Double Delta function potential well

[black_hole]

Consider a one-dimensional system described by a particle of mass m in the presence
of a pair of delta function wells of strength Wo > 0 located at x = L, i.e.
V(x) = -Wo (x + L) - Wo(x - L) This is a rough but illuminating toy model of an electron in the presence of two positive.
charges located at x = L.

(a) Derive a transcendental equation for the allowed eigenenergies of any bound states.
Express your result in terms of the dimensionless quantities go = mLWo/hbar^2 and ε = κL where E = -hbar^2*κ^2 / 2m is the (negative) energy of the bound state.

(b) Solve your transcendental equation(s) graphically / numerically to identify all
bound state energy eigenvalues. How many bound states exist? Does the number
depend on go?

(c) Plot all bound state energy eigenfunctions for go = 0.1, go = 0.5 and go = 10.

(d) How does the energy of the most tightly bound state vary as you vary L? Include
a plot (with axes and units labeled) which shows the energy as a function of L.
Note: You do not have to do this analytically; this is most easily done numerically
using Mathematica.

(e) Suppose we place the particle in the lowest energy bound state. Do the two delta
functions want to be close together or far apart? Plot the induced force between
the delta functions as a function of L. Again, best done numerically.

(f) Use the above to suggest a plausible explanation for why H⁺₂
is a stable molecule.

(g) How does the splitting between levels change as you increase the separation between the wells? Why does the 2rd excited state have the same number of nodes inside each well as the 3rd excited state, but not the same number as the 4th?

So I think I did a). I got ε = go(1-e^(-2ε)) and ε = go(1+e^(-2ε)). I'm not suite sure what b) is. I thought it was 2. The thing I'm getting hung up on c) -g) bc I can't use mathematica to its full 'potential' I guess. Help please? How do I proceed.

 

[mark726]

a) To derive the transcendental equation, we can use the Schrodinger equation:
-Ĥ ψ(x) = E ψ(x)
where Ĥ is the Hamiltonian operator and ψ(x) is the wave function.
In this case, the Hamiltonian is given by:
Ĥ = -ħ^2/2m * d^2/dx^2 + V(x)
Substituting in the potential V(x) given in the problem, we get:
Ĥ = -ħ^2/2m * d^2/dx^2 - Wo (x + L) - Wo(x - L)
Using the boundary conditions that the wave function must be continuous and differentiable at the delta function wells, we can write the wave function as:
ψ(x) = A*e^κx + B*e^-κx for x < -L
ψ(x) = C*e^κx + D*e^-κx for -L < x < L
ψ(x) = E*e^κx + F*e^-κx for x > L
where κ = √(2m(E+Wo)/ħ^2) and A, B, C, D, E, and F are constants that can be determined from the boundary conditions.
Applying the boundary conditions, we get:
ψ(-L) = ψ(L) = 0, which gives A = -D and E = -F
ψ'(-L) = ψ'(L), which gives κ*(A+B) = -κ*(E+F)
Using these conditions and simplifying, we get the transcendental equation:
1 + e^(-2κL) = 2ε/go
where ε = κL and go = mLWo/ħ^2. This is the equation for the allowed energy eigenvalues of the bound states.

b) To solve the transcendental equation, we can use a graphical or numerical approach. Graphically, we can plot the left and right sides of the equation and find the points where they intersect, which will give us the values of ε. Alternatively, we can use a numerical method such as Newton's method to find the roots of the equation.
The number of bound states will depend on the value of go. For small values of go, there will be only one bound state, while for