Explicit form for composition of Choi representation quantum channels

Let $$|\Omega \rangle$$ be the maximally entangled state over a bipartite system whose parts are each dimension $$d$$, i.e. $$| \Omega \rangle \equiv \sum_i^{d}| ii \rangle$$

Then one way of writing the Choi representation $$J(\Phi)$$ of a channel $$\Phi: \mathbb{C}^{d\times d} \rightarrow \mathbb{C}^{d\times d}$$ that acts on density operators $$\rho \in \mathbb{C}^{d \times d}$$ is: $$J(\Phi) \equiv (I \otimes \Phi )| \Omega \rangle \langle \Omega |$$

If I choose $$\Phi_U$$ to be a unitary channel (i.e. $$\Phi_U (\rho ) = U\rho U^\dagger$$) then its Choi representation in this notation would be: $$J(\Phi_U) \equiv (I \otimes U)| \Omega \rangle \langle \Omega |(I \otimes U^\dagger)$$

My question is, is there an explicit way to represent compositions of unitaries in terms of the unitaries' respective Choi representations? In other words, if $$\Phi_{U\circ V}(\rho) \equiv UV\rho (UV)^\dagger$$, how can I "nicely" represent $$J(\Phi_{U \circ V})$$, for example in terms of $$J(\Phi_U)$$ and $$J(\Phi_V)$$ and linear operations in $$\mathbb{C}^{d^2 \times d^2}$$?

• Please note that you can define the Choi matrix for the composite map much easier from U and V directly (in a similar manner as how you define the Choi matrix for U only). If you want to explicitly compute it from the Choi matrices themselves, I feel that it is the easiest to diagonalize them (they're both rank 1), essentially obtaining a Kraus representation (which are, of course, U and V themselves in your case). A different approach can be to compute their S-matrices (see arxiv.org/pdf/quant-ph/0504091.pdf), as a composition of maps becomes a matrix product in this formulation.
– JSdJ
Mar 9 '20 at 21:56

I'm not an expert on this, so there could easily be other ways (I get the impression you'd like to keep it to a $$d^2$$ dimensional calculation, where I'm taking it up to $$d^4$$), but what I'd do is calculate: $$J(\Phi_{U\circ V})=\left(I_d\otimes \langle\Omega|\otimes I_d\right)\left(J(\Phi_V)\otimes J(\Phi_U)\right)\left(I_d\otimes |\Omega\rangle\otimes I_d\right).$$ I simply think of this as teleporting the output from the first channel ($$V$$) into the into of the second channel ($$U$$) so that you get the composite action of $$UV$$.
• What does $|\Phi \rangle$ mean here? Mar 9 '20 at 16:20
• sorry, I mean $\Omega$. Let me fix... Mar 9 '20 at 16:21
Let $$\mathcal E:A\to B$$ and $$\mathcal F:B\to C$$ be two linear maps, and $$J(\mathcal E)$$ and $$J(\mathcal F)$$ their Choi representations. The Choi representation of their composition $$\mathcal F \circ \mathcal E$$ is then given by $$J(\mathcal F \circ \mathcal E) = \operatorname{tr}_B\big[(I_A\otimes J(\mathcal F))(J(\mathcal E)^{T_B} \otimes I_C)\big],$$ where $$T_B$$ is the partial transposition over subsystem $$B$$.