= Solution
The <Born rule> uses the squared <modulus> of the <wavefunction>, not its real part. Write $\psi=A[\sin x+i(1-\cos x)]$. Then
$$
|\psi|^2=|A|^2\bigl[\sin^2x+(1-\cos x)^2\bigr]
=2|A|^2(1-\cos x).
$$
<Wavefunction normalization> requires
$$
1=2|A|^2\int_0^{2\pi}(1-\cos x)\,dx=4\pi|A|^2,
\qquad |A|^2=\frac1{4\pi}.
$$
Integrating the <probability density> over the required interval gives the dimensionless <probability>
$$
\boxed{P=\frac1{2\pi}\int_0^{\pi/2}(1-\cos x)\,dx
=\frac{\pi/2-1}{2\pi}
=\frac14-\frac1{2\pi}\simeq0.09085.}
$$
Thus approximately \b[$9.08\%$] of <position measurements> fall in this interval.
Back to article page
