need to be invoked because the derivation of $(1)$ did not rely on $H_H$ at all. Here is the attempt:
$$ \frac{d \sigma^-_H}{dt} = -\frac{i}{\hbar} [\sigma^-_H, H_H] $$
\begin{align} U^{\dagger}\frac{d \sigma^-_H}{dt}U &= -\frac{i}{\hbar} U^{\dagger}[\sigma^-_H, H_H]U = -\frac{i}{\hbar}U^{\dagger}\left(\sigma^-_H H_H - H_H \sigma^-_H\right)U\\ &= -\frac{i}{\hbar}\left(\sigma^-_H H_H U - U^{\dagger} H_H \sigma^-_H e^{+i\omega t} \right)\\ &= -\frac{i}{\hbar} \left( \sigma^-_H U^{\dagger}H_HU e^{+i\omega t} - U^{\dagger} H_H U \sigma^-_H e^{+i\omega t}\right)\\ &= -\frac{i}{\hbar}[\sigma^-_H,U^{\dagger} H_H U] e^{+i\omega t}\\ &= -\frac{i}{\hbar}[\sigma^-_H, H_B] e^{+i\omega t} \end{align}
Where I've inserted a factor of $U U^{\dagger} =1$ in one location and defined $H_B = U^{\dagger} H_H U$. This would give me the correct answer if $[\sigma^-_H, H_B] = [\sigma^-_H, H_H]$ or likewise if $H_B = H_H$ or $[H_H,U] =0$ but in general I don't think any of these things are necessarily true. I'm surprised by these conditions because as I said before the derivation of $(1)$ didn't require the introduction of $H_H$ at all.
Context
Just a note that I'm working this out in the context of another question which I recently asked here (which got almost zero attention..): Two Level Atom Rotating Frame. The answer to this question would also answer that one for me.
UPDATE
Ok I had an important realization that heavily complicates this problem. It comes from the following confusing notation. Consider
$$ |e(t) \rangle_H = U_H^{\dagger}(t) |e(t) \rangle_S $$
Here $U_H$ is the operator which induces the transformation FROM the Schrodinger picture TO the Heisenberg picture. I've explicitly indicated the time depedence of all kets and operators. The problem is that, if we assume that $|e\rangle_H = |e(t)\rangle_H$ then it is NOT the case that $\sigma^z_H = \left(\left(|e\rangle \langle e|\right)_H - \left(|g\rangle \langle g| \right)_H \right)$ or $\sigma^-_H = \left(|g\rangle \langle e| \right)_H$. This is because (for $\sigma^-$) we have
$$ \sigma^-_S = \left(|g(0) \rangle \langle e(0)|\right)_S $$
That is $\sigma^-_S$ is defined in terms of $|g(0)\rangle_S$ and $|e(0)\rangle_S$ RATHER THAN $|g(t)\rangle_S$ and $|e(t) \rangle_S$. This should have been clear because we know $\sigma^-_S$ should be time independent and we know that $\sigma^-_H$ should be time dependent. This means $\sigma^-_S$ should be defined in terms of Schrodinger kets at fixed time and $\sigma^-_H$ should be defined in terms of the Heisenberg transformation of those kets at fixed times (recalling that it is only the time evolved Schrodinger kets which become time independent after the Heisenber transformation).
In particular the upshot of this is that it causes the argument of the exponential expression for $U$ to be time dependent, a fact which I was not appreciating above. In particular this means the expression I gave for $\frac{dU}{dt}$ is over simplified and any conclusions drawn from that expression are suspect.
I am now trying to attempt the problem keeping in mind this time dependence but it is now much messier. In particular I am now messing around with Baker-Hausdorffformulas to manipulate the exponentials. It looks like this will lead to the right answer but I am not so familiar with these manipulations and could use any tips.
I'm hoping the answer will follow from these Baker-Hausdorff manipulations and the commutation relations between $\sigma^-$ and $\sigma^z$ (which hold in any "picture").
I will post an update if I can clean up my calculations.
