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I'm following the course Mathematical Methods of Quantum Information Theory by Reinhard Werner. In lecture 6, he gave a derivation of Choi-Jamiolkowski isomorphism, and I'm struggling to understand some parts of it.

Here's the proof, in which I added more details for rigor.

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Let $H_A,H_B$ be finite-dimensional Hilbert spaces, and add an ancillary Hilbert space $H_R$ such that $\dim(H_R)=\dim(H_A)=d$. Let $\sigma\in H_R\otimes H_A$ be a pure state, so that $\sigma=|\Omega\rangle\langle\Omega|$, where $|\Omega\rangle\in H_R\otimes H_A$ is given by: $$ |\Omega\rangle=\sum_{i=1}^{d}c_i|ii\rangle, \tag{1} $$ where $|ii\rangle=|i\rangle_R \otimes |i\rangle_A\in H_R\otimes H_A$, being $(|i\rangle)_{i=1}^d$ an orthonormal basis of $H_R$ and $H_A$, with $c_i\geq0$.

Now, we define the quantum channel $\epsilon:B(H_A)\to B(H_B)$ by $\sigma\mapsto\epsilon(\sigma):=|\alpha\rangle$. Here $B(H)$ denote the set of bounded operators acting on an underlying Hilbert space $H$. We define the Choi matrix $\rho_\epsilon\in B(H_A\otimes H_B)$, corresponding to the channel $\epsilon$, by $$ \rho_\epsilon:=(I_R\otimes\epsilon)(|\Omega\rangle\langle\Omega|). \tag{2} $$

By complete positivity of $\epsilon$, $\rho_\epsilon$ must be positive. Then, we can expand $\rho_\epsilon$ as follows: Choosing the basis $( |i \rangle )_{i=1}^d$ of $H_A$ and $(| \alpha \rangle )_{i=1}^d$ of $H_R$, we construct the bases $(| i \rangle \langle j|)_{i=1}^d$ of $B(H_A)$ and $(| \alpha \rangle \langle \beta |)_{i=1}^d$ of $B(H_B)$. Thus, we can express $\rho_\epsilon$ as $$ \rho_\epsilon:=\sum_{i,j}^d \sum_{\alpha,\beta}^d (I_R\otimes\epsilon)(\sigma) |i\rangle\langle j| \otimes |\alpha\rangle\langle\beta|. \tag{3} $$ By taking the trace of (3), we obtain \begin{align} \text{Tr}\left( \sum_{i,j,\alpha,\beta}^d (I_R\otimes\epsilon)(\sigma) |i\rangle\langle j| \otimes |\alpha\rangle\langle \beta| \right) &= \sum_{i,j,\alpha,\beta}^d \text{Tr}((I_R\otimes\epsilon)(|\Omega\rangle\langle\Omega|) |i\rangle\langle j| \otimes |\alpha\rangle\langle \beta|) \tag{4.1} \\ &= \sum_{i,j,\alpha,\beta}^d \text{Tr}(|\Omega\rangle\langle\Omega| |i\rangle\langle j| \otimes \epsilon(|\alpha\rangle\langle \beta|)) \tag{4.2} \\ &= \sum_{i,j,\alpha,\beta}^d \langle\Omega| |i\rangle\langle j| \otimes \epsilon(|\alpha\rangle\langle \beta|) |\Omega \rangle \tag{4.3} \\ &= \sum_{i,j,\alpha,\beta}^d \sum_{k,l}^d \overline{c_k}c_l \langle kk| |i\rangle\langle j| \otimes \epsilon(|\alpha\rangle\langle\beta|) |ll\rangle \tag{4.4} \\ &= \sum_{i,j,\alpha,\beta}^d \sum_{k,l}^d \overline{c_k}c_l \langle k|i \rangle \langle j|l \rangle \langle k| \epsilon(|\alpha\rangle\langle\beta|) |l\rangle \tag{4.5} \\ &= \sum_{i,j,\alpha,\beta}^d \sum_{k,l}^d \overline{c_k}c_l \delta_{ki} \delta_{jl} \langle k| \epsilon(|\alpha\rangle\langle\beta|) |l\rangle \tag{4.6} \\ &= \sum_{i,j,\alpha,\beta}^d \overline{c_i}c_j \langle i| \epsilon(|\alpha\rangle\langle\beta|) |j\rangle \tag{4.7} \end{align}

Since for each $i,j,\alpha,\beta\in\{1,\ldots,d\}$, we have $\text{Tr}(\rho_\epsilon |i\rangle\langle j| \otimes |\alpha\rangle\langle \beta|) = \langle j\beta|\rho_\epsilon|i\alpha\rangle$, the coefficients to describe $\rho_\epsilon$ are in one-to-one correspondence with those of $\epsilon$.

