A quantum particle is confined in a one-dimensional space of \(0 \le x \le 2\). Consider a normalized wavefunction of the particle as
\[ \psi(x) = \sqrt{p/5}\,[1 + \cos(\pi x/2)] \sin(\pi x/2). \]
Which of the following options gives the correct value of \(p\)?
IIT JAM - 2026,
Answer: C
Normalisation gives \(p = 4\).
The wavefunction must satisfy \(\int_0^2 |\psi|^2 dx = 1\). Substitute \(u = \pi x/2\), so \(dx = \frac{2}{\pi}du\) and \(u\) runs from 0 to \(\pi\):
\[ \begin{aligned} 1 &= \frac{p}{5} \cdot \frac{2}{\pi} \int_0^{\pi} I(u)\, du \\ I(u) &= (1 + \cos u)^{2} \sin^{2} u \end{aligned} \]
Expanding, \((1 + \cos u)^2 \sin^2 u = \sin^2 u + 2\cos u \sin^2 u + \cos^2 u \sin^2 u\). Over 0 to \(\pi\) these integrate to \(\frac{\pi}{2}\), 0 and \(\frac{\pi}{8}\):
\[ \begin{aligned} 1 &= \frac{p}{5} \cdot \frac{2}{\pi} \cdot \frac{5\pi}{8} \\ &= \frac{p}{4}, \quad p = 4 \end{aligned} \]
The middle term vanishes because \(\sin^3 u\) is the same at both limits. The values 1, 2 and 3 would leave the total probability at a quarter, a half or three quarters.
Please log in or register to report this question.