Statistical Mechanics Lecture notes — Baruch Horovitz and class of 2007

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Statistical Mechanics Lecture notes — Baruch Horovitz and class of 2007
```Statistical Mechanics
Lecture notes — Baruch Horovitz and class of 2007
Contents
1. Ensemble Theory
2
1a. Thermodynamics (Review)
2
1b. Micro Canonical Ensemble (MCE)
5
1c. Ensemble theory - generalities
12
1d. Canonical Ensemble (CE)
14
1e. Grand Canonical Ensemble (GCE)
19
2. Quantum Statistical Mechanics
23
2a. Ensembles for ideal quantum gases
23
2b. Ideal Bose gas
28
2c. Ideal Fermi gas
34
2d. Non- Ideal gases
36
3. Phase Transitions
39
3a. First order transitions
39
3b. Second order phase transitions – Mean Field theory
45
3c. Exact results
46
3d. Landau’s Theory for second order phase transitions
49
4. Non- equilibrium
53
4a. Kinetic theory and Boltzmann’s equation:
53
4b. Brownian motion
56
4c. Fluctuation Dissipation Theorems (FDT)
60
4d. Onsager’s Relations
68
Appendix: Langevin’s equation from a Hamiltonian
72
1
Reference books:
R. K. Pathria, Statistical Mechanics
Landau & Lifshitz, Statistical Physics
K. Huang, Statistical Mechanics
S.-k Ma, Statistical Mechanics
F. Mohling, Statistical Mechanics Methods and Applications
G. H. Wannier, Statistical Physics
1. ENSEMBLE THEORY
1a. Thermodynamics (Review)
Macroscopic state: Set of measurable ”coordinates” of a system with many (→ ∞) degrees
of freedom, e.g. volume V , number of particles N , energy E.
Equilibrium: A macroscopic state that is uniquely determined by a small number of external
”forces”, e.g. pressure P , chemical potential µ, temperature T .
Pairs of force f and coordinate x generate work δW = f dx (δW is not necessarily an exact
differential, i.e. equation may not be integrated to yield a state function W).
E.g. δW = −P dV , δW = µdN
Microscopic degrees of freedom contain ”heat” energy Q, with S the coordinate, T the force.
δQ = T dS.
Thermodynamic limit: N → ∞, V → ∞ with N/V →const. Macroscopic state, equilibrium
etc. are defined in this limit. Define: Extensive variables that increase with N , e.g. V, E, S
and intensive variables that are constants in the thermodynamic limit, e.g. N/V, P, µ.
First Law: Two ways to exchange energy, work or ”heat”: δE = δQ + δW ; heat δQ is due
to microscopic degrees of freedom. There exists an adiabatic process for which δQ = 0.
Entropy S is a thermodynamic coordinate - proof:
2
Adiabatic process δQ = 0. Equilibrium determines a
curve E(x) from dE = f dx. Between curves change is
nonadiabatic, σ is an integration constant.
E = E(x, σ). Curves do not cross since f =
∂E
∂x
is unique in equilibrium. Also E(x) is single
valued ⇒ E(x, σ) is monotonic in σ. For constant x, δQ = dE = ( ∂E
) dσ
∂σ x
dE = f dx + τ dσ
)
τ = ( ∂E
∂σ x
σ is not unique - can choose σ̄ = A(σ) with A(σ) monotonic. Choose σ which is extensive
σ → S ∼ N , and define temperature T =
∂E
∂S
as intensive. Assumption in proof: only one σ
coordinate.
⇒ dE = T dS − P dV + µdN
H
E = E(S, V, N ) is a (single valued) state function, dE = 0. This is the first law.
R
H
Heat is not single valued - TdS depends on how V, N change along the path ( δQ 6= 0).
Second Law: In a closed system S(t) increases with time. Exchange dE1 = −dE2 in two
subsystems:
S = S1 + S2
dS
dS1 dE1
dS2 dE2
=
+
=
dt
dE1 dt
dE2 dt
1
1
−
T1 T2
dE1
>0
dt
⇒ energy flows from high to low T.
In equilibrium S is maximal ⇒ T1 = T2
dS =
1
P
µ
dE + dV − dN
T
T
T
If volumes of subsystems exchange dV1 = −dV2 with T1 = T2 (hence
1
2
) dV1 + ( ∂S
) dV2 = ( PT1 −
dS = ( ∂S
∂V1 E,N
∂V2 E,N
P2
)dV1
T
∂
∂E
terms vanish),
⇒ P1 = P2 equilibrium.
Particle exchange dN1 = −dN2
dS = (∂S1 /∂N1 )E,V dN1 + (∂S2 /∂N2 )E,V dN2 = (µ1 /T − µ2 /T )dN1
⇒ µ1 = µ2
chemical equilibrium.
T dS > dE +P dV −µdN = δQ irreversible process, i.e. S increases more than its equilibrium
change.
A process in which the energy is changed only by slow variation of external conditions so
3
that at every instant the system is in equilibrium. Furthermore, the system is ”thermally
isolated” - no energy transfer except the external condition.
Adiabatic process is reversible: expand in dλ/dt, λ external condition (e.g. volume)
dS
dt
= a + b dλ
+ 12 c( dλ
)2
dt
dt
a = 0: equilibrium condition
b = 0: since
⇒
dS
dλ
dS
dt
> 0 cannot depend on sign of
= 12 c dλ
+ ..... → 0 when
dt
dλ
dt
dλ
dt
→ 0.
⇒ dS = 0 in adiabatic process.
Note: A reversible process in a closed system is an adiabatic process and S is constant. A
reversible process in an open system is a process for which T dS = dE + P dV − µdN , i.e.
heat exchange with a reservoir is allowed and dS 6= 0.
Thermodynamic Functions
F = E − TS
Helmholtz free energy
dF = −SdT − P dV + µdN ⇒ F (T, V, N )
If an additional variational parameter is present F is minimized to reach dF = 0.
T, V, N fixed: dF = dE − T dS ≤ 0, for < 0 the process is irreversible.
G = E − TS + PV
Gibbs free energy
dG = −SdT + V dP + µdN ⇒ G(T, P, N )
T, P, N fixed: dG = dE − T dS + P dV < 0 and G is minimized.
Ω̃ = F − µN
dΩ̃ = −SdT − P dV − N dµ ⇒ Ω̃(T, V, µ)
T, V, µ fixed: dΩ̃ = dE − T dS − µdN < 0 and Ω̃ is minimized.
Extensiveness: λE = E(λS, λV, λN )
∂
| :
∂λ λ=1
E = T S − P V + µN
F = −P V + µN
G = µN
Ω̃ = −P V
4
(1)
1b. Micro Canonical Ensemble (MCE)
Basic assumption:
S = kB ln Ω(E)
Boltzman’s constant: kB = 1.38 · 10−16 erg/deg.
Ω(E) is the number of microstates for a given N, V, E, and states have equal weight :
Ω(E) =
X
δE−H{r}
{r}
{r} are microscopic degrees of freedom of the N particles.
We shall now prove extensivity and maximum property: Consider two subsystems, neglect
their interaction (justified since the number of particles near the surface is small, or surface
energy bulk energy),
H{r1 , r2 } = H1 {r1 } + H2 {r2 }
X
Ω(E) =
δE−H1 {r1 }−H2 {r2 }
{r1 ,r2 }
=
XX
δE1 −H1 {r1 }
E1 {r1 }
X
δE−E1 −H2 {r2 } =
X
E1
{r2 }
S1 (E1 ) + S2 (E − E1 )
exp
kB
Since Si ∼ Ni are large look for maximum in E1 (equivalent to steepest descent),
∂
(S1 (E1 ) + S2 (E − E1 )) |E¯1 = 0
∂E1
∂S1
∂S2
=
⇒ T1 = T2
∂E1
∂E2
Expand to 2nd order
1
1
(E1 − Ē1 )2 −
(E2 − Ē2 )2
S1 (E1 ) + S2 (E2 ) = S1 (Ē1 ) + S2 (Ē2 ) −
2k1
2k2
2 ∂( T1 )
1
∂ S
1
=
−
= −
= 2
2
ki
∂Ei Ei =Ēi
∂Ei
T Ci
∂Ei
Ci =
∂T V,N
X
−1
1
S(E)/kB
S1 (E¯1 )+S2 (E¯2 )
2
2
e
= e
exp
(E1 − Ē1 ) −
(E2 − Ē2 )
2k1
2k2
E
1
Probability that E1 6= Ē1 is Gaussian, h(E1 − Ē1 )2 i1/2 ∼
5
√
k1 ∼
√
N
Dominant term is E1 = Ē1 with Ē1 such that T1 = T2 .
If ∆ is the spacing of E1 levels (e.g. ∆ ∼ V −2/3 in ideal gas)
Z
X
dE1
1
1
2
→
exp −(
+
)(E1 − Ē1 )
∆
2k1 2k2
E1
1/2
1
k1 k2
=
2π
∆
k1 + k2
⇒ S(E) = S1 (Ē1 ) + S2 (Ē2 ) + O(ln N )
→ Extensivity, ie. Ω(E) is dominated by one term: Ω(E) = Ω1 (Ē1 )Ω2 (Ē2 )
(Note: The only function f (Ω) which is additive when Ω = Ω1 Ω2 is f ∼ ln Ω )
Energy fluctuation ∼
√
√
ki = T CV .
