Calculating energy as a function of lattice spacing

For this post I will attempt to write out expression for $E_q$ in terms of even and odd $\phi_0$

Odd wave functions $\phi(x-mc)_{odd} = \begin{cases} -D e^{\kappa (x-mc)}, & x < mc-a/2 \\ A sin(kx), & mc-a/2 < x < mc+a/2 \\ D e^{-\kappa (x-mc)}, & x > mc+a/2 \end{cases}$

Even wave functions $\phi(x-mc)_{even} = \begin{cases} D e^{\kappa (x-mc)}, & x < mc-a/2 \\ B cos(kx), & mc-a/2 < x < mc+a/2 \\ D e^{-\kappa (x-mc)} & x > mc+a/2 \end{cases}$

Just like the First Homework we can construct a hamiltonian but this time


for periodic finite square wells. In this example I choose c to be the distance between two successive wells

The depth is denoted by $V=-1500eV$

it is characterized by $V_0(x-mc) = \begin{cases} 0, & x < mc-a/2 \\ V, & mc-a/2 < x < mc+a/2 \\ 0, & x > mc+a/2 \end{cases}$

The hamiltonian is:

$H=\sum_{n=-\infty}^{\infty}V_0(x-nc)+{p^2}/{2m}=H_0+\sum_{n\neq 0}V_0(x-nc)$

The test wave function is $\psi_q=\sum_{m=-\infty}^{\infty}e^{imqc}\phi_0(x-mc)$

The S.E. doting with $<\phi_0|$ can yield an expression for $E_q$:

$< \phi_0 (x) | E_q | \psi_q  > = <\phi_0(x)|H|\psi_q> \Leftrightarrow$ 

$< \phi_0 (x) | E_q | \psi_q  > = <\phi _0(x)|  H_0 +  \sum\limits_{n \neq 0} V_0(x-cn)    | \psi _q  > \Leftrightarrow$   

$E_q=E + \frac{ <\phi_0(x) | \sum_{n \neq 0} V_0(x-cn)| \psi_q>}{<\phi_0(x)|\psi_q>}$

Now time to evaluate the denominator $I_{den}=<\phi_0|\psi_q>$ :

Ignoring other terms because we are always dotting with $\phi_0(x)$

  $I_{den}=+...+<\phi_0|e^{-iqc}\phi_0(x+c)>+<\phi_0|\phi_0>+<\phi_0|e^{iqc}\phi_0(x-c)>+...$

$I_{den}\simeq 1+2cos(qc)\int \phi_0(x) \phi_0(x+c) dx$ 

this holds for both even and odd 

Now to deal with the numerator $I_{num}$

In the next steps I assume c>a and by drawing the integrals I get that the following contribute to I

note that the properties of the potential make $<\phi_0|V_0(x-c)|\phi_0(x)>=0$

and other terms zero respectively but the terms m=-1 n=-1 and m=1 n=1 yield

$I_{num}\simeq 2cos(qc)<\phi_0|V_0(x-c)|\phi_0(x-c)>$

I will very soon post the final expressions for $I_{deno} , I_{num}$ but I wrote one that llustrates the form that I got for odd $\phi_0$  

$\frac{I_{deno}-1}{2cos(qc)}=\int_{-\infty}^{-a/2}\frac{D^2}{e^{kc}}*e^{2kx}dx-\int_{-a/2}^{a/2}ADsin(kx)e^{kx}dx-\int_{a/2}^{-a/2+c}D^2*e^{-kc}dx+$

$\int_{-a/2+c}^{a/2+c}ADsin(k(x-c))*e^{-kx}dx+\int_{a/2+c}^{\infty}D^2e^{-2kx}e^{kc}dx$

for the above, I tried calculating it but seemed too hard so maybe its ok to leave it equal 1 ?

For $I_{num}$ , keeping in mind that the range of the integral is dictated by the range of non-zero potential

For odd $I_{num}=2cos(qc)*\int_{c-a/2}^{c+a/2}De^{-k(x-c)}*V*Asin(k(x-c))dx \Leftrightarrow$

$I_{num}=-2V*D*Acos(qc)*e^{kc}((1/2+i/2) e^{-c k} (sin((1/2+i/2) a k)-sinh((1/2+i/2) a k)))*1/k \Leftrightarrow$
$I_{num}=-2V*D*Acos(qc)((1/2+i/2) (sin((1/2+i/2) a k)-sinh((1/2+i/2) a k)))*1/k$

For even $I_{num}=2cos(qc)*\int_{c-a/2}^{c+a/2}De^{-k(x-c)}*V*Bcos(k(x-c))dx \Leftrightarrow$

$I_{num}=2V*B*Dcos(qc)*e^{kc}((1/2-i/2) e^{-c k} (sin((1/2+i/2) a k)+sinh((1/2+i/2) a k)))*1/k \Leftrightarrow$
$I_{num}=2V*B*Dcos(qc)((1/2-i/2) (sin((1/2+i/2) a k)+sinh((1/2+i/2) a k)))*1/k$

Reflections: 

The c dependence for the numerator has disappeared!

Does the significance of the width come in on the denominator ?

 Did I leave out significant integral contributions in the denominatior?

Is the assumption c>a ok ?

I welcome questions and remarks!