What is the intuition behind the Choi-Jamiolkowski isomorphism? It says that with every superoperator $\mathbb{E}$ we can associate a state given by a density matrix

$$ J(\mathbb{E}) = (\mathbb{E} \otimes 1) (\sigma)$$

where $\sigma = \sum_{ij} | ii \rangle \langle jj |$ is the density matrix of some maximally entangled state $\sum_{i} | ii \rangle$.

And then the action of the superoperator is equal to

$$\mathbb{E}(\rho) = \operatorname{tr}_2(J(\mathbb{E}) \cdot 1 \otimes \rho^T).$$

What is the point of this? How does one use this in practice? Is it to simulate the action of the channel $\mathbb{E}$ by first preparing a specific state? I really don't understand the intuition behind this concept.


2 Answers 2


The intuition

Let us consider a channel $\mathcal E$, which we want to apply to a state $\rho$. (This could equally well be part of a larger system.) Now consider the following protocol for applying $\mathcal E$ to $\rho$:

  1. Denote the system of $\rho$ by $A$. Add a maximally entangled state $|\omega\rangle=\tfrac{1}{\sqrt{D}}\sum_{i=1}^D|i,i\rangle$ of the same dimension between systems $B$ and $C$:

  2. Now project systems $A$ and $B$ on $|\omega\rangle$:

    [This can be understood as a teleportation where we have only consider the "good" outcome, i.e., where we don't have to make a (generalized) Pauli correction on $C$, see also the discussion.]
    Our intuition on teleportation (or a simple calculation) tells us that we now have the state $\rho$ in system $C$:

  3. Now we can apply the channel $\mathcal E$ to $C$, yielding the desired state $\mathcal E(\rho)$ in system $C'$:

    However, steps 2 and 3 commute (2 acts on $A$ and $B$, and 3 acts on $C$), so we can interchange the ordering and replace 2+3 by 4+5:

  4. Apply $\mathcal E$ to $C$, which is the right part of $|\omega\rangle$:

    This results in a state $\eta=(\mathbb I\otimes \mathcal E) (|\omega\rangle\langle\omega|)$, which is nothing but the Choi state of $\mathcal E$:

    (This is the original step 3.)

  5. We can now carry out the original step 3: Project $A$ and $B$ onto $|\omega\rangle$:

    Doing so, we obtain $\mathcal E(\rho)$ in $C'$:

Steps 4 and 5 are exactly the Choi-Jamiolkowski isomorphism:

  • Step 4 tells us how to obtain the Choi state $\eta$ for a channel $\mathcal E$
  • Step 5 tells us how we can construct the channel from the state

Going through the math readily yields the expression for obtaining $\mathcal E$ from $\mathcal \eta$ given in the question: $$ \begin{align*} \mathcal E(\rho) &= D\langle \omega|_{AB}\rho_A\otimes \eta_{BC}|\omega\rangle_{AB}\\ &= \sum_{i,j} \langle i|\rho_A|j\rangle_{A} \langle i|_B\eta_{BC} |j\rangle_B \\ & = \mathrm{tr}_B[(\rho_B^T\otimes \mathbb I_C) \eta_{BC}]\ . \end{align*} $$


The intuition above is closely linked to teleportation-based quantum computing and measurement based quantum computing. In teleportation-based computing, we first prepare the Choi state $\eta$ of a gate $\mathcal E$ beforehand, and subsequently "teleport through $\eta$", as in step 5. The difference is that we cannot postselect on the measurement outcome, so that we have to allow for all outcomes. This is, depending on the outcome $k$, we have implemented (for qubits) the channel $\mathcal E(\sigma_k \cdot \sigma_k)$, where $\sigma_k$ is a Pauli matrix, and generally $\mathcal E$ is a unitary. If we choose our gates carefully, they have "nice" commutation relations with Pauli matrices, and we can account for that in the course of the computation, just as in measurement based computing. In fact, measurement based computing can be understood as a way of doing teleportation based computation in a way where in each step, only two outcomes in the teleportation are allowed, and thus only one Pauli correction can occur.


In short, the Choi-Jamiolkowski isomorphism allows to map many statements about states to statements about channels and vice versa. E.g., a channel is completely positive exactly if the Choi state is positive, a channel is entanglement breaking exactly if the Choi state is separable, and so further. Clearly, the isomorphism is very straightforward, and thus, one could equally well transfer any proof from channels to states and vice versa; however, often it is much more intuitive to work with one or the other, and to transfer the results later on.

  • $\begingroup$ When you evaluate $\mathcal{E}(\rho) = \langle \omega|_{AB} \rho_{A} \otimes \eta_{BC} | \omega \rangle_{AB}$, there is no guarantee that this state is normalized, right? In particular if I decompose the $AB$ system into the bell basis, in general there will be components along the other bell states. $\endgroup$
    – r_phys
    Dec 2, 2022 at 2:36
  • $\begingroup$ I see, thanks! At first glance I didn't quite get why you were even guaranteed to have any projection along $| \omega \rangle$, but I guess that the way the Choi state is built, you will always have this "$1/2$ projection". Thanks again $\endgroup$
    – r_phys
    Dec 6, 2022 at 10:32

This is how I have understood it and perhaps you will find it helpful:

Suppose you have a map (channel) $\Phi$ which acts on a system $A$. If $A$ exists in the state $\rho$ we can write,

$$\Phi(\rho) = \Phi(\rho_{ij}|i\rangle\langle j|) = \rho_{ij}\Phi(|i\rangle\langle j|)$$

Where the last step above follows because quantum mechanics is a linear theory. This means that knowing the matrices $\Phi(|i\rangle\langle j|)$ for each $i$ and $j$ helps us define the action of the map on any general density matrix and thus helps us define the map itself.

Note: $\Phi(|i\rangle\langle j|)$ above is a physically meaningless quantity because $|i\rangle\langle j|$ is not, in general, a valid density matrix. For now, let it just stand for one of the matrices that represent the map $\Phi$ and what it can mean physically we will see later.

Now suppose you have two systems of the same dimension as $A$. You have $A \otimes B$ which has been prepared in the Choi state given by $|\Psi\rangle = \Sigma_i |i_Ai_B\rangle $. Let us consider the action of the map $\Phi \otimes I$ (which is a valid transformation map) on this bipartite system.

$$\Phi \otimes I(\Sigma_{ij}|i_Ai_B\rangle\langle j_Aj_B|) = \Sigma_{ij}\Phi(|i_A\rangle\langle j_A|)|i_B\rangle\langle j_B| $$

And suppose you are able to physically perform the measurement $\langle i_B|\sigma |j_B\rangle$ on the above state what you get is $\Phi(|i\rangle\langle j|)$ itself.

Thus everything about $\Phi$ is encoded in the state $\Phi \otimes I(|\text{Choi_state}\rangle)$ and vice versa.

  • $\begingroup$ It's not a valid density matrix so no quantum state will ever have that form. It's meaningless to ask how do we apply the map on such a matrix then. $\Phi(|0\rangle\langle 1 |)$ is just a physically meaningless matrix. $\endgroup$ Jul 28, 2018 at 10:31
  • 1
    $\begingroup$ I don't get what you're trying to say. Sure, the operators $\Phi(|i\rangle\!\langle j|)$ characterise the channel, but how does this connect with the Choi representation? $\endgroup$
    – glS
    Sep 3, 2020 at 13:27

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