Yes, you are. There’s nothing to worry about. I’m just writing this blog post because there are some people trying to convince you otherwise, and I don’t want you to get trolled by them.

First of all, what is the Choi-Jamiołkowski isomorphism? It is one way to write a linear map $\mathcal E$ as a matrix $\mathfrak{C}(\mathcal E)$ so that the matrix is positive semidefinite if and only the map is completely positive. This is very convenient if you want to find out whether $\mathcal E$ is completely positive, or if you’re doing semidefinite programming. If all you want is to write the linear map as a matrix, you’re better off using a representation with a more convenient formula for applying it to a quantum state.

The usual way to define it is

\[\mathfrak{C}(\mathcal E) := \sum_{ij} \ket{i}\bra{j} \otimes \mathcal E(\ket{i}\bra{j}), \]and I assume this is what you’re doing. Yes, it is fine really, there’s nothing wrong with it. What Leifer, Allen et al., and Hoffreumon et al. complain about it is that it’s basis dependent. Which is true: if instead of the basis $\{ \ket{i} \}$ in the above formula you use the basis $\{ U\ket{i}\}$ for some unitary $U$, the resulting matrix does depend on $U$. To see that just let $U$ apply to $\ket{i}$ a phase that depends on $i$.

But so what? It will be a different matrix, but still a perfectly valid representation for $\mathcal E$. And why are you changing the basis anyway? I don’t think anybody ever did the Choi-Jamiołkowski isomorphism in a basis other than the computational one. Nevertheless, since the choice of basis clearly doesn’t matter for the isomorphism, perhaps it’s more elegant to use an isomorphism that doesn’t depend on it.

Leifer proposes using

\[\mathfrak{J}(\mathcal E) := \sum_{ij} \ket{j}\bra{i} \otimes \mathcal E(\ket{i}\bra{j})\]instead. It’s easy to see that this is indeed basis-independent, but it has a fatal flaw: it doesn’t represent completely positive maps as positive semidefinite matrices. For example, the representation of the identity map will have a negative eigenvalue. Since this was the main point of using the Choi-Jamiołkowski isomorphism in the first place, no thanks, I’ll pass.

Allen et al. and Hoffreumon et al. promise to give us the best of both worlds, an isomorphism which is basis-independent and represents completely positive maps as positive semidefinite matrices, by defining instead

\[\mathfrak{B}(\mathcal E) := \sum_{ij} \ket{i}_{A^*}\bra{j} \otimes \mathcal E(\ket{i}_A\bra{j}), \]where $\{\ket{i}_A\}$ is a basis for the primal Hilbert space $A$, and $\{\ket{i}_{A^*}\}$ the dual basis, belonging to the dual Hilbert space $A^*$, the space of linear functionals on $A$.

Now, what is this thing? What I understand as the dual of $\ket{i}$ is simply its bra $\bra{i}$, which would make this isomorphism the same thing that Leifer proposed. After a long discussion with Marco Túlio and Oreshkov, I finally understood what they meant: one can indeed write it as Leifer’s version of the isomorphism, but the operator $\ket{j}\bra{i}$ on the first tensor subsystem has to be understood as a map taking linear functionals to linear functionals, that is, it acts on bras to the left, being defined as

\[ (\bra{\psi})\ket{j}\bra{i} := \langle \psi | j \rangle \bra{i},\]or, to use their notation,

\[ \ket{i}_{A^*}\bra{j}(\ket{\psi}_{A^*}) := \langle \psi | j \rangle \ket{i}_{A^*}.\]Ok, this works, but it is quite inconvenient; I don’t want to use operators that act on the left or change the rules of the bra-ket notation in order to make calculations with this thing. Can’t we redefine this isomorphism so that it works in the usual way?

There is an easy way around it: remember that $\langle \psi | j \rangle = \overline{\langle j | \psi \rangle}$, so if we define these funny vectors as the complex conjugate of the usual vectors, $\ket{\psi}_{A^*} := \overline{\ket{\psi}_A}$, then linear algebra is back to normal.

