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I have some questions about the Choi-Jamiolkowski isomorphism.

I remind how it can be defined. First, we define $|\mathcal{I}_{H_0} \rangle \rangle \in H_0 \otimes H_0$

$$|\mathcal{I}_{H_0} \rangle \rangle = \sum_i |i i \rangle$$

Where the family $\{ |i\rangle \}_i $ is an orthonormal basis of $H_0$.

C.J isomorphism: pure case

Let $X \in \mathcal{L}(H_0, H_1)$, the C.J isomorphism consist in associating to this linear operator, a kets $|X \rangle \rangle \in H_1 \otimes H_0 $:

$$ |X \rangle \rangle \equiv ( X \otimes I_{H_0} ) |\mathcal{I}_{H_0} \rangle \rangle$$

A little calculation shows that the isomorphism simply consists in performing the following mapping:

$$ \sum_{i,j} \langle i | X | j \rangle |i \rangle \langle j | \leftrightarrow \sum_{i,j} \langle i | X | j \rangle |i, j \rangle \rangle$$

Basically we replace: $|i\rangle \langle j | \in \mathcal{L}(H_0,H_1) \leftrightarrow |i,j \rangle \rangle \in H_1 \otimes H_0$

C.J isomorphism: mixed case

The mixed case consist in associating linear operator to quantum map. Let $\mathcal{M} \in \mathcal{L}(\mathcal{L}(H_0),\mathcal{L}(H_1))$. We define the Choi matrix $M \in \mathcal{L}(H_1 \otimes H_0)$ such that:

$$\mathcal{C}(\mathcal{M}) \equiv (\mathcal{M} \otimes \mathcal{I}_{H_0}) (|\mathcal{I}_{H_0} \rangle \rangle \langle \langle \mathcal{I}_{H_0} | )$$

We notice the strong analogy with the definition of the Choi isomorphism for the pure case.

My questions

  • Is there an analog "easy" definition for the Choi isomorphism in the mixed case as there were in the pure case with $|i \rangle \langle j| \leftrightarrow |i,j \rangle \rangle$ ? Or should we always take the "brute force" definition to calculate the isomorphism ?
  • Why is this isomorphism important ? I have seen this post which talks about the intuition from quantum teleportation (I did not really understand the answer, I would need to dig in). But my question is more related on why this isomorphism is often used and not another one to perform calculation.
  • Are there typical relationship between the pure and mixed case ? I managed to prove that if $\mathcal{U}(\rho)=U \rho U^{\dagger}$ is a unitary map, then $\mathcal{C}(\mathcal{U})=|U \rangle \rangle \langle \langle U |$. But are there some other usefull relationship ?
  • (related to my second question): I checked and it seems that the mixed isomorphism is different from the "natural" isomorphism we would consider to associate matrix to quantum map. Basically where the inputs (density matrices) would be replaced by vector: $|i \rangle \langle j | \rightarrow |i,j\rangle$ and the quantum map would then just be a matrix eating vector and giving vector. The resulting matrix seem to not be the same as the mixed case choi matrix. Would you confirm ?
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    $\begingroup$ About your last point: indeed the "natural" isomorphism gives a different matrix than the Choi-Jamiołkowski isomorphism. The key property of the Choi-Jamiołkowski isomorphism is that if $\mathcal M$ is a completely positive map, then $\mathcal C(\mathcal M)$ is a positive semidefinite matrix. This is very convenient, and doesn't hold for the "natural" isomorphism, and it is the reason everybody uses the Choi-Jamiołkowski isomorphism instead. $\endgroup$ – Mateus Araújo Aug 18 '20 at 9:48
  • $\begingroup$ related: quantumcomputing.stackexchange.com/a/11624/55 $\endgroup$ – glS Aug 19 '20 at 15:14
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    $\begingroup$ each post should ask about one single issue/question. Not doing so makes the post less useful and the answers harder to retrieve afterwards $\endgroup$ – glS Aug 19 '20 at 15:16
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Is there an analog "easy" definition for the Choi isomorphism in the mixed case as there were in the pure case with |i⟩⟨j|↔|i,j⟩⟩ ? Or should we always take the "brute force" definition to calculate the isomorphism ?

Basically, you can use a trick to get some intuition for the Choi matrix that works on both pure and mixed states. We know that any quantum channel $\Lambda$ should be a linear map - therefore, if you define the action of $\Lambda$ on a basis of $\mathcal{H}_{0}$, everything that $\Lambda$ can do to any state $\rho$ is encoded into that information.

