# Show that the Choi of a tensor product is the tensor product of the Chois

I have the following problem. Let $$N:L(H_A)\rightarrow L(H_A)$$ be a quantum superoperator. The quantum state corresponding to this operator (via Choi-Jamiolkowski Isomorphism) is $$\Gamma_A^{N}=id\otimes N(\rho)$$, where $$\rho$$ is density matrix of the maximally entangled state and id is the identity operator. This state is called Choi state. Consider two quantum superoperators $$N:L(H_A)\rightarrow L(H_A)$$ and $$M:L(H_B)\rightarrow L(H_B).$$ Show that Choi state of tensor product superoperator $$N\otimes M$$ is: $$\Gamma^{N\otimes M}_{AB}=\Gamma^{N}_A\otimes \Gamma^{M}_B$$ This problem is taken from Principles of Quantum Communication Theory by Sumeet Khatri and Mark M. Wilde. It is problem 4.6 from this textbook.

Edit. I tried to do this the following way. $$\Gamma^{N\otimes M}_{AB}=\mathbb{I_{AB}}\otimes(N\otimes M)\sum_{iji'j'}|ij\rangle|ij\rangle\langle i'j'|\langle i'j'|=\\ \sum_{iji'j'}|ij\rangle\langle i'j'|\otimes N(|i\rangle\langle i'|)\otimes M(|j\rangle\langle j'|)$$ On the other hand: $$\Gamma^{N}_A\otimes \Gamma^{M}_B=\sum_{ij}\mathbb{I}\otimes N(|ii\rangle \langle jj|)\otimes \sum_{i'j'}\mathbb{I}\otimes M(|i'i'\rangle \langle j'j'|)=\\= \sum_{iji'j'}|i\rangle\langle j|\otimes N(|i\rangle\langle j|)\otimes |i'\rangle\langle j'|\otimes M(|i'\rangle\langle j'|)$$ At this point I tried relabelling but I always encountered the problem that on left-hand side N and M was at 2 and 4 position in tensor product and on right-hand side they are at 3 and 4 position.

• What have you tried? Feb 29 at 19:34
• @Rammus I edited the post to show my reasoning. Mar 1 at 13:01
• They are equal up to ordering of the Hilbert spaces, so you will need to swap the order so they are the same. Mar 1 at 14:10