SME
Thermal & Statistical Physics · Problem type
A quantum oscillator: Z, mean quanta and level shares
Use this for “A molecule’s vibration has ħω = 0.12 eV. At 600 K, find Z, the mean number of quanta, the mean energy and the share in the ground level.”
Chart
The bars add to 1.
⟨n⟩ ≥ 0 × 0.6321 + 1 × 0.2325 + 2 × 0.0856 + 3 × 0.0315 + 4 × 0.0116 + 5 × 0.0067 = 0.5781
The last bar holds every value from 5 up, counted here as 5: the true mean is a little higher.
This is an example. Type your own values to start a new problem.
Z
=
1
1
−
e
−
eV
÷
k
B
K
=
x
Reduced spacing ħω ÷ kB T
calculated
⟨n⟩
Mean quanta
calculated
⟨E⟩
Mean energy
calculated · eV
p₀
Share at n = 0
worked out (not typed)
p₁
Share at n = 1
worked out (not typed)
p₂
Share at n = 2
worked out (not typed)
p₃
Share at n = 3
worked out (not typed)
p₄
Share at n = 4
worked out (not typed)
p₅₊
Share at n ≥ 5
worked out (not typed)
C ÷ kB
Heat capacity over kB
worked out (not typed)
Formulas
x
=
ħω
8.617 ×
10
-5
×
T
1
=
0.02585
8.617 ×
10
-5
× 300
Z
=
1
1 −
e
−
x
1.582
=
1
1 −
e
-1
⟨n⟩
=
1
e
x
− 1
0.582
=
1
e
1
− 1
⟨E⟩ = ⟨n⟩ × ħω
0.01505 = 0.582 × 0.02585
p₀
=
1
−
e
−
x
0.6321
=
1
−
e
-1
p₁
=
(1
−
e
−
x
)
×
e
−
x
0.2325
=
(1
−
e
-1
)
×
e
-1
p₂
=
(1
−
e
−
x
)
×
e
−2 ×
x
0.08555
=
(1
−
e
-1
)
×
e
−2 × 1
p₃
=
(1
−
e
−
x
)
×
e
−3 ×
x
0.03147
=
(1
−
e
-1
)
×
e
−3 × 1
p₄
=
(1
−
e
−
x
)
×
e
−4 ×
x
0.01158
=
(1
−
e
-1
)
×
e
−4 × 1
p₅₊
=
e
−5 ×
x
0.006739
=
e
−5 × 1
C
kB
=
x
2
×
e
x
(
e
x
− 1)²
0.9207
=
1
2
×
e
1
÷
(
e
1
− 1)
2
Assumptions
•
One quantum oscillator with levels nħω, n = 0, 1, 2, …, measured from the ground level (the ½ħω zero point cancels in every share).
•
In contact with a reservoir at T: level n’s share is e−nx ÷ Z, with x = ħω ÷ kBT and kB = 8.617 × 10⁻⁵ eV/K.
•
Z is a geometric series: 1 + e−x + e−2x + … = 1 ÷ (1 − e−x). An Einstein solid is 3N such oscillators.
•
The last bar is every level from 5 up together, e−5x, so the bars add to 1.
Step-by-step
WE KNOW
ħω = 0.02585 eV, T = 300 K
FIND
reduced spacing ħω ÷ kB T (x), partition function (Z), mean quanta (⟨n⟩), mean energy (⟨E⟩), share at n = 0 (p₀), share at n = 1 (p₁), share at n = 2 (p₂), share at n = 3 (p₃), share at n = 4 (p₄), share at n ≥ 5 (p₅₊), heat capacity over kB (C ÷ kB)
Step 1 · Find reduced spacing ħω ÷ kB T (x)
Use x = ħω ÷ (8.617 × 10⁻⁵ × T)
The spacing in units of kBT: below 1 many levels are lit, well above 1 almost only the ground level.
x
=
0.02585
8.617 ×
10
-5
× 300
x
=
0.02585
0.02585
x
=
0.999961
Step 2 · Find partition function (Z)
Use Z = 1 ÷ (1 − e⁻ˣ)
Add
the
factors
e
−nx
of
every
level:
a
geometric
series,
first
term
1
and
ratio
e
−
x
.
Z
=
1
1 −
e
−0.999961
Z
=
1
1 − 0.3679
Z
=
1
0.6321
Z
=
1.582
Step 3 · Find mean quanta (⟨n⟩)
Use ⟨n⟩ = 1 ÷ (eˣ − 1)
The
mean
of
n
over
the
shares
(1
−
e
−
x
)
e
−nx
:
the
Planck
formula
for
the
mean
number
of
quanta.