Similarly, separate to subsystems with V = V1 + V2 or N = N1 + N2 to obtain dominant
term at V̄i (hence P1 = P2 ) or at N̄i (hence µ1 = µ2 ).
Ideal Gas (no interactions)
Classical: r → (p, q). For one particle energy (p): Ωcl ∼
R
d3N q
R
P
i
(pi )=E
d3N p ∼ V N .
Note missing dimensional prefactor.
P
∂ ln Ω
=N
=
kB T
∂V
V
N,E
Quantum: r → quantum numbers. Use periodic boundary conditions, e.g. eipx L/~ = 1 ⇒
px = Lh nx , nx integer −∞ < nx < ∞
=
h2
(n2x
2mL2
+ n2y + n2z )
Ω(E) is the number of solutions to
P3N
r=1
n2r =
2m
EV 2/3
h2
⇒ S(N, V, E) = S(N, V 2/3 E)
In an adiabatic process V 2/3 E = const, P = −
⇒ P V 5/3 = const
6
∂E
∂V N,S
=
2E
3V
Ω(E) irregular - No. of points on surface of 3N sphere.
P
Instead Σ(N, V, E) = E 0 ≤E Ω(N, V, E 0 ) has reasonable E → ∞ limit.
∆
2
Replace Ω by Γ = number of microstates with E −
<energy< E +
Ω → Γ(N, V, E; ∆) = ∆
∆
2
∂Σ
∂E
Σ is the volume of 3N sphere with radius R = (2mV 2/3 E/h2 )1/2
Volume of n sphere: Vn = Cn Rn with surface area dVn = nCn Rn−1 dR
Z
∞
2
e−x dx =
√
π
−∞
Z
∞
Z
...
e−
Pn
i=1
x2i
Y
−∞
dxi = π n/2
i
In spherical coordinates:
Z
∞
e
−r2
nCn r
n−1
dr =
n
0
⇒ Vn =
Σ(N, V, E) =
2
!Cn = π n/2
π n/2 Rn
(n/2)!
π 3N/2
3N Σ
(2mV 2/3 E/h2 )3N/2 ⇒ Γ = ∆
(3N/2)!
2E
N → ∞ : ln N ! = N ln N − N
" 3/2 #
V 4πmE
3N
∆
3
ln Γ = N ln 3
+ N + ln
+ ln
h
3N
2
2
E
∆ ∼ level spacing: nr → nr + 1 for some r yields E → E + ∆
2m
2nr + 1 = 2 ∆V 2/3
h
⇒
0 ≤ nr <
∼ N −2/3 < ∆ <∼ N 1/6
⇒
7
2m
EV 2/3
2
h
1/2
ln ∆/E = O(ln N )
∼ N 5/6
⇒ ln Γ = ln Σ + O(ln N )
Volume near surface dominates total volume for N 1. Value of ∆ is not important.
"
⇒ S(N, V, E) = kB ln Γ = N kB ln
T =
∂E
∂S
V
h3
=
N,V
4πmE
3N
3/2 #
3
+ N kB
2
2E
3N kB
3
1
3
E = N kB T ⇒ mhv 2 i = kB T
2
2
2
∂E
3
Cv =
= N kB
∂T N,V
2
2E
∂E
=
P =−
⇒ P V = N kB T
∂V N,S 3V
∂(E + P V )
5
Cp ≡
= N kB
∂T
2
N,P
P
∂V
∂T
N,P
is excess work at constant P .
⇒
Cp
5
=
Cv
3
S is not extensive. Even mixing gases 1,2 with equal T, n leads to Stotal 6= S1 + S2
The ”mixing” entropy is positive:
N kB ln V − N1 kB ln V1 − N2 kB ln V2 = kB
V
V
+ N2 ln
>0
N1 ln
V1
V2
But this must be reversible !
Quantum mechanics - indistinguishability. Classical limit (h → 0) should give Σ →
V
3
5
2πmkB T
⇒ S(N, V, E) = N kB ln + N kB
+ ln
N
2
3
h2
Sackur Tetrude eq. : S is extensive.
Classical derivation ...
X
cl
1
=
w0
Z
d
3N
Z
d3N p =
q
P
i
p2i <2mE
8
1 N
V C3N (2mE)3N/2
w0
1
Σ
N!
w0 is chosen with
P −−−−→ P
h→0
cl
2mE
1
C3N ( 2 )3N/2 ⇒ w0 = N !h3N
N!
h
is the Gibbs correction and h is the volume of one state in the p,q space.
X
Where
1
N!
=
Ex: evaluate w0 for harmonic oscillator.
Equipartition
xi = q i
or
pi
R
∂H
i≡
∂xj
hxi
...
R
E− 21 ∆<H<E+ 12 ∆
R
...
∂H
xi ∂x
dω
j
R
dω
E− 21 ∆<H<E+ 12 ∆
where dω =
d3N qd3N p
.
N !h3N
Noting that
R
R ∂H
∂
∆ ∂E
. . . xi ∂x
dω
j
0<H<E
R
R
=
∂
. . . dω
∆ ∂E
0<H<E
∂E
∂xj
= 0 and integrating by parts
Z
Z
Z
Z
Z
Z
(2)
∂
xj
(H−E)dω = . . . dxk6=j [xi (H−E)] (1) −δij . . . (H−E)dω
numerator = . . . xi
xj
∂xj
0<H<E
0<H<E
Assume H(xj ) → ∞ and monotonic, e.g.
(1,2)
At boundaries, xj
p2
0<H<E
or at walls of container.
2m
(2)
(1)
H(xj ) = H(xj ) = E
, all energy is in xj
∂H
therefore, the first expression on the right side vanish. Returning to hxi ∂x
i we have:
j
hxi
∂H
i = δij
∂xj
∂
∂E
R
R
R
R
. . . (E − H)dw
. . . dw
0<H<E
0<H<E
R
R
R
= δij ∂ R
=
∂
.
.
.
dw
.
.
.
dw
∂E
∂E
0<H<E
0<H<E
∂
∂E
δ
kB δij
R ij R
= δi,j kB T
= ∂S
ln . . . dw
( ∂E )N,V
0<H<E
concluding:
∂H
i = δi,j kB T
hxi ∂x
j
xj → ∞
with H(xj ) monotonic unbounded.
E.g. (Specific examples):
hpi
xi = xj = p i
hqi
xi = xj = q i
∂H
i = hpi q˙i i = kB T
∂pi
∂H
i = −hqi ṗi i = kB T
∂qi
for quadratic Hamiltonians (usually at high T)
X
H=
Aj p2j + Bj qj2
j
9
for such systems we clearly have:
X ∂H
∂H
1
(pj
+ qj
) = 2H ⇒ hHi = f kB T
∂pj
∂qj
2
j
f=No. of degree of freedom ≡ number of quadratic terms in H.
Each harmonic term in the quadratic Hemiltonian makes a contribution of 21 kT towards the
internal energy of the system and hence a contribution of 12 k towards the specific heat Cv .
Example: molecule with m atoms
C.M: 3 translations + 3 rotations
non co-linear molecule
3 translations + 2 rotations
co-linear
(To understand the significance of colinearity, note that quantum levels
classical continuum if
~2
I
~2 `(`+1)
2I
kB T . However, if I → 0 as in a colinear case, at kB T form a
~2
I
only
the single ground state is relevant)
3m coordinates ⇒ no. of vibrations = 3m-6
or = 3m-5
non-collinear
collinear
p2x
2
Translation ( 2m ), rotation ( L2Ix ) - 1 quadratic term
2
px
Vibration ( 2m
+
mω 2 x2
)
2
1
hHi = [6 + 2(3m − 6)]N kB T = (3m − 3)N kB T
2
1
hHi = [5 + 2(3m − 5)]N kB T = (3m − 5/2)N kB T
2
non − collinear
collinear
Virial Theorem (Clausius 1870) ...
The Virial of a system is by definition, the sum of the products of the coordinates of the
various particles and the representative forces acting on them:
V≡
3N
X
i=1
hqi ṗi i = −3N kB T
E.g. ideal gas in box:
10
ṗ 6= 0 only from walls at qi = r
~
p~˙ = −P ds
where: P is pressure, ds is surface element and p~ is
momentum hitting ds (of all particles)
V=
X
r
~r · p~˙ = −P
I
s
~ = −P
~r · ds
Z
~ · ~r)dv = −3P V ⇒ P V = N kB T
(∇
Consider now classical particles i, j with 2-body interaction
u(ri,j ) : rij = |ri − rj |.
X
i
∂ X
p~˙i = −
u(|ri − rj |)
∂~
ri j
r~i · p~˙i = −
X
i,j
r~i ·
∂u(r)
∂r2 ∂u
=
∂~r
∂~r ∂r2
∂
u(|ri − rj |),
∂~
ri
Sum over pairs:
X
i
r~i · p~˙i =
X
i<j
∂u
∂(ri − rj )2
∂(rj − ri )2
− 2 [~
ri ·
+ rj
] + wall pressure =
∂rij
∂~
ri
∂ r~j
2
2rij
}|
{
X
X ∂u z
∂u
2~
r
·
(~
r
−
~
r
)
+
2~
r
·
(~
r
−
r
~
)]
=
−
rij
−
=
[
i
i
j
j
j
i
2
∂r
∂r
ij
ij
i<j
i<j
The net contribution, arising from all the
1
∂u
1
N (N − 1)h−r i = − N 2
2
∂r
2
Z Z
N (N −1)
2
pairs of particles, (N 1):
∂u
N2
r12
g(r1 − r2 )d3 r1 d3 r2 /V 2 = −
∂r12
2V
g(r) pair distribution function
N2
V=−
2V
Z
∞
r
0
∂u
g(r)4πr2 dr − 3P V = −3N kB T
∂r
2πn
P V = N KB T [1 −
3kB T
∞
Z
0
∂u(r)
g(r)r3 dr]
∂r
Also:
3
4πn
E = N KB T [1 +
2
3kB T
Z
u(r)g(r)r2 dr]
where the 1st term in the square brackets stands for the Kinetic energy.
11
Z
r
∂u
g(r)4πr2 dr
∂r
1c. Ensemble theory - generalities
ρ(p, q, t)d3N qd3N p is the probability of microstate(p,q), with normalization
Z
ρ(p, q, t)d3N qd3N p = 1
E=const
Average: hf i =
R
f (p, q)ρ(p, q, t)d3N qd3N p
Equilibrium: all observables
∂<f >
∂t
= 0 ⇒
∂ρ
∂t
p(0),q(0) in the 6N dimensional phase space, follow trajectory and average on time.
Ergodic theorem: Long time average = ensemble average, i.e. probability concept ρ(p, q, t)
is equivalent to mechanics.
Liouvill’s theorem
p(t), q(t) satisfy Hamilton’s equation and define a velocity in phase space
~v = (q̇, ṗ)
∂
Net flow of states from volume ω, with surface σ, is = − ∂t
Z
R
Z
ρdω
∂
⇒
ρ(~v · n̂)dσ =
div(ρ~v )dω = −
∂t
ω
ω
true for any ω ⇒. Continuity:
∂ρ
∂t
Z
ρdω
ω
+ div(ρ~v ) = 0.
⇒ Consider ρ(qi , pi , t) for a collection of states qi (t), pi (t)
X ∂ q̇i ∂ ṗi ∂ρ X ∂ρ
∂ρ
+
q̇i +
ṗi + ρ
+
=0
∂t
∂qi
∂pi
∂qi ∂pi
i
i
∂ q̇i
∂H
∂ ṗi
=
=−
∂qi
∂qi ∂pi
∂pi
and therefore
∂ q̇i
∂qi
+
∂ ṗi
∂pi
dρ
∂ρ
=
+ [ρ, H]poisson = 0
dt
∂t
Local density as viewed on moving points, is constant, hence
dρ
dt
No. of states is conserved ⇒ incompressible ”fluid”.
Equilibrium:
X
∂ρ
=0⇒
∂t
i
∂ρ
∂ρ
q̇i +
ṗi
∂qi
∂pi
12
=0
= 0.
solutions:

 const E − 1 ∆ < H(p, q) < E + 1 ∆ microcanonical
2
2
ρ(p, q) =
0
otherwise
In general ρ can depend on constants of motion, e.g. H(p, q).
ρ[H(q, p)] → canonical ensemble
In microcanonic ρ(p, q, t) =
1
Ω
S = −kB ln ρ = −kB
Z
ρ(p, q, t) ln ρ(p, q, t)d3N pd3N q
defines entropy also in other ensembles, in equilibrium. Is this form of S valid at nonequilibrium? but then
⇒
dS
=0
dt
which violates the second law.
Arrow of time
Consider volume expansion I→II. For N = 1020 , volume increases by factor 2.
20
The increase in number of states is 210 !! all states of I evolve into II, but only fraction
1
21020
of states in II evolve into I. Probability suggests that S(t) increases.
Just probability is not sufficient:
Consider I→II→III, increasing volumes. States in II most probably go to III, but where do
states in II come from - most probably also from III !? not I?
Time reversal invariance: a state x1 in I evolves into x2 in II. Now reverse all momenta in
x2 → R~x2 .
R~x2 is a state in II. It’s time evolution yields x1 in I, i.e. it is possible to find a state that
lowers entropy.
The difficulty with R~x2 is that it must be prepared accurately; a minute perturbation causes
instability (chaos, the butterfly effect). Nature does not allow ”perfect aiming” with accuracy
20
∼ 210 .
13
⇒ need probability + stability, i.e. information on nearby trajectories, ”coarse grained”,
then entropy can increase with time.
Hypothesis: Universe started with low entropy, low entropy radiation from the sun produces
low entropy food etc...
1d. Canonical Ensemble (CE)
Consider a system that is embedded in a larger ”heat
bath”. Energy exchange is allowed, i.e. Er is not fixed.
r is point in phase space of the system.
Reservoir energy ER , Er << E0
ER + Er = E0 = const
probability of state r Pr ∼ ΩR (ER ), i.e. the number of states of the reservoir.
[note: No. of all states with energy Er = Ωr (Er )ΩR (ER )]
⇒ Pr ∼ ΩR (E0 − Er )
∂ ln ΩR
|E0 Er + · · · = ln ΩR (E0 ) − βEr
∂E
e−βEr
P
Pr =
−βEr
re
ln ΩR (E0 − Er ) = ln ΩR (E0 ) −
β=
1
kB T
is determined by the reservoir. The partition function is defined by
ZN (V, T ) =
X
e−βEr
r
Define F BY ZN = e−βF (T,V,N ) . Identify F:
∂ X −βEr +βF X −βEr +βF
∂F
0=
e
=
e
−Er + F + β
∂β r
∂β V
r
⇒ F (T, V ) − E(T, V ) − T
∂F
∂T
=0
V
solution of this differential equation is the free energy F = E − T S.
P
Alternative proof: Assume dominant energy in sum r :
ZN (V, T ) =
X
r
e−βEr =
X
E
14
Ω(E)e−βE =
X
E
eS(E)/kB −βE
dominant term at E ∗ :
∂S(E)
1
∂
(S(E)/kB − βE) = 0 ⇒
|E ∗ =
∂E
∂E
T
E ∗ is such that the MCE at E ∗ has the temperature T.
S(E ∗ )/kB − βE ∗ = −βFM CE
where FM CE is the free energy of the MCE defined with variable E ∗ , V, N .
Note: S(E ∗ (T, V, N ), V, N ) defines F (T, V, N ).
Fluctuation near E ∗ (expanding S around E ∗ ):
S(E)
S(E ∗ )
1 ∂ 2 S ∗
− βE =
− βE +
(E − E ∗ )2 + ... ,
2
kB
kB
2kB ∂E E ∗
where,
∂(1/T )
1
1
1
∂ 2S
=
=
−
=
−
,
∂E 2
∂E
T 2 ∂E/∂T
T 2 CV
and T, CV are for MCE.
ZN (V, T ) = e−βFM CE
X
e−(E−E
∗ )2 /2k T 2 C
B
V
E
⇒
Gaussian.
The weight of E 6= E ∗ decreases rapidly with width,
p
√
h(E − E ∗ )2 i
1
kB T 2 CV
=
∼√
→ 0.
∗
∗
E
E
N
To find the partition function, we replace the Sum with Integral,
Z
X
1
dE,
→
∆
E
where ∆ is the energy level spacing (as above ∆ ∝ N a → ln ∆ ∼ ln N ), and
√
Z
e−βFM CE
e−βF 2πkB T 2 CV
−E 2 /2kB T 2 CV
ZN (V, T ) =
e
dE =
= e−βFCE ,
∆
∆
⇒
FCE = FM CE + O(ln N ),
where FCE , FM CE ∼ O(N ).
Insensitivity of thermodynamic results to type of ensemble due to:
1. Ω(E) ∼ e(...)N ∼ e(...)E → ∞: exponential increase.
15
2. Thermodynamic limit: E, N → ∞.
E.g. for ideal gas, Ω(E)e−βE ∼ e(3/2)N ln E−βE with maximum at E ∗ = 32 N kB T . For E 6= E ∗ ,
Ω(E)e−βE practically vanishes.
Note: The energy fluctuations can also be identified by evaluating the specific heat:
P
−βEr
2
−1
∂
−1 ∂
2
r Er e
P
=
,
CV =
+
hEi
hEi =
−
E
−βE
2
2
r
∂T
kB T ∂β
kB T
re
where
(E − hEi)2
=
E 2 − hEi2
= kB T 2 CV ∼ N.
Examples:
• Classical systems: the general expression for the partition function,
Z
1
ZN (V, T ) =
e−βH(p,q) d3N qd3N p,
3N
N !h
where, N ! is Gibbs normalization, and h corresponds to a volume of one state in the
q, p phase space.
• Ideal gas: H =
P
i
p2i /2m,
VN
ZN (V, T ) =
N !h3N
Z
∞
e
−βp2 /2m
0
2
4πp dp
N
1
=
N!
V
λ3
N
,
where,
h
λ ≡ √
,
the thermal wavelength.
2πmkB T
√
This λ corresponds to a deBroiglie wavelength for momentum ≈ mkB T i.e. λ ∼
p
√
h/ hp2 i = h/ mkB T .
F = −kB T ln ZN = N kB T ln N λ3 /V − 1 ,
∂F
N kB T
P = −
=
,
∂V N,T
V
∂F
S = −
= N kB ln V /N λ3 + 5/2 ,
∂T N,V
∂F
= kB T ln N λ3 /V.
µ =
∂N T,V
16
as in MCE,
• Consider N noninteracting molecules with internal energies i , ni molecules at level i
are indistinguishable.
N

ZN (V, T ) =
X
{ni }
=
=
P

1
e−β i i ni 

n0 !n1 !...
Z
e−βp
|
2 /2m

d3 pd3 q 

{z
}
center of mass
!N
VN
N !λ3N
X
e−βi
,
using multinomial expansion
i
1
(Z1 (V, T ))N ,
N!
Z1 ≡
V X −βi
e
λ3 i
partition of one molecule.
Note: The multinomial expansion is:
!N
X
ai
= N!
i
X
{ni }
1
an1 1 an2 2 ... ,
n1 !n2 !...
where the sum on distributions {ni } is restricted by
P
i
ni = N .
• Diatomic gas:
H =
ε0
|{z}
electron
Z = e−βε0
+ ~ω(ν + 1/2) + ~2 k(k + 1)/2I ,
|
{z
}
{z
}
|
∞
X
rotation
vibration
∞
X
X
2
−β~ω(ν+1/2)
e
(2k + 1)e−β~ k(k+1)/2I
(2s + 1) ,
ν=0
|s
k=0
{z
}
nonidentical atoms
where s is the spin of the two atoms.
T →∞
Z
∞
2 2
2ke−β~ k /2I dk = 2I/β~2 ,
Zrot = →
0
Z
1
2
2
Zrot−class = 2 e−β(Mx +My )/2I dMx dMy dθx dθy = 2I/β~2 ,
h
The two atom axis is ẑ and dθx dθy = dΩ → 4π is the solid angle.
If the two atoms are identical, we need QM:
There is a constraint: s+k needs to be even. [Orbital exchange has (−)k , spin exchange
has (−)s−s1 −s2 , hence both Fermion and Boson symmetries are obeyed for s + k even].
17
Spin
k
degeneracy = 2s + 1
H2 s = 1 odd
3
orthohydrogen
s = 0 even
1
parahydrogen
D2 s = 2 even
5
s = 1 odd
3
s = 0 even
1
H2
Zrot
=
3
X
(2k + 1)e−β~
2 k(k+1)/2I
k odd
D2
Zrot
=
3
X
→
X
(2k + 1)e−β~
2 k(k+1)/2I
,
k even
... + 6
k odd
T →∞
+
X
...
k even
1 2I X
(2s + 1).
2 β~2 s
In the last equation, the
1
2
is the classical reduction of angular integration by factor of
2, for identical atoms.
Equipartition:
R ∂H −βH
dω
xi ∂xj e
∂H
R
=
,
xi
∂xj
e−βH dω
x(2)
Z
j
1
−βH −βH
numerator =
dxi6=j xi e
dωδij .
(1) + β e
xj
E
D
(1,2)
∂H
= δij kB T .
In general, the energy H(xj ) → ∞ at the boundaries of xj ⇒ xi ∂x
j
Z
∂H −βH
−1
xi
e
dω =
∂xj
β
Z
E.g. ideal gas:
∂ p~i 2
pix
∂pix 2m
Extreme relativistic: H =
P
i
1 2
p
= kB T,
m ix
X p~i 2 3
E =
= N kB T.
2m
2
i
=
c |~
pi |,
∂H
∂ q 2
cpix
q̇ix =
=
c pix + p2iy + p2iz =
,
∂pix
∂pix
|~
pi |
*
+
*
+
X
X
3
p~i q~˙i =
c |~
pi |
= hHi = 3N kB T (> N kB T ),
2
i
i
as the power of p smaller, the energy is higher for given temperature.
18
1e. Grand Canonical Ensemble (GCE)
System in heat bath and particle reservoir.
Er ER ,
Nr NR
Er + ER = E0 = Const.
Nr + NR = N0 = Const.
The combined system is microcanonic.
r is a point in phase space of the system, which now includes all Nr ( N0 ). The probability
of a point r with Er , Nr is Pr
Pr ∝ ΩR (N0 − Nr , E0 − Er )
ln Pr ≈ ln ΩR (N0 , E0 ) −
∂ ln ΩR
∂ ln ΩR
|N0 Nr −
|E0 Er
∂N
∂E
= ln ΩR (N0 , E0 ) + βµNr − βEr
β = 1/kB T , µ is imposed by the resevoir
⇒ Pr = eβµNr −βEr /L
The grand partition function is
L(µ, T, V ) =
X
r
eβµNr −βEr ≡ e−β Ω(µ,T,V )
e
e :
Identify Ω
e
∂ X βµNr −βEr +β Ωe X
e + β ∂Ω
µNr − Er + Ω
e
=
0=
∂β r
∂β
r
!
Pr
e
e − T ∂Ω
= µN − E + Ω
∂T
The energy E(µ, T, V ) shows that the solution to this differential equation is the theromoe = F − µN = E − µN − T S = E − µN + T ∂ Ωe
dynamic potential, since Ω
∂T
P
Alternatively, if N is dominated by one term
L ≈ eβµN
X
r,N f ixed
e = F − µN = −P V
e−βEr = eβµN −βF ⇒ Ω
19
Define fugacity ζ = eβµ
L(ζ, T, V ) =
∞
X
ζ N ZN (V, T )
N =0
So that the weight of each N = ζ N ZN /L.
N =ζ
E=−
∂
ln L(ζ, T, V )|T,V
∂ζ
∂
ln L(ζ, T, V )|ζ,V
∂β
etc.
Ideal gas: with internal energies i for each particle,
Z1 =
ZN =
V X −βi
V
e
=
a(T )
λ3 i
λ3
1 N
Z ⇒
N! 1
L=
X
ζN
N
1 N
Z = eζZ1
N! 1
PV
P λ3
V
= ζZ1 = eβµ 3 a(T ) ⇒ µ = kB T ln
kB T
λ
kB T a(T )
N =ζ
PV
nλ3 nλ3 1
∂
ln L(ζ, V, T ) = ζZ1 =
⇒ µ = kB T ln(
) −→ −∞
∂ζ
kB T
a(T )
Particle (N) fluctuations:
Weight W (N ) = eβµN −βF (N,V,T ) /L
Maximum at N such that µ =
∂F
| ,
∂N N
i.e. µcan =
∂F
∂N
equals µ if N is chosen at this maximum,
N = N.
For maximum need
∂2F
|
∂N 2 N
> 0:
F (N, V, T ) = N f (v)
For fixed V ,
∂
∂N
=
−v ∂
,
N ∂v
v=
V
N
hence
∂F
∂f
∂ 2F
v2 ∂ 2f
= f (v) − v
⇒
=
∂N
∂v
∂N 2
N ∂v 2
Since P (v) = − ∂f
⇒γ≡
∂v
∂2F
|
∂N 2 N̄
2
= − vN
∂P
∂v
(or =
∂µ
)
∂N
we need ( ∂P
)| < 0 for W (N ) being
∂v T
a maximum ( this is Van Hove’s theorem, Huang first edition p.321 - general proof; Huang
second edition p.206 - for the special case of hardcore interaction).
Physically obvious: if P > Pext the net force increases V ; now if
more and equilibrium (P = Pext ) is further away.
20
∂P
∂V
> 0 P increases even
CE and GCE are equivalent by this thermodynamic stability criterion ( ∂P
)| < 0.
∂v T
Note however, that at a 1st order phase transition, e.g. the gas-liquid transition, ∂P/∂v = 0
and GCE is not equivalent to CE, hence large N fluctuations are expected.
− 12 βγ(N −N )2
W (N ) ≈ W (N̄ )e
1
⇒ ∆N 2 2 =
s
√
kB T N
∼
N
v 2 (−∂P/∂v)
√
1
∆N 2 2 /N ∝ 1/ N → 0
Energy fluctuations:
E=
X
r
Er Pr = −
∂ ln L
∂β
ζ,V
2
∂
∂E
2 ∂E
ln
L
=
−
|ζ,V
⇒ ∆E 2 = E 2 − E =
=
k
T
B
∂β 2
∂β ζ,V
∂T
ζ,V
ζ,V
∂E (N (ζ, T, V ) , T, V )
∂E
∂E
∂N
∂E
∂N
=
+
= CV +
∂T
∂T N,V
∂N T,V ∂T ζ,V
∂N T,V ∂T ζ,V
∂S
∂E (S (N, T, V ) , V, N )
=µ+T
dE = T dS − P dV + µdN ⇒
∂N
∂N T,V
T,V
!
∂
∂µ
∂F
=µ−T
=µ−T
(*)
∂N ∂T N,V
∂T N,V
1
E2 =
L
∂ 2L
∂β 2
2
T,V
Chain rule
( ∂x
) ( ∂y ) ( ∂z )
∂y z ∂z x ∂x y
= −1
prove by dx = ( ∂x
) dy + ( ∂x
) dz = 0 which yields ( ∂y
) .
∂y z
∂z y
∂z x
∂N
∂µ
∂N
∂µ
∂N
µ
=
+
=−
+
∂µ T,V ∂T ζ
∂µ T,V ∂T N,V
∂µ T,V T
ζ,V
µ,V
"
#
1 ∂N
∂µ
1 ∂N
∂E
=
µ−T
=
from (*)
T ∂µ T,V
∂T N,V
T ∂µ T,V ∂N T,V
"
#2
∂E
∂N
⇒ ∆E 2 = kB T 2 CV + kB T
∂µ T,V
∂N T,V
"
#2
∂E
= (∆E 2 )can + ∆N 2
> (∆E 2 )can
∂N T,V
∂N (µ(T, ζ), T, V )
∂T
∂N
∂T
21
√
1
(∆E 2 ) 2 /E ≈ 1/ N → 0 except at Phase transitions.
Summary: Thermodynamics:
dE = T dS − P dV + µdN
E(S, V, N )
E − T S = F (T, V, N )
dF = −SdT − P dV + µdN
e
F − µN = Ω(T,
V, µ)
e = −SdT − P dV − N dµ
dΩ
Statistical Mechanics:
Ω=
X
1 = eS(E,V,N )/kB
M CE
r (E,N f ixed)
Z=
X
e−βEr = e−F (T,V,N )/kB T
CE
r (N f ixed)
L=
X
eβµNr −βEr = e−Ω(T,V,µ)/kB T
e
r
22
GCE
2. QUANTUM STATISTICAL MECHANICS
A system of N particles is described by a wave function ψ(r1 , r2 , ...). Since the particles are
indistinguishable, upon exchange of particles ri ↔ rj the wave function acquires a phase eiθ
ψ(· · · , ri , · · · , rj , · · · ) = eiθ ψ(· · · , rj , · · · , ri , · · · ).
This particle exchange is equivalent to a rotation by π in the relative coordinate ri − rj .
Exchanging the particles i, j twice is equivalent to a 2π rotation, which in 3-dimensions is
equivalent to the identity operator [a circle on a sphere can be smoothly deformed into a
point]. Hence e2iθ = 1 and there are two types of particles in nature:
θ = 0 symmetric ψ for bosons (Bose Einstein statistics (BE))
θ = π antisymmetric ψ for fermions (Fermi Dirac statistics (FD)).
In particular fermions obey pauli’s exclusion principle [antisymmetric wavefunction at ri = rj
vanishes]. Note also the spin-statistics connection, i.e. integer spin particles are bosons, half
integer particles are fermions. Note also that in 2-dimensions other statistics are allowed;
since the 2π rotation is now a circle with a singular point ri = rj at its center, the circle
cannot be deformed into an identity operation. In 3-dimension one ”escapes in the 3rd
dimension” avoiding the ri = rj singularity.
We also define a ”‘Boltzman statistics”’ (MB) where only the particles at the same energy
level are indistinguishable, i.e. for each energy level we add a factor of
1
.
n !
2a. Ensembles for ideal quantum gases
Micro Canonical Ensemble
We group the different states into energy levels
i with degeneracy of gi . Each level is occupied
with ni particles. For the statistical treatment
we assume gi , ni >> 1.
Ω(N, V, E) =
X0
{ni }
W ({ni })whereW ({ni }) =
23
Y
i
W (i) for a given distribution{ni }
The prime on the sum denotes the following constraints:
X
ni = N
X
and
i
ni i = E
i
For bosons, each energy level can be occupied with ni particles within gi folders with gi − 1
divisions. Hence we choose ni particles from ni + gi − 1 objects, W (i) =
(ni +gi −1)!
ni !(gi −1)!
and for
all energy levels we have
WB.E. =
Y (ni + gi − 1)!
ni !(gi − 1)!
i
For fermions we choose ni sites from the gi available ones, so
WF.D. =
Y
i
gi !
ni !(gi − ni )!
gi ≥ ni >> 1
For the Boltzman statistics any particle can occupy any of the states up to the Gibbs
correction ( n1i ! ) for the indistinguishability of the particles in each energy level
WM.B. =
Y (gi )ni
i
ni !
The Gibbs factor correctly accounts for permutations of particles in differen quantum states,
but also unnecessarily corrects for particles that are in the same quantum state where correction is not necessary (e.g. one symmetric state only). Therefore ni ! is an ”over-correction”
and is valid when the density is low ni << gi and there is a small chance of all particles
being in the same quantum state. This is also reflected in the possible situation gini /ni ! < 1.
E.g., consider two particles in two states n=g=2:
WBE = 3
WF D = 1
WM B =
1
|aai, |bbi, √ (|abi + |bai)
2
1
√ (|abi − |bai)
2
22
=2
2!
1
2
of |aai or of |bbi is ad hoc.
The entropy of the quantum gas will be


0
X
S = kB ln 
W ({ni }).
{ni }
24
We now find the distribution n∗i that maximizes S. Using the method of lagrange multipliers
we demand
"
#
X
δ ln W ({ni }) − α
i
ni − β
X
ni i = 0.
i
with α, β to be determined below. Rewrite our 3 cases as
X
gi
ni
gi
− a − ln 1 − a
ln (W ({ni })) =
ni ln
ni
a
gi
i
where a is defined by a = −1 (BE), a = +1 (FD) and a → 0 (MB).
Performing the variation yields
X gi
ln
− a − α − βi δni = 0
∗
n
i
i
so that S is maximized by the distribution
1
ni ∗
= α+βi
+a
gi
e
The value of S at its maximum is therefore
S/kB = ln W ({n∗i }) =
X
n∗i (α + βi ) +
i
gi
ln 1 + ae−α−βi
a
The coefficients α and β are determined by the constraints on N and E:
−µ
1
∂S
∂α
∂β
=
=N
+α+E
kB T
kB ∂N E,V
∂N E,V
∂N E,V
"
#
X gi −ae−α−βi
∂α
∂β
+
+ i
=α
a 1 + ae−α−βi
∂N E,V
∂N E,V
since the last sum is −N
∂α
∂N E,V
∂S
∂E
∂β
.
∂N E,V
The coefficient β is identified by
∂α
∂β
=N
+E
+β
∂E N,V
∂E N,V
N,V
"
#
X gi −ae−α−βi ∂α ∂β
+
+ i
=β
−α−βi
a
1
+
ae
∂E
∂E
N,V
N,V
i
1
1
=
kB T
kB
−E
The equation of state is obtained by
1X
S
µN − E
PV
gi ln(1 + ae−α−βi ) =
+
=
a i
kB
kB T
kB T
25
so that finally

P
(µ−i )/kB T

∓k
T
g
ln
1
∓
e
B
i





PV =
P −α−βi
P ∗


kB T
gi e
= kB T
ni = kB T N





a = ∓1 





a=0
.






Canonical and Grand-Canonical ensembles
In the GCE there are no constraints on N, E. Hence we label by i all quantum numbers
that completely specify a state so that gi = 1 and
WBE {ni } = 1
WF D {ni } = 1 (if all ni = 0,1) or 0 otherwise
Y 1
WM B {ni } =
ni !
i
notice that WF D < WM B < WBE . E.g. with N = 2 particles in two states
1
|aai, |bbi, √ (|abi + |bai)
2
1
√ (|abi − |bai)
2
BE:
FD:
1
2 |aai,
MB:
1
2 |bbi,
|abi
3 states
1 state
2 states.
P
Consider briefly the CE constrained with i ni = N ,
X0
P
ZN =
W ({ni })e−β i ni i
{ni }
and for the MB distribution
X0 Y N !
1
−βi ni
ZN =
e
= using multinomial expansion =
n
!
N
!
i
i
{ni }
!N
X
e−βi
i
1
1
=
(Z1 (V, T ))N
N!
N!
WM B {ni } = N N /N ! for N particles in N states]. For the
P
BE and FD statistics one has to proceed to the GCE, to avoid the
ni = N constraint.
[Note: with i → 0 we have
L(ζ, V, T ) =
∞
X
N =0
P
ni
N
ζ ZN (V.T ) =
∞ X Y
X
0
N =0 {ni }
ζe−βi
ni
with ζ = eβµ
i
here the first sum determines the constraint N on the second sum. However, since anyway
we some on all occupations, we can sum them on each level independently,
"
#
X
Y
X
n
n
n
0
1
i
L=
ζe−β0
ζe−β1
··· =
ζe−βi
n0 ,n1 ,...
i
26
ni
now using
P
n
xn =
1
1−x
we get the result for each of the distributions
Q
1

BE

i 1−ζe−βi







Q
L=
FD
1 + ζe−βi
i








P

ln L = ζZ1 = ζ i e−βi M B










using
P xn
n!
.








x 
=e
The GCE relation is
ln L =
1X
PV
=
ln 1 + aζe−βi
kB T
a i
and recall a = −1 for BE, a = 1 for FD and a → 0 for MB. For N, E we have :
X
∂(ln L)
1
N =ζ
=
1 βi
∂ζ
e +a
V,T
i ζ
E=−
∂ ln L
∂β
=
ζ,V
X
i
i
1 βi
e +
ζ
a
The mean occupation number is given by
P
1 XX
ni ζ N e−β j nj j
L N
{nj }
1 ∂ ln L
1
1
=−
= 1 β
= β(i −µ)
β
∂i ζ,T,j6=i
e
+a
e i +a
ζ
hni i =
The figure illustrates the various cases. For FD, if µ > and low temperatures the mean
occupation tends to 1. For the BE case, the condition µ < must be valid for all to
27
maintain hni i ≥ 0; this leads (see below) to BE condensation. The MB distribution is valid
i −µ
kB T
>> 1 for all i , hence µ < 0 and |µ| kB T or ζ 1.
P
The condition on the density is N ∼
= ζ i e−βi = ζ λV3 which implies high temperature or
only when hni i << 1 which means
low density limit. Note that both BE and FD approach MB at this limit of ζ 1.
2b. Ideal Bose gas
Here i = p~2 /2m and the summation index is i = p~.
P
A note on momentum summations: The sum p actually stands for a sum on integers
nx , ny , nz that define px , py , pz via periodic boundary conditions. E.g. eipx Lx /~ = 1 with Lx
the length in the x direction, px =
X
2π~
n ,
Lx x
Z
hence for any function f (p)
Lx Ly Lz
V
f (p)d3 p = 3
3
h
h
f (p) =
nx ,ny ,nz
Z
f (p)d3 p
In radial coordinates we then have
Z
4π ∞ 2
1
P
2
=− 3
p ln(1 − ζe−βp /2m )dp − ln(1 − ζ)
kB T
h 0
V
4π
N
= 3
V
h
Z
∞
p
0
2
−1
1 ζ
1 βp2 /2m
e
−1
dp +
ζ
V 1−ζ
The p = 0 term in the original sum is singled out. It is the number of particles N0 =
ζ
1−ζ
in
the single state p = 0. [Equivalently, need ζe−βi < 1 to allow convergence for L.]
The crucial property of bosons is ζ ≤ 1 so as to keep all state occupations hnp i ≥ 0. For
a given density the integral term decreases with T for a fixed ζ; to keep N/V fixed ζ must
increase, but it is bound by ζ = 1. The integral is bound at ζ = 1, hence below some Tc the
integral becomes < N/V and a macroscopic part of N must populate the p = 0 state. This
is Bose Einstein condensation.
The correction in the pressure equation is negligible:
1−ζ =
1
1
1
always
⇒ − ln(1 − ζ) = ln(< N0 > +1) → 0
1+ < N0 >
V
V
The boson thermodynamics are then given by
P
1
= 3 g5/2 (ζ)
kB T
λ
28
N
1
< N0 >
= 3 g3/2 (ζ) +
V
λ
V
Z ∞ n−1
∞
x dx X ζ `
1
=
gn (ζ) =
Γ(n) 0 ζ1 ex − 1
`n
`=1
For small ζ we use the expansion to eliminate ζ:
∞
X
PV
⇒
=
a` (nλ3 )`−1
N kT
`=1
(a` = virial coefficient)
1
a1 = 1, a2 = − √ = −0.176, a3 = −0.0033 .....
4 2
For specific heat:
∂ V g5/2 (ζ)
∂
3
PV
3
2
E = −(
ln L)ζ,V = kB T
= kB T
= PV
3
∂β
∂T
λ
2
kB T
2
ζ,V
since the virial expansion for P V has T (λ3 )`−1 ∼ T 5/2−3`/2 terms.
∞
CV
1
∂E
3 ∂ PV
3 X 5 − 3`
=
a` (nλ3 )`−1
=
=
Nk
N k ∂T N,V
2 ∂T N k N,V
2 `=1 2
3
= (1 + 0.0884nλ3 + 0.0066(nλ3 )2 + ...)
2
Cv
CV = T
∂S
∂T
T →0
N,V
−→ 0
hence CV looks roughly as:
T
g 32 HΖL
If nλ3 < g3/2 (1) = ζ( 32 ) = 2.612
0 < ζ < 1 and
If nλ3 > g3/2 (1)
<N0 >
V
< N0 >
=0
V
=n−
g 3 (1)
2
λ3
2.612
>0
1
29
Ζ
with ζ = 1 and finite density at the single
= 0 level.
Transition at nλ3
2π~2
2
m[g3/2 (1)] 3
n2/3
= g 3 (1) ⇒ kTc
2
=
When ζ = 1, < N1 > is negligible since 1 ∼ h2 /V 2/3
< N1 >
1 1
1 V 2/3 V →∞
=
∼
−→ 0
V
V eβ1 −1
V βh2
<N0 >N
At T ≤ Tc ,
<N0 >
N
=1−
=1−
T
Tc (n)
g3/2 (1)
λ3 n
3/2
=1−
1
nc (T )
n
TTc
1
Two fluid concept: < N0 > condensate,
g 3 (1)
2
λ3
normal component.


 13 g5/2 (ζ) n < nc (normal)
λ
P
=

kT
 13 g5/2 (1) n > nc (condensed phase)
λ
5
5
Pc ∼ T 2 ∼ n 3
P
all excess particles beyond nc occupy p = 0,
< N0 >= N − Nc (T ), even at n → ∞ with
no contribution to P . Along AB
∂P
∂n
5
Pc ~ n 3
A
B
= 0, two
phases coexist with phase A that has n → ∞.
30
1n
P
5
Pc ~T 2
No
This corresponds to a first order phase tran-
realization
 Condensate on line
sition, 1/n jumps by ∆(1/n) = 1/nc as pressures increases.
Normal
T
Note that there is no realization of P > Pc (T ). The equation of state defines a 2-dimensional
surface in P, T, n; the figure is a projection of this surface on the P, T plane. There are no
points on the surface that project onto the region P > Pc (T ). Consider a fixed n and an
initial high temperature where P = nkB T and P Pc (T ). Now as T decreases P approaches
Pc (T ) at Tc , and upon further decrease of T , P stays on the critical curve P (T ) = Pc (T ) for
all T < Tc . This feature is an artifact of the ideal gas; once hard core repulsion is added, at
sufficiently high density (in a constant T trajectory) the pressure will start to increase.
Clausius Clapeyron:
dPc
∆S/N
=
dT
∆( n1 )
∆S is the jump in entropy. We will evaluate S directly:


 5 g5/2 (ζ)
− ln ζ
S
E
µ
5 PV
µ
PV
2 nλ3
+
−
=
−
=
=
g (1) N −<N0 >

N kB
N kB T
N kB T
kT
2 N kB T
kB T
 5 g5/2 (1)
= 52 g5/2
2 nλ3
N
3/2 (1)
T > Tc
T ≤ Tc
At T < Tc , S ∼ N − < N0 >= Nnorm , no entropy for the condensate. Above the Pc (T )
line S/N → 0, while just below this line
N=
V
S
5kB g5/2 (1)
g
(1)
⇒
=
,
3/2
λ3
N
2 g3/2 (1)
∆(1/n) = 1/nc = λ3 /g3/2 (1)
Consider now the derivative
dPc
5 g5/2 (1)
= k
dT
2
λ3
obtained from Pc =
This proves the Clausius Clapeyron relation.
31
kT
g5/2 (1)
λ3
Note that ∆S implies a ”Latent heat” T ∆S ⇒ 1st order transition.
Cv Nk
1.925
Specific heat (T < Tc ):
3V
d T
Cv
15 ζ( 52 )
=
g5/2 (1) ( 3 ) =
∼ T 3/2
Nk
2N
dT λ
4 nλ3
At T > Tc ,
1.5
dg3/2 (ζ)
g1/2 (ζ) ζ→1
dCv
∼
=
−→ ∞
dT
dζ
ζ
T
Tc
The shape of CV (T ) is similar to the ”λ transition” of 4 He, which looks like the letter λ.
For 4 He by this theory Tc = 3.13◦ K, by experimental results Tc = 2.19◦ K.
Note: In a real superfluid there is a finite critical velocity vc below which there is no dissipation to the flow. This implies that at momentum k and frequency ω the current states
have excitations ω = vk (note the transformation to the moving frame evtd/dx ) which cannot
decay into the excitations ω(k) of the system at rest, hence vc = min{ω(k)/k}. The ideal
gas is therefore not a real superfluid since excitation energies are k = ~2 k 2 /2m, so that
vc = 0 at k → 0.
With repulsive weak interaction g
1/2
one has at k → 0 k = ~k VNmg
,
Εk
and critical current is finite either
Ω0
from this limit or from a ”roton”
minimum (see figure) so that
k
vc = ω0 /k0 .
k0
Photons in thermal equilibrium correspond to bosons with µ = 0 since there is no conservation law for photons, hence N minimizes F , i.e. ∂F/∂N = 0 = µ.
Photon spectrum is ~ωk = ~ck with |k| = k. Photons have two polarization states, = ±1.
Z=
X
{nk ,}
e−β
P
n,
~ωk nk,
=
∞
YX
k, n=0
32
e−β~ωk n =
Y
k
2
1 − e−β~ωk
ln Z = −2
P
k
ln(1 − e−β~ωk ), same as the GCE L with µ = 0. (Note F = Ω̃ when µ = 0).
hnk i =
−1 ∂
2
ln Z =
β ∂(~ωk )
exp (β~ωk ) − 1
n(ω) is the number of photons with the frequency ω.
Z ∞
Z
Z
X
V
4πV
ω 2 ω
2
n(ω)dω
hnk i =
4πk hnk idk =
( ) d( )hnk i ≡
3
3
(2π)
(2π)
c
c
0
k
u(ω) = ~ωn(ω) =
ω3
~
exp (β~ω) − 1 π 2 c3
This is the Planck’s formula for energy density.
The total energy
Z ∞
4
π 2 kB
E
~ωn(ω)dω =
=
T4
3 c3
V
15~
0
cV ∝ T 3 is unbounded as ω → ∞ can contribute.
R is the rate of radiation through a small hole in a cavity. We need to average only velocities
with vz > 0
hvz ivz >0 = c
Z
π/2
Z
cos θd(cos θ)/
π
d(cos θ) =
0
0
c
4
4
π 2 kB
T4
60~3 c2
Ec
=
≡ σT 4
V 4
where σ is Stefan’s constant (1879) σ = 5.670 · 10−8 Watt/(meter)2 (Kelvin)4 .
⇒R=
We use periodic boundary quantization where ~n has integer entries,
~c2π|~n|
∂k
1 k
k = ~c|~k| =
⇒
=−
1/3
V
∂V
3V
Therefore
P =
1 ∂
1
1
4σ 4
ln Z =
Σk ~ωk hnk i ⇒ P V = E ⇒ P =
T
β ∂V
3V
3
3c
(for nonreletavistic bosons k ∼ V −2/3 ⇒
Note that
(
∂k
∂V
=
−2 k
3 V
⇒ P V = 23 E).
∂P
)T = 0 ⇒ h∆n2 i → ∞
∂V
If the photon had a mass m ,then at kB T mc2 Stefan’s constant σ would change by a
factor 3/2, inconsistent with experiment ⇒ photon mass = 0 (i.e. less than the experimental
kB T /c2 .)
Phonons
33
Consider N atoms in a lattice that have 3N vibration modes, labeled by phonons with
wavevector ~k and 3 polarizations. In the Debye model: ω = c|~k|, ω < ωm
Z
Z ωm
3ω 2
V
2
f
(ω)dω,
f
(ω)
=
3N = Σk 3 = 3
4πk
dk
≡
V
(2π)3
2π 2 c3
0
Integrating f (ω) yields the cutoff frequency ωm = c(6π 2 n)1/3 ; the Debye temprature is TD :
kB TD = ~ωm .
Z=
X
e(−β
P
i
~ωi ni )
=
ni
3N
Y
i=1
1
1 − e−β~ωi
where i = {~k, polarization}.
hni i = −
1 ∂
1
ln Z =
β ∂(~ωi )
exp (β~ωi ) − 1
X
∂
E=−
ln Z =
~ωi hni i =
∂β
i
E=


3N kB T (1 −

3N kB T π4 (
5
3 TD
8 T
T 3
)
TD
Z
ωm
~ω
0
f (ω)
dω =
exp (β~ω) − 1
T TD
+ ...)
+ O(exp (− TTD ))
T TD
Experiments on specific heat confirm that noninteracting phonons are normal modes of solids
at low T .
2c. Ideal Fermi gas
X
PV
= ln L =
ln (1 + ζ exp (−βp2 /2m))
kT
p,spin
Where ζ = exp (βµ) and
hnp i =
so that N =
P
1
1 βp2 /2m
e
ζ
p,spin hnp i.
34
+1
P
4π
= 3g
kB T
h
∞
Z
dpp2 ln (1 + ζ exp (−βp2 /2m)) =
0
g
f5/2 (ζ)
λ3
where g = 2s + 1 is the number of spin states
Z ∞
1
g
4π
dpp2 1 βp2 /2m
= 3 f3/2 (ζ)
n= 3g
~
λ
e
+ 1)
0
ζ
1
fn (ζ) =
Γ(n)
Z
0
∞
xn−1
dx
1
exp (x) + 1
ζ
∞
X
ζl
=
(−)l+1 n
l
l=1
0 < ζ < ∞ (unlike bosons!)
For nλ3 1 or ζ 1:
nλ3 /g = ζ − ζ 2 /23/2 + ...,
1
[hnp i → nλ3 exp (−βp ) is MB form]
g
P
g
1
=
(ζ − ζ 2 /25/2 + ...) = 1 + 5/2 nλ3 + ...
3
nkB T
nλ
g2
This is the virial expansion. Consider next nλ3 1
1 3
4
π2
nλ = √ [(ln ζ)3/2 + (ln ζ)−1/2 + ...] + O(1/ζ)
g
8
3 π
Where ζ → eβF , so that by comparing T → 0 terms
~2 6π 2 n 2/3
F =
(
)
2m g
To confirm the T = 0 result hnp i →
1
exp (β(p −F ))−1
so that all the states
with p < F are occupied. Therefore
V
N =g 3
h
confirming the result F =
~2
2m
Z
d3 p =
p <F
gV 4π 3
p ⇒ F = p2F /2m
(2π~)3 3 F
2
( 6πg n )2/3 .
Expansion at kB T F :
µ = F [1 −
π 2 kB T 2
(
) + ...]
12 EF
35
5 kB T 2
3
3
) + ...] = P V.
E = N F [1 + (
5
12 EF
2
The first term is g
P
|p|<pF
p2 /2m. The second term corresponds to a fraction kB T /F of
excited particles with energy kB T , hence excess energy is ∼ T 2 and cV ∼ T .
At T=0 the pressure is due to occupied states that have p~ 6= 0.
2d. Non- Ideal gases
Atoms have on average hdipoleitime = 0 (time average is also equivalent to ensemble average).
However, the dipole moment of one atom induces a dipole on a neighboring atom such that
hdipole · dipoleitime 6= 0. This energy is ∼ − r16 .
Lennard - Jones potential:
U = [( br0 )12 − 2( br0 )6 ]
b0 ≈ 2 − 5 Å,
≈ (1 − 40) · 10−22 Joule
≈ (1 − 40) · 0.1◦ K
U
b0
0
r
−ε
Virial expansion:
Z
P
P
1
−β[ i p2i /2m+ i<j U (ri −rj )]
3N
3N
Z =
d
pd
re
N !h3N
Z
P
1
= Z0 N
d3N re−β i<j U (ri −rj )
V
Z
P
1
d3N r[(e−β i<j U (ri −rj ) − 1) + 1]
= Z0 N
V
where Z0 is the non interacting partition function.
36
1
For low density, each pair contributes independently:
R 3 3
Z = Z0 [ N (N2−1) ( d rV1 d2 r2 e−βU (r1 −r2 ) − 1) + 1]
1−e−β U(r)
0
r
Z
1
N2
F = F0 − kB T 2 (e−βU (r1 −r2 ) − 1)d3 r1 d3 r2
2
V
Z
1
(1 − e−βU (r) )d3 r
B(T ) ≡
2
N2
F = F0 + kB T
B(T )
V
∂F
N kB T
N
P = −
=
(1 + B(T ))
∂V
V
V
1
assuming convergence (U 6= r ).
>0
<0
zZ }| { Z
z }| {
b0
∞
B ≈
... +
... = b − a/T
b0
0
N
N
N kB T
(1 + b) − a( )2
P =
V
V
V
N kB T
N 2
≈
− a( ) ⇒ Van der Waals equation
V − bN
V
)2 , is the pressure reduction
where V −bN is the excluded volume and the second term, −a( N
V
from the attractive part of the interaction.
GCE formulation
Low density → low fugacity ζ
X
L=
ζ N ZN (V, T ) = 1 + ζZ1 + ζ 2 Z2 + ...
N
for both classical and quantum theories define partition functions for N = 1, 2 particles:
1
PV
ln L = ζZ1 + ζ 2 (Z2 − Z12 ) + ... =
2
kB T
∂
1
N = ζ (ln(L)) = ζZ1 + 2ζ 2 (Z2 − Z12 )
∂ζ
2
eliminate ζ ⇒
2
PV
N
1 2
N
= N−
(Z2 − Z1 ) ≡ N 1 + B
kB T
Z1
2
V
1
1 2
B = −V 2 (Z2 − Z1 )
Z1
2
37
Evaluate now Z1 , Z2 : For Z1 there are no effects of interactions or of quantum statistics, i.e.
Z1 =
X
e−βp
2 /2m
=
p
V
λ3
h
λ= √
2πmkB T
Consider next Z2 = Tr e−βH for N = 2, first the classical problem:
Z
2
Z
1
1
3 −βp2 /2m
Z2 =
d pe
· 6 d3 r1 d3 r2 e−βU (r1 −r2 )
2!
h
Z
V
=
· d3 re−βU (r)
6
2λ
R
From the expression above for B we obtain B(T ) = − 21 (e−βU (r1 −r2 ) − 1)d3 r1 d3 r2 as in the
previous derivation.
Consider next the quantum problem with the 2-particle Hamiltonian
H=
−~2
~2 2
~2
∇2r1 + ∇2r2 + U (r1 − r2 ) = −
∇R − ∇2r + U (r)
2m
4m
m
where r = r1 − r2 , R = 21 (r1 + r2 ). The eigenvalues of H are
P2
4m
+ En where P is the center
of mass momentum, hence
Z2 =
X
P
e−βP
2 /4m
·
X
e−βEn =
n
X
23/2
e−βEn
V
·
λ3
n
Note that En contain information on statistics, i.e. En is restricted to eigenfunctions that
are symmetric in r for bosons, or to antisymmetric ones for fermions.
