Perfect discrimination of nonorthogonal quantum states with posterior classical partial information

The indistinguishability of nonorthogonal pure states lies at the heart of quantum information processing. Although the indistinguishability reflects the impossibility of measuring complementary physical quantities by a single measurement, we demonstrate that the distinguishability can be perfectly retrieved simply with the help of posterior classical partial information. We demonstrate this by showing an ensemble of nonorthogonal pure states such that a state randomly sampled from the ensemble can be perfectly identified by a single measurement with the help of postprocessing of the measurement outcomes and additional partial information about the sampled state, i.e., the label of the subensemble from which the state is sampled. When an ensemble consists of two subensembles, we show that the perfect distinguishability of the ensemble with the help of postprocessing can be restated as a matrix-decomposition problem. Furthermore, we give the analytical solution for the problem when both subensembles consist of two states.

Figure 1

Indistinguishability of nonorthogonal pure states in a QKD-like protocol. First, the sender randomly chooses label $X\in \{A,B\}$ and encodes his secret bit in a basis state of ${\mathbb{S}}^{\left(X\right)}$. Second, the eavesdropper intercepts the state transmitted from the sender and measures it. She cannot identify the transmitted state perfectly even if she can process her measurement outcomes with label $X$.

Figure 2

The region of $(P_{21},P_{22})$ for perfectly distinguishable $({\mathbb{S}}^{\left(A\right)},{\mathbb{S}}^{\left(B\right)})$ with the help of postprocessing when $P_{11}=P_{12}=1/4$, shown by the white region. The example shown in Table 1 resides on the boundary of perfectly distinguishable pairs. Note that since ${\mathbb{S}}^{\left(B\right)}$ is an indexed set of orthonormal vectors, $(P_{21},P_{22})$ cannot be in the dark gray region for any $({\mathbb{S}}^{\left(A\right)},{\mathbb{S}}^{\left(B\right)})$.