Semi-infinite potential well

  • #1
StillAnotherDave
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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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Answers and Replies

  • #2
StillAnotherDave
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Any thoughts folks?
 
  • #3
hutchphd
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Sketch the first few eignstates for each case...
 
  • #4
StillAnotherDave
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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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  • #5
PeroK
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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.
 
  • #6
StillAnotherDave
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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.
 
  • #7
PeroK
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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?
 
  • #8
StillAnotherDave
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The general solutions? Yes. That was the first part of the question which I didn't upload:

1584783048128.png

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

For an infinite well:
1584783424427.png
 
  • #9
PeroK
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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.
 
  • #10
StillAnotherDave
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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.
 
  • #11
StillAnotherDave
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So ... ? :rolleyes:
 
  • #12
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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  • #13
StillAnotherDave
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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.
 
  • #14
PeroK
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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.
 
  • #15
StillAnotherDave
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Thanks! I'll try to work this out tomorrow morning and see how I do.

Appreciate the help.
 
  • #16
PeroK
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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.
 
  • #17
StillAnotherDave
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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?
 
  • #18
PeroK
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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}##?
 
  • #19
StillAnotherDave
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If so, comparing E to E' doesn't give the required factor... what have I got wrong?
 
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  • #20
Remember that ##k## is necessarily positive. ##k = \frac{\sqrt{2mE}}{\hbar}##
 
  • #21
PeroK
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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##.
 
  • #22
StillAnotherDave
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Sorry, I corrected.

It's
$$k=\frac{-\pi}{4a}=\frac{3\pi }{4a}$$
 
  • #24
StillAnotherDave
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Yea ... just getting to grips with the latex formatting.
 
  • #25
StillAnotherDave
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Got there in the end... thanks for the help!
 
  • #26
PeroK
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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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  • #27
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!
 
  • #28
PeroK
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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.
 

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