Symmetric finite square double potential well I'm having trouble evaluating the Hamiltonian that comes with a symmetric finite square double potential well.
What I want to do is calculate the overlap Integrals contained in the Hamiltonian Matrix of the system when both wells do not lie infinitely far apart.
This is how my Potential looks like:
Region I ( x < -(d/2 + a):
$$V(x) = V_1 $$ 
Region II ( -(d/2 +a) < x < -d )
$$V(x) = 0, $$
Region III ( -d/2 < x < 0 )
$$V(x) = V_1, $$
Region III ( 0 < x < d/2 )
$$V(x) = V_2$$
Region IV ( d/2 < x < d/2 + a)
$$V(x) = 0$$
Region V (x > d/2 + a)
$$V(x)=V_2$$
Sorry I couldnt figure out how to present this better. To make things easier, let $V_1=V_2$.
The coordinate system is in the middle of region III.
I have two states for the separated wells:
$$ (T+V_1) |\psi_1> = \epsilon_1 |\psi_1 > $$
$$ (T+V_2) |\psi_2> = \epsilon_2 |\psi_2 > $$
Considering the total Hamiltonian:
$$ H= T + V_1 + V_2 $$
I get:
$$ 
         \begin{bmatrix}
         <\Psi_1|H|\Psi_1> & <\Psi_1|H|\Psi_2>\\         
         <\Psi_2|H|\Psi_1> & <\Psi_2|H|\Psi_2> \\
         \end{bmatrix}
 $$
Now how do I go about calculating the overlap Terms $ <\Psi_1|H|\Psi_2>$ and $<\Psi_2|H|\Psi_1>$ ?
I know that 
$ <\Psi_1|H|\Psi_2> = \int_{R} \Psi_1^* (T+V_1+V_2)\Psi_2$
Now comes the tricky part for me. I use the Wave functions I got from solving the Schrödinger Equation and split the integral in several parts because I have defined the functions to be different in each region.
Is this the correct way to do this?
My Integral would have a term like this for Region I then:
$$\int_{-\infty}^{-(d/2 + a)} \Psi_1^* (T+V_1+V_2)\Psi_2$$ 
where we should have $\Psi_1 = Aexp(\alpha x)$ but what about $\Psi_2$ ? I know that in the "positive" part of region III we have $\Psi_2 = Aexp(\alpha x)$ as well. But can I use that function in the Integral that is representing Region I (as above) ?
(Im not even entirely sure if $\Psi_2$ should represent the wave function in the second well but from what I'vea read online that's how you go about solving a double well problem like this. By considering both isolated wells and then bringing them together)
I need some advice. How do I calculate the overlap (which could be considered to be zero when both wells are "isolated") and which wave functions do I use?
 A: Assuming typing error in region II (should be -(d/2 +a) < x < -d/2) the potential looks like this: 

If it's not what your potential well looks like maybe you would like to post a picture.
The solutions to this problem are obtained by solving one dimensional time independent Schrodinger equation (TISE):
$\hat{H}\Psi(x)=E\Psi(x)$
Where $\hat{H}$ is Hamiltonian, $\Psi$ is an eigen function and $E$ an eigen value. We can further expand the hamiltonian operator
$\hat{H}=\hat{T}(x)+V(x)=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\,,$
where $\hbar=h/(2\pi)$ is reduced Planck constant.
If you are not sure how to solve the TISE please see finite potential well. If you never did that I strongly recomend that you find it in your text book and first understand the solution of a single finite well. If you don't know what is Schrodinger equation or how the hamiltonian is defined in QM I recomend starting with that. I will describe a general process of solution for the double well. The full solution is rather lengthy because there are many cases. Please let me know which section you don't understand and I'll expand it.
The energy of the particle is $E$. Assuming $V_1<V_2$. We divide the solution into 3 cases:
a)$E<V_1$ (discrete)
b)$V_1<E<V_2$ (discrete)
c)$V_2<E$ (continuous)
For transition probabilities you probably don't need to bother with the discrete spectrum.
In the later I'll use:
$$
V(x)=
\begin{cases} 
V_1&x < -(\frac{d}{2} + a)\\
0&-(\frac{d}{2} + a) < x < -\frac{d}{2}\\
V_1& -\frac{d}{2} < x < 0\\
V_2& 0 < x < \frac{d}{2}\\
0&\frac{d}{2} < x < \frac{d}{2} + a\\
V_2&x > \frac{d}{2} + a\\
\end{cases}
$$
We can rewrite the TISE:
$$\frac{-\hbar^2}{2m}\frac{d^2\Psi(x)}{d^2x}+V(x)\Psi(x)=E\Psi(x)$$
$$\frac{d^2\Psi(x)}{d^2x}+\frac{2m(E-V(x))}{\hbar^2}\Psi(x)=0$$
We need to do the folowing part separately for a)$E<V_1$ and b)$V1<E<V2$
Let's take case a)
Let's take a look at region I.
In this region $E<V(x)=V_1$. We define wave vector
$$k_I=\frac{\sqrt{2m(V_1-E)}}{\hbar}$$
(We want $k_I$ to be real number)
We solve the equation:
$$\frac{d^2\Psi_I(x)}{d^2x}-k_I^2\Psi_I=0$$
which leads to solution:
$$\Psi_I=A_Ie^{k_Ix}+B_Ie^{-k_Ix}$$
This applies also for regions IIIa,IIIb,V because in all these regions E

Let's take a look at region II. 
In this region $E>V(x)=0$
We define wave vector
$$k_{II}=\frac{\sqrt{2m(E-0)}}{\hbar}$$
We solve the equation:
$$\frac{d^2\Psi_{II}(x)}{d^2x}-k_{II}^2\Psi_{II}=0$$
which leads to solution:
$$\Psi_{II}=A_{II}\sin(x)+B_{II}\cos(x)\,,$$
Similarly for region IV.
Case b) is analogical to a). If $V_1=V_2$ then you only need to calculate case a)

Coefficients
You can calculate the numbers $A_I,B_I,A_{II},B_{II}$ from boundary conditions:
$\Psi_I(-\infty)=0$=>\Psi_I(x)=
$\Psi_I(-(d/2 + a))=\Psi_II(-(d/2 + a))$
$\frac{d\Psi_I}{dx}(-(d/2 + a))=\frac{d\Psi_II}{dx}(-(d/2 + a))$
$\vdots$
$\Psi_V(\infty)=0$
This system of equations has non-zero solution only for certain values of $E$. You can calculate the deterinant in similar manner as for the finite potential well and find values of $E$ for which the determinant is zero. This CAN'T BE SOLVED ANALYTICALLY = you need computer for this step.
Hamiltonian matrix (Overlap)
You can't define ${H=T+V_1+V_2}$ there is no such thing as total hamiltonian. The hamiltonian is:
$$\hat{H}=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)$$
You just plug this in to your integral and get for example:
$$H_{12}=\int_{−(d/2+a)}^{−∞}Ψ_1(x)\frac{-\hbar^2}{2m}\frac{d^2}{dx^2(x)}Ψ_2(x)+\Psi_1V_1Ψ_2 dx$$
Please notice that $\Psi_I$ relates to region while $\Psi_1$ relates to energy level.
