How do you integrate integrate ψ1(x)?

[maxbye3]

Homework Statement

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?

Homework 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)) .

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.

 

[thebigstar25]

I hope i will be helpful (but don't completely count on what I am 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 don't 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) .. :)

 

[vela]

I am confused how to integrate ψ1(x) or ψ2(x) i.e. is it as simple as (ψ1(x)*x)/2?

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