# How do you integrate integrate ψ1(x)?

1. Feb 19, 2010

### maxbye3

1. The problem statement, all variables and given/known data
Normalise Ψ(x,0) (use the fact that both ψ1 and ψ2 are stationary states).
Will this wave function be normalised at any later time, t > 0?

2. Relevant equations
A particle in an infinite square well of width a has as its initial wave function an even mixture of the first two stationary states,
Ψ(x, 0) = A(ψ1(x) + ψ2(x)) .

3. The attempt at a solution
I get the concept of normalisation I would integrate the square of A(ψ1(x) + ψ2(x)) between 0 and a and equate it to 1.
I am confused how to integrate ψ1(x) or ψ2(x) i.e. is it as simple as (ψ1(x)*x)/2?
Perhaps I can say ψ1(x) = the time-independent Schr ̈odinger equation for the infinite square well un=(2/L)^0.5sin(nx/L)

Anyway any help would be deeply appreciated, thanks.

2. Feb 19, 2010

### thebigstar25

I hope i will be helpful (but dont completely count on what im going to say)..

well, the whole idea of normalization is to determine that "A" ,and that can be accomplished through taking the integral of square (A(ψ1(x) + ψ2(x)) between 0 and a and equate it to 1) as you stated .. I dont think you would need to break them , you can do the integration as the following:

integral(from 0 to a) [of] : square(A(ψ1(x) + ψ2(x)) dx

// ;square(A(ψ1(x) + ψ2(x)) = A^2(ψ1(x))^2 + A^2(ψ2(x))^2 + 2A^2 ψ1(x)ψ2(x)

taking the integral(from 0 to a) [of] A^2(ψ1(x))^2 dx = A^2 integral((ψ1(x))^2 dx , since the integral of ((ψ1(x))^2 dx =1 , then the first term is A^2

doing the same thing with the second term you will also get A^2

the third term as can be seen is an integral of both integral ψ1(x)ψ2(x) , this one will give zero (why?) you can verify this by making use of orthognality ..

so finally you will end up with : A^2 +A^2 =1 >> 2A^2=1 >> A^2=1/2 >> A=sqrt(1/2)..

then the normalised function : sqrt(1/2)*(ψ1(x) + ψ2(x))

hopefully , this answers the part of the question which asks to normalize Ψ(x,0) .. :)

3. Feb 19, 2010

### vela

Staff Emeritus
How did you come up with that? What do you get when you square $\psi$?