⟨n⟩
=
1
e
0.999961
− 1
⟨n⟩
=
1
2.718 − 1
⟨n⟩
=
1
1.718
⟨n⟩ = 0.582012
Step 4 · Find mean energy (⟨E⟩)
Use ⟨E⟩ = ⟨n⟩ × ħω
Each quantum carries ħω, so the mean energy above the ground level is the mean number of quanta times ħω.
⟨E⟩ = 0.582012 × 0.02585
⟨E⟩ = 0.01505 eV
Step 5 · Find share at n = 0 (p₀)
Use p₀ = 1 − e⁻ˣ
Level
0’s
factor
is
1,
so
its
share
is
1
Z
=
1
−
e
−
x
.
p₀
=
1
−
e
−0.999961
p₀
=
1
−
0.3679
p₀
=
0.6321
Step 6 · Find share at n = 1 (p₁)
Use p₁ = (1 − e⁻ˣ) × e⁻ˣ
Level
1’s
factor
over
Z
:
1
Z
=
1
−
e
−
x
,
times
e
−
x
.
p₁
=
(1
−
e
−0.999961
)
×
e
−0.999961
p₁
=
(1
−
0.36789)
×
e
-1
p₁
=
0.63211
×
e
-1
p₁
=
0.63211
×
0.36789
p₁
=
0.2325
Step 7 · Find share at n = 2 (p₂)
Use p₂ = (1 − e⁻ˣ) × e−2 × x
Level
2’s
factor
over
Z
:
1
Z
=
1
−
e
−
x
,
times
e
−2
x
.
p₂
=
(1
−
e
−0.999961
)
×
e
−2 × 0.999961
p₂
=
(1
−
0.36789)
×
e
−1.9999
p₂
=
0.63211
×
e
−1.9999
p₂
=
0.63211
×
0.13535
p₂
=
0.08555
Step 8 · Find share at n = 3 (p₃)
Use p₃ = (1 − e⁻ˣ) × e−3 × x
Level
3’s
factor
over
Z
:
1
Z
=
1
−
e
−
x
,
times
e
−3
x
.
p₃
=
(1
−
e
−0.999961
)
×
e
−3 × 0.999961
p₃
=
(1
−
0.3679)
×
e
-3
p₃
=
0.6321
×
e
-3
p₃
=
0.6321
×
0.04979
p₃
=
0.03147
Step 9 · Find share at n = 4 (p₄)
Use p₄ = (1 − e⁻ˣ) × e−4 × x
Level
4’s
factor
over
Z
:
1
Z
=
1
−
e
−
x
,
times
e
−4
x
.
p₄
=
(1
−
e
−0.999961
)
×
e
−4 × 0.999961
p₄
=
(1
−
0.3679)
×
e
-4
p₄
=
0.6321
×
e
-4
p₄
=
0.6321
×
0.01832
p₄
=
0.01158
Step 10 · Find share at n ≥ 5 (p₅₊)
Use p₅₊ = e−5 × x
Levels
5
and
up:
(1
−
e
−
x
)(
e
−5
x
+
e
−6
x
+
…)
adds
to
e
−5
x
,
whatever
the
first
five
leave.
p₅₊
=
e
−5 × 0.999961
p₅₊
=
e
-5
p₅₊ = 0.006739
Step 11 · Find heat capacity over kB (C ÷ kB)
Use C ÷ kB = x² × eˣ ÷ (eˣ − 1)²
d⟨E⟩/dT for one oscillator (Einstein’s heat capacity), in kB: near 1 when kBT ≫ ħω, falling to 0 as the oscillator freezes into its ground level.
C
kB
=
0.999961
2
×
e
0.999961
(
e
0.999961
− 1)²
C
kB
=
1
2
×
e
1
÷
(2.7182 − 1)
2
C
kB
=
1
2
×
e
1
1.7182
2
C
kB
=
0.99992
×
2.7182
2.9522
C
kB
=
2.718
2.9522
C
kB
=
0.9207
Check
Put the numbers back in. Both sides should match.
0.999961
=
0.02585
8.617 ×
10
-5
× 300
✓
1.582
=
1
1 −
e
−0.999961
✓
0.582012
=
1
e
0.999961
− 1
✓
0.01505 = 0.582012 × 0.02585 ✓
0.6321
=
1
−
e
−0.999961
✓
0.2325
=
(1
−
e
−0.999961
)
×
e
−0.999961
✓
0.08555
=
(1
−
e
−0.999961
)
×
e
−2 × 0.999961
✓
0.03147
=
(1
−
e
−0.999961
)
×
e
−3 × 0.999961
✓
0.01158
=
(1
−
e
−0.999961
)
×
e
−4 × 0.999961
✓
0.006739
=
e
−5 × 0.999961
✓
0.9207
=
0.999961
2
×
e
0.999961
(
e
0.999961
− 1)²
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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