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 $\omega_0+\omega \gg |\omega_0-\omega|$ ?

  • 3
    $\begingroup$ if $\omega_0 + \omega \gg |\omega_0 - \omega|$ then $\frac{1}{\omega_0 + \omega} \ll \frac{1}{ |\omega_0 - \omega|}$ $\endgroup$ Commented Mar 16, 2018 at 19:46
  • 1
    $\begingroup$ The averages of the two numerators over time is the same, so it's the denominators that really matter. $\endgroup$
    – Red Act
    Commented Mar 16, 2018 at 19:53
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Mar 16, 2018 at 19:54
  • 2
    $\begingroup$ 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). $\endgroup$
    – DanielSank
    Commented Mar 16, 2018 at 19:56
  • 1
    $\begingroup$ @far.westerner Erm, well... to be honest, the particular quantum mechanics textbook referenced in this question is, in my humble opinion, terrible. $\endgroup$
    – DanielSank
    Commented Mar 16, 2018 at 21:22

1 Answer 1


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.

However, while we're here, it's important to note that there's a wide array of important real-world situations where the RWA can fail to capture important parts of the physics. It's a great tool, and much of the time you can build enormous theoretical edifices within its confines while keeping excellent agreement with experiment, but it can indeed break. (Though countering that, as Peter Knight once said, if you're looking to study physics beyond the RWA, you'd better think carefully about whether there are other approximations which will break before it and which you should investigate first.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.