# Approximations with the Finite Square Well

• doggydan42
In summary,The homework statement is asking for the standard square well potential. Using the relation between ##\xi## and ##\eta##, and the approximation for ##\eta##, the approximate values of ##\xi## and ##\eta## to leading order in ##z_0## can be found. If one of these is right, how would they choose one?
doggydan42

## Homework Statement

Consider the standard square well potential
$$V(x) = \begin{cases} -V_0 & |x| \leq a \\ 0 & |x| > a \end{cases}$$
With ##V_0 > 0##, and the wavefunctions for an even state
$$\psi(x) = \begin{cases} \frac{1}{\sqrt{a}}cos(kx) & |x| \leq a \\ \frac{A}{\sqrt{a}}e^{-\kappa |x|} & |x| > a \end{cases}$$

where we have included the ##\frac{1}{\sqrt{a}}## prefactor to have consistent units for ##\psi##, and A is a constant required by continuity at x = a.

For the finite square well, recall that
$$\eta = ka, \xi = \kappa a, k^2 = \frac{2m(V_0-|E|)}{\hbar^2}, \kappa^2 = \frac{2m|E|}{\hbar^2}$$

and
$$\xi^2 + \eta^2 = z_0^2, \xi = \eta tan(\eta)$$

We want to have a better understanding of the limit as ##V_0 \rightarrow \infty##, and understand why the discontinuity in ##\psi'## in the infinite well does not give trouble. Keeping m and a constant as we let ##V_0## grow large is the same thing as letting ##z_0## grow large, where
$$z_0^2 = \frac{2ma^2V_0}{\hbar^2}$$

Part A:
Consider the ground state of the potential. In the limit of large ##z_0##, compute the approximate values of ##\eta## and ##\xi## to leading order in ##z_0##. Your answers should have no trigonometric functions in them. (Hint: You will need to approximate the relations between ##\xi## and ##\eta##. Think about the range you expect ##\eta## to lie in.)

##\eta = ##
##\xi = ##
##A = ##

## Homework Equations

These are given in the problem statement.

## The Attempt at a Solution

My main issue was with the approximations.
I was thinking of how to approximate the tangent.

I was also thinking of possibly using the average of the range ##\eta## should lie in, which I believe is 0 to ##z_0##, but that does not seem to be right.

What would be the best way to go about approximating ##\eta## and ##\xi##?

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doggydan42 said:
I was also thinking of possibly using the average of the range ηη\eta should lie in, which I believe is 0 to z0z0z_0, but that does not seem to be right.
You have the relation ##\xi^2 + \eta^2 = z_0^2##. The righthand side grows large as ##V_0## becomes large. How do you get the lefthand side grow large?

vela said:
You have the relation ##\xi^2 + \eta^2 = z_0^2##. The righthand side grows large as ##V_0## becomes large. How do you get the lefthand side grow large?

I would expect at least one of ##\eta## or ##\xi## to grow large. Given the relation between ##\eta## and ##\xi##, ##\eta## would have to grow large?

Good question. Do you have a reason for choosing ##\eta## over ##\xi##, or are you just guessing?

vela said:
Good question. Do you have a reason for choosing ##\eta## over ##\xi##, or are you just guessing?
Graphically, it looks like if ##\xi## increases, ##\eta## could take many different values. That was my main reason, but there is probably a more concrete one. Also, if you plug in the relation between ##\eta## and ##\xi## into the circlular equation, the graph shows that as ##\eta## increases ##z_0## generally increases.

If ##\eta## is large, then ##k = \eta/a## is large. What does that mean? Does that make sense for, say, the ground state?

vela said:
If ##\eta## is large, then ##k = \eta/a## is large. What does that mean? Does that make sense for, say, the ground state?
That would not make sense, since the cosine function would have a large frequency. So ##\xi## grows large while ##\eta## remains small?

If so, how would I get rid of the tangent function in the approximation?

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I tried plugging in the function:

$$z_0^2 = \eta^2 + (\eta tan(\eta))^2 = \eta^2sec^2(\eta)$$
since ##z_0## is positive.
$$z_0 = \eta sec(\eta)$$
So using the relationship between ##\xi## and ##\eta##,
$$\xi = \eta tan(\eta) = z_0 sin(\eta)$$

By the graphs of ##\eta## and ##\xi##, ##\eta## should be just less than ##\pi/2##.
Since as ##z_0## goes to infinity, ##\eta## approaches a value, I was thinking that I might be able to approximate
$$\eta = \frac{\pi]{2} - z_0^{-1}$$
or
$$\eta = \frac{\pi]{2} +z_0^{-1}$$

If one of these is right, how would I choose one?

