2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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
    Nov 6, 2023 at 18:01

1 Answer 1

3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.