Normalisation of a wavefunction for a particle in a box

Box is from 0 to L
Prob Density is |A|^2 sin^2(npix/L)

So you'd get the integral of that from 0 to L is 1

And the answer is A = Root(2/L)

But i can't figure how to get that? =S

Heeelp. I swear i used to understand this...

Reply 1

F1 fanatic

mc_watson87

\int_0^L A^2sin^2(n\pi x/L)dx=1

sin^2(n \pi x/a) = 1/2(1-cos(n \pi x/(2L) )

integrate up and the cos will disappear when you stick in the limits, leaving you with:

A^2[x/2]_0^L=1

from which point it's obvious. :smile:

Reply 2

mc_watson87

Cheers dude! Man guess i'm gunna have to learn sum trig for my QM exam tomorrow. =\

Reply 3

F1 fanatic

mc_watson87

Cheers dude! Man guess i'm gunna have to learn sum trig for my QM exam tomorrow. =\

that's one of the only tricks that you need to use for QM though, that and integrating the cos/sin etc themselves. Depending on the level of the stuff a knowledge of orthogonality and even/odd functions can come in handy too.

Good luck with your exam mate :smile: