SME
Thermal & Statistical Physics
Partition functions
Use this for “A defect has two states 0.02 eV apart in a crystal at 150 K. What share of the defects is in the upper state, and what are Z and the mean energy?”
Chart
The bars add to 1.
E(X) = 0 × 0.8737 + 1 × 0.1263
E(X) = 0.1263
This is an example. Type your own values to start a new problem.
Z
=
1
+
e
−
eV
÷
k
B
K
=
x
Reduced gap ε ÷ kB T
calculated
p₀
Ground share
calculated
p₁
Upper share
calculated
⟨E⟩
Mean energy
calculated · eV
F
Free energy
calculated · eV
Formulas
x
=
ε
8.617 ×
10
-5
×
T
1.934
=
0.05
8.617 ×
10
-5
× 300
Z
=
1
+
e
−
x
1.1445
=
1
+
e
−1.934
p₀
=
1
Z
0.8737
=
1
1.1445
p₁
=
e
−
x
Z
0.1263
=
e
−1.934
1.1445
p₀
+
p₁
=
1
0.8737 + 0.1263 = 1
⟨E⟩
=
ε
×
p₁
0.006315 = 0.05 × 0.1263
F
=
−8.617 ×
10
-5
×
T
×
ln(
Z
)
−0.00349
=
−8.617 ×
10
-5
×
300
×
ln(1.1445)
Assumptions
•
The system trades energy with a reservoir at T (the canonical ensemble); kB = 8.617 × 10⁻⁵ eV/K.
•
A state’s share is its Boltzmann factor e−E/kBT over Z, the sum of the factors of every state.
•
Two single levels (no degeneracy): the ground level at 0 and the upper one at ε.
•
The free energy F = −kBT ln Z; at low T, Z is near 1 and almost every system is in the ground level.
Step-by-step
WE KNOW
ε = 0.05 eV, T = 300 K
FIND
reduced gap ε ÷ kB T (x), partition function (Z), ground share (p₀), upper share (p₁), mean energy (⟨E⟩), free energy (F)
Step 1 · Find reduced gap ε ÷ kB T (x)
Use x = ε ÷ (8.617 × 10⁻⁵ × T)
The gap measured in units of kBT, the typical thermal energy: every share below depends on this ratio alone.
x
=
0.05
8.617 ×
10
-5
× 300
x
=
0.05
0.02585
x
=
1.9342
Step 2 · Find partition function (Z)
Use Z = 1 + e⁻ˣ
Add
the
Boltzmann
factors
of
the
two
levels:
e
0
=
1
for
the
ground
level,
e
−
ε
/kBT
for
the
upper
one.
Z
=
1
+
e
−1.9342
Z
=
1
+
0.1445
Z
=
1.1445
Step 3 · Find ground share (p₀)
Use p₀ = 1 ÷ Z
The ground level’s factor is 1, so its share is 1 over the sum of the factors.
p₀
=
1
1.1445
p₀
=
0.8737
Step 4 · Find upper share (p₁)
Use p₁ = e⁻ˣ ÷ Z
The upper level’s Boltzmann factor over the sum of the factors.
p₁
=
e
−1.9342
1.1445
p₁
=
0.14454
1.1445
p₁
=
0.1263
Step 5 · Find mean energy (⟨E⟩)
Use ⟨E⟩ = ε × p₁
Average
the
energies
by
their
shares:
0
×
p₀
+
ε
×
p₁
.
⟨E⟩ = 0.05 × 0.1263
⟨E⟩ = 0.006315 eV
Step 6 · Find free energy (F)
Use F = −8.617 × 10⁻⁵ × T × ln(Z)
The
free
energy
follows
from
Z
alone:
F
=
−kBT
ln
Z
,
never
above
0.
F
=
−8.617 ×
10
-5
×
300
×
ln(1.1445)
F
=
(−8.617
×
10
-5
)
×
300
×
0.13497
F
=
−0.025851
×
0.13497
F
=
−0.00349
eV
Check
Put the numbers back in. Both sides should match.
1.9342
=
0.05
8.617 ×
10
-5
× 300
✓
1.1445
=
1
+
e
−1.9342
✓
0.8737
=
1
1.1445
✓
0.1263
=
e
−1.9342
1.1445
✓
0.8737 + 0.1263 = 1 ✓
0.006315 = 0.05 × 0.1263 ✓
−0.00349
=
−8.617 ×
10
-5
×
300
×
ln(1.1445)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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