E vs q for All States

Below are my general solutions to the energies for an infinite set of square wells, as well as my specific solutions to the E1 and E3 states.

Here are my constants:

Well width: a = 2b = 1 Angstrom 

Well Depth: \(V_0\) = 250 eV

Well Spacing (from one center to another): c = 1.75 Angstroms (chosen because it causes essentially no difference in E1, but a very large difference in E3)

E1:

E1 = -226 eV  , k = 2.51 Angstroms^(-1) , \(\kappa\) = 7.7 Angstroms^(-1)

                        D = 1.26 Angstroms^(-1/2), B = 0.39 Angstroms^(-1/2)

E3:

 E3 = -50 eV  , k = 7.24 Angstroms^(-1) , \(\kappa\) = 3.62 Angstroms^(-1)

                        D = 1.14 Angstroms^(-1/2), B = -1.01 Angstroms^(-1/2)

\( E_q^1 = -226 + 0.46cos(1.75q) \)

\( E_q^3 = -50 + 32cos(1.75q)      \)

Here are the steps to get here and general equations for all states, in case I got anything wrong:

Following from previous discussions we have

\( E_q = E_0 + 2cos(qc)I_{11}^m \)

\( I_{11}^m = < \phi_0 (x) | V (x-c) | \phi_0 (x-c)> \)

\( < \phi_0 (x) | V (x-c) | \phi_0 (x-c)> = \int_{c-b}^{c+b} \phi_0^* (x) V_0 (x-c) \phi_0  (x-c) dx \)

\(  = V_0 \int_{c-b}^{c+b} \phi_0^* (x) \phi_0  (x-c) dx \)

\(  \phi_0(x) \)  between c-b and c+b is the exponentially falling tail of the wave  function centralized at x = 0, and is the same for even and odd functions.

\(\phi_0 (x-c)\) between c-b and c+b is the sinusoidal wave function centered at x=c, and varies from even to odd.

\(  = V_0 \int_{c-b}^{c+b} e^{- \kappa x} \phi_0  (x-c) dx \)

odd \(  = V_0 D \int_{c-b}^{c+b} e^{- \kappa x} sin(k(x-c)) dx \)

even \(  = V_0 D \int_{c-b}^{c+b} e^{- \kappa x} cos(\kappa(x-c)) dx \)

odd \( \frac{V_0 D A e^{-\kappa c}}{k^2 + \kappa^2} [ k(e^{2b\kappa} - 1)   cos(bk) -  \kappa(e^{2b\kappa} + 1)   sin(bk)   ]   \)

even \( \frac{V_0 D B e^{-\kappa c}}{k^2 + \kappa^2} [ \kappa(e^{2b\kappa} - 1)   cos(bk) +  k(e^{2b\kappa} + 1)   sin(bk)   ]   \)



odd \( E_q = E_0 + \frac{V_0 D A e^{-\kappa c} cos(qc)}{k^2 + \kappa^2} [ k(e^{2b\kappa} - 1)   cos(bk) -  \kappa(e^{2b\kappa} + 1)   sin(bk)   ]   \)

even \( E_q = E_0 + \frac{V_0 D B e^{-\kappa c} cos(qc)}{k^2 + \kappa^2} [ \kappa(e^{2b\kappa} - 1)   cos(bk) +  k(e^{2b\kappa} + 1)   sin(bk)   ]   \)

\( E_q^1 = -226 + 327,050e^{-7.7c}cos(cq) \)

\( E_q^3 = -50 + 17,885e^{-3.62c}cos(cq)      \)

Does this look reasonable?