Infinite Square Well - Quantum Mechanics

[benjy1]

SPECIFICALLY SEE POST 8 AND AFTER PLEASE


Hi so that I can get the help for the specific problem I am working on I will set the question up and include all the steps that I can get and work out. The end question will be about quantized energy levels. This is for a maths module.

I am working on infinite wells and particularly on a question that will tend to the energy levels of an infinite well.

v(x) = infinity x<0 and x>2a , V for a<x<2a and 0 for 0<x<a

and for the energy E>V i can work out the two wave functions

Psi=Asin(kx) for 0<x<a where k^2=2mE/h^2 (i don't know how to put h bar)


and Psi= Bsin(K(2a-x)) for a<x<2a where K^2=2m(E-V)/h^2

I use the boundary condition at x=a to find the relation between K and k.

Its that the derivative and the wave functions are equal.

I have

1. Asin(ka)=Bsin(Ka)
2.Akcos(ka)=-BKcos(Ka)

So from this I can get

3. Btan(Ka)/K=-Btan(ka)/k

So now is where I am stuck. I need to solve this where V tends to zero, so this means it will be like an infinite well.
So if V tends to 0 then K tends to k.
so how do I solve 3? I can't cancel the B because B=0 might be a solution.

So what I am thinking is that it is either B=0 or tan(ka)=0

so for tan(ka)=0 we have sin(ka)=0 so k=npi/a

For B=0 then from 1 and 2 we get different solutions. I don't want A=0 so that is ruled out.

I need to find the quantized energy levels. These must be the same as for the infinite well case i.e E=(h*pi*n)^2/8*m*a^2.

To do this k has to be equal to n*pi/2a.

So any advice on what to do further please? Thanks

 

[benjy1]

55 views and no1 can help or suggest something? PLEASE let me know if you want me to scan the question and my work I can do that. Thats the only way to include all my work done on here.

 

[vela]

Hi so that I can get the help for the specific problem I am working on I will set the question up and include all the steps that I can get and work out. The end question will be about quantized energy levels. This is for a maths module.

I am working on infinite wells and particularly on a question that will tend to the energy levels of an infinite well.

v(x) = infinity x<0 and x>2a , V for a<x<2a and 0 for 0<x<a

and for the energy E>V i can work out the two wave functions

Psi=Asin(kx) for 0<x<a where k^2=2mE/h^2 (i don't know how to put h bar) and Psi= Bsin(K(2a-x)) for a<x<2a where K^2=2m(E-V)/h^2

I use the boundary condition at x=a to find the relation between K and k.

Its that the derivative and the wave functions are equal.

I have

1. Asin(ka)=Bsin(Ka)
2.Akcos(ka)=-BKcos(Ka)

So from this I can get

3. Btan(Ka)/K=-Btan(ka)/k

So now is where I am stuck. I need to solve this where V tends to zero, so this means it will be like an infinite well.
So if V tends to 0 then K tends to k.
so how do I solve 3? I can't cancel the B because B=0 might be a solution.

I didn't work this out on paper, but I'm pretty sure if you let B=0, the entire wavefunction will be 0, which isn't a solution you're interested in. So you can go ahead and assume B is non-zero and cancel it.

So what I am thinking is that it is either B=0 or tan(ka)=0

so for tan(ka)=0 we have sin(ka)=0 so k=npi/a

For B=0 then from 1 and 2 we get different solutions. I don't want A=0 so that is ruled out.

I need to find the quantized energy levels. These must be the same as for the infinite well case i.e E=(h*pi*n)^2/8*m*a^2.

To do this k has to be equal to n*pi/2a.

So any advice on what to do further please? Thanks

 

[benjy1]

IF B=0 then its either A=0 or sin(ka)=0 or cos(ka)=0 as I see it from equations 1 and two..

I will probably scan my work and if I can figure out how to post it on here I will..

 

[vela]

IF B=0 then its either A=0 or sin(ka)=0 or cos(ka)=0 as I see it from equations 1 and two.

If B=0, then it's either A=0 or sin(ka)=0 and cos(ka)=0. You can't satisfy both equations at the same time, so A has to vanish.

 

[benjy1]

If B=0, then it's either A=0 or sin(ka)=0 and cos(ka)=0. You can't satisfy both equations at the same time, so A has to vanish.

Yes you are right there. But then I have the case that tan(ka)=-tan(Ka) and we have that K tends to k. so then we solve sin(ka)=0 which is k=n*pi/a.

But this does not satisfy the energy levels of the infinite well which is

E=((h*pi)^2/8ma^2)*n^2

 

[vela]

If you let k=K, you have

A sin ka = B sin ka
A cos ka = -B cos ka

You know what the solutions are supposed to be, so figure out how they manage to satisfy both conditions. You should find you're overlooking half of the solutions.

 

[benjy1]

If you let k=K, you have

A sin ka = B sin ka
A cos ka = -B cos ka

You know what the solutions are supposed to be, so figure out how they manage to satisfy both conditions. You should find you're overlooking half of the solutions.

