Quantum state discrimination

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In quantum information science, quantum state discrimination refers to the task of inferring the quantum state that produced the observed measurement probabilities.

More precisely, in its standard formulation, the problem involves performing some POVM ${\displaystyle (E_{i})_{i}}$ on a given unknown state ${\displaystyle \rho }$, under the promise that the state received is an element of a collection of states ${\displaystyle \{\sigma _{i}\}_{i}}$, with ${\displaystyle \sigma _{i}}$ occurring with probability ${\displaystyle p_{i}}$, that is, ${\displaystyle \rho =\sum _{i}p_{i}\sigma _{i}}$. The task is then to find the probability of the POVM ${\displaystyle (E_{i})_{i}}$ correcting guessing which was state was received. Since the probability of the POVM returning the ${\displaystyle i}$-th outcome when the given state was ${\displaystyle \sigma _{j}}$ has the form ${\displaystyle {\text{Prob}}(i|j)=\operatorname {tr} (E_{i}\sigma _{j})}$, it follows that the probability of successfully determining the correct state is ${\displaystyle P_{\rm {success}}=\sum _{i}p_{i}\operatorname {tr} (\sigma _{i}E_{i})}$.^[1]

Helstrom Measurement[]

The discrimination of two states can be solved optimally using the Helstrom measurement.^[2] With two states ${\displaystyle \{\sigma _{0},\sigma _{1}\}}$ comes two probabilities ${\displaystyle \{p_{0},p_{1}\}}$ and POVMs ${\displaystyle \{E_{0},E_{1}\}}$. Since ${\displaystyle \sum _{i}E_{i}=I}$ for all POVMs, ${\displaystyle E_{1}=I-E_{0}}$. So the probability of success is:

${\displaystyle P_{success}=p_{0}\operatorname {tr} (\sigma _{0}E_{0})+p_{1}\operatorname {tr} (\sigma _{1}E_{1})=p_{0}\operatorname {tr} (\sigma _{0}E_{0})+p_{1}\operatorname {tr} (\sigma _{1}I-\sigma _{1}E_{0})=p_{0}-\operatorname {tr} [(p_{0}\sigma _{0}-p_{1}\sigma _{1})E_{0}]}$

To maximize the probability of success, the trace needs to be minimized. That's accomplished when ${\displaystyle E_{0}}$ is a projector on the positive eigenspace of ${\displaystyle p_{0}\sigma _{0}-p_{1}\sigma _{1}}$.^[2]

Pretty Good Measurement[]

For distinguishing more than two states, the Pretty Good Measurement (PGM), also known as the square root measurement is not optimal, but it does pretty well.^[3] In PGM, ${\displaystyle \{E_{i}\}=\{S^{-{\frac {1}{2}}}p_{i}\sigma _{i}S^{-{\frac {1}{2}}}\}}$ [2] where ${\displaystyle S=\sum _{i}p_{i}\sigma _{i}}$. This makes ${\displaystyle P_{success}=\sum _{i}p_{i}\operatorname {tr} (\sigma _{i}\{S^{-{\frac {1}{2}}}p_{i}\sigma _{i}S^{-{\frac {1}{2}}}\})}$.^[3]

For example, consider ${\displaystyle \sigma _{0}=|+\rangle \langle +|}$, ${\displaystyle \sigma _{1}=|-\rangle \langle -|}$, ${\displaystyle \sigma _{2}={\frac {1}{2}}(|+\rangle \langle +|+|-\rangle \langle -|}$ where probabilities are ${\displaystyle p_{0}={\frac {1}{6}}}$, ${\displaystyle p_{1}={\frac {1}{3}}}$ and ${\displaystyle p_{2}={\frac {1}{2}}}$. This makes ${\displaystyle S={\frac {1}{6}}\sigma _{0}+{\frac {1}{3}}\sigma _{1}+{\frac {1}{2}}\sigma _{2}={\begin{bmatrix}{\frac {1}{12}}&{\frac {1}{12}}\\{\frac {1}{12}}&{\frac {1}{12}}\end{bmatrix}}+{\begin{bmatrix}{\frac {1}{6}}&-{\frac {1}{6}}\\-{\frac {1}{6}}&{\frac {1}{6}}\end{bmatrix}}+{\begin{bmatrix}{\frac {1}{4}}&0\\0&{\frac {1}{4}}\end{bmatrix}}={\begin{bmatrix}{\frac {1}{2}}&0\\0&{\frac {1}{2}}\end{bmatrix}}}$ and ${\displaystyle S^{-{\frac {1}{2}}}={\begin{bmatrix}{\sqrt {2}}&0\\0&{\sqrt {2}}\end{bmatrix}}}$.

${\displaystyle E_{0}={\begin{bmatrix}{\sqrt {2}}&0\\0&{\sqrt {2}}\end{bmatrix}}{\begin{bmatrix}{\frac {1}{12}}&{\frac {1}{12}}\\{\frac {1}{12}}&{\frac {1}{12}}\end{bmatrix}}{\begin{bmatrix}{\sqrt {2}}&0\\0&{\sqrt {2}}\end{bmatrix}}={\begin{bmatrix}{\frac {3}{8}}&1\\{\frac {3}{8}}&{\frac {3}{8}}\end{bmatrix}}]}$

${\displaystyle E_{1}={\begin{bmatrix}{\sqrt {2}}&0\\0&{\sqrt {2}}\end{bmatrix}}{\begin{bmatrix}{\frac {1}{6}}&-{\frac {1}{6}}\\-{\frac {1}{6}}&{\frac {1}{6}}\end{bmatrix}}{\begin{bmatrix}{\sqrt {2}}&0\\0&{\sqrt {2}}\end{bmatrix}}={\begin{bmatrix}{\frac {1}{3}}&-{\frac {1}{3}}\\-{\frac {1}{3}}&{\frac {1}{3}}\end{bmatrix}}}$

${\displaystyle E_{2}={\begin{bmatrix}{\sqrt {2}}&0\\0&{\sqrt {2}}\end{bmatrix}}{\begin{bmatrix}{\frac {1}{2}}&0\\0&{\frac {1}{2}}\end{bmatrix}}{\begin{bmatrix}{\sqrt {2}}&0\\0&{\sqrt {2}}\end{bmatrix}}={\begin{bmatrix}{\frac {1}{2}}&0\\0&{\frac {1}{2}}\end{bmatrix}}}$.

With these projectors, ${\displaystyle P_{success}={\frac {1}{6}}\operatorname {tr} [{\begin{bmatrix}{\frac {1}{2}}&{\frac {1}{2}}\\{\frac {1}{2}}&{\frac {1}{2}}\end{bmatrix}}{\begin{bmatrix}{\frac {3}{8}}&1\\{\frac {3}{8}}&{\frac {3}{8}}\end{bmatrix}}]+{\frac {1}{3}}\operatorname {tr} [{\begin{bmatrix}{\frac {1}{2}}&-{\frac {1}{2}}\\-{\frac {1}{2}}&{\frac {1}{2}}\end{bmatrix}}{\begin{bmatrix}{\frac {1}{3}}&-{\frac {1}{3}}\\-{\frac {1}{3}}&{\frac {1}{3}}\end{bmatrix}}]+{\frac {1}{2}}\operatorname {tr} [{\begin{bmatrix}{\frac {1}{2}}&0\\0&{\frac {1}{2}}\end{bmatrix}}{\begin{bmatrix}{\frac {1}{2}}&0\\0&{\frac {1}{2}}\end{bmatrix}}]=0.65}$

References[]

External links[]

• Interactive demonstration about quantum state discrimination