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Quantum Mechanics

Angular momentum and spin

Use this for “An electron in a d orbital has ℓ = 2. For mℓ = −2, find the size of L, its z-part and the angle between L and the z-axis.”

Diagram

z−2ħ−ħ0ħ2ħ|L| = 2.449ħLz = ħθ = 65.91°
|L|
=
√
2 × 3
ħ
=
2.449ħ,
and
its
z-part
is
mħ
=
ħ.
cos
θ
=
1
2.449
,
so
θ
=
65.91°
from
the
z-axis.
2ℓ + 1 = 5 cones; L is never straight along z, and its x- and y-parts stay unknown round the cone.
This is an example. Type your own values to start a new problem.
cos
θ
°
=
mℓ
(
+
1)
|L|
Size of L
calculated · ħ
Lz
z-part of L
calculated · ħ
states
Number of states
calculated

Formulas

|L|
=
√(
ℓ
×
(
ℓ
+
1))
2.4495 = √(2 × (2 + 1))
Lz = mℓ
1 = 1
θ
=
cos⁻¹(mℓ
÷
|L|)
65.91
=
cos⁻¹(
1
2.4495
)
states
=
2
×
ℓ
+
1
5 = 2 × 2 + 1

Assumptions

•
Only |L| and one component, Lz, are known together; Lₓ and Lᵧ spread round a cone about z.
•
mℓ runs over the whole numbers from −ℓ to ℓ, so there are 2ℓ + 1 states. Sizes are in units of ħ.
•
|mℓ| < √(ℓ(ℓ + 1)), so L never points straight along z. An s state (ℓ = 0) has no L at all.

Step-by-step

WE KNOW
ℓ = 2, mℓ = 1
FIND
size of L (|L|), z-part of L (Lz), angle from the z-axis (θ), number of states (states)
Step 1 · Find size of L (|L|)
Use |L| = √(ℓ × (ℓ + 1))
The
size
of
L
in
ħ
is
the
square
root
of
ℓ
(
ℓ
+
1),
always
a
little
more
than
ℓ
.
|L| = √(2 × (2 + 1))
|L|
=
√
2 × 3
|L|
=
√
6
|L| = 2.4495 ħ
Step 2 · Find z-part of L (Lz)
Use Lz = mℓ
The z-part of L is mℓ units of ħ.
Lz = 1 ħ
Step 3 · Find angle from the z-axis (θ)
Use θ = cos⁻¹(mℓ ÷ |L|)
The z-part over the size is the cosine of the angle from z.
θ
=
cos⁻¹(
1
2.4495
)
θ
=
65.91°
Step 4 · Find number of states (states)
Use states = 2 × ℓ + 1
Count
mℓ
from
−
ℓ
to
ℓ
:
2
ℓ
+
1
values.
states = 2 × 2 + 1
states = 4 + 1
states = 5
Check
Put the numbers back in. Both sides should match.
2.4495 = √(2 × (2 + 1)) ✓
1 = 1 ✓
65.91
=
cos⁻¹(
1
2.4495
)
✓
5 = 2 × 2 + 1 ✓
Topic check
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