$\renewcommand{\ket}[1]{\left\lvert #1 \right\rangle}$ 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.


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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.

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  • $\begingroup$ Nice @DanielSank. But I just wonder as what would the transformation $R(t) = U(t)^\dagger$ 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! $\endgroup$ – Seeker Oct 5 '17 at 21:36
  • $\begingroup$ @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 $\endgroup$ – DanielSank Oct 5 '17 at 21:41
  • $\begingroup$ Well, backward temporal evolution is something which doesn't go down the throat easily. I mean, it is hard to visualize this transformation. $\endgroup$ – Seeker Oct 5 '17 at 21:46
  • $\begingroup$ @Seeker really? You can't imagine a car rolling backwards, or a spinning top reversing direction? What about moving your hand to the left, and then to the right? $\endgroup$ – DanielSank Oct 5 '17 at 21:48
  • $\begingroup$ 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. $\endgroup$ – Seeker Oct 5 '17 at 22:24

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