Questions about changing frame in QM and rotating wave approximation

[Malamala]

Hello! I have a 3 level system with energies $E_0$, $E_1$ and $E_2$ and 2 time varying electric fields, the first one connects $E_0$ and $E_1$ while the second one connects $E_0$ and $E_2$ with Rabi frequencies $\Omega_1$ and $\Omega_2$ and the frequencies of the electric fields are $\omega_1$ and $\omega_2$ such that $|E_0-E_1-\omega_1|<<|E_0-E_1|$ and $|E_0-E_2-\omega_2|<<|E_0-E_2|$. So my Hamiltonian looks like:

$$
\begin{pmatrix}
E_0 & \Omega_1\cos\omega_1 t & \Omega_2\cos\omega_2 t\\
\Omega_1\cos\omega_1 t & E_1 & 0\\
\Omega_2\cos\omega_2 t t & 0 & E_2
\end{pmatrix}
$$
If I had only $E_0$, $E_1$, I would use this matrix:

$$
U = \begin{pmatrix}
1 & 0\\
0 & e^{i\omega_1 t}
\end{pmatrix}
$$
to go in a frame rotating with the electric field and remove all the explicit time dependence from the problem. But I am not sure what to do for a 3 level system (i.e. can I remove all the time dependence?). In principle I can use this for a 3 level system:

$$
U = \begin{pmatrix}
1 & 0 & 0\\
0 & e^{i\omega_1 t} & 0\\
0 & 0 & e^{i\omega_2 t}
\end{pmatrix}
$$
which would give a time-independent Hamitlonian (and that is a valid unitary transformation), but I am not sure this is ok physically. Basically, if I use this, the 2 different excited states seem to be in different rotating frames, which is confusing to me. What is the right way to proceed? Can I actually go to a time independent frame that makes physical sense?

 

[thomsj4]

Let us choose an operator

$$
U(t) = \begin{pmatrix}
1 & 0 & 0 \\
0 & e^{i\omega t} & 0 \\
0 & 0 & e^{i\omega t}
\end{pmatrix}
$$

where

$$
\omega = \frac{\omega_1 + \omega_2}{2}.
$$

Transform the state vector as

$$
|\tilde{\psi}(t)\rangle = U^\dagger(t)|\psi(t)\rangle
$$

and the Hamiltonian as

$$
\tilde{H}(t) = U^\dagger(t)H(t)U(t) - iU^\dagger(t)\frac{dU(t)}{dt}.
$$

The original Hamiltonian is

$$
H(t) = \begin{pmatrix}
E_0 & \Omega_1\cos(\omega_1 t) & \Omega_2\cos(\omega_2 t) \\
\Omega_1\cos(\omega_1 t) & E_1 & 0 \\
\Omega_2\cos(\omega_2 t) & 0 & E_2
\end{pmatrix}.
$$

First, compute the term

$$
iU^\dagger(t)\frac{dU(t)}{dt}.
$$

Since

$$
\frac{dU(t)}{dt} = \begin{pmatrix}
0 & 0 & 0 \\
0 & i\omega e^{i\omega t} & 0 \\
0 & 0 & i\omega e^{i\omega t}
\end{pmatrix},
$$

we obtain

$$
iU^\dagger(t)\frac{dU(t)}{dt} = \begin{pmatrix}
0 & 0 & 0 \\
0 & -\omega & 0 \\
0 & 0 & -\omega
\end{pmatrix}.
$$

Next,

$$
U^\dagger(t)H(t)U(t)
$$

amounts to multiplying the off-diagonal driving terms by exponentials of $\pm i\omega t$. Thus,

$$
U^\dagger(t)H(t)U(t) = \begin{pmatrix}
E_0 & \Omega_1\cos(\omega_1 t)e^{-i\omega t} & \Omega_2\cos(\omega_2 t)e^{-i\omega t} \\
\Omega_1\cos(\omega_1 t)e^{i\omega t} & E_1 & 0 \\
\Omega_2\cos(\omega_2 t)e^{i\omega t} & 0 & E_2
\end{pmatrix}.
$$

