# The width of a finite potential well

Tags:
1. Nov 6, 2016

### TheSodesa

1. The problem statement, all variables and given/known data
An electron is enclosed in a potential well, whose walls are $V_0 = 8.0eV$ high. If the energy of the ground state is $E = 0.50eV$, approximate the width of the well.

Answer: $0.72nm$

2. Relevant equations
For an electron in a potential well, whose energy is less than the height of the potential walls, outside the well the time-independent Schrödinger equation becomes:

\Psi ''(x) = \frac{2m}{\hbar^2}(V_0 - E) \Psi (x) = \alpha^2 \Psi (x),

where

\alpha^2 = \frac{2m}{\hbar} (V_0 - E) > 0

Inside the well, where $V(x) = 0$, it is the familiar:

\Psi ''(x) = -k^2 \Psi (x), k^2 = \frac{2mE}{\hbar^2}

By doing a whole bunch of math (by requiring that $\Psi$ be continuously differentiable at the potential walls and solving the resulting linear system with the assumption that it has more than one solution, meaning $det() = 0$), we and up with the result

\frac{\sin (kL)}{\cos (kL)} = \tan (kL) = \frac{\alpha}{k}.

Substituting $\alpha$ from $(2)$ and $k$ from $(3)$, we get

\tan (kL) = \sqrt{\frac{V_0 - E}{E}}

3. The attempt at a solution
Now my idea was to use $(5)$ to solve for $L$ as follows:
$$\tan (kL) = \sqrt{\frac{V_0 - E}{E}}\\ \iff\\ kL = \arctan (\sqrt{\frac{V_0 - E}{E}})\\ \iff\\ L = k^{-1}\arctan (\sqrt{\frac{V_0 - E}{E}}) = \sqrt{\frac{2mE}{\hbar^2}}^{-1} \cdot \arctan (\sqrt{\frac{V_0 - E}{E}})\\ = \sqrt{\frac{2(9.109 \cdot 10^{-31}kg)(0.5 \cdot 1.6022 \cdot 10^{-19}J)}{(\frac{6.626\cdot 10^{-34}Js}{2\pi})^2}}^{-1} \cdot\\ \arctan (\sqrt{\frac{8eV - 0.5eV}{0.5eV}})\\ = 2.084 751 \cdot 10^{-8}m,$$
which is a bit off. Any idea what I'm doing wrong?

Last edited: Nov 6, 2016
2. Nov 6, 2016

3. Nov 6, 2016

### TSny

I grabbed a QM text off the shelf to check your equation (5). It appears to me that this equation is for a well that extends from x = -L to x = +L. So, L in (5) would be half the width of the well.

Also, make sure your calculator is in the radian mode for taking the arctan.

4. Nov 7, 2016

### TheSodesa

Yeah, it was the radian thing, and the fact that the well goes from -L to L. The online material for our book (Tipler, Modern Physics, itself glosses over this part, the mathematics in particular) actually uses the letter $a$ in place of $L$, but I figured I could just change the place of the origin without doing anything to make up for it, so to speak.

Thanks again.

Last edited: Nov 7, 2016
5. Nov 7, 2016

### TheSodesa

Apparently there are even and odd solutions. The online material for our book states, that the $-\cot(ka)$ actually corresponds to the odd case.

6. Nov 7, 2016

Yes?

7. Nov 9, 2016

### TheSodesa

I as just noting that the link (the part of the page) you provided used $-\cot$ instead of $\tan$. At the time I didn't notice I could scroll up on that page, where the even solution could be found as well.

8. Nov 9, 2016

### Simon Bridge

Yes? Where are you up to in your solution?

9. Nov 13, 2016

### TheSodesa

Gah, sorry. I just fell of the map for a week, didn't I? I basically forgot to use radians in my calculation and also forgot, that the formula applies to a potential well from $-a$ to $a$, not $0$ to $L$.

Should I have posted my final numerical answer in this thread? I guess that would have been in good housekeeping...

The answer I got was $\approx 0.73nm$.