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You have ##z_0 \cos\eta = \eta##. Expand the cosine to first order about ##\eta=\pi/2##. Then you can solve for ##\eta## in terms of ##z_0##.

vela said:
You have ##z_0 \cos\eta = \eta##. Expand the cosine to first order about ##\eta=\pi/2##. Then you can solve for ##\eta## in terms of ##z_0##.

That makes a lot of sense; I forget about taylor series.

This gave me

$$\eta = \frac{\pi z_0}{2(1+z_0)}$$

Plugging that into the circular equation for I get
$$\xi = z_0 \frac{\sqrt{4(1+z_0)^2-\pi^2}}{2(1+z_0)}$$

Do these look right?

Yup, that's what I got here.

vela said:
Yup, that's what I got here.
So for finding A, by continuity at a

$$Ae^{-\kappa a} = cos(ka) \Rightarrow A = cos(\eta)e^{\xi} = (\frac{\eta}{z_0})e^{\xi} =(\frac{\pi}{2(1+z_0)}) e^{z_0 \frac{\sqrt{4(1+z_0)^2-\pi^2}}{2(1+z_0)}}$$

The problem statement says to express the results to leading order in ##z_0##, so you'll want to do a little more work on your expressions for ##\eta## and ##\xi##.

vela said:
The problem statement says to express the results to leading order in ##z_0##, so you'll want to do a little more work on your expressions for ##\eta## and ##\xi##.
So,
$$\xi = \eta tan(\eta) = \eta \frac{sin\eta}{cos\eta}$$
$$sin\eta \approx 1,cos\eta \approx \frac{\pi}{2}-\eta = \frac{\pi}{2}(1-\frac{z_0}{1+z_0}) = \frac{\pi}{2}(\frac{1}{1+z_0})$$
Plugging these into ##\xi##:
$$\xi = (\frac{\pi z_0}{2(1+z_0)})\frac{1}{\frac{\pi}{2(1+z_0)}} = z_0$$

But this would give ##\eta = 0## in the circular equation.

I was also thinking that maybe
$$\eta = \frac{\pi}{2}(1-z_0^{-1})$$
but I could only justify this graphically.

Use your expressions from post 10:
$$\eta = \frac{\pi z_0}{2(1+z_0)} = \frac{\pi}{2}\frac{1}{1+(1/z_0)} = \cdots$$
$$\xi = z_0 \sqrt{1-\left[\frac{\pi}{2(1+z_0)}\right]^2} = \cdots$$

vela said:
Use your expressions from post 10:
$$\eta = \frac{\pi z_0}{2(1+z_0)} = \frac{\pi}{2}\frac{1}{1+(1/z_0)} = \cdots$$
$$\xi = z_0 \sqrt{1-\left[\frac{\pi}{2(1+z_0)}\right]^2} = \cdots$$
For ##\eta##, if I expand the terms, I get
$$\eta = \frac{\pi}{2}(1-\frac{1}{z_0})$$.

For ##\xi##, I was thinking
$$\xi = z_0-\frac{\pi^2}{8}z_0^{-1}$$

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For A, I would then get

$$A = cos(\eta)e^\xi = (\frac{\pi}{2}-\eta)e^\xi = \frac{\pi}{2}z_0^{-1}e^{z_0}$$

Do I need to simplify anymore?

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## What is a finite square well?

A finite square well is a simplified model used in quantum mechanics to study a particle's behavior within a potential energy barrier. It consists of a potential energy function that is constant within a finite range and infinite outside of that range.

## What is the significance of approximations with the finite square well?

Approximations with the finite square well allow us to understand the behavior of particles within a potential energy barrier, which has important applications in various fields such as nuclear physics, solid state physics, and quantum computing.

## How are approximations with the finite square well used in real-life scenarios?

One common real-life application of approximations with the finite square well is in the study of radioactive decay. The decay of a nucleus can be modeled by a particle moving within a potential energy barrier, making the finite square well a useful tool for understanding this process.

## What are the limitations of using a finite square well approximation?

One limitation is that it is a simplified model and does not account for all the complexities of a real potential energy barrier. Additionally, it assumes that the potential energy within the well is constant, which may not always be the case in real-life scenarios.

## Can the finite square well approximation be extended to other potential energy functions?

Yes, the finite square well can be generalized to other potential energy functions, such as the harmonic oscillator potential or the Coulomb potential. This allows for a more comprehensive understanding of particle behavior in different types of barriers.

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