Question

Quantum measurement and qubits.

Answer 1

A qubit is either in the state\[|u\rangle=\cos\left(\frac\pi4-\frac x2\right)|0\rangle+\sin\left(\frac\pi4-\frac x2\right)|1\rangle\]or\[|v\rangle=\cos\left(\frac\pi4+\frac x2\right)|0\rangle+\sin\left(\frac\pi4+\frac x2\right)|1\rangle\]and we want to determine which is the case by measuring it. One of the following two measurements is optimal in terms of the probability of success.

Measurement I:
Measure in the\[|u\rangle=\cos\left(\frac\pi4-\frac x2\right)|0\rangle+\sin\left(\frac\pi4-\frac x2\right)|1\rangle\]\[|u^\perp\rangle=-\sin\left(\frac\pi4-\frac x2\right)|0\rangle+\cos\left(\frac\pi4-\frac x2\right)|1\rangle\]basis and interpret the outcome \(u\) as \(|u\rangle\) and \(u^\perp\) as \(|v\rangle\).

Measurement II:
Measure in the standard basis and interpret \(0\) as \(|u\rangle\) and \(1\) as \(|v\rangle\).

The probability of success is defined as \(\frac12p_u+\frac12p_v\), where\[p_u=\text{Pr[we output }|u\rangle\text{|the qubit was in the state}|u\rangle]\]and\[p_v=\text{Pr[we output }|v\rangle\text{|the qubit was in the state}|v\rangle]\]