The paper which I'm reading mentions the following:

The Hamiltonian,

$$ \frac{\varepsilon}{2}\sigma_z + \sum_k \omega_k b_k^\dagger b_k + \sum_k (\sigma^+ g_k^* b_k + \sigma^- g_k b_k^\dagger), $$

is the same as the Hamiltonian:

$$ \frac{\varepsilon}{2}\sigma_z + \sum_k \omega_k b_k^\dagger b_k + \sigma_x \sum_k (g_k^* b_k + g_k b_k^\dagger), $$

except that the non-rotating wave approximation terms have been dropped.

With non-rotating wave approximation terms now included, we get:

$$ \frac{\varepsilon}{2}\sigma_x + \sum_k \omega_k b_k^\dagger b_k + \sigma_z \sum_k (g_k^* b_k + g_k b_k^\dagger). $$

I'm not sure what's the (non-) rotating wave approximation (RWA) and how it gives the second and the third Hamiltonians under the given conditions. I do get that we'll get the Hamiltonians in the second and the third equations by applying the relevant rotation operators.

If someone could provide the details, especially where the RWA is made and when are where are the non-RWA


1 Answer 1


This looks to me like a Jaynes-Cummings Hamiltonian, describing the interaction of a two-level atom with a quantized light field. For one polarization of the light field (k=1) the total Hamiltonian is made up of the atoms hamiltonian, the field hamiltonian and the interaction hamiltonian $$ H_{atom} = \frac{\hbar\omega_a}{2}\sigma_z\\ H_{field} = \hbar\omega_{field} \,a^\dagger a\\ H_{int} = \frac{\hbar\Omega}{2}(\sigma_+ + \sigma_-)(a^\dagger + a),$$ where $\sigma_x = (\sigma_+ + \sigma_-)/2$ and $a^\dagger$ and $a$ are the creation and annihilation operators of the light field. In this case, the trick is to look at the time evolution of the interaction hamiltonian in the interaction picture. The operators will evolve as follows $$ a_{int}(t) = a e^{-i\omega_{field}t} \\ a^\dagger_{int}(t) = a^\dagger e^{+i\omega_{field}t} \\ \sigma_-(t) = \sigma_{-} \, e^{-i\omega_at}\\\sigma_+(t) = \sigma_{+} \,e^{+i\omega_at}, $$ thus the interaction hamiltonian becomes $$ H_{int}^I = \frac{\hbar\Omega}{2}\left( \sigma_+a^{\dagger}e^{i\left(\omega_a+\omega_{field}\right)t}+\sigma_+ae^{i\left(\omega_a-\omega_{field}\right)t}\\+\sigma_-a^{\dagger}e^{i\left(-\omega_a+\omega_{field}\right)t}+\sigma_-ae^{-i\left(\omega_a+\omega_{field}\right)t} \right).$$ What is now referred to as the "rotating wave approximation" is to ommit the quickly oscillating terms in $H^I_{int}$, i.e. the terms oscillating with $\omega_a + \omega_{field}$ instead of $\left| \omega_a-\omega_{field}\right|$. When you transform back to the Schrödinger picture this leaves you with $$H_{int} = \frac{\hbar\Omega}{2}\left(\sigma_+a + \sigma_-a^\dagger\right), $$ coresponding to the first of the formulae you posted.

  • $\begingroup$ Why do you have terms of the form $\sigma_{+}a^\dagger$ and $\sigma_{-}a$ in $H_{int}$ in the Schordinger picture. These aren't included in the first expression. Am I missing out on something? $\endgroup$ Jul 15, 2016 at 17:40
  • $\begingroup$ If you factor out the not approximated version of $H_{int}$ in Schrödinger picture, it is $H_{int} = \frac{\hbar\Omega}{2}\left(\sigma_+ a^\dagger + \sigma_-a^\dagger + \sigma_+ a + \sigma_- a\right)$. Is it that what you mean? $\endgroup$ Jul 15, 2016 at 17:44
  • $\begingroup$ Yes. The first two terms don't make sense. In the population decay model, we assume that the transition from the excited to the ground state is cause by a light quanta, modeled by the a mode being excited by the relevant creation operator and vice verse. I assume this is the reason these two terms aren't included in the first expression as well. $\endgroup$ Jul 15, 2016 at 17:47
  • $\begingroup$ This therms are included in your second hamiltonian when you insert $\sigma_x = (\sigma_+ + \sigma_-)/2$. Then you apply the rotating wave approximation and only the terms that you mention are left, which allow this interpretation. $\endgroup$ Jul 15, 2016 at 17:54
  • $\begingroup$ So maybe I am missing out on the larger picture. Could you please tell me why these terms were included in the Hamiltonian in the first place since the only meaningful interpretation can be given to only the last two terms? $\endgroup$ Jul 15, 2016 at 17:58

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