Consider an $N$-level Hamiltonian with energies $\omega_1...\omega_N$ with coupling drives at frequencies $f_{i,j}$ which couple the $i$ and $j$-th levels (not necessarily resonantly, so $f_{i,j} \neq \omega_j - \omega_i$).

This Hamiltonian looks something like

$$H = \sum_{i}^{N} \omega_i \vert i\rangle \langle i \vert + \sum_{i,j} \Omega_{i,j}\cos(f_{i,j} t) \vert i\rangle \langle j \vert + h.c.$$ Or as a matrix

$$H= \begin{bmatrix} \omega_1 & \Omega_{1,2}\cos(f_{1,2} t) & \Omega_{1,3}\cos(f_{1,3} t) & \dots\\ \Omega^*_{1,2}\cos(f_{1,2} t)& \omega_2& \Omega_{2,3}\cos(f_{2,3} t) & \dots \\ \Omega_{1,3}^*\cos(f_{1,2} t) & \Omega^*_{2,3}\cos(f_{2,3} t)&\omega_3 & \dots\\ \vdots & \vdots &\vdots & \ddots \end{bmatrix}$$

The rotating wave approximate to $H$, denoted $H_{RWA}$ would involve first replacing the $\cos(f_{i,j} t)$ terms with $e^{i f_{i,j} t}$

$$H\approx \tilde{H}=\begin{bmatrix} \omega_1 & \Omega_{1,2}e^{if_{1,2} t} & \Omega_{1,3}e^{if_{1,3} t}& \dots\\ \Omega^*_{1,2}e^{-if_{1,2} t}& \omega_2& \Omega_{2,3}e^{if_{2,3} t} & \dots \\ \Omega_{1,3}^*e^{-if_{1,2} t} & \Omega^*_{2,3}e^{-if_{2,3} t}&\omega_3 & \dots\\ \vdots & \vdots &\vdots & \ddots \end{bmatrix}$$

The second step is a matter of convenience which is to come up with an appropriate unitary matrix $U$ to get rid of the explicit time dependent with

$$H_{RWA}= U^{\dagger} \tilde{H} U + iU^{\dagger}\partial_t U$$


Is there an algorithm for going through the frequencies appearing in the Hamiltonian $H$ and coming up with the appropriate rotating frame transform $U$ and obtain $H_{RWA}$ from $H$?

It is easy enough to get $U$ with the RWA when you have two levels, but it is unclear to me how to do it systematically for larger $N$.

  • $\begingroup$ Dropping one of the exponents is already the final result of RWA, isn't it? You start by going to the interaction picture, using unitary U=exp(i*H0*t), where the diagonal is zero. The off-diagonal terms will get the w0 phase, and the ones in which the frequencies are summed are rapidly oscillating, so we neglect them. Going back to Schroedinger's picture, you'll get your second Hamiltonian. That's already RWA Hamiltonian. $\endgroup$
    – EvgeniyZh
    Sep 3, 2023 at 8:15
  • $\begingroup$ @EvgeniyZh I think that's correct, applying $U$ involves no approximation on its own. However, it seems to virtually always be the case in RWA that the goal is to end up with a time independent Hamiltonian, so the optional step of $U$ is always done in practice. I guess I don't understand how to systemically find $U$ which gives a time independent result (or nearly time independent at least) $\endgroup$ Sep 3, 2023 at 9:07
  • $\begingroup$ You could parametrize $U$ with some anti-symmetric matrix $U = e^{iGt}$ and solve the resulting linear equations. Yet, there are more equations than variables (you have diagonal terms for $F$ but not for $G$), maybe that's what you mean by "nearly"? $\endgroup$
    – EvgeniyZh
    Sep 3, 2023 at 10:01
  • $\begingroup$ Actually, you have diagonal terms too, since G needs to be Hermitian (or iG skew-Hermitian) so you should be able to solve the system unless there are some accidental degeneracies $\endgroup$
    – EvgeniyZh
    Sep 3, 2023 at 11:07
  • $\begingroup$ @EvgeniyZh is there an algorithm for solving that? It looks rather messy. It's relatively straightforward for a diagonal $U$, but the general solution would not become time independent only with a diagonal $U$, some off-diagonals are needed $\endgroup$ Sep 3, 2023 at 16:57


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