Explicit form for composition of Choi representation quantum channels

Question

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 $\mathbb{C}^{d^2 \times d^2}$?

Answer 1

Yes, there is a nice way to represent this composition, valid for any linear map, not only unitaries. It is so useful that it has a name, link product.

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 $B$.

Answer 2

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 $UV$.