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Context

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, which actually involves no approximation and is for convenience, is to come up with an appropriate unitary matrix $U$ which gets rid of all (or almost all?) time dependence with

$$H_{RWA}= U^{\dagger} \tilde{H}U + i(\partial_t U^{\dagger}) U$$

Question

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 $H, \tilde{H}$?

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

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  • $\begingroup$ You're confusing two separate concepts: 1) the rotating frame and 2) the rotating wave approximation. Your $U$ is not a "rotating wave transform" (I've never heard that term before). Rather, it's just a change of frame. $\endgroup$
    – DanielSank
    Sep 3, 2023 at 2:29
  • $\begingroup$ @DanielSank thank you. Perhaps I used the wrong term, but is it not true that you have two ingredients: replace the real sinusoid with one one complex exponential, then a unitary transform to get rid of the time dep. My question is about getting $U$ properly $\endgroup$
    – KF Gauss
    Sep 3, 2023 at 2:46

1 Answer 1

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A partial answer to your question is the following. Consider a unitary transformation of the form $$ U(t) = \text{diag}\left(e^{ix_1t},\,e^{ix_2t},\, ... \right), $$ the Hamiltonian, written in the rotating frame, is $U^{\dagger}(t) H U(t) + iU^{\dagger}(t)\partial_t U(t)$. A diagonal element at position $i$ is time independent and given by $\omega_i-x_i$; while an off-diagonal element at position $ij$ (with $j>i$) is time dependent, and it reads $$ \Omega_{ij} e^{i(f_{ij} - x_i + x_j)t}. $$ In order to make all of these terms time independent, we get a set of equation of the form $$ x_i - x_j = f_{ij}. $$

This is a set of (at most) $N(N-1)/2$ equations with only $N$ variables, which means that it can't be solved in general. However, under specific circumstances, it can be solved. In fact notice that if one of the couplings $\Omega_{ij}$ vanishes, then we have one less equation to solve, so the number of independent equations is $N(N-1)/2 - n$, where $n$ is the number of couplings that are set to 0.

For example, if $N=2$, then $N(N-1)/2 = 1$ and we have one equation and two variables: $x_1 - x_2 = f_{12}$, which has infinite solutions, for example $x_1 = f_{12}$ and $x_2=0$. Another example is the case $N=3$ with, say, $\Omega_{12}=0$. In this case, the equation $x_1-x_2=f_{12}$ doesn't hold and we just get $$ \left\{ \begin{array}{c} x_1 - x_3 = f_{13} \\ x_2 - x_3 = f_{23} \end{array} \right. $$ which has infinite solutions, for example $x_3=0$, $x_2 = f_{23}$ and $x_1 = f_{13}$.

Of course this is not the most general answer to your question, but at least it paves the way to a first attempt of solution. Hope this helps! :)

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  • $\begingroup$ Thank you! There was a typo in my equation, the time derivative for $U$ was in the wrong place, which changes a few of the signs. Anyways, so would you think that in general it should not be possible to find a $U$ which gets rid of all time dep of the RWA form, even if you allow for off diagonal terms? $\endgroup$
    – KF Gauss
    Sep 3, 2023 at 14:00
  • $\begingroup$ Or is there some theorem that RWA Hamiltonians of the form in my question $\tilde{H}$ can always be written in a time dep form after a unitary transformation? $\endgroup$
    – KF Gauss
    Sep 3, 2023 at 14:37
  • $\begingroup$ Unfortunately I don't know what happens when allowing off diagonal terms in $U(t)$. However the trick shown here can be used in some of the most relevant real life situations. $\endgroup$
    – Matteo
    Sep 3, 2023 at 15:02

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