Normalization of a wavefunction problem

[hellomister]

Homework Statement

Normalize sin ((n*pi*x)/L) where x is between 0 and L and n is a positive integer

Homework Equations

integral (psi*psi)dx=1
N^2 integral sin ((n*pi*x)/L)dx =1

I don't really understand if this integral is correct, what is the complex conjugate of the wavefunction?

Can i just integrate my wavefunction and say tht is psi*psi?

The Attempt at a Solution

N^2 integral sin ((n*pi*x)/L)dx =1

-N^2 Lcos ((n*pi*x/L))/n*pi=1
I just integrated the wavefunction and then solved for N

 

[Matterwave]

The complex conjugate of a real number or function is just that number/function itself. You should be squaring your function. You can't just integrate your function alone, it has to be multiplied by itself.

 

[hellomister]

Ohh, thanks! I was very confused.
So then I would square my function and then integrate and solve for N?
then my normalized wavefunction would be N times the original function?

 

[Matterwave]

Yes. I don't know why you have a N squared though. There's no reason for N to be squared. N is just some as yet undetermined constant. Squaring it will just confuse you.

 

[vela]

The usual way it's done is you write

$$\psi_n(x) = N \sin{\left(\frac{n\pi x}{L}\right)$$

where N is the yet unknown normalization constant. Then you plug this wavefunction into

$$1=\int_0^L \psi_n^*(x)\psi_n(x) dx$$

and solve for N. Since both $\psi_n^*(x)$ and $\psi_n(x)$ contain N, you get $N^2$ in the equation.

 

[Matterwave]

Ah right, it's been a while...I suppose N^2 would be appropriate. Otherwise, you'd square-root N. (I always use A instead of N which adds to my confusion...)

 

[hellomister]

Sorry about that, i just saw it in my textbook and copied it down, but Vela's explanation really cleared it up! thanks both of you guys. Sorry I am a newbie when it comes to this stuff... and I'm rusty on math.

Also would you guys know anything about guassian integrals? if i was trying to integrate e^(-2ax^2)?