Chapter 2
Schrödinger equation for
central potentials
In this chapter we extend the concepts and methods introduced in the previous
chapter for a one-dimensional problem to a specific and very important class
of three-dimensional problems: a particle of mass m under a central potential
V (r), i.e. depending only upon the distance r from a fixed center. The Schrö-
dinger equation we are going to study in this chapter is thus
" #
h̄2 2
Hψ(r) ≡ − ∇ + V (r) ψ(r) = Eψ(r). (2.1)
2m
The problem of two interacting particles via a potential depending only upon
their distance, V (|r1 − r2 |), e.g. the Hydrogen atom, reduces to this case, with
m equal to the reduced mass of the two particles.
The general solution proceeds via the separation of the Schrödinger equation
into an angular and a radial part. In this chapter we consider the numerical
solution of the radial Schrödinger equation. A non-uniform grid is introduced
and the radial Schrödinger equation is transformed to an equation that can still
be solved using Numerov’s method introduced in the previous chapter.
2.1 Variable separation
Let us introduce a polar coordinate system (r, θ, φ), where θ is the polar angle,
φ the azimuthal one, and the polar axis coincides with the z Cartesian axis.
After some algebra, one finds the Laplacian operator in polar coordinates:
1 ∂ ∂ 1 ∂ ∂ 1 ∂2
2
∇ = 2 r2 + 2 sin θ + (2.2)
r ∂r ∂r r sin θ ∂θ ∂θ r2 sin2 θ ∂φ2
It is convenient to introduce the operator L2 = L2x + L2y + L2z , the square of the
angular momentum vector operator, L = −ih̄r × ∇. Both L ~ and L2 act only
on angular variables. In polar coordinates, the explicit representation of L2 is
" #
1 ∂ ∂ 1 ∂2
2 2
L = −h̄ sin θ + . (2.3)
sin θ ∂θ ∂θ sin2 θ ∂φ2
16
, The Hamiltonian can thus be written as
h̄2 1 ∂ ∂ L2
H=− r2 + + V (r). (2.4)
2m r2 ∂r ∂r 2mr2
The term L2 /2mr2 has a classical analogous: the radial motion of a mass having
classical angular momentum Lcl can be described by an effective radial potential
V̂ (r) = V (r) + L2cl /2mr2 , where the second term (the “centrifugal potential”)
takes into account the effects of rotational motion. For high Lcl the centrifugal
potential “pushes” the equilibrium position outwards.
In the quantum case, both L2 and one component of the angular momentum,
for instance Lz :
∂
Lz = −ih̄ (2.5)
∂φ
commute with the Hamiltonian, so L2 and Lz are conserved and H, L2 , Lz have
a (complete) set of common eigenfunctions. We can thus use the eigenvalues of
L2 and Lz to classify the states. Let us now proceed to the separation of radial
and angular variables, as suggested by Eq.(2.4). Let us assume
ψ(r, θ, φ) = R(r)Y (θ, φ). (2.6)
After some algebra we find that the Schrödinger equation can be split into an
angular and a radial equation. The solution of the angular equations are the
spherical harmonics, known functions that are eigenstates of both L2 and Lz :
Lz Y`m (θ, φ) = mh̄Y`m (θ, φ), L2 Y`m (θ, φ) = `(` + 1)h̄2 Y`m (θ, φ) (2.7)
(` ≥ 0 and m = −`, ..., ` are integer numbers).
The radial equation is
" #
h̄2 1 ∂ ∂Rn` h̄2 `(` + 1)
− 2
r2 + V (r) + Rn` (r) = En` Rn` (r). (2.8)
2m r ∂r ∂r 2mr2
In general, the energy will depend upon ` because the effective potential does;
moreover, for a given `, we expect to find bound states with discrete energies
and we have indicated with n the corresponding index.
Finally, the complete wave function will be
ψn`m (r, θ, φ) = Rn` (r)Y`m (θ, φ) (2.9)
The energy does not depend upon m. As already observed, m classifies the
projection of the angular momentum on an arbitrarily chosen axis. Due to
spherical symmetry of the problem, the energy cannot depend upon the orien-
tation of the vector L, but only upon his modulus. An energy level En` will
then have a degeneracy 2` + 1 (or larger, if there are other observables that
commute with the Hamiltonian and that we haven’t considered).
