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I've been reading the following lecture notes in which two isomorphisms are defined:

  • vectorization: $\text{vec}:B(H_A,H_B)\to H_A\otimes H_B$ and
  • the Choi-Jamiołkowski isomorphism: $C:B(B(H_A), B(H_B )) \to B(H_A \otimes H_B )$ where $H_A, H_B$ are Hilbert spaces and $B(H)$ is the space of bounded linear operators on the Hilbert space $H.$

The graphical representation for $\text{vec}(A)$ is drawn on page 4, and reproduced here for convenience:

this

I think I understand it, but I'm struggling to draw a similarly intuitive picture of $C(L)$ for $L\in B(B(H_A), B(H_B))$

I'm aware that they are effectively equivalent, but I'd like to draw a diagram that would let me see immediately e.g. the identity listed as Lemma 1.3: $C(L) = (id_A\otimes L) (\omega_{d_A})$ (for $\omega_{d_A}$ the density matrix describing the $d_A$ dimensional maximally entangled state, or generalized Bell state) as well as similar such calculations.

Put simply, if $\text{vec}(A)$ is given by the picture above, what is $C(L)?$

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    $\begingroup$ (OP has cross-posted this on phys.SE: link) $\endgroup$
    – Frederik vom Ende
    Commented Aug 13 at 5:25
  • $\begingroup$ > for $\omega_{d_A}$ the $d_A$ dimensional maximally mixed density matrix Do you mean maximally entangled state? If so, it's the same picture, you just add a second arc $\supset \subset$ to get the density matrix for the maximally entangled state. $\endgroup$
    – Refik Mansuroglu
    Commented Aug 13 at 6:21
  • $\begingroup$ related: quantumcomputing.stackexchange.com/q/11580/55, quantumcomputing.stackexchange.com/a/21168/55 (see picture in last part of the answer) $\endgroup$
    – glS ♦
    Commented Aug 13 at 10:36
  • $\begingroup$ @RefikMansuroglu Yes, that's also in the handout I linked. What I'm nusure about is how to draw the rest of the picture. How do I represent $C$ and $C(L)?$ $\endgroup$
    – Luke
    Commented Aug 13 at 10:54

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You can use the Kraus operators $K_i$ of $L$ and write

Choi matrix

EDIT: Of course, this requires the existence of Kraus operators, which is equivalent to $L$ being completely positive.

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