What is the Choi matrix of the $H$ gate?

I see the definition of Choi matrix is:

The (unnormalized) maximally entangled bipartite state between a quantum system $$S$$ and an ancilla system $$A$$ is $$|\psi\rangle=\sum_{k=1}^d|k\rangle_A|k\rangle_S$$ , where $$\{|k\rangle\}_{k=1}^d$$ represents an orthonormal basis. For a quantum process $$\mathcal{E}$$ acting only on the system $$S$$ of $$|\psi\rangle$$, the output state is given by $$\Upsilon_{\mathcal{E}}=(\mathcal{I} \otimes \mathcal{E})(|\psi\rangle\langle\psi|)=\sum_{k, l=1}^d|k\rangle\langle l| \otimes \mathcal{E}(|k\rangle\langle l|),$$ which is called the Choi matrix of the process $$\mathcal{E}$$.

And the Hadamard gate is $$H=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right]$$, with the corresponding Choi matrix $$\Upsilon_H=(\mathcal{I} \otimes \mathcal{H})(|\psi\rangle\langle\psi|)=\frac{1}{2}\left[\begin{array}{cccc} 1 & 1 & 1 & -1 \\ 1 & 1 & 1 & -1 \\ 1 & 1 & 1 & -1 \\ -1 & -1 & -1 & 1 \end{array}\right],$$ but How can I get it from the definition of Choi matrix?

• generally speaking, a map of the form $\Phi(X)=UXU^\dagger$ has Choi $\sum_{ij} \Phi(E_{ij})\otimes E_{ij}$ equal to $\operatorname{vec}(U)\operatorname{vec}(U)^\dagger$ (that is, $uu^\dagger$ with $u$ the vectorization of $U$)
– glS
Commented Nov 6, 2023 at 18:01

Let $$\rho$$ be your state.

Let $$\mathcal{E}$$ be the Hadamard map.

$$\therefore \mathcal{E}(\rho) = H \rho H^{\dagger} = H \rho H\;.$$

Let $$\Upsilon_{\mathcal{E}}$$ be the Choi matrix. For this case, $$d=2$$.

($$d$$ is the dimension of the system on which this super-operator $$\mathcal{E}$$ acts.)

$$\therefore\Upsilon_{\mathcal{E}} = \sum_{k,l=0}^{1,1} |k\rangle\langle l| \otimes \mathcal{E}|k\rangle \langle l| \,.$$

So you need to calculate

$$\Upsilon_{\mathcal{E}} = \bigg( |0\rangle\langle 0| \otimes \mathcal{E} \big(|0\rangle \langle 0| \big)\bigg) + \bigg( |0\rangle\langle 1| \otimes \mathcal{E}\big(|0\rangle \langle 1| \big)\bigg) +\bigg( |1\rangle\langle 0| \otimes \mathcal{E}\big(|1\rangle \langle 0| \big)\bigg) +\bigg( |1\rangle\langle 1| \otimes \mathcal{E}\big(|1\rangle \langle 1| \big)\bigg) \,,$$

which is

$$\Upsilon_{\mathcal{E}} = \bigg( |0\rangle\langle 0| \otimes |+\rangle \langle +| \bigg) + \bigg( |0\rangle\langle 1| \otimes |+\rangle \langle -| \bigg) +\bigg( |1\rangle\langle 0| \otimes |-\rangle \langle +| \bigg) +\bigg( |1\rangle\langle 1| \otimes |-\rangle \langle -| \bigg) \,.$$

In response to OP's comment to this answer:

How to get this equation for Choi Matrix?

As you said, we start with a maximally entangled bipartite state $$|\psi \rangle$$, upto some normalization. If $$\mathcal{E}$$ is a $$d^2 \times d^2$$ operator, then each part of this bipartite state is of dimension $$d$$ i.e., Anciallas are of $$d$$ dimension, and systems is also of $$d$$ dimension. Then, we do nothing to the ancilla and apply the $$\mathcal{E}$$ process to your system.

$$|\psi \rangle = \sum_{k=0}^{d-1} |k \rangle_A |k \rangle_S\,.$$

So let $$\mathcal{M} = |\psi\rangle \langle \psi|$$.

\begin{align} \mathcal{M} &= |\psi\rangle \langle \psi|\,,\\ &= \bigg(\sum_{k=0}^{d-1} |k \rangle_A |k \rangle_S \bigg)\bigg(\sum_{l=0}^{d-1} \langle l|_A \langle l|_S\bigg)\,,\\ &= \sum_{k=0}^{d-1}\sum_{l=0}^{d-1} \bigg(|k \rangle_A |k \rangle_S \langle l|_A \langle l|_S \bigg)\,,\\ &= \sum_{k,l=0}^{d-1,d-1} |k\rangle _A \langle l| \otimes |k\rangle _S \langle l|\,. \end{align} So now, the Choi matrix is

$$\Upsilon_{\mathcal{E}} = \big(\mathcal{I} \otimes \mathcal{E}\big) \mathcal{M}\;.$$

Simplifying the above expression: \begin{align} \Upsilon_{\mathcal{E}} &= \big(\mathcal{I} \otimes \mathcal{E}\big) \mathcal{M}\;,\\ &= \big(\mathcal{I} \otimes \mathcal{E}\big) \bigg(\sum_{k,l=0}^{d-1,d-1} |k\rangle _A \langle l| \otimes |k\rangle _S \langle l| \bigg) \,,\\ &= \sum_{k,l=0}^{d-1,d-1} \big(\mathcal{I} \otimes \mathcal{E}\big) \bigg( |k\rangle _A \langle l| \otimes |k\rangle _S \langle l| \bigg)\,,\\ &= \sum_{k,l=0}^{d-1,d-1} \mathcal{I} \big(|k\rangle _A \langle l|\big) \otimes \mathcal{E} \big(|k\rangle _S \langle l| \big)\,,\\ &= \sum_{k,l=0}^{d-1,d-1} |k\rangle _A \langle l| \otimes \mathcal{E} \big(|k\rangle _S \langle l| \big)\,. \end{align} And this is how you end up with the equation for the Choi matrix you have written.

• Thanks, I see. I also want to ask why $\Upsilon_{\mathcal{E}}=(\mathcal{I} \otimes \mathcal{E})(|\psi\rangle\langle\psi|)=\sum_{k, l=1}^d|k\rangle\langle l| \otimes \mathcal{E}(|k\rangle\langle l|)$ Commented Nov 3, 2023 at 12:08
• @Karry No problem! Also, I have edited my answer. Let me know if that helps. Commented Nov 3, 2023 at 13:17
• So clear, I see. Thanks! : ) Commented Nov 4, 2023 at 3:17