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.
Best Answer
I) First, I am doing it analytically,
Solutions of Schrodinger Eqution:
$\psi_I(x; 0<x\le l)=A \sin k_1 x+B \cos k_1 x \Rightarrow \psi_I(x)=A \sin k_1 x$ (after using the BC: $\psi(0)=0$)
$\psi_{II}(x; l<x\le 2l)=C \sin k_2 x+D \cos k_2 x \Rightarrow \psi_{II}(x)=C (\sin k_2 x - $ $\tan k_22l \cos k_2x)$ (after using the BC: $\psi_{II}(2l)=0)$.
where $k_1=\sqrt(e); k_2=\sqrt(e-V_0)$ (I have taken $\hbar^2=1=2m$)
Now matching the solutions at $l$: $\psi_{I}(l)=\psi_{II}(l),$ and $\psi_{I}'(l)=\psi_{II}'(l),$ This eliminates the coefficients $A$ and $C$.
We get the following transcendental Eq.,
$\frac{\tan k_1l}{k_1l}=\frac{\sin k_2l-\tan k_2 2l \cos k_1 l}{k_2l(\cos k_2l+\tan k_2 2l \sin k_2 l)} \Rightarrow \frac{\tan (k_1(e)l)}{k_1(e)l}-\frac{\sin (k_2(e)l)-\tan (k_2(e) 2l) \cos (k_1(e) l)}{k_2(e)l [\cos (k_2(e)l)+\tan (k_2(e) 2l) \sin (k_2(e) l)]} {=}0$
Example: Let's take $l=1$ and $V_0=2.$
First two roots of this equation are: $\textbf {3.367, 10.944} $; These are the first two eigenvalues.
II) Numerically, There is a nice way to do it in Mathematica,
Here, I am integrating the Schrodinger equation from ( with BCs:$ \psi(0)=0,\psi'(0)=1)$ $0$ to $2l$, then looking for zeros of $\psi(2l,e)$.
Here I have taken $l=1$ and $V_0=2.$
The above plot is for "sy2[2 l, e]" ( i.e. $\psi(2l,e)$). The first two roots are {e $\rightarrow$ $\textbf {3.36726, 10.9443} $}; These are first two eigenvalues.