Comparing Energy Levels in Semi-Infinite and Infinite Potential Wells

[StillAnotherDave]

Homework Statement

How do I compare energy values between an infinite and semi-finite potential well?

Relevant Equations

E=h^2k^2/2m

Hello folks,

A bit stumped with the following question:

Consider a potential well with an infinite wall at x=o and a finite wall at x=a. The height at x=a is such that U₀=2E₁' where E₁' is the energy of the particle's n=1 state in this semi-infinite well.

How can one show that E₁' is lower that E₁ (the energy at n=1 for an infinite well) by a factor of 9/16?

 

[StillAnotherDave]

Any thoughts folks?

 

[hutchphd]

Sketch the first few eignstates for each case...

 

[StillAnotherDave]

As a starting point, the n=1 state energy E₁ of a particle with mass m in an infinite well is given by:

$$^{E_{1}}=\frac{h^2}{8mL^2}$$​

So the sticking point is how to calculate an energy value for n=1 E₁' when the barrier at x=a is U₀=2E₁', such that:

$$\frac{^{E_{1}'}}{E_{1}}=\frac{9}{16}$$

 

[PeroK]

Any thoughts folks?

I would solve the SDE! I don't know a quicker way. There may be one, of course.

 

[StillAnotherDave]

I would solve the SDE! I don't know a quicker way. There may be one, of course.

Please explain if you can. This is a year one physics question - i.e. first exposure to this material for me, so I'm having difficulty connecting the dots.

 

[PeroK]

Please explain if you can. This is a year one physics question - i.e. first exposure to this material for me, so I'm having difficulty connecting the dots.

Are you familiar with the solutions to the infinite and finite square well problems?

 

[StillAnotherDave]

The general solutions? Yes. That was the first part of the question which I didn't upload:

For the semi-finite well:
Ψ(x)=Asin(kx) [0<x<a]
Ψ(x)=De^−κx [x>a]

For an infinite well:

 

[PeroK]

This is precisely why you are supposed to include all the relevant information about a problem. And, why you had to bump for any ideas.

 

[StillAnotherDave]

It seems so. In my defense, the question as to what exactly is or is not relevant to making the question intelligible vs providing unhelpful background isn't always straightforward.

 

[StillAnotherDave]

So ... ?

 

[etotheipi]

For the semi-finite well:
Ψ(x)=Asin(kx) [0<x<a]
Ψ(x)=De^−κx [x>a]

Once you get the relation in part f), you should be able to substitute the new bit of information $U_0 = 2E_{1}^ {'}$ back into the previous parts of the question to get something useful out for $k$. I'm trying to be vague because I think you've nearly got it.

Then you just need to find out how to get $E$ from $k$, for the infinite and finite cases! It might help to know that $k$ is a wavenumber, i.e. $k = \frac{2\pi}{\lambda}$...

 

[StillAnotherDave]

Once you get the relation in part f), you should be able to substitute the new bit of information $U_0 = 2E_{1}^ {'}$ back into the previous parts of the question to get something useful out for $k$. I'm trying to be vague because I think you've nearly got it.

Then you just need to find out how to get $E$ from $k$, for the infinite and finite cases! It might help to know that $k$ is a wavenumber, i.e. $k = \frac{2\pi}{\lambda}$...

Hmm ... many thanks! I'll reflect on that and try to figure it out. Watch this space.

 

[PeroK]

Hmm ... many thanks! I'll reflect on that and try to figure it out. Watch this space.

I think I can make sense of this question. First, we have the ground state energy of the infinite square well:
$$E = \frac{\pi^2 \hbar^2}{2ma^2}$$
Next, we have a half-infinite half-finite well. The ground state energy, $E'$, for this satisfies the equations:
$$k^2\hbar^2 = 2mE' \ \text{and} \ \kappa^2 \hbar^2 = 2m(U_0 - E')$$
And, you are given that $U_0 = 2E'$.

The first step, therefore, is find an equation for $k$ and $\kappa$ from this.

Next, you are given another equation for $k$ and $\kappa$:
$$\tan (ka) = -\frac{k}{\kappa}$$

This should allow you to find $ka$.

