Question #174223

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


1
Expert's answer
2021-03-23T11:16:08-0400

1=ΨΨdx=A201sin2(2πx)dx=A2201(1cos(4πx)dx=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=

=A22(x14πsin(4πx))x=0x=1=A22,    =\frac{A^2}{2}(x-\frac{1}{4\pi}sin(4\pi x))\vert_{x=0}^{x=1}=\frac{A^2}{2},\implies

A=2,A=\sqrt2,

Ψ=2sin(2πx)\Psi=\sqrt2 sin(2\pi x)


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS