f(x,y) = (1-x)² + 100(y-x²)² ,
-(1/2)f'' -(1/r)f = εf ,
The bound s-state wave-function satisfies this equation and the two boundary conditions,
f(r → 0) = r-r², (prove this)
f(r → ∞) = 0 .
These two boundary conditions can only be satisfied for certain discrete (negative) values of the energy.
Since one cannot integrate numerically to ∞ one substitutes ∞ with a reasonably large number, rmax, such that it is much larger than the typical size of the hydrogen atom but still managable for the numerical inregrator (say, rmax = 10 Bohr radii),
f(rmax)=0 .
Let Fε(r) be the solution (to be found numericall
via gsl_odeiv
) to our differential equation with energy
ε and initial condition Fε(r → 0)=r-r².
Generally, for a random negative ε, this solution will not satisfy the
boundary condition at rmax. It will only be satisfied when
ε is equal one of the bound state energies of the system.
Now define an auxiliary function
M(ε) ≡ Fε(rmax) .
The shooting method is then equivalent to finding the (negative) root of the equation
M(ε) = 0 .
Find the lowest root, ε0, of the equation
M(ε) = 0
for, say, rmax=8. Plot the resulting
function and compare with the exact result (which es
ε0=-1/2, f0(r)=re-r –
check this by inserting ε0 and f0(r)
into the Schredinger equation above).
(Optional) Investigate the convergence of the solution towards the exact result as function of rmax.
(Optional) Try also to use a more precise boundary condition for bound states (which have ε<0),
f(r → ∞) = r e-kr , (prove this)