Preliminary Part for Homework

Here is what I got as result for the preliminary part:

The ground state energy, n=1, of a finite well with width a = 0.6 Å:

\(E_n = \frac{\hbar ^2 \pi^2}{2ma^2} n^2 \)



\(E_1  = 104 eV \)

To fit four bound states in our well it does not have to be deeper than \( 4^2 * 104 = 1664 eV \). In the calculations below I use \( V_0 = 1500 eV \). I set \( V(x) = 0 \) at the top of the well and the bottom is located at \( -V_0\) which then gives \( E_1\) the new value \( E_1 = - 1396 eV \) (We should however obtain \( E_1\) below, which we don't. See my comment in the comment section). 

We solve the Schrödinger equation and look at the even and odd solutions respectively, with the center of the well set at x = 0 and to make it less messy I set \( \frac{a}{2} = b \).

Even states are described by:

\( \phi_{even} = \begin{cases} D e^{\kappa x}, & x<-b \\ B cos(kx), & |x|<b \\ D e^{-\kappa x} & x>b \end{cases} \) 

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

\( k = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \)

Fitting those by making the function and its derivative continuous at the well boundary gives us the following equation:

\( \sqrt{\frac{-E}{E+V_0}} = tan \left(\sqrt{\frac{2mb^2}{\hbar^2}(E+V_0)} \right) \)

And we can write:

\( \frac{2mb^2}{\hbar^2} = \frac{2*9.1*10^{-31}kg*(0.3*10^{-10}m)^2}{(1.1*10^{-34}J*s)^2} * {1.6*10^{-19}} = 0.02eV^{-1} \)

(The term \(1.6*10^{-19} \) converts the \(J^{-1}\) into \(eV^{-1}\))

From here WolframAlpha (P = E in the equation, Wolfram didn't like E) did the job and gave: \( E_3 = -733 eV \).

The odd states are described by:

\( \phi_{odd} = \begin{cases} -D e^{\kappa x}, & x<-b \\ A sin(kx), & |x|<b \\ D e^{-\kappa x}, & x>b \end{cases} \) 

Which gives the following equation and energies:

\( -\sqrt{\frac{E+V_0}{-E}} = tan \left( \sqrt{0.02eV^{-1}(E+V_0)} \right) \)

Again Wolfram Alpha, with E = P.

\( E_2 = -1152 eV \), \( E_4 = -204 eV \)

The four bound states are: \( E_1 = -1396 eV\), \( E_2 = -1152 eV\), \( E_3 = -733 eV\) and \( E_4 = -204 eV\).

If you want the derivations I can take pictures of my notes and send by email, just let me know. Also let me know if there are mistakes, questions, anything! 

Addendum on calculating numbers: (added 1-20)
For calculating k and kappa, one can multiply through by c to get:

\( \kappa = \sqrt{\frac{2mc^2(-E)}{\hbar^2 c^2}} \)

\( k = \sqrt{\frac{2mc^2(E+V_0)}{\hbar^2c^2}} \)

where,

\(mc^2 = 0.511 \times 10^6 \) eV

and

\(\hbar c = 1970\) eV-A

Then I would suggest calcuating A, B and D from integrals where everything is in Angstroms (with A, B and D in Angstrom^{-1/2}.  Does that make sense?