# Semi-infinite potential well

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?

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StillAnotherDave
Any thoughts folks?

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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:

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

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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.

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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:

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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 ... ?

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}##...

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PeroK
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.

etotheipi
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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.

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

Appreciate the help.

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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.

etotheipi
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?

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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?

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

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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##.

etotheipi
StillAnotherDave
Sorry, I corrected.

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

etotheipi
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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!

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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.

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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!