Writing the Jaynes-Cumming model in QuTip

Question

I want to write the Jaynes-Cumming Hamiltonian in QuTip.
$$H = hbar omega_{C} a^dagger a + frac12 hbar omega_{a} sigma_{z} + hbar lambda (sigma_{+} a + sigma_{-} a^dagger)$$
I assume $hbar = 1$, and we know $a, a^dagger$ are the field (cavity) operators, and $sigma_{z}, sigma_{+}, sigma_{-}$ are the atomic operators; and they commute. $lambda$ is the coupling constant.
I assume that the # of Fock basis states that I choose is N. Now, to write the Hamiltonian using qutip I would use the following statements,
$a$: a = tensor(qeye(2), destroy(N)) 
$a^dagger$: a.dag() 
$a^dagger a$: a.dag() * a
$sigma_{z}$: sz = tensor(sigmaz(), qeye(N))
$sigma_{+} a$: sp = tensor(sigmap(), qeye(N)) * a 
$sigma_{-} a^dagger$: sp.dag() 
$sigma_{+} a + sigma_{-} a^dagger$: sp + sp.dag()
But in the QuTip documentation the interaction term has been written as,

sm = tensor(destroy(2), qeye(N)); Hint = a.dag() * sm + a * sm.dag()

Could someone explain why they use the previous statement but not the one I would have assumed?

chrysaor4 · Accepted Answer

Usually we say $|0rangle$ is the ground state and is at the north pole, but sometimes people choose the convention that since the north pole is ``higher" than the south pole on the Bloch sphere, that $|0rangle$ should be the excited state instead.  Thus, the qubit raising operator (with basis ordering $|0rangle, |1rangle$) is
begin{align}
sigma_+ = begin{bmatrix} 0 & 1  0 & 0 end{bmatrix}
end{align}
which actually looks like the lowering operator in the first convention.  QuTiP is treating the qubit as a two-level oscillator ($|0rangle$ is ground), and you are using $|0rangle$ as excited.

Writing the Jaynes-Cumming model in QuTip

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