Writing the Jaynes-Cumming model in QuTip

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?

• You write $\sigma_{+} a$: sp = tensor(sigmap(), qeye(N)) * a and then two lines later you add the two exchange terms, multiplying sp again by $a$, even though you defined it to include $a$ already? – chrysaor4 Apr 2 at 19:17
• @chrysaor4: Typo. Corrected. – sbp Apr 2 at 19:27
• What part of the QuTiP equation is confusing you exactly? – chrysaor4 Apr 2 at 19:30
• @chrysaor4: Like I mentioned that interaction part. – sbp Apr 2 at 19:31

Usually we say $$|0\rangle$$ 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 $$|0\rangle$$ should be the excited state instead. Thus, the qubit raising operator (with basis ordering $$|0\rangle, |1\rangle$$) 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 ($$|0\rangle$$ is ground), and you are using $$|0\rangle$$ as excited.
• The ordering is not an issue as the atomic and field (cavity) part either way commutes. Yes, one needs to be careful when one is interested in the partial traces. But the major issue here is why do they write tensor(destroy(2), qeye(N)). $\sigma_{+}$ is not destroy(2). This part is unclear to me. – sbp Apr 2 at 19:39
• Ah, editing now. Your question is about $a$, not the interaction term. – chrysaor4 Apr 2 at 19:42
• No. It is the sm term in the interaction Hamiltonian. – sbp Apr 2 at 19:46