How to describe the evolution of a density matrix using the Choi matrix?

Question

How do I apply the Choi matrix on a Density matrix.

Say my process is a Hadamard gate, and my input state is the ground state on 1 qubit (qubit id 0).

$U = H = \dfrac{1}{\sqrt{2}} \begin{bmatrix}1&1\\1&-1\end{bmatrix}$

Thus, to do ancilla assisted QPT, I need 1 ancilla qubit (I took it as the MSB, qubit id 1).

So,:

$\rho_{in} = |0\rangle \langle 0| = \begin{bmatrix}1&0\\0&0\end{bmatrix}$

Considering the full system (along with ancilla), the:

$\rho_{in}^{sys} = |00\rangle \langle 00| = \begin{bmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$

The (normalized) Choi matrix for the quantum circuit with Hadamard of Qubit 0, is:

$\rho_{choi}^{sys} = \begin{bmatrix}0.25&0.25&0.25&-0.25\\0.25&0.25&0.25&-0.25\\0.25&0.25&0.25&-0.25\\-0.25&-0.25&-0.25&0.25\end{bmatrix}$

This can be found either using Qiskit, or solving $(I \otimes H)(|\Omega \rangle \langle \Omega |)(I \otimes H)^\dagger$, where $|\Omega \rangle$ is the Bell pair $\dfrac{1}{\sqrt{2}} (|00\rangle +|11\rangle )$

Considering I am applying a Hadamard on the $\rho_{in}$, the output density matrix:

$\rho_{out} = H|0\rangle \langle 0|H^{\dagger} = \begin{bmatrix}0.5&0.5\\0.5&0.5\end{bmatrix}$

How exactly do I recover this back from the $\rho_{choi}^{sys}, \rho_{in}^{sys}$ and partial trace as given in the Qiskit link https://qiskit.org/documentation/stubs/qiskit.quantum_info.Choi.html?

Answer 1

You want to calculate $$ \rho_{out}=2\text{Tr}_0(\rho_{in}^T\otimes I\cdot\rho^{sys}_{choi}). $$

Answer 2

For any map $\Lambda(\cdot)$ with Choi representation $\rho_{C}$ defined as you are doing it (i.e. the channel on the second biparition of the maximally entangled state), the output $\rho_{out} = \Lambda(\rho_{in})$ can be calculated as:

$$ \rho_{out} = \mathrm{tr}_{1}\big[\rho_{C}\big(\rho_{in}^{t} \otimes I\big)\big] $$

where the trace is over the ancilla space and $\rho_{in}^{t}$ is the (non-complex) transpose of $\rho_{in}$.

Sometimes rather than $I$ the maximally mixed state $I/d$ is used, but all that changes is that there's an extra factor of $d$ - just renormalize the output state as $\rho_{out} \rightarrow \frac{\rho_{out}}{\mathrm{tr}[\rho_{out}]}]$ and you should be all set.

Btw, usually the Choi matrix is defined with the channel on the first bipartition (e.g. in your case the first qubit), and not the second qubit. In your case the Choi matrix doesn't really change though, but strictly speaking you then get $\big(I \otimes \rho_{in}^{t}\big)$ and the trace over the other bipartition.

Answer 3

Let $\Phi$ be a channel acting on a state $\rho$ (or more generally, a map acting on a linear operator; we don't actually need restrict to CPTP maps and states for these calculations). Let $J(\Phi)$ be the Choi representation of $\Phi$, i.e. $$J(\Phi)\equiv \sum_{ij} \Phi(E_{ij})\otimes E_{ij},$$ where $E_{ij}\equiv|i\rangle\!\langle j|$. Denote with $J(\Phi)_{ij,k\ell}$ the matrix elements of $J(\Phi)$ (remember that $J(\Phi)$ is a bipartite operator/state. We then have $$\Phi(\rho)_{ij} = \sum_{k\ell}J(\Phi)_{ik,j\ell}\rho_{k\ell}.$$ It is worth stressing that $J(\Phi)_{ik,j\ell}$ here is defined as $$J(\Phi)_{ik,j\ell} \equiv \langle i,k|J(\Phi)|j,\ell\rangle = \langle i|\Phi(E_{k\ell})|j\rangle.$$ If you want a component-free expression, you can equivalently write the above as $$\Phi(\rho) = \operatorname{Tr}_2[J(\Phi)(I\otimes \rho^T)].$$