Lets pick the most straightforward basis (actually, it's not really a basis for $\mathcal{H}_{0}$ that we use but for the space of $d\times d$ matrices, where $d=2^{n}$): $$\{|i\rangle\langle j|\}\,\,\,\,\forall i,j \in \{1,...,d\},$$ this gives us $d^{2}$ elements. All information about $\Lambda$ is encoded into the following $d^{2}$ elements:

$$\{\Lambda\left(|i\rangle\langle j|\right)\}\,\,\,\,\forall i,j \in \{1,...,d\}.$$

The Choi matrix the above $d^{2}$ elements, but then not as a list, but we put the $(i,j)$-th element of that list (which is a $d\times d$ matrix) into a new matrix as the $(i,j)$-th element: we get a $d \times d$ block matrix where each entry is itself a $d \times d$ matrix: $$\rho_{\mathrm{Choi}} = \begin{bmatrix} \Lambda\left(|0\rangle\langle 0|\right) & \Lambda\left(|0\rangle\langle 1|\right) & \dots & \Lambda\left(|0\rangle\langle d-1|\right) \\ \Lambda\left(|1\rangle\langle 0|\right) & \Lambda\left(|1\rangle\langle 1|\right) & \dots & \Lambda\left(|1\rangle\langle d-1|\right) \\ \vdots & \vdots & \ddots & \vdots \\ \Lambda\left(|d-1\rangle\langle 0|\right) & \Lambda\left(|d-1\rangle\langle 1|\right) & \dots & \Lambda\left(|d-1\rangle\langle d-1|\right) \\ \end{bmatrix}. $$

From this point of view, it is pretty clear that the Choi matrix is a clever way of writing down the information encoded into $\Lambda$.

The important thing to notice is that the above matrix is always a valid density matrix, which is obvious from your definition.

Why is this isomorphism important ? I have seen this post which talks about intuition from quantum teleportation (I did not really understand the answer, I would need to dig in). But my question is more related on why this isomorphism is often used and not another one to perform calculation.

The Choi-Jamiolkowski isomorphism states that having $d^{2}$ elements in the second equation is always enough (and that the resulting Choi matrix is always a state), which makes it important just for that. Moreover, it is often easier to work with density matrices rather than maps, and some properties that one needs or looks for in maps nicely translate into the 'normal' properties of density matrices for the Choi matrix.

  • The process fidelity of a map $\lambda$ is the same as the state fidelity of it's Choi matrix.
  • The CP constraint of $\Lambda$ translates nicely into the positive-semidefiniteness of the Choi matrix.

However, it is important that not all properties translate nicely. The TP constraint on $\Lambda$ is a stronger constraint than $\rho$ just having unit trace. This also means that not all $d^{2} \times d^{2}$ density matrices represent a $d\times d$ map!

People have been using (at least theoretically) the isomorphism to perform quantum process tomography by actually performing the (easier) quantum state tomography. However, because of the above reason of the TP constraint, not all methods of state tomography give valid 'isomorphism-correct' quantum states - something which I've seen gone wrong in a paper or two!

Are there typical relationship between the pure and mixed case ? I managed to prove that if U(ρ)=UρU† is a unitary map, then C(U)=|U⟩⟩⟨⟨U|. But are there some other usefull relationship?

I believe that the first answer answers this question as well. If you have some other representation of $\Lambda$ (e.g. a Kraus representation, it's $\chi$ matrix or a Stinespring dilation) there exist nice ways of translating one representation into another, but I guess that this is best asked as a separate question (and for the $\chi \leftrightarrow \rho_{\mathrm{Choi}}$ translation I already have an answer on a previous question; the Kraus decomposition should be really straightforward as well).

(related to my second question): I checked and it seems that the mixed isomorphism is different from the "natural" isomorphism we would consider to associate matrix to quantum map. Basically where the inputs (density matrices) would be replaced by vector: |i⟩⟨j|→|i,j⟩ and the quantum map would then just be a matrix eating vector and giving vector. The resulting matrix seem to not be the same as the mixed case choi matrix. Would you confirm ?

This is indeed not the same as the Choi matrix. Rather, this would be a representation sometimes just called the superoperator representation. It isn't used that often though, because both the CP and the TP constraints (especially the TP) are really finicky. I'd advice you to stay away from this representation unless you really know what you're doing. These matrices are often invertible, but that is a big no-no physically speaking.

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  • $\begingroup$ Thank you for your answer. About your first point, I am not sure to see the graphical trick you talk about. What you explains is just that we can write down the matrix in the computational basis. But is there something more to understand ? $\endgroup$ – StarBucK Aug 17 '20 at 21:23
  • $\begingroup$ (1/2)I don't fully agree that it's 'just' writing the matrix in the computational basis, allthough it's really close. The point I was trying to make is that the topleft $d \times d$ subblock of the Choi matrix is $\Lambda\left(|0\rangle\langle 0|\right)$, the one next to it is $\Lambda\left(|0\rangle\langle 1|\right)$, the $i,j$-th subblock is $\Lambda\left(|i\rangle\langle j|\right)$, etc. Maybe the word graphical is not a very good choice (I edited it). $\endgroup$ – JSdJ Aug 18 '20 at 9:18
  • $\begingroup$ (2/2) To some extend, there is not that much more to understand - the Choi matrix is a nifty way of representing the action of the map $\Lambda$ on the computational basis. The CJ isomorphism tells us that this representation always is a valid state. $\endgroup$ – JSdJ Aug 18 '20 at 9:19
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    $\begingroup$ That's definitely correct - the statement holds only for quantum channels, which are by definition CPTP. In the context of open quantum systems we just don't really think about non-CP maps all that much, and the constraints on the Choi matrix are easily adopted to allow for trace-decreasing maps. $\endgroup$ – JSdJ Aug 18 '20 at 9:27
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    $\begingroup$ Great thank you very much. $\endgroup$ – StarBucK Aug 18 '20 at 9:27

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