# The Choi-Jamiołkowski isomorphism: you’re doing it right

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.

This entry was posted in Uncategorised. Bookmark the permalink.