Lecture 27:Orbital Angular Momentum

Phy851 Fall 2009

The General Theory of Angular Momentum

• Starting point:– Assume you have three operators that

satisfy the commutation relations:

– Let:

• Conclusions:– Simultaneous eigenstates of J2 and Jz exist– They must satisfy:

– Where the quantum numbers take on thevalues:

yx

zyx

iJJJ

JJJJ

±=

++=

±

2222

yxz

xzy

zyx

JiJJ

JiJJ

JiJJ

h

h

h

=

=

=

],[

],[

],[

mjmmjJ

mjjjmjJ

z ,,

,)1(, 22

h

h

=

+=

1,)1()1(, ±±−+=± mjmmjjmjJ h

jjjjjm

j

,1,,2,1,

,3,2

5,2,

2

3,1,

2

1,0

−+−+−−=

=

K

K

Orbital Angular Momentum

• For orbital angular momentum we have:

• So that:

• In coordinate representation we have:

PRLrrv

×=

xyz YPXPL −=

€

v L →−ih

r r ×

r ∇ ( )

φθθ

φθ

∂

∂

∂

∂

∂

∂=∇×

sin

1100det

rrr

r

eee

rr

rrr

rr

θφθ φθ ∂

∂+

∂

∂−= ee

rr

sin

1

∂

∂+

∂

∂−−=

θφθ φθ eeiLrr

hr

sin

1

The z-component of L

• Lz is defined by:

LeL zz

rr⋅=

∂

∂+

∂

∂−−→

θφθ φθ eeiLrr

hr

sin

1

xer

zer

θ rr

θθθθ eee rz

rrrsincos −=

θer

rer

LeLeLe rz

rrrrrr⋅−⋅=⋅ θθθ sincos

φ∂∂

−= hiLz

0=⋅Lerrr

φθθ ∂

∂−=⋅sin

1Lerr

Laplacian

• In spherical coordinates, the Laplacian isgiven by:

• In QM, the Kinetic Energy obeys:

• Now from Classical Mechanics, we know that:

• By comparison, we see that we must have:

2

2

2222

22

sin

1sin

sin

11

φθθθ

θθ ∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂=∇

rrrrrr

2

22

22..

Mr

L

M

pEK r +=

( )rMM

Pr

rhrψψ 2

22

22∇−=

( )rLrr

hr

ψφθθ

θθθ

ψ

∂

∂+

∂

∂

∂

∂−=

2

2

222

sin

1sin

sin

1

( )rr

rrMr

rhψ

φθθθ

θθ

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂=

2

2

22

2

2

sin

1sin

sin

1

2

Changing coordinate system and/orbasis

• Coordinate representation in sphericalcoordinates:

• Note that the angular momentum operatordidn’t depend on r or ∂/∂r:

• Decompose as tensor product:

• Alternate basis sets:

• Can be combined to give four basis choices:

( ) ( )Ω⊗= θφθφ

Rrr

0],[ =RLr

0],[ =rPLr

rR→r

iPr ∂

∂−→ h

( ) ( )Rr

Rpr ↔

( ) ( )ΩΩ↔ mlθφ

€

rθφ{ }

€

r,θ,φ{ }

€

pr,θ,φ{ }

€

r,l,m{ }

€

pr,l,m{ }

φθ ,, →ΦΘθφθ φθ ∂

∂+

∂

∂−→ eeL

rrr

sin

1

These two are the most commonly used

Allowed quantum numbers for orbitalangular momentum

• In coordinate representation, the eigenvalueequation for Lz becomes:

• Which has the solution:

• The wave-function must be single-valued, sothat:

• Integer m values occur only for integer lvalues

• Therefore half-integer l values are forbiddenfor the case of orbital angular momentum

mmmi ,,,, lhlh φθφθφ

=∂

∂−

φθφθ imemm ,0,,, ll =

12 =πimeK,3,2,1,0 ±±±=m

Kl ,3,2,1,0=

mmmLz ,, lhl =

Spherical Harmonics

• The transformation coefficients from angularto angular momentum representation arecalled `Spherical Harmonics’– Denoted as:

– In Mathematica:

• Some properties:

• Any function of θ and φ can be expandedonto Spherical Harmonics:

),(,, φθφθ mYm ll =

SphericalHarmonicY[l,m,_,_]

€

sinθdθ dφ0

2π

∫0

π

∫ Ylm (θ,φ)( )

∗Y ′ l

′ m (θ,φ) = δl, ′ l δm, ′ m

€

Ylm (θ,φ)( )

∗= −1( )mYl

−m (θ,φ)

€

Ylm (θ,φ) = −1( )m

2l +1( ) l −m( )!4π l + m( )!

Plm (cosθ)eimφ

€

f θ,φ( ) = cl,mm=−l

l

∑l= 0

∞

∑ Ylm (θ,φ)

cl,m = sinθdθ dφ0

2π

∫0

π

∫ Ylm (θ,φ)( )

∗f (θ,φ)

€

l,m ′ l , ′ m = δl, ′ l δm, ′ m

Orbitals

• We call the different l states ‘orbitals’

l=0 ‘S’ orbitals

m=0

l=1 ‘P’ orbitals

m=-1,0,1

l=2 ‘D’ orbitals

m=-2,-1,0,1,2

l=3 ‘F’ orbitals

m=-3,-2,-1,0,1,2,3 etc…

• The number of sub-orbitals (m-states) isgiven by:

12 += llN

€

Y00(θ,φ) =

14π