# 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, 2020 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, 2020 at 7:11
• So do I. Anyway it answers partially. Thus, thanks a lot! May 6, 2020 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 by Norbert Schuch. 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): $$$$\rho_{\mathrm{Choi}} = \big(\Lambda \otimes I\big) |\Omega\rangle\langle\Omega|.$$$$ As the Choi matrix is a state, it must be positive semidefinite (corresonding the the CP constraint) and must have unit trace (necessary but not sufficient for the TP constraint).

The process- or $$\chi$$-matrix comes from the fact that we can write our map as a double sum: $$$$\Lambda(\rho) = \sum_{m,n} \chi_{mn}P_{m}\rho P_{n}^{\dagger},$$$$ where $$\{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: $$$$\begin{split} \rho_{\mathrm{Choi}} &= \sum_{m,n} \chi_{m,n} |P_{m}\rangle\rangle\langle\langle P_{n}|, \\ \chi_{m,n} &= \langle\langle P_{m} | \rho_{\mathrm{Choi}} |P_{n}\rangle\rangle, \end{split}$$$$ where $$|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, 2020 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, 2020 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, 2021 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, 2021 at 11:19