Simple harmonic oscillator
The SHO has non-degenerate levels with n = (n + 21 )h̄ω for n = 0, 1, 2, . . ..
a. The internal partition function is
ζ(T )
∞
X
=
n=0
∞
X
=
e−βn
e−β(n+1/2)h̄ω
n=0
= e−βh̄ω/2
∞
X
e−βh̄ωn ,
n=0
but
so
1
= 1 + x + x2 + · · · for 0 ≤ x < 1,
1−x
1
−βh̄ω/2
ζ(T ) = e
.
1 − e−βh̄ω
Thus
ln ζ(T ) = −βh̄ω/2 − ln 1 − e−βh̄ω
and
eint (T )
= −
=
=
∂ ln ζ
∂β
h̄ωe−βh̄ω
1
h̄ω +
2
1 − e−βh̄ω
1
h̄ω
h̄ω + βh̄ω
.
2
e
−1
b. The internal heat capacity is
cint
V (T )
∂eint
∂β ∂eint
=
∂T
∂T ∂β
1
h̄ωeβh̄ω
(h̄ω) − βh̄ω
=
−
kB T 2
(e
− 1)2
2
h̄ω
e−h̄ω/kB T
= kB
.
kB T
(1 − e−h̄ω/kB T )2
=
c. As is often the case, it is easier to derive this result than to interpret it.
Low temperature behavior: kB T h̄ω; βh̄ω 1; e−βh̄ω ≈ 0, so
2
h̄ω
cint
(T
)
≈
k
e−h̄ω/kB T .
B
V
kB T
As in the Schottky case, this is very flat at T = 0.
1
High temperature behavior: kB T h̄ω; βh̄ω 1; e−βh̄ω ≈ 1. But if we just leave it like this, we get
≈ 0(1/0), which is not helpful. We need a better approximation for e−βh̄ω .
cint
V (T )
Let
x≡
h̄ω
−→ 0.
kB T
Then
e−x = 1 − x + 21 x2 − 16 x3 + O(x4 )
so
e−x
(1 − e−x )2
1 − x + 12 x2 − 16 x3 + O(x4 )
2
= kB x 2
1 − 1 − x + 21 x2 − 16 x3 + O(x4 )
1 − x + 21 x2 − 16 x3 + O(x4 )
= kB x2
2
x2 1 − 12 x + 61 x2 + O(x3 )
1 − x + 21 x2 + O(x3 )
= kB 2 .
1 − 21 x + 16 x2 + O(x3 )
cint
V (T )
= kB x2
This result is good enough to tell us that
when
T −→ ∞,
x −→ 0,
so
cint
V −→ kB ,
the classical equipartition result.
d. But we can also go on to get the leading order quantal corrections to equipartition:
1 − x + 21 x2 + O(x3 )
int
cV (T ) = kB 2 .
1 − 21 x + 16 x2 + O(x3 )
But
1
= 1 − 2z + 3z 2 − 4z 3 + O(z 4 ),
(1 + z)2
where in our case
z = − 21 x + 16 x2 + O(x3 ).
So
cint
V (T )
i
h
2
= kB 1 − x + 21 x2 + O(x3 ) 1 − 2 − 12 x + 61 x2 + O(x3 ) + 3 − 21 x + 16 x2 + O(x3 ) + O(x3 )
= kB 1 − x + 12 x2 + O(x3 ) 1 + x − 13 x2 + 3 14 x2 + O(x3 ) + O(x3 )
5 2
= kB 1 − x + 12 x2 + O(x3 ) 1 + x + 12
x + O(x3 )
5 2
= kB 1 + (x − x) + ( 12
x + 12 x2 − x2 ) + O(x3 )
1 2
= kB 1 − 12
x + O(x3 ) .
The leading order correction is negative, so as the temperature goes down, the heat capacity will start off
by falling below the equipartition line cint
V (T ) = kB .
2
e. So the graph is
cVint(T)
kB
0
hω/kB
T
3