Therefore, the transformed Hamiltonian becomes

$$
\tilde{H}(t) = \begin{pmatrix}
E_0 & \Omega_1\cos(\omega_1 t)e^{-i\omega t} & \Omega_2\cos(\omega_2 t)e^{-i\omega t} \\
\Omega_1\cos(\omega_1 t)e^{i\omega t} & E_1 - \omega & 0 \\
\Omega_2\cos(\omega_2 t)e^{i\omega t} & 0 & E_2 - \omega
\end{pmatrix}.
$$

This single rotating frame captures part of the oscillatory phase of both fields without forcing each excited state into a separate rotating frame, leaving behind only the differences

$$
\omega_1 - \omega
$$

and

$$
\omega_2 - \omega
$$

in the exponentials when we use

$$
\cos(\omega_1 t)e^{-i\omega t} = \frac{1}{2}\left(e^{i(\omega_1-\omega)t} + e^{-i(\omega_1+\omega)t}\right)
$$

(and similarly for $\omega_2$). We may then invoke a two-frequency rotating-wave-like approximation if $\omega_1 \approx E_1 - E_0$, $\omega_2 \approx E_2 - E_0$, and $\omega$ is chosen to capture the main portion of both transitions, leaving smaller detunings

$$
\Delta_1 = \omega_1 - \omega
$$

and

$$
\Delta_2 = \omega_2 - \omega.
$$

If $\Delta_1$ and $\Delta_2$ are not too large, the rapidly oscillating counter-rotating terms can be truncated, yielding an effectively time-independent or slowly varying block of the Hamiltonian. The procedure is mathematically rigorous because it uses a globally well-defined unitary transformation of the Hilbert space and allows us to keep a single frame for all excited states while still systematically treating any leftover time dependence via perturbation expansions or by focusing on the resonant (slow) components of the dynamics. Physically, it can be viewed as “splitting the difference” between the two fields, thereby simplifying the system without introducing separate rotating frames for each excited state. This resolves the confusion of placing each level in a different frame, and in the regime of near resonance, we can reduce the problem to a slowly varying effective Hamiltonian that accurately reproduces Rabi oscillations for both transitions.

 

[renormalize]

...leaving smaller detunings

$$
\Delta_1 = \omega_1 - \omega
$$

and

$$
\Delta_2 = \omega_2 - \omega.
$$

If ...

Asking @thomsj4 for a personal favor: could you dramatically reduce your use of carriage returns so I don't have to wade through vast empty spaces (and risk snow blindness )?

Try this for example:
"...leaving smaller detunings$$\Delta_1 = \omega_1 - \omega$$and$$\Delta_2 = \omega_2 - \omega.$$If ..."
which renders as:
...leaving smaller detunings$$\Delta_1 = \omega_1 - \omega$$and$$\Delta_2 = \omega_2 - \omega.$$If ...

Or better yet: put small, unnumbered equations like those above inline:
"...leaving smaller detunings $\Delta_1 = \omega_1 - \omega$ and $\Delta_2 = \omega_2 - \omega.$ If ..."
giving:
...leaving smaller detunings $\Delta_1 = \omega_1 - \omega$ and $\Delta_2 = \omega_2 - \omega.$ If ...

Much more readable.
Thanks!

 

[Malamala]

Let us choose an operator

$$
U(t) = \begin{pmatrix}
1 & 0 & 0 \\
0 & e^{i\omega t} & 0 \\
0 & 0 & e^{i\omega t}
\end{pmatrix}
$$

where

$$
\omega = \frac{\omega_1 + \omega_2}{2}.
$$

Transform the state vector as

$$
|\tilde{\psi}(t)\rangle = U^\dagger(t)|\psi(t)\rangle
$$

and the Hamiltonian as

$$
\tilde{H}(t) = U^\dagger(t)H(t)U(t) - iU^\dagger(t)\frac{dU(t)}{dt}.
$$

The original Hamiltonian is

$$
H(t) = \begin{pmatrix}
E_0 & \Omega_1\cos(\omega_1 t) & \Omega_2\cos(\omega_2 t) \\
\Omega_1\cos(\omega_1 t) & E_1 & 0 \\
\Omega_2\cos(\omega_2 t) & 0 & E_2
\end{pmatrix}.
$$

