What is the meaning of rotating the frame in quantum mechanics?

Question

I want to ask about the meaning of rotating the frame in quantum mechanics.

Many papers write the rotating frame and I'm not sure what the meaning of it.

Answer 1

It is essentially equivalent to going to the “interaction picture” with respect to some hamiltonian. For instance, let's say we have a Hamiltonian that can be broken down to two pieces-

$H = H_0 + V(t)$

In the Schrodinger picture, we know that any state $|\psi (t)\rangle$ will evolve under the Schrodinger equation given by $i\hbar\dot{|\psi (t)\rangle} = H |\psi (t)\rangle$. In the Heisenberg picture, we consider the states to be stationary, however, the operators evolve with time. The interaction picture is somewhat in between, where we consider the states to evolve under some part of the hamiltonian ($V(t)$), and the operators to evolve under the rest ($H_0$). You can mathematically show the evolutions to be like this-

$i\hbar\dot{|\psi (t)\rangle_I} = V_I(t) |\psi (t)\rangle_I$

$i\hbar\dot{A_I} = [A_I, H_0]$

where $A_I = U^{\dagger}A U$ and $U = \exp{(-\frac{iH_0t}{\hbar})}$

With this processes, we have basically simplified some parts of the evolution of the system (evolution due to $H_0$) and focus on the more interesting parts ($V(t)$). This is similar to the case where a classical object is going through a complex motion along with some simple rotation, and you can basically focus on the complex motion by going to the rotating frame with the same frequency.

I'll give you one simple example to understand this. Let's say we have a hamiltonian like this-

$H = \hbar \omega \sigma_z + \hbar \frac{\Omega}{2} (\exp(i\omega t) + \exp(-i\omega t))\sigma_x$.

If we take $H_0 = \hbar \omega \sigma_z$ and $V(t) = \hbar \frac{\Omega}{2} (\exp(i\omega t) + \exp(-i\omega t))\sigma_x$ and go to the interaction picture with $H_0$, we can show that the term $V(t)$ in the interaction picture will be- $\hbar \frac{\Omega}{2} \sigma_x (1+ \exp(2i\omega t))$. Usually, the $\exp{(2i\omega t)}$ term implies a faster rotating term, which we can safely ignore as it violates energy conservation. Thus, in the rotating frame, our state only evolves with the simple $\hbar \frac{\Omega}{2}\sigma_x$, which makes the calculations a lot easier.

Answers in this thread- What is the "interaction picture" or "rotating frame" in quantum mechanics? discusses the underlying math nicely.