## 7.8 State-channel duality

Suppose that \mathcal{H} is of dimension d and \mathcal{H}' is of dimension d', and pick a basis for each space. Then any linear map \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H'}) can be completely characterised by its action on the d^2-many basis matrices |i\rangle\langle j| of \mathcal{B}(\mathcal{H}), where i,j\in\{1,2\ldots,d\}, i.e. for any density operator \rho on \mathcal{H} we have \mathcal{E}(\rho) = \mathcal{E}\left(\sum_{ij}\rho_{ij} |i\rangle\langle j|\right) = \sum_i\rho_{ij}\mathcal{E}(|i\rangle\langle j|). \tag{$\natural$} We can now tabulate all the (d'\times d') matrices \mathcal{E}(|i\rangle\langle j|) in \mathcal{H}' by forming a bigger (dd'\times dd') block matrix in \mathcal{H}\otimes\mathcal{H}':

We call this block matrix \widetilde{\mathcal{E}}\in\mathcal{B}(\mathcal{H}\otimes\mathcal{H}') the Choi matrix of \mathcal{E}.

The Choi matrix is essentially another way of representing a linear map \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H'}): if you are given the Choi matrix \widetilde{\mathcal{E}} of \mathcal{E} and you want to evaluate \mathcal{E}(\rho), then you simply follow Equation (\natural), taking the values of \mathcal{E}(|i\rangle\langle j|) from the Choi matrix. This can be formally written as

The Choi matrix \widetilde{\mathcal{E}} of a linear map \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H'}) satisfies \mathcal{E}(\rho) = d(\operatorname{tr}\otimes\mathbf{1})\left[(\rho^T\otimes\mathbf{1})\widetilde{\mathcal{E}}\right] for all density matrices \rho, where d=\dim\mathcal{H}.

The expression above may look baffling to an untrained eye, but this is often the case when we turn something conceptually obvious into a precise and compact mathematical notation. In order to gain some intuition here, recall that, for matrices A and B, \operatorname{tr}A^T B = \sum_{ij} A_{ij}B_{ij}. If we take A and B to be the block matrices \rho\otimes\mathbf{1} and \widetilde{\mathcal{E}}, respectively, then we can use this to show that (\operatorname{tr}\otimes\mathbf{1})\left[(\rho^T\otimes\mathbf{1})\widetilde{\mathcal{E}}\right] = \frac{1}{d}\sum_i\rho_{ij}\mathcal{E}(|i\rangle\langle j|).

This state-channel duality thus gives us a one-to-one correspondence between linear maps \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') and matrices \widetilde{\mathcal{E}} acting on the tensor product \mathcal{H}\otimes\mathcal{H}'. This correspondence is sometimes called the Choi–Jamiołkowski isomorphism, and we discuss it further in the appendix of the same name.

Mathematically, it is not too surprising that the matrix elements of an operator on a tensor product can be reorganised and reinterpreted as the matrix elements of an operator between operator spaces. What is interesting, and perhaps not so obvious, however, is that the positivity conditions for maps correspond exactly to conditions on their Choi matrices under this correspondence. In order to see this, let us express the Choi matrix as the result of \mathbf{1}\otimes\mathcal{E} acting on the maximally entangled state in \mathcal{H}\otimes\mathcal{H}:

The Choi matrix \widetilde{\mathcal{E}} of a linear map \mathcal{E}\colon\mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H'}) is given by \widetilde{\mathcal{E}} = (\mathbf{1}\otimes\mathcal{E})|\Omega\rangle\langle\Omega| = \frac{1}{d} \sum_{ij} |i\rangle\langle j|\otimes\mathcal{E}(|i\rangle\langle j|) where |\Omega\rangle=\frac{1}{\sqrt d}\sum_{i=1}^d|i\rangle|i\rangle is the maximally entangled state in \mathcal{H}\otimes\mathcal{H}, and where d=\dim\mathcal{H}.

In this form, the Choi–Jamiołkowski isomorphism gives us a simple necessary and sufficient condition for a linear map to be a quantum channel:

State-channel duality: \mathcal{E} is a quantum channel if and only if \widetilde{\mathcal{E}} is a density matrix.

One direction of this claim is immediate: we already know that any quantum channel \mathcal{E} is, in particular, a completely positive map, and so \mathbf{1}\otimes\mathcal{E} maps density matrices to density matrices, whence \widetilde{\mathcal{E}}=(\mathbf{1}\otimes\mathcal{E})|\Omega\rangle\langle\Omega| is a density matrix, since |\Omega\rangle\langle\Omega| is a density matrix. The other direction is less obvious. If \widetilde{\mathcal{E}} is a density matrix, then it can be written as a mixture of pure states |\psi_k\rangle\langle\psi_k| with probabilities p_k: \widetilde{\mathcal{E}} = \sum_k|\widetilde{\psi}_k\rangle\langle\widetilde{\psi}_k|, where |\widetilde{ \psi}_k\rangle=\sqrt{p_k}|\psi_k\rangle are (non-normalised) vectors, and any such vector can be written as |\widetilde{ \psi}_k\rangle = (\mathbf{1}\otimes E_k)|\Omega\rangle for some operator E_k.

