Consider Hilbert spaces $\mathcal{X}, \mathcal{Y}$. For any quantum channel $\mathcal{E}_{\mathcal{X}\rightarrow \mathcal{Y}}$, the bipartite Choi state $J(\mathcal{E}) \in L(\mathcal{Y}\otimes\mathcal{X})$ is given by

$$J(\mathcal{E}) = (\mathcal{E}\otimes I)\sum_{a,b} \vert a\rangle\langle b\vert\otimes\vert a\rangle\langle b\vert$$

It is also possible to show (see here for example) that the trace preserving condition of the map $\mathcal{E}$ is equivalent to the following condition on its Choi state

$$\text{Tr}_\mathcal{Y}J(\mathcal{E}) = I_\mathcal{X} \tag{1}$$

The proof is easy - it relies on noticing that the trace preserving condition implies that $\mathcal{E}(\vert a\rangle\langle b\vert) = \delta_{a,b}$ due to the trace preserving condition. Meanwhile, positive-semidefiniteness of $J(\mathcal{E})$ corresponds to complete positivity of $\mathcal{E}$.

If $J(\mathcal{E})$ is a density matrix but does not fulfill (1) is it still related to physical quantum channels (i.e. completely positive and trace preserving maps) in some way? Specifically, since completely positive maps correspond to positive semidefinite Choi matrices, does imposing that $\text{Tr}(J(\mathcal{E})) = 1$ (in addition to positive semidefiniteness) give us any condition on the map $\mathcal{E}$?


1 Answer 1


It is exactly what you say: A "Choi state" which is only positive semi-definite corresponds to a completely positive map which is not necessarily trace preserving. Fixing the trace of the Choi state rescales the "success probability" when implementing the CP map via the isomorphism - if you wish, it rescales the trace of all output states $\mathcal E(\rho)$ by a constant factor. (Note that deviating from the local trace condition $\mathrm{tr}_Y J = I$, but keeping the total trace $\mathrm{tr} J =1$, implies that there will be inputs with outputs with trace larger than one.)


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