What is the difference between the Interaction picture (Dirac Picture) and a rotating reference frame? In David McIntyre's Quantum Mechanics, we examine an electron within a magnetic field
$$\vec{B}=B_o \hat{z}+B_1[\cos(\omega t)\hat{x}+\sin(\omega t)\hat{y}]$$
The Hamiltonian is then time-dependent and in matrix form it is as follows
$$H=\frac{\hbar}{2}\begin{bmatrix}\omega_0& \omega_1e^{-i\omega t}\\\omega_1e^{i\omega t}&-\omega_0\end{bmatrix}=H_0+H_t(t)$$
where $\omega_0=\frac{eB_0}{m}$ and $\omega_1=\frac{eB_1}{m}$. McIntyre then says that the solution to the Schrödinger equation can be written using the energy basis of the time-independent part of the Hamiltonian (this is the usual basis: $|\uparrow\rangle_z=(1,0)$and $ |\downarrow\rangle_z=(0,1)$) as
$$|\psi(t)\rangle=c_+(t)|\uparrow\rangle_z +c_-(t)|\downarrow\rangle_z=(\,c_+(t)\,,\,c_-(t)\,) \tag{1}$$
Now as far as I am aware, the above solution is in the Schrödinger picture where $|\psi(t)\rangle_s=e^{-iHt/\hbar}|\psi(t=0)\rangle$. Now I have learnt in class that the state in the interaction picture is defined as
$$
|\psi(t)\rangle_I=e^{iH_0t/\hbar}|\psi(t)\rangle \tag{2}
$$
where the exponent only includes the time-independent part of the Hamiltonian. My issue is that McIntyre then says that we can simplify the problem by transforming the state in equation (1) to the rotating frame which he says (without justification) yields
$$
|\psi(t)_*\rangle=c_+(t)e^{i\omega t/2}|\uparrow\rangle_z +c_-(t)e^{-i\omega t/2}|\downarrow\rangle_z=(\,c_+(t)e^{i\omega t/2}\,,\,c_-(t)e^{-i\omega t/2}\,) \tag{3}
$$
Now I have read that the rotating frame is related to (if not entirely equivalent to) the interaction picture. This leads me to think that equation (3) is the state vector in the interaction picture for this problem. However, how can this be if the interaction picture is defined as in equation (2)? If we are to believe equation (2), then should we not get something like
$$
|\psi(t)_*\rangle=(\,c_+(t)e^{i\omega_1 t/2}\,,\,c_-(t)e^{-i\omega_1 t/2}\,)
$$
where we've used $\omega_1$ instead of $\omega$. In fact, why should we not use $\omega_0$ considering in equation (3), it is the unperturbed time-independent Hamiltonian that is present while $\omega$ is associated with the energy due to the time-dependent part?
 A: Yes, the interaction picture is a rotating frame, but not every rotating frame is "the" interaction picture. Therefore we do not use equation (2) to find (3) but rather some [yet-to-be-proven-to-be] useful transformation
$$
\begin{pmatrix}c_+^R(t)\\c_-^R(t)\end{pmatrix}=\exp\left(i \begin{pmatrix}1&0\\0&-1\end{pmatrix}\omega t\right)
\begin{pmatrix}c_+^S(t)\\c_-^S(t)\end{pmatrix}.
$$ Note that we have not said anything about what functional form the coefficients $\begin{pmatrix}c_+^S(t)\\c_-^S(t)\end{pmatrix}$ actually take - we have only specified how they related to the "rotated" coordinates $\begin{pmatrix}c_+^R(t)\\c_-^R(t)\end{pmatrix}$. You have tried to guess the answer for what the coefficients should look like, but that is exactly what is hard to do because we are solving a problem with a time-dependent Hamiltonian!
Another way of looking at this is by considering the transformation to act on the basis states:
$$
|\uparrow\rangle_z\to |\uparrow(t)\rangle_z\equiv e^{i\omega t/2}|\uparrow\rangle_z ,\quad |\downarrow\rangle_z\to |\downarrow(t)\rangle_z\equiv e^{-i\omega t/2}|\downarrow\rangle_z.
$$ If we write the Hamiltonian in bra-ket notation, dropping the $z$ subscript, it looks like
\begin{align}
\frac{2H}{\hbar}=\omega_0|{\uparrow}\rangle\langle{\uparrow}|-\omega_0|{\downarrow}\rangle\langle{\downarrow}|
+\omega_1e^{-i\omega t}|{\uparrow}\rangle\langle{\downarrow}|
+\omega_1e^{i\omega t}|{\downarrow}\rangle\langle{\downarrow}|.
\end{align}  We can substitute in our new definitions of the states to see that (remembering to take all of the complex conjugates)
\begin{align}
\frac{2H}{\hbar}=\omega_0|{\uparrow(t)}\rangle\langle{\uparrow(t)}|-\omega_0|{\downarrow(t)}\rangle\langle{\downarrow(t)}|
+|{\uparrow(t)}\rangle\langle{\downarrow(t)}|
+|{\downarrow(t)}\rangle\langle{\downarrow(t)}|.
\end{align} Do you see the trick? Using this redefinition, we have made the Hamiltonian appear to be time-independent! It is only time-independent in this funky basis in which the states depend on time, but that time dependence is easy to sort out at the end of any calculation. This final form of the Hamiltonian justifies the choice of rotating frame. In other contexts, such as "the" interaction picture, another final form of the Hamiltonian justifies a different choice of rotating frame  like the one you suggested using $\exp(i H_0 t)$.
