Schrodinger equation for one dimensional square well

[stigg]

Homework Statement

the question as well as the hint is shown in the 3 attachments

Homework Equations

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

 

[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 \psi(x) to be normalized?

 

[stigg]

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

 

[genericusrnme]

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?

 

[stigg]

no i do not

 

[genericusrnme]

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?

 

[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 \int Sin[n \pi x] Sin[m \pi x] dx = 0 if m \ne n

 

[stigg]

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

 

[genericusrnme]

|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

 

[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

 

[genericusrnme]

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

 

[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)

 

[genericusrnme]

Okay, so what are we going to do with this?
What do we need to set this equal to and what do we then need to solve for?

 

[stigg]

to normalize it it has to be set equal to 1 and we need to find A, also to fix my integral above, i did it out wrong and it is actualy equal to A^2 +(19/20) so that means if i set it equal to 1 that A = \sqrt{1/20}

 

[stigg]

with A solved for and pulgged back in the function is now normalized, so the next part of the question asks me to find the possible results of measures of the energy and what are the respective probabilities of obtaining each result

 

[genericusrnme]

Do you know how to do that?
Do you know what the energy eigenstates are for the particle in a box?

 

[stigg]

i do not, i had trouble with finding eignenstates in a previous problem as well, my notes are too vague to be of any use.

 

[genericusrnme]

Well what better time to get some practice in at deriving the energy eigenstates for the particle in a box problem

All we need to do is solve the energy eigenvalue equation;
H \psi = E \psi

With the Hamiltonian as

H_{inside} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}

And we set \psi = 0 everywhere outside of the box, since the potential energy is taken to be infinite there, this gives us the boundary conditions which give us the quantisation of energy!

This gives us the simple second order ODE for \psi

- \frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = E \psi (x)

Which can be rewritten as

\frac{d^2 \psi(x)}{dx^2} = -\frac{2\ m\ E}{\hbar^2} \psi (x)

The allowed solutions to this give you the energy eigenstates and eigenvalues.
You should then be able to rewrite your \Psi(x) as a sum of energy eigenstates \psi_n (x)


Once you've done this, do you know how you would go about finding out what the probabilities for measuring each energy eigenvalue are?

 

[stigg]

i am taking differential equations at the moment so i am not overly familiar with solving a second order DE such as this

 

[genericusrnme]

Oh, well I'll just go ahead and say it, it's pretty common knowledge, the solutions to
\frac{d^2 f(x)}{dx^2} = -a^2 f(x)
are simply

f(x)=A\ Sin(a\ x)+ B\ Cos(a \ x)

 

[stigg]

so in the case of my function, -a² = -2mE/\hbar²

so i would get ψ(x)= Asin(x\sqrt{}2mE/\hbar²) + Bcos(x\sqrt{}2mE/\hbar^2[/SUP)⁾

 

[genericusrnme]

\Psi(x) = A\ Sin(\sqrt{\frac{2mE}{\hbar^2}}x) + B\ Cos( \sqrt{\frac{2mE}{\hbar^2}}x)

Yes, you are correct, now you just need to apply the boundary conditions, that is \psi(0)=\psi(a)=0

I'll give you a hint here, you should be trying to write E in terms of a natural number n

 

[stigg]

wheni apply the boundary conditions i find that B=0 and therefore am left with Psi(x) = A\ Sin(\sqrt{\frac{2mE}{\hbar^2}}x), by a natural number do you mean e? because i had seen before that Psi(x) = A\ Sin(\sqrt{\frac{2mE}{\hbar^2}}x) can turn into some function Psi(x) = Ae^someting but i do not know if that applies here

 

[genericusrnme]

A natural number is just an integer which is greater than or equal to 1
And yes, you're on the right tracks setting B=0, you just need to work out what constraints you need to place on E now