7.3 Orbital angular momentum
7.3.1 Polar coordinates
These coordinates are particularly useful if the potential
depends only on the distance from center: V(r) = V(r)
e.g. Coulomb potential etc.
z
θ
r
y
ϕ
x
x = x(r, θ, ϕ)
y = y(r, θ, ϕ)
z = z(r, θ, ϕ)
In the Schrödinger picture we have to transform
H =−
¤2
2m
∂2
∂x 2
+
∂2
∂y 2
+
∂2
∂z 2
+ V(r)
Transformation needs some efforts of partial differentiation - we only give the results:
2
2
¤ ∇2 = − ¤
− 2m
2m
1 ∂
r 2 ∂r
r 2 ∂r∂
2
+ L 2
2mr
(first term: radial energy - last term: rotational energy)
with: L 2 = −¤ 2
∂
1
sin θ ∂θ
∂
sin θ ∂θ
+
∂2
1
2
sin θ ∂ϕ 2
L 2 is here just an abbreviation but we guess:
L 2 is the square of the angular momentum operator!
2
We can write H = H rad + L 2 + V(r)
2mr
which depends on r, θ, ϕ so that we have to solve:
Hψ nlm (r, θ, ϕ) = E nlm ψ nlm (r, θ, ϕ)
Ansatz (separation): ψ nlm (r, θ, ϕ) = R n (r)Y lm (θ, ϕ)
7.3.2 Orbital angular momentum: definition
Alternative approach to angular momentum:
11
L = r×p
3
hence in quantum mechanics: L = r × −i¤∇
= −i¤
ex
ey
ez
x
y
z
∂
∂x
∂
∂y
∂
∂y
this has to be expressed in polar coordinates we only calculate one component as an example
L z = −i¤ x ∂y∂ − y ∂x∂
now, since: ϕ = ϕ(x, y, z) with x = r sin θ cos ϕ and y = r sin θ sin ϕ
ì
∂
∂ϕ
=
∂x ∂
∂ϕ ∂x
+
∂y ∂
∂ϕ ∂y
∂z ∂
∂ϕ ∂z
+
=
∂
∂
= −r sin θ sin ϕ ∂x
+ r sin θ cos ϕ ∂y
+0
∂
∂
= −y ∂x
+ x ∂y
=
∂
∂ϕ
∂
∂
= −y ∂x
+ x ∂y
∂
∂
hence L z = −i¤ ∂ϕ
in complete analogy to p x = −i¤ ∂x
so called pairs of canonic conjugate coordinates:(L z and ϕ) or (p x and x)
The components L x and L y are more complicated.
Without proof we communicate that:
L 2 = L 2x + L 2y + L 2z = -¤ 2
∂
1
sin θ ∂θ
∂
sin θ ∂θ
+
∂2
1
2
sin θ ∂ϕ 2
as before (alternative - formal - derivation of orbital angular momentum).
7.3.3 Eigenvalues and eigenfunctions of angular momenta
7.3.3.1 The z - component
L z Φ(ϕ) = § z Φ(ϕ)
∂
with L z = −i¤ ∂ϕ
Φ = Ce
∂ Φ + § Φ = 0 (this is the Eigenvalue equation)
ì i¤ ∂ϕ
z
i §z ϕ
¤
C = normalization constant
Now we have to apply some physics: which values of § z are meaningfull?
!
Φ must be single valued! i.e. Φ(0) = Φ(2π)
!
ì e0 = e
i § z 2π
¤
this is only possible if
then e
i § z 2π
¤
§z
¤
= m is an integer (m = 0, ±1, ±2, . . . . )
= e i2πm = 1
L z Φ m (ϕ) = m¤Φ m (ϕ)
hence: eigenvalues of L z are § z = m¤ with m = 0, ±1, ±2, . . . .
Φ m = C m exp(imϕ)
we call m the magnetic or projection quantum number
!
Orthonormality: δ mm v =
Φ m |Φ m v
= C ∗m C m v ∫ 0 exp(−imϕ) exp im v ϕ dϕ =
2π
12
for m ≠ m v
0
= C ∗m C m v
Φ m |Φ m v
2π for m = m v
!
hence |C m | 2 2π = 1 ì C m =
1
2π
note phase convention (C m is real)!
Final result:
L z Φ m = m¤Φ m : Φ m =
exp(imϕ) with m = 0, ±1, ±2, . . . .
1
2π
7.3.3.2 Components in x and y direction
For the L x and L y components the calculations are more complicated but in principle trivial. Without proof
we note:
L z and L x and L y are not (pairwise) simultaneously measurable (i.e. do not commute):
L x L y ≠ L y L x and L x L z ≠ L z L x and L y L z ≠ L z L y
This can be shown exactly but is also plausible, since each component depends on different combinations
of differential operators of ϕ and θ.
7.3.3.3 Magnitude of angular momentum
we had
L 2 = -¤ 2
∂
1
sin θ ∂θ
∂
sin θ ∂θ
+
∂2
1
2
sin θ ∂ϕ 2
Eigenvalue equation for L 2 :
L 2 Y(θ, ϕ) = λ 2 Y(θ, ϕ)
Ansatz: Y(θ, ϕ) = Θ(θ)Φ(ϕ) and we try the eigenfunctions of L z : Φ m = C m exp(imϕ)
insert into eigenvalue equation:
−¤ 2
∂
1
sin θ ∂θ
ì −¤ 2
sin θ
∂
1
sin θ ∂θ
∂Θ(θ)
∂θ
sin θ ∂Θ
∂θ
Φ m (ϕ) +
−
1 Θ(θ)
sin 2 θ
m2 Θ
sin 2 θ
−m 2 Φ m (ϕ) = λ 2 Θ(θ)Φ m (ϕ)
= λ2Θ
Without proof we report that physically reasonable solutions: finite and single valued for 0 ≤ θ ≤ π
exist only for eigenvalues:
λ 2 = §(§ + 1)¤ 2 with § = 0, 1, 2, . . . and m = 0, ±1, ±2, . . . . ±§
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