Computational Physics - Department of Physics

228 7 Eigensystems

To solve the Schrödinger equation as a matrix diagonalization problem, let us study the
radial part of the Schrödinger equation. The radial part of the wave function,R(r), is a solution
to

−

h ̄^2

2 m

(

1

r^2

d

dr

r^2

d

dr

−

l(l+ 1 )

r^2

)

R(r)+V(r)R(r) =E R(r).

Then we substituteR(r) = ( 1 /r)u(r)and obtain

−

h ̄^2

2 m

d^2

dr^2

u(r)+

(

V(r)+

l(l+ 1 )

r^2

h ̄^2

2 m

)

u(r) =E u(r).

We introduce a dimensionless variableρ= ( 1 /α)rwhereαis a constant with dimension length
and get

− ̄

h^2

2 mα^2

d^2

dρ^2

u(r)+

(

V(ρ)+

l(l+ 1 )

ρ^2

̄h^2

2 mα^2

)

u(ρ) =E u(ρ).

In the example below, we will replace the latter equation with that for the one-dimensional
harmonic oscillator. Note however that the procedure whichwe give below applies equally
well to the case of e.g., the hydrogen atom. We replaceρwithx, take away the centrifugal
barrier term and set the potential equal to

V(x) =

1

2

kx^2 ,

withkbeing a constant. In our solution we will use units so thatk=h ̄=m=α= 1 and the
Schrödinger equation for the one-dimensional harmonic oscillator becomes

−

d^2

dx^2 u(x)+x

(^2) u(x) = 2 E u(x).
Let us now see how we can rewrite this equation as a matrix eigenvalue problem. First we
need to compute the second derivative. We use here the following expression for the second
derivative of a functionf
f′′=f(x+h)−^2 f(x)+f(x−h)
h^2
+O(h^2 ),
wherehis our step. Next we define minimum and maximum values for the variablex,Rmin
andRmax, respectively. With a given number of steps,Nstep, we then define the stephas
h=
Rmax−Rmin
Nstep

.

If we now define an arbitrary value ofxas

xi=Rmin+ih i= 1 , 2 ,...,Nstep− 1

we can rewrite the Schrödinger equation forxias

−

u(xk+h)− 2 u(xk)+u(xk−h)

h^2

+x^2 ku(xk) = 2 E u(xk),

or in a more compact way

−

uk+ 1 − 2 uk+uk− 1

h^2

+x^2 kuk=−

uk+ 1 − 2 uk+uk− 1

h^2

+Vkuk= 2 E uk,