# Approximations with the Finite Square Well

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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Staff Emeritus
Homework Helper
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?

doggydan42
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?

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

doggydan42
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.

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

doggydan42
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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doggydan42
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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Staff Emeritus
Homework Helper
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##.

doggydan42
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?

Staff Emeritus
Homework Helper
Yup, that's what I got here.

doggydan42
Yup, that's what I got here.
So for finding A, by continuity at a

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

Staff Emeritus
Homework Helper
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##.

doggydan42
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.

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

doggydan42
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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doggydan42
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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