I am not sure that I know exactly what you mean but this is what i think.

Do the 2 equations mean that sin and cos have to be zero as u said. So what if i have k=npi/2a for even for sin and k=npi/2a for odd for the cos?

Im not sure what u mean I am missing half of the solutions.

Thank you for your help btw i forgot to say

 

[vela]

In post 6, you set sin ka = 0. That means the wave function is equal to 0 at x=a, so the condition sin ka = 0 corresponds to the solutions with a node at x=a. You should know, however, that there is another set of solutions where x=a is an antinode. You were missing those solutions. As you've guessed, those solutions arise when cos ka = 0.

 

[benjy1]

In post 6, you set sin ka = 0. That means the wave function is equal to 0 at x=a, so the condition sin ka = 0 corresponds to the solutions with a node at x=a. You should know, however, that there is another set of solutions where x=a is an antinode. You were missing those solutions. As you've guessed, those solutions arise when cos ka = 0.

I haven't hear of node and antinode before so although I understand that we have two sets of solutions I'm not sure if that is sufficient to just say that those are the solutions. Is my reasoning correct to write down that k=n*pi/2a. Is it just a result to say that there is an antinode,and how do we know that it is for cos ka=0?(other than the reasoning I had)


The question has an extension part which I was stuck on because I am stuck on this part. If you have any suggestions on how I should tackle the next step I would appreciate it as I do not know.

The next step is

Investigate possible energy levels if c^2=2ma^2*V/h^2 where c=1 and the more general case c>0. I have to use MAPLE to do the graphs,thats not the part I have difficulty with as I know how to do that, what I can't see is what graphs I need ie the relations between K and k.

Here K does not equal to k. I think the idea is to get two function or a function involvin K and k and plot it to find the energy levels but I don't know how to find the relations for K and k.

Thank you very much I am now going to bed so I will reply as soon as I wake up.

 

[vela]

I haven't hear of node and antinode before so although I understand that we have two sets of solutions I'm not sure if that is sufficient to just say that those are the solutions. Is my reasoning correct to write down that k=n*pi/2a. Is it just a result to say that there is an antinode,and how do we know that it is for cos ka=0?(other than the reasoning I had)

Well, even if you're not already familiar with the infinite square well solutions, you should be able to deduce how to get all of the solutions just from the math. I'm not clear on how you deduced that cos ka=0. As long as you can explain how you know sin ka=0 and cos ka=0 leads to solutions, that's all you need to do. (And, of course, say what those solutions are.)

 

[benjy1]

Well, even if you're not already familiar with the infinite square well solutions, you should be able to deduce how to get all of the solutions just from the math. I'm not clear on how you deduced that cos ka=0. As long as you can explain how you know sin ka=0 and cos ka=0 leads to solutions, that's all you need to do. (And, of course, say what those solutions are.)

Thats the problem I don't know though. I am not sure why coska=0. I just used the equation in your post 7. Thats why I asked what if that is enough of a reasoning or is there a way to conclude that cos ka=0?

 

[vela]

Thats the problem I don't know though. I am not sure why coska=0. I just used the equation in your post 7. Thats why I asked what if that is enough of a reasoning or is there a way to conclude that cos ka=0?

The first equation can be rewritten as

$$(A-B)\sin ka = 0$$

For it to hold, you have to have A-B=0 or sin ka=0. Consider the two cases separately. Using the second equation, what can you conclude about A, B, and cos ka in each case?

If you're familiar with linear algebra, you could also take this alternate approach. You can express the two equations as the matrix equation

$$\begin{bmatrix}\sin ka & -\sin ka \\ \cos ka & \cos ka\end{bmatrix}\begin{bmatrix} A \\ B\end{bmatrix} = 0$$

For this equation to have a non-trivial solution, the determinant of the matrix must vanish.

 

[benjy1]

The first equation can be rewritten as

$$(A-B)\sin ka = 0$$

For it to hold, you have to have A-B=0 or sin ka=0. Consider the two cases separately. Using the second equation, what can you conclude about A, B, and cos ka in each case?

If you're familiar with linear algebra, you could also take this alternate approach. You can express the two equations as the matrix equation

$$\begin{bmatrix}\sin ka & -\sin ka \\ \cos ka & \cos ka\end{bmatrix}\begin{bmatrix} A \\ B\end{bmatrix} = 0$$

For this equation to have a non-trivial solution, the determinant of the matrix must vanish.

Ok. So if A=B then from the second equation we have that A=B=0 so we don't want that so we must have coska=0. So this gives the solutions we want.
I also worked out from the determinant and we get sinka*coska=0 so this is what we want. Thank you very much

Does anyone know about the extension part of the question??

Investigate possible energy levels if c^2=2ma^2*V/h^2 where c=1 and the more general case c>0. I have to use MAPLE to do the graphs,thats not the part I have difficulty with as I know how to do that, what I can't see is what graphs I need ie the relations between K and k.

Here K does not equal to k. I think the idea is to get two function or a function involvin K and k and plot it to find the energy levels but I don't know how to find the relations for K and k.