# Comparing Energy Levels in Semi-Infinite and Infinite Potential Wells

• StillAnotherDave
In summary: 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
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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Any thoughts folks?

Sketch the first few eignstates for each case...

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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StillAnotherDave said:
Any thoughts folks?
I would solve the SDE! I don't know a quicker way. There may be one, of course.

PeroK said:
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.

StillAnotherDave said:
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?

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:

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.

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.

So ... ?

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

Appreciate the help.

StillAnotherDave said:
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
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?

StillAnotherDave said:
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}##?

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}##

StillAnotherDave said:
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
Sorry, I corrected.

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

etotheipi
StillAnotherDave said:
Sorry, I corrected.

It's
$$k=\frac{-\pi}{4}=\frac{3\pi }{4}$$
You mean?
$$ka=\frac{3\pi }{4}$$

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

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

StillAnotherDave said:
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
PeroK said:
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!

etotheipi said:
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.

## 1. What is a semi-infinite potential well?

A semi-infinite potential well is a concept in quantum mechanics where a particle is confined to a certain region and cannot escape due to the potential energy barrier at one end being infinitely high.

## 2. How is a semi-infinite potential well different from a finite potential well?

A finite potential well has a potential energy barrier at both ends, while a semi-infinite potential well only has a barrier at one end. This means that in a semi-infinite potential well, the particle is confined to a smaller region and has a higher probability of being found near the barrier.

## 3. What is the significance of a semi-infinite potential well in quantum mechanics?

The semi-infinite potential well is a simplified model used in quantum mechanics to understand the behavior of particles in confined spaces. It helps in understanding concepts such as quantization of energy levels and wave-particle duality.

## 4. How is the energy of a particle in a semi-infinite potential well calculated?

The energy of a particle in a semi-infinite potential well is calculated using the Schrödinger equation and boundary conditions. The energy levels are quantized and depend on the width and depth of the potential well.

## 5. Can a particle in a semi-infinite potential well tunnel through the potential barrier?

Yes, there is a small probability that a particle in a semi-infinite potential well can tunnel through the potential barrier. This phenomenon is known as quantum tunneling and is a fundamental aspect of quantum mechanics.

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