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 $E_2$?


1 Answer 1


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.


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