Now, my doubt arises because I expanded $\rho_\epsilon$ with sums as in $(3)$; in contrast, Prof. Werner omitted them. So, my question is if this derivation is correct. Any comment or suggestion is welcomed.

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1 Answer 1

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It's easier to work with matrix units $E_{ij} = |i\rangle \langle j|$. In particular, we have $$ |\Omega\rangle \langle \Omega| = \sum_{i,j}c_i\overline{c_j} E_{ij} \otimes E_{ij}\tag{1}\,. $$

The Choi map $\Phi$ that gives us a state from a channel is $$ \begin{align}\rho_\epsilon &= \Phi(\epsilon) \tag{2.1}\\&= (I_R \otimes \epsilon)(|\Omega\rangle \langle \Omega|) \tag{2.2}\\&= \sum_{i,j}c_i\overline{c_j} \cdot E_{ij} \otimes \epsilon(E_{ij})\tag{2.3}\,.\end{align} $$

Note that $\epsilon(E_{ij})$ is a linear combination of all matrix units $E_{kl}$. In total $\epsilon$ can be described by $d^4$ parameters. They are independent if $\epsilon$ is any linear map (not necessarily a quantum channel).

When we compute the expectation with $|k\rangle\langle l| \otimes |\alpha\rangle\langle \beta| = E_{kl} \otimes E_{\alpha\beta}$ we get $$\begin{align} {\rm Tr}(\rho_\epsilon E_{kl} \otimes E_{\alpha\beta}) &= {\rm Tr}\bigg(\sum_{i,j}c_i\overline{c_j} E_{ij} \otimes \epsilon(E_{ij}) \cdot E_{kl} \otimes E_{\alpha\beta}\bigg)\tag{3.1} \\&= {\rm Tr}\bigg(\sum_{i,j}c_i\overline{c_j} E_{ij}E_{kl} \otimes \epsilon(E_{ij})E_{\alpha\beta}\bigg)\tag{3.2}\\&= \sum_{i,j}c_i\overline{c_j}\cdot{\rm Tr}(E_{ij}E_{kl})\cdot{\rm Tr}\big(\epsilon(E_{ij})E_{\alpha\beta}\big) \tag{3.3}\\&= c_l\overline{c_k}\cdot{\rm Tr}\big(\epsilon(E_{lk})E_{\alpha\beta}\big)\tag{3.4}\,.\end{align} $$

Numbers ${\rm Tr}(\epsilon(E_{kl})E_{\alpha\beta})$ over $\alpha,\beta$ are exactly the coefficients in the linear decomposition of $\epsilon(E_{kl})$ over the basis $\{E_{\alpha\beta}\}$ in the space of matrices since $$ \epsilon(E_{kl}) = \sum_{\alpha,\beta} {\rm Tr}\big(\epsilon(E_{kl})E_{\alpha\beta}\big)E_{\alpha\beta}^T.\tag{4} $$

If $c_k=1/\sqrt{d}$ for all $k$ then there is a reverse formula $$\begin{align} \epsilon(X) &= \Phi^{-1}(\rho_\epsilon)(X) \tag{5.1}\\&= {\rm Tr}_R\big((X^T\otimes I_B)\rho_\epsilon\big)\tag{5.5}\,, \end{align} $$ which gives a channel from a state. It's enough to prove it for $X = E_{kl}$ since both $\epsilon$ and ${\rm Tr}_R$ are linear functions. And it's not hard to generalize it for non-zero $c_k$.

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  • $\begingroup$ Thank you, great answer! I just have one further question: In your equation (4), wouldn't it be $\epsilon(E_{kl}) = \sum_{\alpha,\beta} {\rm Tr}\big(\epsilon(E_{kl})E_{\alpha\beta}^T\big)E_{\alpha\beta}$? Using the trace inner product in $B(H_B)$. $\endgroup$
    – Manuel E
    Nov 14, 2023 at 17:58
  • $\begingroup$ It's the same thing. $E_{\alpha\beta}^T = E_{\beta\alpha}$ and since we sum over all possible values of $\alpha, \beta$ it doesn't matter. $\endgroup$
    – Danylo Y
    Nov 14, 2023 at 20:12

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