# Derivation of Choi-Jamiolkowski isomorphism

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.

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.

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 $$c_k$$.

• 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 $B(H_B)$. Commented Nov 14, 2023 at 17:58
• 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. Commented Nov 14, 2023 at 20:12