In this tutorial, we are going to show how to make use of qustop to
calculate the optimal probability of distinguishing a state from an ensemble of
quantum states when Alice and Bob are allowed to use global (positive)
measurements on their system.
The optimal probability with which Bob can distinguish the state he is given may be obtained by solving the following semidefinite program (SDP).
\begin{align*}
\textbf{Primal:} \quad & \\
\text{maximize:} \quad & \sum_{i=0}^n p_i \langle M_i,
\rho_i \rangle \\
\text{subject to:} \quad & \sum_{i=0}^n M_i = \mathbb{I}_{\mathcal{A} \otimes
\mathcal{B}},\\
& M_1, \ldots, M_n \in \text{Pos}(\mathcal{A} \otimes \mathcal{B}).
\end{align*}
\begin{equation}
\begin{aligned}
\textbf{Dual:} \quad & \\
\text{minimize:} \quad & \text{Tr}(Y) \\
\text{subject to:} \quad & Y - \rho_i \in \text{Pos}(\mathcal{A} \otimes \mathcal{B}),
\quad \forall i = 1, \ldots, n, \\
& Y \in \text{Herm}(\mathcal{A} \otimes
\mathcal{B}).
\end{aligned}
\end{equation}
The qustop package solves either of these two optimization problems depending on whether the
optimal measurements are required.
For the special case of distinguishing between two states, the probability of optimally distinguishing is exactly
\text{opt}_{\text{POS}}(\eta) = \frac{1}{2} + \frac{1}{4} \left\lVert \eta(0) - \eta(1) \right\rVert_1
where \left\lVert \cdot \right\rVert_1.
A result of [tWalgate00] shows that any two orthogonal pure states can be distinguished perfectly. This result actually applies to LOCC measurements and is a stronger claim than just for positive measurements, but since \text{opt}_{\text{LOCC}}(\eta) \leq \text{opt}_{\text{POS}}(\eta) is true for any ensemble \eta, it also holds for positive measurements.
For example, consider the two orthogonal pure states
| \psi_0 \rangle = \sqrt{\frac{3}{4}} | + \rangle + \sqrt{\frac{1}{4}} | - \rangle
\quad \text{and} \quad
| \psi_1 \rangle = \sqrt{\frac{1}{4}} | + \rangle - \sqrt{\frac{3}{4}} | - \rangle.
Since | \psi_0 \rangle and | \psi_1 \rangle are pure and mutually orthogonal, they are able to be perfectly distinguished.
.. literalinclude:: ../examples/opt_dist/positive/min_error/two_pure_states.py :language: python :linenos: :start-after: # along with this program. If not, see <https://www.gnu.org/licenses/>.
Consider now the following two mixed states
| \phi_1 \rangle = \sqrt{\frac{3}{4}} |+ \rangle + \sqrt{\frac{1}{4}} |- \rangle
\quad \text{and} \quad
| \phi_2 \rangle = \sqrt{\frac{1}{4}} |+ \rangle + \sqrt{\frac{3}{4}} |- \rangle.
The following code sample shows that the closed-form equation matches the result obtained from
qustop, however, since they are mixed states and not pure, we are not able to perfectly
distinguish them.
.. literalinclude:: ../examples/opt_dist/positive/min_error/two_mixed_states.py :language: python :linenos: :start-after: # along with this program. If not, see <https://www.gnu.org/licenses/>.
On the note of orthogonality, if the ensemble of states provided consist of all mutually orthogonal states, then it is possible to distinguish with perfect probability in this special case.
As a prototypical example, consider the four Bell states
\begin{equation}
\begin{aligned}
| \psi_0 \rangle = \frac{| 00 \rangle + | 11 \rangle}{\sqrt{2}}, &\quad
| \psi_1 \rangle = \frac{| 01 \rangle + | 10 \rangle}{\sqrt{2}}, \\
| \psi_2 \rangle = \frac{| 01 \rangle - | 10 \rangle}{\sqrt{2}}, &\quad
| \psi_3 \rangle = \frac{| 00 \rangle - | 11 \rangle}{\sqrt{2}}.
\end{aligned}
\end{equation}
.. literalinclude:: ../examples/opt_dist/positive/min_error/bell_states.py :language: python :linenos: :start-after: # along with this program. If not, see <https://www.gnu.org/licenses/>.
If there are more than two states and those states are not mutually orthogonal, no closed-form equation is known to exist, so we resort to solving the SDP.
For instance, consider the following set of three non-mutually-orthogonal states
\begin{equation}
\begin{aligned}
| \phi_1 \rangle = \sqrt{\frac{3}{4}}| + \rangle + \sqrt{\frac{1}{4}} | - \rangle, &\quad
| \phi_2 \rangle = \sqrt{\frac{1}{4}}| + \rangle + \sqrt{\frac{3}{4}} | - \rangle, \\
| \phi_3 \rangle = \sqrt{\frac{1}{2}}| + \rangle + \sqrt{\frac{1}{2}} | - \rangle.
\end{aligned}
\end{equation}
.. literalinclude:: ../examples/opt_dist/positive/min_error/three_nonorthogonal_states.py :language: python :linenos: :start-after: # along with this program. If not, see <https://www.gnu.org/licenses/>.
The optimal probability with which Bob can distinguish the state he is given unambiguously may be obtained by solving the following semidefinite program (SDP).
\begin{align*}
\textbf{Primal:} \quad & \\
\text{maximize:} \quad & \sum_{i=0}^n p_i \langle M_i,
\rho_i \rangle \\
\text{subject to:} \quad & \sum_{i=0}^{n+1} M_i = \mathbb{I}_{\mathcal{A} \otimes
\mathcal{B}},\\
& \langle M_i \rho_j \rangle = 0 \quad
\forall i \not= j = 1, \ldots, n \\
& M_1, \ldots, M_n \in \text{Pos}(\mathcal{A} \otimes \mathcal{B}).
\end{align*}
As an example, consider the set of three non-mutually-orthogonal states we considered earlier
\begin{equation}
\begin{aligned}
| \phi_1 \rangle = \sqrt{\frac{3}{4}}| + \rangle + \sqrt{\frac{1}{4}} | - \rangle, &\quad
| \phi_2 \rangle = \sqrt{\frac{1}{4}}| + \rangle + \sqrt{\frac{3}{4}} | - \rangle, \\
| \phi_3 \rangle = \sqrt{\frac{1}{2}}| + \rangle + \sqrt{\frac{1}{2}} | - \rangle.
\end{aligned}
\end{equation}
The probability to distinguish amongst these states unambiguously gives a value of zero, while the minimum-error case, as we saw, gave a value of 3/4.
.. literalinclude:: ../examples/opt_dist/positive/unambiguous/three_nonorthogonal_states.py :language: python :linenos: :start-after: # along with this program. If not, see <https://www.gnu.org/licenses/>.
| [tWalgate00] | Walgate, J., Short, A. J., Hardy, L., & Vedral, V "Local distinguishability of multipartite orthogonal quantum states." Physical Review Letters 85.23 (2000): 4972. |