## Schrodinger equation for one dimensional square well

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

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

## Schrodinger equation for one dimensional square well

Well, do you know what it would mean for $\psi(x)$ to be normalized?

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

 Quote by stigg normalizing ψ(x) means to determine if ∫ |Ψ(x)|^2 dx = 1 right?
That is correct
Do you know what it means if two functions are orthogonal?

 no i do not
 Okay, orthogonality is what the hint is describing two functions, f and g are orthogonal if $\int f*g\ dx = 0$ the hint is just restating that each of the Sins are orthogonal to the other ones, that is; $\int Sin(\pi x)Sin(2 \pi x) dx = 0$ etc Using this, how do you think you should proceed in determining A?
 do i expand the equation and cancel out the terms with Sin(πx)Sin(2πx) because they will integrate to 0?
 Yes, show me what you get
 does ∫Sin(πx)Sin(3πx)dx=0 and ∫Sin(2πx)Sin(3πx)dx=0 ?
 yes, if you work the integral out yourself you'll find that $\int Sin[n \pi x] Sin[m \pi x] dx = 0$ if $m \ne n$
 alright so then in that case i will only be dealing with the sin2 functions and therefore |Ψ(x)|^2= (1/10a)sin2($\pi$x/a)+(aA2/a)sin2(2$\pi$x/a)+(9/5a))sin2(3$\pi$x/a)

 |Ψ(x)|^2= (1/10a)sin2(πx/a)+(aA2/a)sin2(2πx/a)+(9/5a))sin2(3πx/a)
That is not exactly true, the terms aren't zero on their own, what IS true is that

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

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

 Quote by 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 -$\infty$ to$\infty$
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

 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)

 Tags quantum, quantum well, schrodinger equation, wave equation