How do I calculate the initial alpha for an iterative process in a finite well?

[bluestar]

I am having a devil of a time trying to successfully perform an iterative process for a particle’s energy in a finite well. The equation describing the energy for a finite well is transcendental thus requiring a graphical or numerical solution. Although, the graphical solution is straight forward I choose to pursue the numeric solution.

The only Internet resource I have been able to find that addresses the subject in specific detail is a page at the hyperphysics site:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pfbox.html#c1
(Page down about 4 pages to “Numerical Solution for Ground State”

The specific problem is that I am not able to determine how to calculate the initial alpha or ‘attenuation coefficient’ as he calls it. If you wish to tackle my example here are the numbers you can plug into the input boxes and it automatically calculates the rest of the values.

Well Length: .39E-9 m or .39 nm
Electron Mass: 9.1E-31 kg
Infinite well ground state for n=1: .396508E-18 J or 2.474808 eV
Potential Energy U: 1.025393E-17 or 64 eV

The first estimate of the attenuation coefficient is: .401643E11 m
This is the number I can not determine how it is calculated. When I used the formula for alpha provided in the section I do not come anywhere close to his first estimated value.

This next link, at the bottom, provides the energy levels for the first 4 levels of a finite well using the variables described above.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html

Pursuing this example may be the easiest way to help; otherwise, if anyone has a good reference on numerical solutions for finite well energies and if it is in an electronic form I would much prefer to study that document and try to solve the problem myself. Otherwise, any and all help is greatly appreciated.

RON

 

[jtbell]

The first estimate of the attenuation coefficient is: .401643E11 m
This is the number I can not determine how it is calculated. When I used the formula for alpha provided in the section I do not come anywhere close to his first estimated value.

When I plug the given numbers into the formula

\alpha = \sqrt {\frac {2 m (U_0 - E)} {\hbar^2}}

I get the indicated value. Make sure you're using consistent units. I used kg for mass, J for energy, and J-sec for Planck's constant. Also make sure you're using \hbar = h/2\pi, not plain old h.

 

[bluestar]

Many thanks to you, JTBell.

I was using eV for both energy factors. When I used energy in Joules I was accurate to 4 digits and this may be because I only used 3 digits of the Planck constant to calculate Dirac’s Constant. Anyway I am very pleased and greatly relieved to now get it to work. (Wished I has posted this 3 days ago).

This calculation was for the ground state energy and length for n=1 in a finite well. Now I will start looking to calculate the energy and lengths for the next 3 or 4 elevated energy levels. Before I begin trying to figure this out do you know how to determine the next levels? Do I use the n=2 energy level for the infinite well case and then duplicate the steps as we did for the ground state?

Your input is greatly appreciated.
RON

 

[bluestar]

It looks like I might have been premature in thinking the iteration problem had been resolved. When I use the alpha calculated by the internet page .401643 E11 in the formula to calculate the new length I get an incorrect answer. The length is calculated by the following equation and the formula and illustration are on a page up from the Internet calculation sheet.

L-old + 2 / alpha = L-new
Thus,
.39 E-9 + 2 / .401643 E11 = .3904979 E-09 and it should be .439795 E-9

However, if I use the following alpha 4.01643 E09 then I get a correct answer

.39 E-9 + 2 / .401643 E09 = .43979 E-09

So where does the difference of a factor of 2 in the alpha come from?

If I use subsequent alphas to the E09 power then all iterations match the Internet data sheet and converge to a correct number within 3 iterations and with a 3 digit accuracy.