# What is the link between the rotating wave approximation and the algebraic representation of a dynamical system?

In analyzing a system of two coupled oscillators, I noticed a rather interesting correspondence between the so-called "rotating wave approximation" (RWA) for solving differential equations and the structure of the algebraic representation of the matrix form of those equations.

## Background

The system of equations under consideration is $$\frac{d}{dt} \left( \begin{array}{c} a \\ a^* \\ b \\ b^* \end{array} \right) = i \left( \begin{array}{cccc} - \omega_a & 0 & -g & g \\ 0 & \omega_a & -g & g \\ -g & g & - \omega_b & 0 \\ -g & g & 0 & \omega_b \end{array} \right) \left( \begin{array}{c} a \\ a^* \\ b \\ b^* \end{array} \right) \tag{\star} \, .$$ This system is the Hamiltonian version of the system described in a previous question, and can be derived from the Hamiltonian $$H = \omega_1 a^* a + \omega_2 b^* b - g \left( a b - a^* b - a b^* + a^* b^* \right)$$ using the appropriate Hamiltonian equations of motion $$\dot{a} = -i \frac{\partial H}{\partial a^*} \quad \text{and} \quad \dot{b} = -i \frac{\partial H}{\partial b^*}$$ and similarly for $$a^*$$ and $$b^*$$.

The parameter $$g$$ characterizes the strength of coupling between the two oscillators. Note that if $$g=0$$, then we have two uncoupled equations $$\dot a = -i \omega_a a \quad \text{and} \quad \dot b = -i \omega_b b$$ with solutions $$a(t) = a(0) e^{-i \omega_a t} \quad \text{and} \quad b(t) = b(0) e^{-i \omega_b t}$$ representing two uncoupled oscillators.

Now note that the matrix in Equation $$(\star)$$ can be written algebraically as $$-i \sigma_z \otimes \left( g \sigma_x + \frac{d}{2} \sigma_z + \frac{s}{2} \mathbb{I} \right) - g (\sigma_y \otimes \sigma_x)$$ where $$s \equiv \omega_1 + \omega_2$$ and $$d \equiv \omega_1 - \omega_2$$.

## Rotating wave approximation

The rotating wave approximation (RWA) comes essentially from noting that while the time dependences of e.g. $$a$$ and $$b^*$$ have opposite sign and therefore lead to slow oscillation, the time dependences of e.g. $$a$$ and $$b$$ have the same sign and lead to fast oscillation. Therefore, the terms $$ab$$ and $$a^* b^*$$ are dropped from the Hamiltonian. Doing this leads to an equation like $$(\star)$$, but with a matrix given by $$-i \sigma_z \otimes \left( g \sigma_x + \frac{d}{2} \sigma_z + \frac{s}{2} \mathbb{I} \right) \, ,$$ i.e. the $$-g (\sigma_y \otimes \sigma_x)$$ term is gone.

## Question

Why is the rotating wave approximation equivalent to dropping the $$(\sigma_y \otimes \sigma_x)$$ term in the dynamical equation matrix? What intuitive link is there between the argument we gave for the RWA (or any other argument) and the dropping of the algebraic term $$(\sigma_y \otimes \sigma_x)$$?

This question has been psuedo-cross-posted to the Math site in an attempt to get a rigorous and more mathematically oriented answer.

I think it may be easier to understand the system of equations if you swap the ordering of operators in ($$\star$$) to $$(a,b,a^*,b^*)$$. In this case the equation becomes
$$\frac{d}{dt} \left( \begin{array}{c} a \\ b \\ a^* \\ b^* \end{array} \right) = -i H_{\textrm{eff}} \left( \begin{array}{c} a \\ b\\ a^* \\ b^* \end{array} \right) \tag{\star} \, .$$ \begin{align*} H_{\textrm{eff}} & = \sigma_z \otimes \left( g \sigma_x + \frac{d}{2} \sigma_z + \frac{s}{2} \mathbb{I} \right) - i\, g (\sigma_y \otimes \sigma_x) \\ &= \left( \begin{array}{cccc} \omega_1 & g & 0 & -g \\ g & \omega_2 & -g & 0 \\ 0 & g & -\omega_1 & -g \\ g & 0 & -g & -\omega_2 \end{array} \right) \end{align*} The $$i\, g\, \sigma_y\otimes \sigma_x$$ term then corresponds to the block off-diagonal part of the matrix. The first diagonal block has eigenvalues $$\frac{s}{2} \pm \frac{1}{2}\sqrt{d^2 + 4 g^2}$$, where $$s = \omega_1 + \omega_2, d = \omega_1-\omega_2$$. To get the second block's eigenvalues we send $$s\rightarrow-s$$. So for $$s\gg |d|, g$$, these blocks are split in 'energy' by approximately $$s$$.
The 'secular approximation' for this 'Hamiltonian' corresponds to throwing away terms that are 1) small and 2) do not 'conserve energy'. In the case $$s\gg |d|, g$$, the term $$i\, g\, \sigma_y\otimes \sigma_x$$ is small and changes the energy of a 'state' by approximately $$\pm s$$. This is unlike the other term proportional to $$g$$, which although is small does not change the energy of a 'state' by more than $$\sim |g|$$.