# Sinusoidal drive of two level system: why can we ignore one of the two terms?

I am studying the time-dependent perturbation theory from Griffith's Introduction to Quantum Mechanics. The context is a two-level system under a sinusoidal perturbation: $H'(\textbf{r}, t) = V(\textbf{r}) \cos(\omega t)$, where the symbols have their usual meanings.

He gets the following expression (eq. (9.25)):

$$c_b(t) = -\frac{V_{ba}}{2\hbar} \left[ \frac{e^{i(\omega_0+\omega)t} - 1}{\omega_0+\omega} + \frac{e^{i(\omega_0-\omega)t} - 1}{\omega_0-\omega} \right].$$

For $\omega_0+\omega \gg |\omega_0-\omega|$, he says that the second term in the square brackets dominates.

How do we know that the second term in the square brackets dominates from the condition $\omega_0+\omega \gg |\omega_0-\omega|$ ?

• if $\omega_0 + \omega \gg |\omega_0 - \omega|$ then $\frac{1}{\omega_0 + \omega} \ll \frac{1}{ |\omega_0 - \omega|}$ – By Symmetry Mar 16 '18 at 19:46
• The averages of the two numerators over time is the same, so it's the denominators that really matter. – Red Act Mar 16 '18 at 19:53
• Not really. They have magnitudes between $0$ and $2$, which to a physicist is just $\sim 1$. The only time you might think they matter is when the numerator of the second term vanishes, but since they are oscillating in time this can't happen for long, i.e. there is no physical case where the second term is always $0$, so we ignore it. – By Symmetry Mar 16 '18 at 19:54
• FYI the approximation of dropping the second term is called the "rotating wave approximation". For fun, here's a paper about a real life system where the rotating wave approximation is not very good (this is a biased reference because it's my paper). – DanielSank Mar 16 '18 at 19:56
• @far.westerner Erm, well... to be honest, the particular quantum mechanics textbook referenced in this question is, in my humble opinion, terrible. – DanielSank Mar 16 '18 at 21:22

As mentioned in the comments, this is known as the Rotating-Wave Approximation, and it holds because the exponents $(\omega\pm \omega_0)t$ are real, so that the exponentials $|e^{i(\omega\pm \omega_0)t}|\leq 1$ are bounded in modulus, so that $$\omega_0+\omega \gg |\omega_0-\omega| \implies \frac{1}{\omega_0+\omega} \ll \left|\frac{1}{\omega_0-\omega}\right|$$ and the exponentials don't change the picture much.