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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 $(\sigma_y \otimes \sigma_x)$?

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.