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?

  • $\begingroup$ 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 $U$) $\endgroup$
    – glS
    Commented Nov 6, 2023 at 18:01

1 Answer 1


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.

  • $\begingroup$ 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|) $ $\endgroup$
    – karry
    Commented Nov 3, 2023 at 12:08
  • $\begingroup$ @Karry No problem! Also, I have edited my answer. Let me know if that helps. $\endgroup$
    – FDGod
    Commented Nov 3, 2023 at 13:17
  • $\begingroup$ So clear, I see. Thanks! : ) $\endgroup$
    – karry
    Commented Nov 4, 2023 at 3:17

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