I was reading TQI-notes by Watrous where they introduce different representations for quantum channels and wondering how to go from one to the other. I have:

\begin{align} &|\Phi(\rho)\rangle\!\rangle \tag{1} \\ &= K(\Phi)|\rho\rangle\!\rangle \tag{2} \\ &=|\text{tr}_2[(I\otimes \rho^T) J(\Phi)]\rangle\!\rangle \tag{3} \end{align}

where $\Phi$ is the channel, $\rho$ is the density operator of the quantum state input into the channel, $|\rangle\!\rangle$ represents vectorization, $K(\Phi)$ is the normal representation of the channel ($= \sum_i A_i \otimes \bar{A}_i$ where $A_i$ are the Kraus operators), and $J(\Phi)$ is the Choi representation ($=\sum_i |A_i\rangle\!\rangle \langle\!\langle A_i|$) of the channel.


  1. Are the equations correct?
  2. Is there a more straightforward way to rewrite Eq. (3) in terms of $J(\Phi)$? Choi representation seems very convoluted so where can this be useful?
  3. I have omitted the Strinespring representation as I don't have a clear intuition for it but how can it be related to Eq.(1)?

1 Answer 1


The only question which I believe has not been addressed by the comments yet is your question 3: how is the Stinespring representation related to the representation matrix $K(\Phi)$ of $\Phi$? To recall, for every completely positive trace preserving map $\Phi:\mathcal H_A\to\mathcal H_B$ the Stinespring dilation asserts the existence of a Hilbert space $\mathcal H_E$ (with dimension at most $\dim\mathcal H_A\dim\mathcal H_B$) as well as an isometry $V:\mathcal H_A\to\mathcal H_B\otimes\mathcal H_E$ such that $\Phi={\rm tr}_E(V(\cdot)V^*)$. Combining this with the vectorization procedure yields a rather simple form for the representation matrix of $\Phi$:

Theorem. Given any Stinespring dilation of $\Phi$ of the above form it holds that $$ K(\Phi)=\big( {\bf1}_{BB}\otimes\langle\eta|_E \big)\big( {\bf1}_{B}\otimes\mathbb F_{EB}\otimes {\bf1}_{E} \big)\big( \overline{V}\otimes V \big)\tag{1} $$ where $|\eta\rangle_E=\sum_j|j\rangle\otimes|j\rangle$ is the (unnormalized) maximally entangled state on $\mathcal H_{E}\otimes\mathcal H_{E}$, $\overline{(\cdot)}$ stands for entrywise complex conjugation (in the computational basis), and $\mathbb F_{EB}:\mathcal H_{E}\otimes\mathcal H_{B}\to\mathcal H_{B}\otimes\mathcal H_{E}$ is the usual flip operator $x\otimes y\mapsto y\otimes x$.

Before proving this let me note that one can obtain an analogous result for unitary Stinespring dilations, I'll leave that as an exercise to keep things simple.

Proof. The key properties we will need is that ${\rm vec}(|i\rangle\langle j|)=\sum_k|k\rangle\otimes|i\rangle\langle j|k\rangle=|j\otimes i\rangle$ and that $K$ is a (linear) homomorphism, that is, $K(\Phi_1\Phi_2)=K(\Phi_1)K(\Phi_2)$ for all $\Phi_1,\Phi_2$. Thus writing $\Phi$ as the composition of ${\rm tr}_E$ and $V(\cdot)V^*$ means that $K(\Phi)=K({\rm tr}_E)K(V(\cdot)V^*)$. The second factor is simple to evaluate via the well-known identity $| ABC\rangle\!\rangle=(C^T\otimes A)|B\rangle\!\rangle$ for all $A,B,C$: \begin{align*} K(V(\cdot)V^*)|X\rangle\!\rangle&=|VXV^*\rangle\!\rangle\\ &=((V^*)^T\otimes V)|X\rangle\!\rangle\\ &=(\overline V\otimes V)|X\rangle\!\rangle \end{align*} Because $X$ was arbitrary this implies $K(V(\cdot)V^*)=\overline V\otimes V$. Thus in order to show (1) all we have to check is that $K({\rm tr}_E)=({\bf1}_{BB}\otimes\langle\eta|_E )( {\bf1}_{B}\otimes\mathbb F_{EB}\otimes {\bf1}_{E})$, and even this we only have to verify for product vectors (by linearity). Indeed, for all $i,j,k,l=1,2,\ldots$ \begin{align*} K({\rm tr}_E)|i\otimes j\otimes k\otimes l\rangle&=K({\rm tr}_E){\rm vec}(|k\otimes l\rangle\langle i\otimes j|)\\ &={\rm vec}\big({\rm tr}_E(|k\rangle\langle i|\otimes |l\rangle\langle j|)\big)\\ &={\rm vec}(|k\rangle\langle i|\langle j|l\rangle)=\langle j|l\rangle|i\otimes k\rangle\,. \end{align*} On the other hand \begin{align*} ({\bf1}_{BB}\otimes\langle\eta|_E )( {\bf1}_{B}\otimes\mathbb F_{EB}\otimes {\bf1}_{E})|i\otimes j\otimes k\otimes l\rangle&= ({\bf1}_{BB}\otimes\langle\eta|_E )|i\otimes k\otimes j\otimes l\rangle\\ &=\sum_a({\bf1}\otimes{\bf1}\otimes\langle a|\otimes\langle a|)|i\otimes k\otimes j\otimes l\rangle\\ &=|i\otimes k\rangle\underbrace{\sum_a \langle a|j\rangle\langle a|l\rangle}_{=1\text{ if }j=l\text{ and 0 else}}=\langle j|l\rangle|i\otimes k\rangle\,. \end{align*} Thus these expressions are the same so, altogether, we proved (1). $\square$


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