We can then redefine the isomorphism as

\[\mathfrak{B}(\mathcal E) := \sum_{ij} \overline{\ket{i}\bra{j}} \otimes \mathcal E(\ket{i}\bra{j}), \]and it is easy to show that it is indeed independent of the choice of basis $\{\ket{i}\}$: let $\mathcal E$ be the identity map again (by linearity it is enough to consider this case). Its representation is given by

\[\mathfrak{B}(\mathcal I) = \sum_{ij} \overline{U\ket{i}\bra{j}U^\dagger} \otimes U\ket{i}\bra{j}U^\dagger. \]Inserting two resolutions of the identity matrix we get

\begin{align*} \mathfrak{B}(\mathcal I) &= \sum_{ijkl} \overline{U\ket{i}\bra{j}U^\dagger} \otimes \ket{k}\bra{k}U\ket{i}\bra{j}U^\dagger\ket{l}\bra{l} \\

&= \sum_{ijkl} \overline{U\ket{i}\bra{i}U^\dagger\ket{k}\bra{l}U\ket{j}\bra{j}U^\dagger} \otimes \ket{k}\bra{l} \\

&= \sum_{kl} \overline{\ket{k}\bra{l}} \otimes \ket{k}\bra{l},

\end{align*} where we used the fact that $\bra{i}M\ket{j} = \overline{\bra{j}M^\dagger\ket{i}}$.

This isomorphism is not actually basis-independent, though, because it depends on the basis we have chosen to do the complex conjugation in. And to add insult to the injury, the operator you get by choosing some specific basis for the complex conjugation is exactly the same as the operator you get by using this basis to define the usual Choi-Jamiołkowski isomorphism.

So yes, you are doing it right.

I am glad that you take my blog post seriously enough to criticize it.

You should not call it “Leifer’s version of the isomorphism”, but just the Jamiolkowski isomorphism, because it is the version that Jamiolkowski originally proposed. The one you are advocating is the Choi isomorphism. We usually lump them together under the label “Choi-Jamiolkowski” because they are equivalent up to a partial transpose, so for many purposes it does not matter which one you use.

To be fair, there are many reasons for doing the Choi-Jamiolkowski isomorphism, and representing CP maps as positive matrices is only one of them. If I were doing semidefinite programming, you can be sure that I would use the Choi matrix as you have defined it, just like everyone else.

The main point of my blog post was not to make the isomoprhism basis independent, but rather to reduce the mystery of the Choi-Jamiolkowski isomorphism by showing that it is just the natural generalization of the relationship between stochastic dynamics and conditional probability that occurs in classical probability theory. For that, the Jamiolkowski version has some benefits, i.e. you do not need to put partial transposes in the equations, but it is by no means impossible to work with the Choi matrix instead.

When working with the Choi-Jamiolkowski operator of CPT maps, I personally find it very useful to think “I am working with the quantum version of a conditional probability distribution” rather than “I am working with a subset of the set of bipartite quantum states with a funny normalization”. The latter point of view is more common, but I think it obscures some simple facts that are obvious on the first point of view. See my blog post and the paper with Rob Spekkens that it was based on for examples.

Well, I am also glad you took my blog post seriously enough to criticize it. To be fair, I was mostly annoyed by the badly explained version of the isomorphism defended by Allen, Hoffreumon, etc., and took you along for the ride. Also, I wanted to steal your title.

That said, I do think it’s a rather misleading analogy to consider the CJ operator a quantum version of a conditional probability distribution. It doesn’t behave like one, as you explained in your paper. You can’t further conditionalize it, and more generally it doesn’t work for more than two subsystems.

First and foremost, I think the best way to interpret the CJ operator is as the linear map itself. I know, not very deep, but sometimes life is like this. Secondly, I think it is very useful to notice that they are equal to density matrices modulo normalization. You can actually produce them in the lab by applying the CPTP map to one-half of a maximally entangled state! That’s kind of cool, no? There’s no such “physical interpretation” for, say, the constraint matrix that is used in the NPA hierarchy. Not only it gives a warm fuzzy feeling, but it also makes immediately obvious that quantum process tomography is possible.