First, compute the term

$$
iU^\dagger(t)\frac{dU(t)}{dt}.
$$

Since

$$
\frac{dU(t)}{dt} = \begin{pmatrix}
0 & 0 & 0 \\
0 & i\omega e^{i\omega t} & 0 \\
0 & 0 & i\omega e^{i\omega t}
\end{pmatrix},
$$

we obtain

$$
iU^\dagger(t)\frac{dU(t)}{dt} = \begin{pmatrix}
0 & 0 & 0 \\
0 & -\omega & 0 \\
0 & 0 & -\omega
\end{pmatrix}.
$$

Next,

$$
U^\dagger(t)H(t)U(t)
$$

amounts to multiplying the off-diagonal driving terms by exponentials of $\pm i\omega t$. Thus,

$$
U^\dagger(t)H(t)U(t) = \begin{pmatrix}
E_0 & \Omega_1\cos(\omega_1 t)e^{-i\omega t} & \Omega_2\cos(\omega_2 t)e^{-i\omega t} \\
\Omega_1\cos(\omega_1 t)e^{i\omega t} & E_1 & 0 \\
\Omega_2\cos(\omega_2 t)e^{i\omega t} & 0 & E_2
\end{pmatrix}.
$$

Therefore, the transformed Hamiltonian becomes

$$
\tilde{H}(t) = \begin{pmatrix}
E_0 & \Omega_1\cos(\omega_1 t)e^{-i\omega t} & \Omega_2\cos(\omega_2 t)e^{-i\omega t} \\
\Omega_1\cos(\omega_1 t)e^{i\omega t} & E_1 - \omega & 0 \\
\Omega_2\cos(\omega_2 t)e^{i\omega t} & 0 & E_2 - \omega
\end{pmatrix}.
$$

This single rotating frame captures part of the oscillatory phase of both fields without forcing each excited state into a separate rotating frame, leaving behind only the differences

$$
\omega_1 - \omega
$$

and

$$
\omega_2 - \omega
$$

in the exponentials when we use

$$
\cos(\omega_1 t)e^{-i\omega t} = \frac{1}{2}\left(e^{i(\omega_1-\omega)t} + e^{-i(\omega_1+\omega)t}\right)
$$

(and similarly for $\omega_2$). We may then invoke a two-frequency rotating-wave-like approximation if $\omega_1 \approx E_1 - E_0$, $\omega_2 \approx E_2 - E_0$, and $\omega$ is chosen to capture the main portion of both transitions, leaving smaller detunings

$$
\Delta_1 = \omega_1 - \omega
$$

and

$$
\Delta_2 = \omega_2 - \omega.
$$

If $\Delta_1$ and $\Delta_2$ are not too large, the rapidly oscillating counter-rotating terms can be truncated, yielding an effectively time-independent or slowly varying block of the Hamiltonian. The procedure is mathematically rigorous because it uses a globally well-defined unitary transformation of the Hilbert space and allows us to keep a single frame for all excited states while still systematically treating any leftover time dependence via perturbation expansions or by focusing on the resonant (slow) components of the dynamics. Physically, it can be viewed as “splitting the difference” between the two fields, thereby simplifying the system without introducing separate rotating frames for each excited state. This resolves the confusion of placing each level in a different frame, and in the regime of near resonance, we can reduce the problem to a slowly varying effective Hamiltonian that accurately reproduces Rabi oscillations for both transitions.

Thank you for this, it all makes sense. However, I am not sure how to get rid of $\Delta_1$ and $\Delta_2$. For example, I have $\Delta_1 = \frac{\omega_1-\omega_2}{2} = \frac{E_1-E_2}{2}$ assuming the 2 lasers are on resonance. But there is no relationship between $E_1$ and $E_2$ i.e. $E_1-E_2$ can be arbitrarily large or small and I am not sure what to do with this term in general.