A particle of mass m in the infinite square well (of width a) starts out in the state
\Psi(x, 0)= \begin{cases}A, & 0 \leq x \leq a / 2 \\ 0, & a / 2 \leq x \leq a\end{cases} ,
for some constant A, so it is (at t = 0 ) equally likely to be found at any point in the left half of the well. What is the probability that a measurement of the energy (at some later time would yield the value \pi^{2} \hbar^{2} / 2 m a^{2} ?.
A^{2} \int_{0}^{a / 2} d x=A^{2}(a / 2)=1 \Rightarrow A=\sqrt{\frac{2}{a}} .
From Eq. 2.37,
c_{n}=\int \psi_{n}(x)^{*} f(x) d x . (2.37)
c_{1}=A \sqrt{\frac{2}{a}} \int_{0}^{a / 2} \sin \left(\frac{\pi}{a} x\right) d x=\left.\frac{2}{a}\left[-\frac{a}{\pi} \cos \left(\frac{\pi}{p} x\right)\right]\right|_{0} ^{a / 2}=-\frac{2}{\pi}\left[\cos \left(\frac{\pi}{2}\right)-\cos 0\right]=\frac{2}{\pi} .
P_{1}=\left|c_{1}\right|^{2}= (2/ π)^2 = 0.4053 .