7.9 Channel-state duality

Suppose that \dim\mathcal{H}=d and \dim\mathcal{H}'=d', and pick a basis for each space. Now 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_{i,j=1}^d\rho_{ij} |i\rangle\langle j|\right) = \sum_{i,j=1}^d\rho_{ij}\mathcal{E}(|i\rangle\langle j|). \tag{$\natural$} We can now tabulate the (d\times d)-many (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}':

After scaling by a factor of \frac{1}{d}, we call this block matrix \widetilde{\mathcal{E}}\in\mathcal{B}(\mathcal{H}\otimes\mathcal{H}') the Choi matrix139 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'}), since 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. We can write this more formally as follows.

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

The expression above may look baffling at first glance, but this is often the case when we turn something conceptually obvious into more compact mathematical notation. In order to gain some intuition here, recall that, for matrices A and B, \operatorname{tr}A^T B = \sum_{i,j} 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,j}\rho_{ij}\mathcal{E}(|i\rangle\langle j|).

This 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}', known as the Choi–Jamiołkowski isomorphism \mathcal{E}\mapsto\widetilde{\mathcal{E}}.

The correspondence between linear maps \mathscr{B}(\mathcal{H})\to\mathscr{B}(\mathcal{H'}) and operators in \mathscr{B}(\mathcal{H}\otimes\mathcal{H'}), known as the Choi–Jamiołkowski isomorphism (or channel-state duality in the specific setting of quantum information), is another example of a well known correspondence between vectors in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} and operators \mathscr{B}(\mathcal{H}_{\mathcal{A}}^\star,\mathcal{H}_{\mathcal{B}}) or \mathscr{B}(\mathcal{H}_{\mathcal{B}}^\star,\mathcal{H}_{\mathcal{A}}).

Take a tensor product vector in |a\rangle\otimes|b\rangle\in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}. Then it defines natural maps in \mathscr{B}(\mathcal{H}_{\mathcal{A}}^\star,\mathcal{H}_{\mathcal{B}}) and \mathscr{B}(\mathcal{H}_{\mathcal{B}}^\star,\mathcal{H}_{\mathcal{A}}), via \begin{aligned} \langle x| &\longmapsto \langle x|a\rangle|b\rangle \\\langle y| &\longmapsto |a\rangle\langle y|b\rangle \end{aligned} for any linear forms \langle x|\in\mathcal{H}^\star_A and \langle y|\in\mathcal{H}^\star_B. We then extend this construction (by linearity) to any vector in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}. These isomorphisms are canonical: they do not depend on the choice of any bases in the vectors spaces involved.

However, some care must be taken when we want to define correspondence between vectors in \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} and operators in \mathscr{B}(\mathcal{H}_{\mathcal{A}},\mathcal{H}_{\mathcal{B}}) or \mathscr{B}(\mathcal{H}_{\mathcal{B}},\mathcal{H}_{\mathcal{A}}). For example, physicists like to “construct” \mathscr{B}(\mathcal{H}_{\mathcal{B}},\mathcal{H}_{\mathcal{A}}) in a deceptively simple way: |a\rangle|b\rangle \longleftrightarrow |a\rangle\langle b|. Flipping |b\rangle and switching from \mathcal{H}_{\mathcal{B}} to \mathcal{H}^\star_B is an anti-linear operation (since it involves complex conjugation). This is fine when we stick to a specific basis |i\rangle|j\rangle and use the ket-flipping approach only for the basis vectors. This means that, for |b\rangle=\sum_j\beta_j|j\rangle, the correspondence looks like |i\rangle|b\rangle \longleftrightarrow \sum_j \beta_j |i\rangle\langle j| and not like |i\rangle|b\rangle \longleftrightarrow |i\rangle\langle b| = \sum_j \beta^\star_j |i\rangle\langle j|. This isomorphism is non-canonical: it depends on the choice of the basis. But it is still a pretty useful isomorphism! The Choi–Jamiołkowski isomorphism is of this kind (i.e. non-canonical) — it works in the basis in which you express a maximally mixed state |\Omega\rangle=\sum_i|i\rangle|i\rangle.

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. That is, this 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}' descends to a one-to-one correspondence between quantum channels and some specific family of matrices (which we will shortly discuss). In other words, we can classify quantum channels as being exactly those linear maps that have a certain image under the Choi–Jamiołkowski isomorphism! In order to see this, let us express the Choi matrix as the result of \mathbf{1}\otimes\mathcal{E} acting on the maximally mixed state |\Omega\rangle\coloneqq\frac{1}{\sqrt{d}}\sum_{i=1}^d|i\rangle|i\rangle 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}_{d\times d}\otimes\mathcal{E})|\Omega\rangle\langle\Omega| = \frac{1}{d} \sum_{i,j} |i\rangle\langle j|\otimes\mathcal{E}(|i\rangle\langle j|) where d=\dim\mathcal{H}.

Pictorially, we might represent this by something like

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

Channel-state 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, or directly as \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, which can all 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 as140 |\psi\rangle=\mathbf{1}\otimes V|\Omega\rangle where V=\sum_{i,j}V_{ij}|j\rangle\langle i| is an operator from \mathcal{H} to \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_{i,j} \left(|i\rangle\langle j|\otimes\underbrace{\sum_k E_k(|i\rangle\langle j|) E^\dagger_k}_{\mathcal{E}(|i\rangle\langle j|)}\right). \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.

Finally, \operatorname{tr}\widetilde{\mathcal{E}}=1 implies that141 \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.

This establishes the desired isomorphism between states and channels. The equations \mathcal{E}(\rho) = d(\operatorname{tr}\otimes\mathbf{1})\left[(\rho^T\otimes\mathbf{1})\widetilde{\mathcal{E}}\right]. and \widetilde{\mathcal{E}} = (\mathbf{1}\otimes\mathcal{E})|\Omega\rangle\langle\Omega| = \frac{1}{d} \sum_{i,j} |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 can summarise the flow of implications in a diagram: \begin{CD} \mathcal{E} @>{\widetilde{\mathcal{E}}=(\mathbf{1}\otimes\mathcal{E})|\Omega\rangle\langle\Omega|}>> \widetilde{\mathcal{E}} \\@VVV @VVV \\E_k 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 (in particular) that \mathbf{1}\otimes\mathcal{E} takes a maximally mixed 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| (taking 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. Man-Duen Choi was brought up in Hong Kong. He received his Ph.D. degree under the guidance of Chandler Davis at Toronto. He taught at the University of California, Berkeley, from 1973 to 1976, and has worked since then at the University of Toronto. His research has been mainly in operator algebras, operator theory, and polynomial rings. He is particularly interested in examples/counterexamples and 2\times2 matrix manipulations.↩︎

  2. Exercise. Prove this!↩︎

  3. Exercise. Prove this! Hint: you can decompose the trace into a partial trace followed by a trace, and it’s helpful to do so here — \operatorname{tr}\widetilde{\mathcal{E}}=\operatorname{tr}[(\operatorname{tr}\otimes\mathbf{1})\widetilde{\mathcal{E}}].↩︎