Now, any vector |\psi\rangle in \mathcal{H}\otimes\mathcal{H}' can be written as115 |\psi\rangle=\mathbf{1}\otimes V|\Omega\rangle where V=\sum_{ij}V_{ij}|j\rangle\langle i| is an operator from \mathcal{H} to \mathcal{H}', and |\Omega\rangle=\frac{1}{d}\sum_i|i\rangle|i\rangle is a maximally entangled state in \mathcal{H}\otimes\mathcal{H}. (Here, the vectors |i\rangle and |j\rangle form orthonormal bases in \mathcal{H} and \mathcal{H}', respectively).

Using this, we see that \begin{aligned} \widetilde{\mathcal{E}} &= \sum_k|\widetilde{\psi}_k\rangle\langle\widetilde{\psi}_k| \\&= \sum_k (\mathbf{1}\otimes E_k)|\Omega\rangle\langle\Omega| (\mathbf{1}\otimes E^\dagger_k) \\&= \frac{1}{d} \sum_{ij} |i\rangle\langle j|\otimes\underbrace{\sum_k E_k(|i\rangle\langle j|) E^\dagger_k}_{\mathcal{E}(|i\rangle\langle j|)}. \end{aligned} Comparing the last expression on the right-hand size with the definition of \widetilde{\mathcal{E}}, or using the Choi–Jamiołkowski isomorphism, we conclude that \mathcal{E} is of the form \mathcal{E}(\rho) = \sum_k E_k \rho E^\dagger_k. Moreover, \operatorname{tr}\widetilde{\mathcal{E}}=1 implies that116 \sum_k E^\dagger_kE_k=\mathbf{1}. So if \widetilde{\mathcal{E}} is a density operator, then the map \mathcal{E} can be expressed in the Kraus form. Thus \mathcal{E} is a quantum channel, and, therefore, also a CPTP map. We have now established the desired isomorphism between states and channels.

The equations117 \mathcal{E}(\rho) = d(\operatorname{tr}\otimes\mathbf{1})\left[(\rho^T\otimes\mathbf{1})\widetilde{\cal{E}}\right]. and \widetilde{\mathcal{E}} = (\mathbf{1}\otimes\mathcal{E})|\Omega\rangle\langle\Omega| = \frac{1}{d} \sum_{ij} |i\rangle\langle j|\otimes\mathcal{E}(|i\rangle\langle j|) tell us how to obtain the state \widetilde{\mathcal{E}} from the channel \mathcal{E}, and vice versa. We have also shown that quantum channels are exactly the completely positive trace-preserving maps.

We summarise the flow of implications in the following diagram: \begin{CD} \mathcal{E} @>{\widetilde{E}=(\mathbf{1}\otimes\mathcal{E})|\Omega\rangle\langle\Omega|}>> \widetilde{\mathcal{E}} \\@VVV @VVV \\E_k\cdot E_k^\dagger @<<{|\widetilde{\psi}_k\rangle=(\mathbf{1}\otimes E_k)|\Omega\rangle}< |\widetilde{\psi}_k\rangle\langle\widetilde{\psi}_k| \end{CD} We start in the top left corner with a quantum channel \mathcal{E}. This channel is a CPTP map, which means that \mathbf{1}\otimes\mathcal{E} takes a maximally entangled state |\Omega\rangle to a density matrix \widetilde{\mathcal{E}}. This is our first implication: if \mathcal{E} is a quantum channel, then its Choi matrix \widetilde{\mathcal{E}} is a density matrix. The reverse implication goes as follows. The density matrix \widetilde{\mathcal{E}} can be expressed as a mixture of pure states, |\widetilde\psi_k\rangle\langle\widetilde\psi_k| (and this takes us to the bottom right corner in the diagram). We use this mixture decomposition when we construct map the \mathcal{E} from the Choi matrix \widetilde{\mathcal{E}}, via the Choi-Jamiolkowski isomorphism. We notice that \mathcal{E} admits the operator-sum representation, and that each of the pure states in the mixture is associated with a Kraus operator, with |\widetilde{\psi}_k\rangle = (\mathbf{1}\otimes E_k)|\Omega\rangle. All together this gives us the reverse implication: if the Choi matrix \widetilde{\mathcal{E}} is a density matrix, then \mathcal{E} is a quantum channel.

1. Exercise. Prove this!↩︎

2. Exercise. Prove this!↩︎

3. Where \operatorname{tr}\otimes\mathbf{1} acts as (\operatorname{tr}\otimes\mathbf{1})(A\otimes B)\coloneqq(\operatorname{tr}A)\otimes B.↩︎