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

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    $\begingroup$ 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. $\endgroup$ – JSdJ Mar 9 at 21:56
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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 $UV$.

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  • $\begingroup$ What does $|\Phi \rangle $ mean here? $\endgroup$ – forky40 Mar 9 at 16:20
  • $\begingroup$ sorry, I mean $\Omega$. Let me fix... $\endgroup$ – DaftWullie Mar 9 at 16:21

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