I'm trying to understand the process of deriving the master equation in this article. The system in general is composed of two interacting qubits each of which is coupled to their local thermal bath and both of them are coupled to a common bath. The figure below describes the system.

Scheme of the two qubits interacting with thermal baths

In section 2.3.1 the authors neglect the interaction between qubits so the system hamiltonian of the two qubits become:

$H_S = \frac{\omega_1}{2}\sigma_1^z + \frac{\omega_2}{2}\sigma_2^z$

which is already diagonal in the ‘canonical basis’ $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ with eigenvalues respectively $E_0 = -\omega_+/2, E_1 = -\omega_-/2, E_2 = \omega_-/2, E_3 = \omega_+/2$ where $\omega_{\pm}=\omega_1 \pm \omega_2$. In order to derive the jump operators we must decompose the system-bath interaction Hamiltonian such that:

$H_I = \sum_{\beta} A_{\beta}(t) \otimes B_{\beta}(t)$

in which $A_{\beta}(t)$ are the system operators. From the interaction Hamiltonian in the article (equation 4 of the article), we see that the system components of the interaction Hamiltonian are $\sigma_i^x$ and $\sigma_i^z$. The jump operators can be found using (equation 17):

$A_{\beta}(\omega) = \sum_{\epsilon'-\epsilon=\omega} |\epsilon\rangle \langle \epsilon| A_{\beta} |\epsilon'\rangle \langle \epsilon'|$

where $|\epsilon\rangle$ are the basis of the eigenvectors of the system Hamiltonian $H_S$ which I have given above. Using this relation, the authors go on to derive the jump operators, stating that:

In the interaction Hamiltonian we can find the system operators $\sigma_j^x$ and $\sigma_j^z$ coupled to the bath operators, with $j=1, 2$. Their decomposition in terms of jump operators is readily written according to equation (17):

And then they write the jump operators as:

$\sigma_j^x(\omega_j) = \sigma_j^-, \sigma_j^x(-\omega_j) = \sigma_j^+ \rightarrow \sigma_j^x = \sigma_j^- + \sigma_j^+, \sigma_j^z(0) = \sigma_j^z \rightarrow \sigma_j^z = \sigma_j^z$

I didn't understand how they derived this last part. I know that they have used the equation 17 of the article and this is a simple question but for some reason I can't figure out how they did this. For example when I try to find $\sigma_1^x(\omega_1)$ using equation 17, I get:

$\sigma_1^x(\omega_1) = |00\rangle \langle 00|\sigma_1^x |10\rangle \langle 10| + |01\rangle \langle 01|\sigma_1^x |11\rangle \langle 11| = |00\rangle \langle 10|+|01\rangle \langle 11|$

However the text states that $\sigma_1^x(\omega_1) = \sigma_1^-$. I don't know where I have made a mistake. The questions are:

1- What is the step-by-step derivation for the jump operators? Also how do we decide how many transitions are possible and which operators correspond to which transitions?

2- Is it possible to figure out what transitions will we have and to which operators will they correspond before doing these calculations and have a pictorial idea of the problem like figure (b)? I'm asking this because the figure is given before the derivation of the jump operators. So I suspect there is a way to visualize the transitions, the operators corresponding to those transitions and their energies just by looking at the interaction Hamiltonian.

  • 1
    $\begingroup$ Note that the authors are using the standard convention $\sigma_1^-:=\sigma_1^-\otimes\mathbb{I}_2$. This means that $\sigma_1^-=|00\rangle\langle10|+|01\rangle\langle 11|$, which is exactly what you have found. $\endgroup$ Commented Jul 9, 2021 at 10:44

1 Answer 1


I am reading the same paper and here is how to derive their jump operators:

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