SME
Quantum Mechanics · Problem type
Adding two angular momenta
Use this for “A d electron has ℓ = 2 and spin s = ½. Which total j can it have, how many states does each hold, and how many states are there in all?”
Table
One total j (j)
States with this j (gⱼ)
Smallest total j jₘᵢₙ = 1/2
Largest total j jₘₐₓ = 3/2
Number of j values nⱼ = 2
Total states N = 6
This is an example. Type your own values to start a new problem.
j₁
First angular momentum
example
j₂
Second angular momentum
example
jₘᵢₙ
Smallest total j
worked out (not typed)
jₘₐₓ
Largest total j
worked out (not typed)
nⱼ
Number of j values
worked out (not typed)
N
Total states
worked out (not typed)
j
One total j
example
gⱼ
States with this j
calculated
Formulas
jₘᵢₙ
=
|
j₁
−
j₂
|
1
2
=
|1
−
1
2
|
jₘₐₓ
=
j₁
+
j₂
3
2
=
1
+
1
2
nⱼ = jₘₐₓ − jₘᵢₙ + 1
2
=
3
2
−
1
2
+
1
N
=
(2
×
j₁
+
1)
×
(2
×
j₂
+
1)
6
=
(2
×
1
+
1)
×
(2
×
1
2
+
1)
gⱼ
=
2
×
j
+
1
4
=
2
×
3
2
+
1
Assumptions
•
j₁ and j₂ are whole or half numbers (ℓ is whole, a spin s is ½); sizes in units of ħ.
•
The total j runs from |j₁ − j₂| to j₁ + j₂ in steps of 1, and each j holds 2j + 1 states.
•
Coupling only regroups the states: the 2j + 1 add up to (2j₁ + 1)(2j₂ + 1).
Step-by-step
WE KNOW
j₁ = 1, j₂ = 1/2, j = 3/2
FIND
smallest total j (jₘᵢₙ), largest total j (jₘₐₓ), number of j values (nⱼ), total states (N), states with this j (gⱼ)
Step 1 · Find smallest total j (jₘᵢₙ)
Use jₘᵢₙ = |j₁ − j₂|
The two point as nearly opposite as they can: the smaller is taken from the larger.
jₘᵢₙ
=
|1
−
1
2
|
jₘᵢₙ = |0.5|
jₘᵢₙ
=
1
2
Step 2 · Find largest total j (jₘₐₓ)
Use jₘₐₓ = j₁ + j₂
The two line up as nearly as they can: their sizes add.
jₘₐₓ
=
1
+
1
2
jₘₐₓ
=
3
2
Step 3 · Find number of j values (nⱼ)
Use nⱼ = jₘₐₓ − jₘᵢₙ + 1
Count
the
j
values
from
jₘᵢₙ
to
jₘₐₓ
in
steps
of
1,
both
ends
included.
nⱼ
=
3
2
−
1
2
+
1
nⱼ = 2
Step 4 · Find total states (N)
Use N = (2 × j₁ + 1) × (2 × j₂ + 1)
Every
state
of
the
first
with
every
state
of
the
second;
the
2
j
+
1
of
each
j
add
up
to
the
same.
N
=
(2
×
1
+
1)
×
(2
×
1
2
+
1)
N
=
3
×
(1
+
1)
N
=
3
×
2
N
=
6
Step 5 · Find states with this j (gⱼ)
Use gⱼ = 2 × j + 1
Its
mⱼ
run
from
−
j
to
j
in
steps
of
1.
gⱼ
=
2
×
3
2
+
1
gⱼ = 3 + 1
gⱼ = 4
Check
Put the numbers back in. Both sides should match.
1
2
=
|1
−
1
2
|
✓
3
2
=
1
+
1
2
✓
2
=
3
2
−
1
2
+
1
✓
6
=
(2
×
1
+
1)
×
(2
×
1
2
+
1)
✓
4
=
2
×
3
2
+
1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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