Schrodinger equation for one dimensional square well

stigg

1. The problem statement, all variables and given/known data
the question as well as the hint is shown in the 3 attachments


2. Relevant equations



3. The attempt at a solution
i know how to normalize an equation, however i do not understand what the hint is saying, or how to do these integrals, any guidance would be greatly appreciated

Attached Thumbnails

     

genericusrnme

The hint is just telling you that each Sin(a pi x) is orthogonal to the other Sins
I'll give you my hint - all you need to know is how to integrate Sin^2(x)

Do you know how to do the other two problems?

stigg

im not sure exactly what you mean by that, what is the integral i need to take?

genericusrnme

Well, do you know what it would mean for [itex]\psi(x)[/itex] to be normalized?

stigg

normalizing ψ(x) means to determine if ∫ |Ψ(x)|^2 dx = 1 right?

genericusrnme

That is correct
Do you know what it means if two functions are orthogonal?

stigg

no i do not

genericusrnme

Okay, orthogonality is what the hint is describing
two functions, f and g are orthogonal if [itex]\int f*g\ dx = 0[/itex]
the hint is just restating that each of the Sins are orthogonal to the other ones, that is;
[itex]\int Sin(\pi x)Sin(2 \pi x) dx = 0[/itex] etc

Using this, how do you think you should proceed in determining A?

stigg

do i expand the equation and cancel out the terms with Sin(πx)Sin(2πx) because they will integrate to 0?

genericusrnme

Yes, show me what you get

stigg

does ∫Sin(πx)Sin(3πx)dx=0 and ∫Sin(2πx)Sin(3πx)dx=0 ?

genericusrnme

yes, if you work the integral out yourself you'll find that [itex]\int Sin[n \pi x] Sin[m \pi x] dx = 0[/itex] if [itex]m \ne n[/itex]

stigg

alright so then in that case i will only be dealing with the sin² functions and therefore |Ψ(x)|^2= (1/10a)sin²([itex]\pi[/itex]x/a)+(aA²/a)sin²(2[itex]\pi[/itex]x/a)+(9/5a))sin²(3[itex]\pi[/itex]x/a)

genericusrnme

That is not exactly true, the terms aren't zero on their own, what IS true is that

[itex]\int |Ψ(x)|^2dx=\int (1/10a)sin2(πx/a)+(aA2/a)sin2(2πx/a)+(9/5a))sin2(3πx/a)dx[/itex]

stigg

yes yes youre right i jumped the gun on that, now will the limits of my integral be from -a to a or from -[itex]\infty[/itex] to[itex]\infty[/itex]

genericusrnme

Since this is the particle in a box problem, the potential outside of the box is set to infinity and so we make ψ = 0 everywhere outside, so it doesn't matter where you set the limits of integration (as long as the box is contained in them of course) since we pick up exactly 0 from the outside region.
If you look at the problem statement, you'll see that the box isn't -a<x<a, it's 0<x<a

stigg

ah yes my mistake so using 0 to a as the limits of integration i get that it is equal to (1/10a)+(9/5a)+(2A²/a)