# Multi-qubit quantum channels in Pauli-Liouville basis: Tensor product of Pauli-Transfer Matrices?

I would like to verify something, need a sanity check. Are the quantum channels for different qubits in the Pauli-Liouville basis (Pauli Transfer Matrices) also given by a tensor product?

The Kraus matrices of a multi-qubit quantum channel acting non-trivially on a qubit, but trivially on the other qubits, are given as the tensor product of the matrices with identity matrices. For example, the bit-flip channel (with the strength $$p$$) on the qubit A of a two-qubit system (A & B) is given as,

$$\mathcal{E}(\rho_{AB}) = (1-p)\rho_{AB} + p X_A\otimes I_B \rho_{AB} X_A \otimes I_B.$$

Here the first Kraus matrix is trivial (identity), and the second Kraus matrix is the tensor product of the Kraus matrix on the subsystem A with identity on B, $$M_{AB}^{(1)} = M_A^{(1)} \otimes I_B = X_A\otimes I_B$$. When we use the $$n$$ qubit Pauli-Liouville basis $$B_{PL} = \left\{I/\sqrt{2}, X/\sqrt{2},Y/\sqrt{2},Z/\sqrt{2} \right\}^{\otimes n}$$, where each basis element $$P_i$$ is a member of this direct-product of bases, we can write a quantum channel as a $$2^{2n}\times 2^{2n}$$ matrix,

$$R_{ij}(\mathcal{E}) = \text{Tr}\left[P_i\mathcal{E}\left(P_j\right)\right].$$

Each Pauli transfer matrix element of row $$i$$ and column $$j$$ of is given through Hilbert-Schmidt inner product using the bases $$P_i, P_j \in B_{PL}$$. Going back to the bit-flip channel example given above

$$R_{ij}(\mathcal{E};p) = (1-p)\text{Tr}\left(P_i P_j\right) + p \text{Tr}\left(P_i X_A \otimes I_B P_j X_A \otimes I_B \right) \\ = (1-p)\delta_{ij}+ p \text{Tr}\left(\tilde{P}_i^A X_A \tilde{P}_j^A X_A\right)\text{Tr}\left(\tilde{P}_i^B \tilde{P}_j^B\right).$$

Above I decomposed the two-qubit bases in the $$AB$$ space as $$P_i^{AB}=\tilde{P}_i^A\tilde{P}_i^B$$. These Pauli-Liouville bases are normally written in standard quaternary (basis 4) notation such that the first basis $$P_0^{AB}=0.5I^A\otimes I^B$$, can be denoted as $$\tilde{P}_0^A = \tilde{P}_0^B I/\sqrt{2}$$. Similarly the second basis $$P_1^{AB}=0.5I^A\otimes X^B$$, can be denoted as $$\tilde{P}_0^A = I/\sqrt{2}$$, and $$\tilde{P}_0^B X/\sqrt{2}$$, and so on. In the equation above the first term clearly leads to $$2^{4}\times 2^{4}$$ identity matrix as $$P_i$$ and $$P_j$$ are coming from a complete, orthonormal set of basis functions.

And, I "think" (but can not show with a convincing, rigorous argument) the second term in the above equation, $$\text{Tr}(\dots)\text{Tr}(\dots)$$, leads to a tensor product of two matrices. In particular, it leads to the tensor product,

$$R = R_X^A \otimes R_I^B$$

Where $$R_X^A$$ denotes the PTM of single qubit unitary channel of $$X$$ gate on qubit A such that $$R_{X_{ij}}^A = \text{Tr}(\tilde{P}_i X \tilde{P}_j X)$$, and the second term is the PTM of the single qubit identity channel (a 4D identity matrix). Can someone verify this? Can we just generalize PTMs by taking tensor products as well?

Yes, the Pauli transfer matrix of the tensor product of two quantum channel is the tensor product of the Pauli transfer matrices of the individual channels: $$R(\mathcal{E}\otimes\mathcal{F})=R(\mathcal{E})\otimes R(\mathcal{F})$$.
Proof. The element in the $$i$$th row and $$j$$th column of the Pauli transfer matrix of $$\mathcal{E}\otimes\mathcal{F}$$ is \begin{align} R(\mathcal{E}\otimes\mathcal{F})_{ij}=\mathrm{tr}(P_i\circ(\mathcal{E}\otimes\mathcal{F})(P_j))\tag1 \end{align} where $$P_i$$ and $$P_j$$ are Pauli operators on the joint system. By indexing over each subsystem individually, we can write the row index as $$i=i_1i_2$$, so that $$P_i=P_{i_1}\otimes P_{i_2}$$ and similarly for the column index. Then $$(1)$$ becomes \begin{align} R(\mathcal{E}\otimes\mathcal{F})_{ij}&=\mathrm{tr}(P_{i_1}\otimes P_{i_2}\circ(\mathcal{E}\otimes\mathcal{F})(P_{j_1}\otimes P_{j_2}))\tag2\\ &=\mathrm{tr}(P_{i_1}\mathcal{E}(P_{j_1})\otimes P_{i_2}\mathcal{F}(P_{j_2}))\tag3\\ &=\mathrm{tr}(P_{i_1}\mathcal{E}(P_{j_1}))\cdot\mathrm{tr}(P_{i_2}\mathcal{F}(P_{j_2}))\tag4\\ &=R(\mathcal{E})_{i_1j_1}\cdot R(\mathcal{F})_{i_2j_2}\tag5\\ &=\left[R(\mathcal{E})\otimes R(\mathcal{F})\right]_{ij}\tag6 \end{align} as expected.$$\square$$