17
Schrödinger equation for
central potentials
In this chapter we extend the concepts and methods introduced in the previous
chapter for a one-dimensional problem to a specific and very important class
of three-dimensional problems: a particle of mass m under a central potential
V (r), i.e. depending only upon the distance r from a fixed center. The Schrö-
dinger equation we are going to study in this chapter is thus
" #
h̄2 2
Hψ(r) ≡ − ∇ + V (r) ψ(r) = Eψ(r). (2.1)
2m
The problem of two interacting particles via a potential depending only upon
their distance, V (|r1 − r2 |), e.g. the Hydrogen atom, reduces to this case, with
m equal to the reduced mass of the two particles.
The general solution proceeds via the separation of the Schrödinger equation
into an angular and a radial part. In this chapter we consider the numerical
solution of the radial Schrödinger equation. A non-uniform grid is introduced
and the radial Schrödinger equation is transformed to an equation that can still
be solved using Numerov’s method introduced in the previous chapter.
2.1 Variable separation
Let us introduce a polar coordinate system (r, θ, φ), where θ is the polar angle,
φ the azimuthal one, and the polar axis coincides with the z Cartesian axis.
After some algebra, one finds the Laplacian operator in polar coordinates:
1 ∂ ∂ 1 ∂ ∂ 1 ∂2
2
∇ = 2 r2 + 2 sin θ + (2.2)
r ∂r ∂r r sin θ ∂θ ∂θ r2 sin2 θ ∂φ2
It is convenient to introduce the operator L2 = L2x + L2y + L2z , the square of the
angular momentum vector operator, L = −ih̄r × ∇. Both L ~ and L2 act only
on angular variables. In polar coordinates, the explicit representation of L2 is
" #
1 ∂ ∂ 1 ∂2
2 2
L = −h̄ sin θ + . (2.3)
sin θ ∂θ ∂θ sin2 θ ∂φ2
16
, The Hamiltonian can thus be written as
h̄2 1 ∂ ∂ L2
H=− r2 + + V (r). (2.4)
2m r2 ∂r ∂r 2mr2
The term L2 /2mr2 has a classical analogous: the radial motion of a mass having
classical angular momentum Lcl can be described by an effective radial potential
V̂ (r) = V (r) + L2cl /2mr2 , where the second term (the “centrifugal potential”)
takes into account the effects of rotational motion. For high Lcl the centrifugal
potential “pushes” the equilibrium position outwards.
In the quantum case, both L2 and one component of the angular momentum,
for instance Lz :
∂
Lz = −ih̄ (2.5)
∂φ
commute with the Hamiltonian, so L2 and Lz are conserved and H, L2 , Lz have
a (complete) set of common eigenfunctions. We can thus use the eigenvalues of
L2 and Lz to classify the states. Let us now proceed to the separation of radial
and angular variables, as suggested by Eq.(2.4). Let us assume
ψ(r, θ, φ) = R(r)Y (θ, φ). (2.6)
After some algebra we find that the Schrödinger equation can be split into an
angular and a radial equation. The solution of the angular equations are the
spherical harmonics, known functions that are eigenstates of both L2 and Lz :
Lz Y`m (θ, φ) = mh̄Y`m (θ, φ), L2 Y`m (θ, φ) = `(` + 1)h̄2 Y`m (θ, φ) (2.7)
(` ≥ 0 and m = −`, ..., ` are integer numbers).
The radial equation is
" #
h̄2 1 ∂ ∂Rn` h̄2 `(` + 1)
− 2
r2 + V (r) + Rn` (r) = En` Rn` (r). (2.8)
2m r ∂r ∂r 2mr2
In general, the energy will depend upon ` because the effective potential does;
moreover, for a given `, we expect to find bound states with discrete energies
and we have indicated with n the corresponding index.
Finally, the complete wave function will be
ψn`m (r, θ, φ) = Rn` (r)Y`m (θ, φ) (2.9)
The energy does not depend upon m. As already observed, m classifies the
projection of the angular momentum on an arbitrarily chosen axis. Due to
spherical symmetry of the problem, the energy cannot depend upon the orien-
tation of the vector L, but only upon his modulus. An energy level En` will
then have a degeneracy 2` + 1 (or larger, if there are other observables that
commute with the Hamiltonian and that we haven’t considered).
17
