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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.

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  • $\begingroup$ Your initial question does a reasonable job at explaining the situation without requiring the reader to go through several pages of text. This one, unfortunately, doesn't. $\endgroup$ – Emilio Pisanty Aug 8 '18 at 17:37
  • $\begingroup$ Yes, the amount of text has gotten a bit out of hand. Now that I understand things better I think I can consolidate the information and I think I may be close to providing answers myself. What do recommend as per Stack Exchange best practices for me to delete material and clean up the posts? $\endgroup$ – jgerber Aug 8 '18 at 17:43
  • $\begingroup$ That depends on whether you have two core questions or just one and, if you have two distinct core questions, whether you can provide two solid answers to both core questions and fix this one so that it is readable. If you've solved the original, it may be a good idea to delete this one and just answer the initial question. $\endgroup$ – Emilio Pisanty Aug 8 '18 at 17:45
  • $\begingroup$ Ok I wrote down a complete answer to the previous question. I think there is a distinct question here. The previous question was asking about deriving a Hamiltonian whereas this question is asking about relating temporal dynamics in two different frames. I'll quickly answer it and then leave it up as a warning to anyone trying to calculate things when transforming under unitaries generated by time-dependent operators.. $\endgroup$ – jgerber Aug 9 '18 at 8:30
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The claim $(*)$ in the "Short Version" section of the original question is wrong.

The original hypothesis at $(*)$ was arrived at by comparison of $(1)$ and $(2)$, however $(2)$ was derived by erroneously assuming $(| e \rangle \langle e|)_H$ was time independent. As such, I see no reason to believe $(*)$ is in general true. It was the belief that it was true that caused me a lot of confusion.

For a counter example consider

$$ H_H = \sigma^-_H+\sigma^+_H $$

In this case

$$ \frac{d}{dt}\sigma^-_H = -\sigma^z_H $$

So

$$ \frac{d}{dt}\sigma^-_H e^{+i\omega t} = -\sigma^z_H e^{+i\omega t} $$

But $[\frac{d}{dt}\sigma^-_H,U] = 0$ so

$$ U^{\dagger} \frac{d}{dt}\sigma^-_H U = \frac{d}{dt}\sigma^-_H = -\sigma^z_H \neq -\sigma^z_H e^{+i\omega t} $$

This example shows very saliently that when there is a Hamiltonian which causes $(|e \rangle \langle e|)_H$ to evolve the equality can easily break down.

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