Integration in quantum physics?

Question

i m trying to integrate from 0 to a in respect to x the equation I lambda I^2 *sin^2(n*pi*x/a) given that this is equal to 1 i need to find I lambda I^2. i just used the fact that integration of sin^2(x) = x/2 - 1/4sin(x)

thanks for any help

Answer 1

Well you can pull out the lambda^2 since it's a constant, and then just integrate the sin^2 function over x.  http://www.integrals.com can do most integrals for you if you can't find it in a table.  I don't have my table at home, sorry.

Answer 2

this sounds like you are normalizing a wave function

let's say you have:

L^2 Int[sin^2[n pi x/a] dx] = L^2[x/2-(a Sin[(2 n pi x)/a])/(4 n pi)]

substituting your limits:

L^2 a/4 [2-sin(2 n pi)/n pi]=1

assuming you have the quantum condition that n is an integer, the sin term is zero (sin 2 n pi) is zero for all integer values of n, you then have:

a L^2/2=1

or your normalization condition is

L=sqrt[2/a]