Answer to Question #283840 in Quantum Mechanics for Kuttus

Question

A particle of mass m is confined between 0<=x<=l. If px be the momentum, then find px in the ground state

Answer 1

$\psi(x)=\sqrt\frac{{2}}{L}sin\frac{n\pi x}{L}\\n=1 \\grounstate \\\psi(x)=\sqrt\frac{{2}}{L}sin\frac{\pi x}{L}$

$<p_x>=\frac{\smallint_{0}^{L}\psi^* p_x \psi dx}{\smallint_{0}^L\psi\psi^* dx}$

$<p_x>=\frac{{\smallint_{0}^{L}}\sqrt\frac{2}{L}sin\frac{\pi x}{L}\frac{d}{dx}(\sqrt\frac{2}{L}sin\frac{\pi x}{L})dx}{\smallint_{0}^{L}(\sqrt\frac{2}{L}sin\frac{\pi x}{L}\sqrt\frac{{2}}{L}sin\frac{\pi x}{L})dx}$

$<p_x>=\smallint_{0}^{L}\frac{2\pi}{L^2}sin\frac{\pi x}{L}cos\frac{\pi x}{L}dx$

Where

$\smallint_{0}^{L}\sqrt\frac{{2}}{L}sin\frac{\pi x}{L}\sqrt\frac{{2}}{L}sin\frac{\pi x}{L}dx=1$

$<p_x>=\smallint_{0}^{L}\frac{\pi}{L^2}sin\frac{2\pi x}{L}$

$<p_x>=\frac{\pi}{L^2}\smallint_{0}^{L}sin\frac{2\pi x}{L}$

$<p_x>=\frac{\pi}{L^2}(\frac{L}{2\pi})|(\frac{cos2\pi x}{L})|_{0}^{L}$

$<p_x>=\frac{\pi}{L^2}(\frac{L}{2\pi})(cos0-cos{2\pi})$

$<p_x>=\frac{\pi}{L^2}\times\frac{L}{2\pi}(1-1)=0$

$<p_x>=0$