Finally, you may be able to express $E'$ in terms of $a$ and hence compare it with the ground state energy of the infinite square well.

That's what I think you are expected to do.

 

[StillAnotherDave]

Thanks! I'll try to work this out tomorrow morning and see how I do.

Appreciate the help.

 

[PeroK]

Thanks! I'll try to work this out tomorrow morning and see how I do.

Appreciate the help.

I just checked it all comes out. I assumed initially that $k$ had to do with the infinite well and $\kappa$ with the finite well. But that made no sense. Then I realized that both $k$ and $\kappa$ related to the finite well and the $E$ in those equations could more sensibly have been denoted $E'$. Leaving $E$ for the energy of the infinite well.

 

[StillAnotherDave]

This is what I've got so far:

$$tan(ka)=-1$$

So,
$$ka=\frac{-\pi }{4}$$
and
$$k=\frac{-\pi }{4a}$$

If you substitute this into:
$$E^{'}=\frac{k^{2}h^{2}}{2m}$$

You get:

$$E^{'}=\frac{\pi^{2}h^{2}}{16ma^{2}}$$

That's as far as I've got. Does that look right?

 

[PeroK]

This is what I've got so far:

$$tan(ka)=-1$$

So,
$$ka=\frac{\pi }{4}$$

That's as far as I've got. Does that look right?

That is definitely not right!

What is $\tan \frac{\pi}{4}$?

 

[StillAnotherDave]

If so, comparing E to E' doesn't give the required factor... what have I got wrong?

 

[etotheipi]

Remember that $k$ is necessarily positive. $k = \frac{\sqrt{2mE}}{\hbar}$

 

[PeroK]

If so, comparing E to E' does give the required factor... what have I got wrong?

The factor was supposed to be $9/16$. Not $1/16$.

 

[StillAnotherDave]

Sorry, I corrected.

It's
$$k=\frac{-\pi}{4a}=\frac{3\pi }{4a}$$

 

[PeroK]

Sorry, I corrected.

It's
$$k=\frac{-\pi}{4}=\frac{3\pi }{4}$$

You mean?
$$ka=\frac{3\pi }{4}$$

 

[StillAnotherDave]

Yea ... just getting to grips with the latex formatting.

 

[StillAnotherDave]

Got there in the end... thanks for the help!

 

[PeroK]

Sorry, I corrected.

It's
$$k=\frac{-\pi}{4a}=\frac{3\pi }{4a}$$

As an aside, a mathematical subtlety. If you have:
$$\tan ka = -1$$
And you know that $ka > 0$, then:
$$ka \ne \arctan(-1)$$
The function $\arctan$ has a range of $(-\frac \pi 2, \frac \pi 2)$. And $ka$ cannot be in that range. You can, however, simply look for the appropriate value of $ka$ based on the work you did on the problem to see that $ka = \frac{3\pi }{4}$. In other words, it was important that you already knew that $ka$ had approximately this value (to the nearest $\pi$).

Note that if you were looking for a different energy level, you may well have had to choose $ka = \frac{7\pi }{4}$, for example.

 

[etotheipi]

As an aside, a mathematical subtlety. If you have:
$$\tan ka = -1$$
And you know that $ka > 0$, then:
$$ka \ne \arctan(-1)$$
The function $\arctan$ is defined on $(-\frac \pi 2, \frac \pi 2)$. And $ka$ cannot be in that domain. You can, however, simply look for the appropriate value of $ka$ based on the work you did on the problem to see that $ka = \frac{3\pi }{4}$. In other words, it was important that you already knew that $ka$ had approximately this value (to the nearest $2\pi$).

Note that if you were looking for a different energy level, you may well have had to choose $ka = \frac{11\pi }{4}$.

I suppose $ka = \arctan(-1) + n\pi$ where $n \in \mathbb{Z}$ might be a good way of writing it!

 

[PeroK]

I suppose $ka = \arctan(-1) + n\pi$ where $n \in \mathbb{Z}$ might be a good way of writing it!

That's how to write it formulaically, yes.