# How can one derive the POVM to unambiguously discriminate between $|0\rangle$ and $|+\rangle$?

Page 92 of Nielsen and Chuang describes a POVM that can determine if a given state is either $$|0\rangle$$ or $$|+\rangle$$ with no error, but with some chance of an inconclusive result. The POVM is:

$$E_1 = \frac{\sqrt{2}}{1+\sqrt{2}} |1\rangle\langle 1|$$ $$E_2 = \frac{\sqrt{2}}{1+\sqrt{2}} |-\rangle\langle -|$$ $$E_3 = I - E_1 - E_2$$

I see how $$E_1$$ detects $$|+\rangle$$ because it is projecting orthogonal to $$|0\rangle$$, $$E_2$$ works similarly, and that $$E_3$$ is the inconclusive case that occurs the rest of the time.

How can one derive the constant in front of $$E_1$$ and $$E_2$$?

• Dec 22, 2023 at 6:44

The main idea is to make the constants in front of $$E_1$$ and $$E_2$$ as large as possible so that the probability of obtaining an unambiguous result is maximized while still ensuring that $$E_3$$ is positive semidefinite to make your POVM a valid one.

If you set the prefactors in front of $$E_1$$ and $$E_2$$ to be positive constants $$a$$ and $$b$$, you have that

$$E_3 = I - aE_1 - bE_2 = \begin{pmatrix}1 -\frac{b}{2} & \frac{b}{2}\\ \frac{b}{2} & 1 - a -\frac{b}{2} \end{pmatrix}$$

A $$2\times 2$$ positive semidefinite matrix has trace and determinant both nonnegative. The determinant $$D$$ is

$$D = 1 - a - b - \frac{ab}{2}$$

By symmetry, we can choose $$a = b$$ and solving for $$D\geq 0$$ yields the quadratic

$$(a - 2)^2 \geq 2$$

Taking the negative root (otherwise the trace becomes negative), we have $$a \leq 2 - \sqrt{2}$$ which is equal to the constants you see in Nielsen and Chuang.