Question

Psi(x)= Asin(2πx), for what value of A is the wave function normalised

Answer 1

$1=\int \Psi \Psi^* dx=A^2\int _0^1sin^2(2\pi x)dx=\frac{A^2}{2}\int_0^1(1-cos(4\pi x)dx=$

$=\frac{A^2}{2}(x-\frac{1}{4\pi}sin(4\pi x))\vert_{x=0}^{x=1}=\frac{A^2}{2},\implies$

$A=\sqrt2,$

$\Psi=\sqrt2 sin(2\pi x)$