When reading articles on CQED or atomic physics, I often encounter the Jaynes-Cummings Hamiltonian, which in the simplistic form I will write down as:

$$H = \hbar \omega_c a^\dagger a + \hbar \frac{\Omega}{2} \sigma_z + \hbar g(\sigma_+a + \sigma_-a^\dagger) \, .$$

Often one wants to add a drive term, that represents a laser/MW source driving the cavity. As the microwave source is a field oscillating, I would intuitively write it down as

$$H_\text{drive}(t) = \epsilon_\text{drive} \cos({\omega_\text{drive}t}) \, \left( a + a^{\dagger}\right) \, .$$

However, often I find it in the following form:

$$H_\text{drive}(t) = \epsilon_\text{drive} \, \left( e^{i\omega_\text{drive}t} a + e^{-i\omega_\text{drive}t}a^{\dagger} \right) \, .$$

Which is familiar, yet will lead to different results (even though often, for some 'magical' reason, results will be the same).

Can someone explain me what is the difference between the two versions, and what they physically represent?

Note: Since the second form for $H_{drive}$ when written in the rotating reference frame does not lead fast oscillating terms ($2\omega_d$), I guess people use it as the drive term already written in the RWA approximation, but I am not completely sure of that.

  • $\begingroup$ Are you absolutely sure the second form you wrote there is what you saw written? Often we work in the rotating frame, rotating at the drive frequency $\omega_\text{drive}$. However, this would not produce the second form of $H_\text{drive}$ written above, so I'm not sure if there is an error here. $\endgroup$
    – DanielSank
    May 25, 2016 at 5:28

1 Answer 1


Your Hamiltonian is much the same actually. If I write $e^{i\omega t}~=~cos(\omega t) + i sin(\omega t)$ the quoted driving term is $$ H_d = \epsilon \left(a(cos(\omega t) + i sin(\omega t)) + a^\dagger(cos(\omega t) - i sin(\omega t))\right) $$ $$ = \epsilon \left((a + a^\dagger)cos(\omega t) + i(a - a^\dagger)sin(\omega t))\right) $$ You can check that this is self-adjoint. This is however a bit more general in that the two terms are in a sense rotating and counter rotating.


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