# What is the "interaction picture" or "rotating frame" in quantum mechanics?

$\renewcommand{\ket}[1]{\left\lvert #1 \right\rangle}$ In typical quantum mechanics courses, we learn about the so-called "Schrodinger picture" and "Heisenberg picture". In the Schrodinger picture, the equation of motion brings time dependence to the states, $$i \hbar \partial_t \ket{\Psi(t)} = H(t) \ket{\Psi(t)}$$ while in the Heisenberg picture the equation brings time dependence to the operators, $$i \hbar \partial_t A(t) = [A(t), H(t)] \, .$$

When we have a Hamiltonian which can be split into an "easy" part $H_0(t)$$^{[a]} and a time dependent "difficult" part V(t),$$H(t) = H_0(t) + V(t)$$people talk about the "interaction picture" or "rotating frame". What's the difference between the interaction picture and rotating frame, and how do they work?$[a]$: We assume that$H_0(t)commutes with itself at different times. • Feb 16, 2022 at 11:43 ## 1 Answer $$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$$ Basic idea: the rotating frame "unwinds" part of the evolution of the quantum state so that the remaining part has a simpler time dependence. The interaction picture is a special case of the rotating frame. Consider a Hamiltonian with a "simple" time independent part $$H_0$$, and a time dependent part $$V(t)$$: $$H(t) = H_0 + V(t) \, .$$ Denote the time evolution operator (propagator) of the full Hamiltonian $$H(t)$$ as $$U(t,t_0)$$. In other words, the Schrodinger picture state obeys $$\ket{\Psi(t)} = U(t, t_0) \ket{\Psi(t_0)}$$. The time evolution operator from just $$H_0$$ is (assuming $$H_0$$ is time independent, or at least commutes with itself at different times) $$U_0(t, t_0) = \exp\left[ -\frac{i}{\hbar} \int_{t_0}^t dt' \, H_0(t') \right] \, .$$ Note that $$i\hbar \partial_t U_0(t, t_0) = H_0(t) U_0(t, t_0) \, .$$ Define a new state vector $$\ket{\Phi(t)}$$ as $$\ket{\Phi(t)} \equiv R(t) \ket{\Psi(t)}$$ where $$R(t)$$ is some "rotation operator". Now find the time dependence of $$\ket{\Phi(t)}$$: \begin{align} i \hbar \partial_t \ket{\Phi(t)} =& i \hbar \partial_t \left( R(t) \ket{\Psi(t)} \right) \\ =& i \hbar \partial_t R(t) \ket{\Psi(t)} + R(t) i \hbar \partial_t \ket{\Psi(t)} \\ =& i \hbar \dot{R}(t) \ket{\Psi(t)} + R(t) H(t) \ket{\Psi(t)} \\ =& i \hbar \dot{R}(t) R(t)^\dagger \ket{\Phi(t)} + R(t) H(t) R(t)^\dagger \ket{\Phi(t)} \\ =& \left( i \hbar \dot{R}(t) R(t)^\dagger + R(t) H(t) R(t)^\dagger \right) \ket{\Phi(t)} \, . \end{align} Therefore, $$\ket{\Phi(t)}$$ obeys Schrodinger's equation with a modified Hamiltonian $$H'(t)$$ defined as $$H'(t) \equiv i \hbar \dot{R}(t) R(t)^{\dagger} + R(t) H(t) R(t)^\dagger \, . \tag{\star}$$ This is the equation of motion in the rotating frame. Useful choices of $$R$$ depend on the problem at hand. Choosing $$R(t) \equiv U_0(t, t_0)^\dagger$$ has the particularly useful property that the first term in $$(\star)$$ cancels the $$H_0(t)$$ part of the second term, leaving \begin{align} i \hbar \partial_t \ket{\Phi(t,t_0)} = \left( U_0(t, t_0)^\dagger V(t) U_0(t, t_0) \right)\ket{\Phi(t, t_0)} \, . \end{align} which is Schrodinger's equation with effective Hamiltonian $$H'(t) \equiv U_0(t)^\dagger V(t) U_0(t) \, .$$ This is called the interaction picture. It is also known by the name Dirac picture. • Nice @DanielSank. But I just wonder as what would the transformationR(t) = U(t)^\dagger$mean. Since$U(t) = e^{-i H t}$, which means$R(t) = e^{-i H (-t) }$. Does it mean that R(t) takes the wavefunction$\Psi (t)$backward in time! Oct 5, 2017 at 21:36 • @Seeker "Does it mean that$R(t)$takes the wave function$\Psi(t)$backward in time!" Yes, although I'm not sure why that question gets an exclamation mark instead of the usual question mark :-P Oct 5, 2017 at 21:41 • Well, backward temporal evolution is something which doesn't go down the throat easily. I mean, it is hard to visualize this transformation. Oct 5, 2017 at 21:46 • Reversibility of the motions which you are talking about (the car, the top or even one's head) is the spatial reversibility by which I mean you simply change the spatial coordinates. Say a transformation$x \rightarrow -x$. Now that is easy to visualize. But when it comes to$t \rightarrow -t$, it is really not so easy. Could you give me just one such trivial example which takes you from$t$all the way back to$-t\$? Surely, you can't. Oct 5, 2017 at 22:24
• @Lefteris I hope the linked document has been useful (you can email any questions you have). Was this post clear enough to resolve your question? Oct 1, 2021 at 0:39