What is the relationship between Choi and Chi matrix in Qiskit?

I'm struggling with the framework for quantum process tomography on Qiskit.

The final step of such a framework is running fit method of ProcessTomographyFitter class. Documentation states that such function gives a Choi matrix as output. Nevertheless, I'd want the Chi matrix to define the superoperator of a circuit. Specifically, I'm interested in understanding how a 2-qubit circuit affects a single qubit.

Thus, my questions are:

• What is the relationship between Choi and Chi matrix?
• When do they coincide?
• How to obtain Chi from Choi matrix?
• Does this answer your question? How to perform Quantum Process Tomography for three qubit gates?
– glS
May 3 '20 at 14:20
• I feel that that question is not a complete answer to this question, as it does not describe how to obtain the chi matrix from the Choi matrix or when they are the same.
– JSdJ
May 4 '20 at 7:11
• So do I. Anyway it answers partially. Thus, thanks a lot! May 6 '20 at 13:51

( I copied some text from a previous answer of mine)

Defining the Choi and $$\chi$$ matrix

The Choi matrix is a direct result of the Choi-Jamiolkowski isomorphism. Some intuition on what this is can be found in this previous answer. Consider the maximally entangled state $$|\Omega \rangle = \sum_{\mathrm{i}}|\mathrm{i}\rangle \otimes |\mathrm{i}\rangle$$, where $$\{|\mathrm{i}\rangle\}$$ forms a basis for the space on which $$\rho$$ acts. (Note that we thus have a maximally entangled state of twice as many qubits). The Choi matrix is the state that we get when on one of these subsystems $$\Lambda$$ is applied (leaving the other subsystem intact): $$\begin{equation} \rho_{\mathrm{Choi}} = \big(\Lambda \otimes I\big) |\Omega\rangle\langle\Omega|. \end{equation}$$ As the Choi matrix is a state, it must be positive semidefinite (corresonding the the CP constraint) and must be unity trace (corresponding to the TP constraint).

The process- or $$\chi$$-matrix comes from the fact that we can write our map as a double sum: $$\begin{equation} \Lambda(\rho) = \sum_{m,n} \chi_{mn}P_{m}\rho P_{n}^{\dagger}, \end{equation}$$ where $$\{P_{m}\}$$ & $$\{P_{n}\}$$ form a basis for the space of density matrices; we use the Pauli basis $$\{I,X,Y,Z\}^{\otimes n}$$ (thereby omitting the need for the $$\dagger$$ at $$P_{n}$$). The matrix $$\chi$$ now encapsulates all information of $$\Lambda$$; the CP constraint reads that $$\chi$$ must be positive semidefinite, and the trace constraint reads that $$\sum_{m,n}\chi_{mn}P_{n}P_{m} \leq I$$ (with equality for TP).

Computing one from another

From this, we get the following two identities: $$\begin{equation} \begin{split} \rho_{\mathrm{Choi}} &= \sum_{m,n} \chi_{m,n} |P_{m}\rangle\rangle\langle\langle P_{n}|, \\ \chi_{m,n} &= \langle\langle P_{m} | \rho_{\mathrm{Choi}} |P_{n}\rangle\rangle, \end{split} \end{equation}$$ where $$|P_{m}\rangle\rangle$$ is the 'vectorized' version of $$P_{m}$$, which is essentially just the columns of $$P_{m}$$ stacked on top of each other, giving a vector. That answers question 3.

Again I shamelessly 'self-promote': in the first appendix of my thesis I work through proofs of all these relations. The most intuitive way is by using the Kraus decomposition as an intermediary, but it is not needed.

Relationship between the two

From this, you can see that the Choi matrix and the chi matrix do indeed have some relationship. In fact, by choosing either the (qubit)-basis in which we express the Choi matrix, or choosing the (operator)-basis that we associate with the $$\chi$$-matrix, they can be one and the same.

As @AdamZalcman has pointed out in his comment (Thank you!), from the identity $$\chi_{m,n} = \langle \langle P_{m}|\rho_{\mathrm{Choi}}| P_{n}\rangle\rangle$$ we can choose the $$P_{m/n}$$ so that we just select the $$m$$-th row and $$n$$-th column of $$\rho_{\mathrm{Choi}}$$. This works if $$P_{k} = |i\rangle \langle j|$$, with $$k = id + j$$. Since both $$i$$ and $$j$$ run from $$0$$ to $$d-1$$ (indicating the column and row, respectively), this gives exactly $$d^{2}$$ elements.

The same effect can be reached if one expresses the Choi matrix in a different basis, while keeping the $$P_{k}$$ associated with $$\chi_{m,n}$$ the usual Paulis. For the two to coincide then (i.e. $$\chi_{m,n} = \rho_{\mathrm{Choi}}^{m,n}$$), we see that $$\rho_{\mathrm{Choi}}^{m,n}$$ should be expressed in the `vectorized-Pauli-basis' (which is a set of states, i.e. a basis for the Hilbert space!) - this is exactly the Bell basis.

• Thank you! What I believe is missing now is how all this math behind QPT and Quantum Channels is applied in Qiskit. The Qiskit's method works with Pauli basis, and it returns a Choi-matrix, but it seems corresponding to a Chi-matrix. Speaking more in programming language, the method returns a Choi object that is a 4x4 matrix (since I'm interested in the superoperator from one qubit to one qubit for a 2-qubit circuit). From documentation of the Chi object we have that Chi-matrix "is related to the Choi representation by a change of basis of the Choi-matrix into the Pauli basis". May 8 '20 at 12:06
• You're welcome!I am not 100 percent sure what is the problem here anymore. The fit method indeed returns a Choi matrix (at least that is what it did when I used it ~1.5 years ago). If you want to have the chi matrix, you can just calculate it yourself using the equations I provided. Or, if you check the source code for the chi matrix class (qiskit.org/documentation/_modules/qiskit/quantum_info/operators/…), you can pass the Choi matrix to the init function as the 'data' variable; thereby obtaining the chi matrix for that channel.
– JSdJ
May 8 '20 at 12:19
• +1 Very nice answer! I think the statement that the Choi matrix is the $\chi$ matrix when $P_k$ are the Bell basis rather than the Pauli basis is not correct. Note that as you wrote $\chi_{m,n} = \langle\langle P_m|\rho_{Choi}|P_n\rangle\rangle$ which means that for Choi and $\chi$ to coincide we need $P_k$ to vectorize to "one-hot" basis (aka standard basis). Consequently $P_k = |i\rangle\langle j|$ where $i$ and $j$ are determined by $k$ from $k=id + j$ where $d$ is Hilbert space dimension $i,j=0,1,\dots,d-1$ and $k=0,1,\dots,d^2-1$. Jun 21 '21 at 19:54
• @AdamZalcman (A quite delayed) thanks for your comment! That was indeed quite sloppy of me - I'll update the text.
– JSdJ
Aug 12 '21 at 11:19