Computational Physics - Department of Physics

236 7 Eigensystems

−

h ̄^2

2 mα^2

d^2

dρ^2 u(ρ)+

k

2 α

(^2) ρ (^2) u(ρ) =E u(ρ).
We multiply thereafter with 2 mα^2 /h ̄^2 on both sides and obtain
−d
2
dρ^2
u(ρ)+mk
h ̄^2
α^4 ρ^2 u(ρ) =^2 mα
2
h ̄^2
E u(ρ).
The constantαcan now be fixed so that
mk
h ̄^2
α^4 = 1 ,
or
α=

(

̄h^2

mk

) 1 / 4

.

Defining

λ=

2 mα^2

̄h^2

E,

we can rewrite Schrödinger’s equation as

−

d^2

dρ^2

u(ρ)+ρ^2 u(ρ) =λu(ρ).

This is the first equation to solve numerically. In three dimensions the eigenvalues forl= 0
areλ 0 = 3 ,λ 1 = 7 ,λ 2 = 11 ,....
We use the by now standard expression for the second derivative of a functionu

u′′=

u(ρ+h)− 2 u(ρ)+u(ρ−h)

h^2

+O(h^2 ),

wherehis our step. Next we define minimum and maximum values for the variableρ,ρmin= 0
andρmax, respectively. You need to check your results for the energies against different values
ρmax, since we cannot setρmax=∞.
With a given number of steps,nstep, we then define the stephas

h=

ρmax−ρmin

nstep

.

Define an arbitrary value ofρas

ρi=ρmin+ih i= 0 , 1 , 2 ,...,nstep

we can rewrite the Schrödinger equation forρias

−

u(ρi+h)− 2 u(ρi)+u(ρi−h)

h^2

+ρi^2 u(ρi) =λu(ρi),

or in a more compact way

−

ui+ 1 − 2 ui+ui− 1

h^2

+ρi^2 ui=−

ui+ 1 − 2 ui+ui− 1

h^2

+Viui=λui,

whereVi=ρ^2 iis the harmonic oscillator potential. Define first the diagonal matrix element

di=

2

h^2

+Vi,