Download Lecture 25: Introduction to the Quantum Theory of Angular Momentum Phy851 Fall 2009

Lecture 25:
Introduction to the Quantum Theory
of Angular Momentum
Phy851 Fall 2009
Goals
1. Understand how to use different coordinate
systems in QM (cartesian, spherical,…)
2. Derive the quantum mechanical properties
of Angular Momentum
•
Use an algebraic approach similar to what we
did for the Harmonic Oscillator
3. Use the resulting theory to treat spherically
symmetric problems in three dimensions
•
Calculating the Hydrogen atom energy levels
will be our target goal
Motion In 3Dimensions
• For a particle moving in three dimensions,
there is a distinct quantum state for every
point in space.
– Thus each position state is now labeled by a
vector
v
X →R
r
x → r
– Vector operators are really three operators
r
r
r
r
R = X ex + Y e y + Z ez
Scalar Operators
X ,Y , Z
Ordinary
r r Vectors
r
ex , e y , ez
• Coordinate system not unique:
– For example, we could use spherical
coordinates
r
r
R = R er (Θ, Φ )
Scalar Operators
Vector Operator
r
er (Θ, Φ )
R, Θ, Φ
r
r
r
r
er (Θ, Φ ) = sin Θ cos Φex + sin Θ sin Φe y + cos Θez
You can never go wrong with Cartesian
Coordinates
• In all other coordinate systems, the unit
vectors are also operators
– so must be treated carefully
• Eigenstates:
rr rr
Rr =r r
• For each point in space there is a position
eigenstate
• How we want to label these points is up to
us:
r r rr r rr
r
X r = ex ⋅ R r = ex ⋅ r r = x r
r
r
Yr =yr
r
r
Z r =zr
r
r = x, y , z
r
r = r ,θ , φ
r
r = ρ , z, φ
Different ways
to refer to
the same state
Orthogonality
• In three dimensions, the orthogonality
condition becomes:
rr
r
3 r
′
r r = δ ( r − r ′)
– Cartesian coordinates:
€ xyz x ′y ′z′ = δ ( x − x ′)δ ( y − y ′)δ ( z − z′)
– Spherical coordinates:
€
δ (θ − θ ′) δ (φ − φ ′)
rθφ r′θ ′φ ′ = δ ( r − r′)
r sin θ
r
– Mixing coordinates:
€
€
xyz rθφ = δ ( x − r sin θ cos φ )δ ( y − r sin θ sin φ )δ ( z − r cos θ )
Wavefunctions
• Wavefunctions are defined in the usual way
as:
r
r
ψ (r ) = r ψ
r 2
∫ d r ψ(r ) = 1
3
€
ψ ( x, y, z ) = xyz ψ
∞
∞
∞
∫ dx ∫ dy ∫ dz ψ (x, y,z)
−∞
−∞
2
=1
−∞
ψ (r , θ , φ ) = rθφ ψ
€
∞
π
2π
2
∫ dr ∫ r sinθdθ ∫ rdφ ψ (r,θ,φ ) = 1
0
€
0
0
Momentum
• The three-dimensional momentum vector
operator is:
r r
r
r
P = ex Px + e y Py + ez Pz
• The three-dimensional Hamiltonian is
r r
r
P⋅P
H=
+V R
2m
()
• In Cartesian components, this becomes:
€
€
2
r
Pz2
Px2 Py
H=
+
+
+V R
2m 2m 2m
()
 h2  d 2

d2
d2 


(
)
xyz H ψ = −
+ 2 + 2  + V x, y, z  xyz ψ
2

dy
dz 
 2m  dx

or
 h2 2
r
r r
r H ψ = −
∇ + V (r ) r ψ
 2m

Angular Momentum
• We can decompose the momentum operator
onto spherical components as:
r r
r
r
P = er (Θ,Φ) Pr + eθ (Θ,Φ) Pθ + eφ (Θ,Φ) Pφ
r
r
Pr = er (Θ,Φ) ⋅ P
r
r
Pθ = eθ (Θ,Φ) ⋅ P
r
r
Pφ = eφ (Θ,Φ) ⋅ P
• The unit-vector operators are:
€
r
r
r
r
er (Θ, Φ ) = sin Θ cos Φex + sin Θ sin Φe y + cos Θez
r
r
r
r
eθ (Θ, Φ ) = − cos Θ cos Φex − cos Θ sin Φe y + sin Θez
r
r
r
eφ (Θ, Φ ) = − sin Φex + cos Φe y
• From the above decomposition, we are led to
define the Angular momentum as:
r r
r
L = eθ (Θ,Φ) Pθ + eφ (Θ,Φ) Pφ
• This is not the usual definition (L=RxP) but is
equivalent.
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