The drive Hamiltonian is describe with a cos and sin function of the phase but in the reference given QUANTUM COMPUTING WITH NEUTRAL ATOMS (arxiv.org) (page 13)
Those functions are not present.
Why the difference ?

Great question! This really boils down to a choice of representation that doesn’t have affects on the physics. The Rydberg atom Hamiltonian’s interaction part is proportional to PauliZ-PauliZ interactions, since we can write

n_i = \frac{1 - \sigma_z}{2}.

The drive term just needs to be transverse to the interaction (i.e., \sigma_x or \sigma_y). In our demo (Neutral-atom quantum computers) the choice for a transverse field is more general (a linear combination of \sigma_x and \sigma_y), whereas the paper you referenced is choosing a specific value of \phi.

Another way to think about it: both Hamiltonians are a unitary transformation away from each other, where the unitary is a Pauli Z rotation .

Thank you for your answer ! But I’m sorry, I don’t really understand the idea behind the transverse field. Do you think you could explain it or share some ressources so that I could learn about it ?

The transverse field is something that’s there to represent the energy cost of exciting / unexciting your degrees of freedom (small aside: the Rydberg Hamiltonian is extremely similar to the infamous transverse-field Ising model — Transverse-field Ising model - Wikipedia). It’s present in the Rydberg Hamiltonian to quantify the energy difference between a Rydberg atom in its ground state and in an excited state (where \Omega — the thing that is coupled to the transverse field — is called the Rabi frequency).

I wouldn’t get too lost in terminology here — the transverse field is something that quantifies the energy required to excite things .

Ah yeah that’s the formula I’m still a bit confused about :

Why is there a “-” in front of the detuning ?

If the detuning drives the transition between the ground state and the rydberg state, why do we need the \Omega term ?

What does \frac{I_{q}-\sigma^{z}_{q}}{2} represent ? I know it’s either 0 or 1 and that \sigma^{z}_{q} is the Pauli Z operator of the q-th qubit but why do we need that formula ? Why not just \sum_{q\in wires}^{}\sigma^{z}_{q} ? Also why Pauli Z and not X or Y ?

The answer to your third question is that it represents the occupation number (i.e., is the Rydberg atom excited — value of 1 — or not — value of 0). That’s the equation that’s in my first response to you:

n_i = \frac{1 - \sigma^z_i}{2}.

That’s the Rydberg occupation operator .

For your first question, the detuning \delta couples to this term in the Hamiltonian, meaning that for a positive value of \delta, the drive Hamtiltonian favours Rydberg atoms that are “occupied” (excited) — it lowers the energy. The opposite is true if \delta is negative.

If the detuning drives the transition between the ground state and the rydberg state, why do we need the Ω.

The detuning is accounting for the “leftover” / “excess” energy from incident photons. So, the roles of the detuning and the Rabi frequency \Omega are different.