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.)

Write your answer in terms of $z_0$.
$\eta =$
$\xi =$
$A =$

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 $\xi$?

 

[vela]

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?

 

[vela]

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.

 

[vela]

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?

 

[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?

 

[vela]

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?

 

[vela]

Yup, that's what I got here.

 

[doggydan42]

Yup, that's what I got here.

So for finding A, by continuity at a

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

 

[vela]

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.

 

[vela]

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
$$\xi = z_0-\frac{\pi^2}{8}z_0^{-1}$$

 

[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?