38
3. PHASE TRANSITIONS
Order Parameter
Example
Tc (◦ K)
Liquid-gas
density
H2 O
647
Ferromagnetic
magnetization
Fe
1044
Antiferromagnetic
sublattice magnetization
F eF2
78
Bose condensation
superfluid amplitude
Superconductivity
4
He
2
electron pair amplitude
Y Ba2 Ca3 O7
90
Binary fluid
concentration of one fluid
CCl4 − C7 F14
302
Binary alloy
density of one kind on a sublattice
Cu − Zn
739
Ferroelectric
polarization
T riglycine − sulf ate
322
Ferroelastic
q = 0 distortion
Ni − T i
charge density wave
q 6= 0 distortion
N bSe3
59
Metal-Insulator
Percolation
fraction of sites in percolating cluster
Definition: n-th order phase transition has a discontinuity in the n-th derivative of a free
energy. Hence a 1st order transition has a jump in entropy (−∂F/∂T ) which is the latent
heat. A 2nd order transition has a jump in Cv (contains ∂ 2 F/∂T 2 ).
3a. First order transitions
n
M
nl
mid point
(assymetry)
Tc
coexisting domains
ng
T
T
Infinitesimal change in P (liquid-gas on left figures) or in H (magnetic field in a ferromagnet
39
on right figures) leads to a jump in density n (on left) or in magnetization (on right). In
P − T plane (left) or H − T plane (right) first order transition ends at a critical point Tc .
P
H
liquid
triple point
solid
•
↑
↓
critical point
•
↑
↓
↑
↓
•
gas
T
T
Consider the liquid-gas phase transition as a prototype of a 1st order transition
Liquid-gas: 1st transition, no symmetry change.
Solid-liquid: 1st order transition, change in translation symmetry.
Fixing V below Tc leads to liquid/gas coexistence.
Fixing P leads to a jump in V from fully gas phase to fully liquid phase.
40
Clapeyron’s equation
g(P, T ) =
1
G(P, T, N )
N
where G(P, T, N ) is the Gibbs free energy.
∆g(P, T ) = g2 (P, T ) − g1 (P, T )
where gi (T, P ) is formally continued across Tc .
Define s = S/N, v = V /N
∆s = s2 − s1 ;
∂∆g
= −∆s;
∂T P
41
∆v = v2 − v1
∂∆g
= ∆v
∂P T
By the chain rule
∂∆g
∂T
P
∂T
∂P
∆g
∂P
∂∆g
T
= −1
we get
(∂∆g/∂T )P
=−
(∂∆g/∂P )T
∂P
∂T
∆g
=−
∆s
∆v
Along the transition line P (T ) we get ∆g = 0 =const.
dP
∆s
latent heat
=
=
.
dT
∆v
T ∆v
Van-der Waals equation
A rough argument:
Vef f = V − b
a
P = Pkin − 2
V
where b ∼ N is the excluded volume and a ∼ N 2 is a measure of attractive forces among
the molecules of the system; a ∼ N 2 since there are N 2 /V 2 pairs. Hence
a Pkin Vef f = (V − b) P + 2 = N KB T
V
At T > Tc we get one solution for
the equation, and for T < Tc we get
three.
The three roots merge to one
root at inflection point Pc , Vc , Tc
The parameters a, b are sample specific. To identify them in term of Pc , Vc , and Tc rewrite
the equation in the form
h
iV2
a (V − Vc )3 = 0 = (V − b) Pc + 2 − N kB Tc
V
Pc
By identifying each power of V in this form we write a, b in terms of Pc , Vc , Tc , and with
P̄ = P/Pc , T̄ = T /Tc , V̄ = V /Vc we obtain
3
1
8
P̄ + 2
V̄ −
= T̄
3
3
V̄
42
which is the law of corresponding states – all substances have the same equation of state
when their P, V, T are measured in units of the critical point.
∂P
> 0 is unstable so that (∆N )2 → 0.
Note the region with ∂V
Alternative derivation (S.K.Ma p.470)
G0 (T, P, N ) = E − T S + P V
where
3
N
E = N K B T − a1 N
2
V
attraction from neighboring ∼ N/V atoms, and
1 (V − b1 N )N
S = kB ln
N!
λ3N
where b1 N is the excluded volume and λ is the thermal wavelength.
Proper G(T, P, N ) is obtained at minimum with respect to V which is a redundant variable
for G.
3
N2
V − b1 N
G0 (T, P, N ; V ) = N KB T − a1
− kB T N ln
+ N + PV
2
V
N λ3
If
∂G0
∂V
= 0, we get the Van der Waals equation with a = a1 N 2 and b = b1 N .
G0 shows (see figure) that with 3 solutions 1 is metastable and 1 is unstable (max of G0 )
For P1 < P < P2 gas is stable, liquid is metastable.
For P2 < P < P3 liquid is stable, gas is metastable.
43
P2 is when gas and liquid are degenerate G0 (V1 ) = G0 (V2 ).
Maxwell’s construction
In general G0 = F (V, T, N ) + P V where P (V ) = −∂F/∂V . At P2 :
0 = G0 (V2 ) − G0 (V1 ) =
Z
V2
[−P (V ) + P ]dV .
V1
Hence the area between P = P2 line and the P (V ) curve vanishes – this is Maxwell’s
construction.
An alternative derivation: In terms of F (V ) choose V1 and V2 such that parallel tangents
generate one line with
∂F
∂V1
=
∂F
.
∂V2
The line is
∂F
F2 − F1
=
⇒
V2 − V1
∂V1
Z
P2 (V2 − V1 ) = −(F2 − F1 ) =
V2
P (V )dV
V1
which is the same Maxwell’s construction. The line is
realized by coexistence of the two phases with fractions
x and 1 − x, respectively (0 < x < 1),
V = xV1 + (1 − x)V2
Fline = xF1 +(1−x)F2 =
(V − V2 )F1 + (V1 − V )F2
<F
V1 − V2
⇒ Phase separation at P2 since V1 < V < V2 has lower
Fline then F on the continuous curve.
When V is changed carefully to avoid strong fluctuation, metastable phases can persist until they become
44
3b. Second order phase transitions – Mean Field theory
Consider ferromagnetism as a prototype second order phase transition. Localized independent moments ±µ in a magnetic field B lead to magnetization
M =N
µeβµB − µe−βµB
= N µ tanh (βµB) .
eβµB + e−βµB
Mean field theory for interactions: neighbors with magnetization ∼ M/V induce an addi-
, α ∼ interaction strength.
tional field, B → B + α M
V
=⇒
M = N µ tanh βµ B + α M
has M 6= 0 solutions even if B = 0.
V
If T = Tc , 1 =
N 2
µ α/kB Tc
V
Expand near Tc :
Tc
1
M= M−
T
3
3
M
βµα
N µ =⇒ M ∼ (Tc − T )1/2 .
V
Susceptibility at T > Tc : B → 0, M → 0
M
∂M
1
βN µ2
M = N µβµ B + α
∼
=⇒ χ ≡
=
.
2
V
∂B B=0 1 − βN µ α/V
T − Tc
Microscopic Model
Consider two neighbors of spin 1/2, total spin is S = 0, 1 with energy difference which defines
the ”exchange energy” J. The requirement for an antisymmetric wavefunction requires a
symmetric orbital for the singlet S = 0 and antisymmetric one for the triplet S = 1, hence
a large energy difference from the difference Coulomb interactions. Hence J is much larger
45
then that of interacting magnetic dipoles. Consider
~si · ~sj =
1
2
(~si + ~sj )2 − ~s2i − ~s2j = 21 S(S + 1) − 12 ·
3
2

 −3/4 S = 0
=
.
 1/4 S = 1
P
=⇒ Interaction energy = −J 0n.n. ~si ·~sj +const (n.n. is summation on γ nearest neighbors).
P0
means that each pair is summed once.
If the crystal has an easy axis for the spin, ~si → (sz )i , hence two models:
H = −J
0
X
hi,ji
~si · ~sj
H = −J
~s spin operator: Heisenberg model
0
X
σi = ±1: Ising model.
σi σj
hi,ji
Solution of the Ising model by mean field:
H MF = − 21 Jγ hσi
1
2
X
σi
γ no. of nearest neighbors.
i
is needed so that H MF = − 21 N Jγ hσi2 ,
1
Nγ
2
no. of bonds. (e.g. N/2 ”odd” sites each
generates γ bonds.)
P
hσi = hσi i =
βγJhσiσi /2
σi e
eβγJhσiσi /2
σi =±
=⇒ kTc =
βγJhσi
e
P
σj /2
j6=i
{σj6=i }
σi =±
P
P
P
βγJhσi
e
P
j6=i
σj /2
= tanh βJγ 12 hσi
{σj6=i }
1
Jγ.
2
3c. Exact results
Mapping between systems
1. Lattice gas model to Ising
Na atoms occupy sites in lattice with N cells; Naa number of nearest neighbors, each with
energy 0 , g(Na , Naa ) is the number of configurations with a given Na , Naa (a non-trivial
function). Grand partition
LG =
X
Na
ζ Na
X
g (Na , Naa ) eβ0 Naa .
Naa
46
Canonical partition of Ising has the same form:
N+ no. of + spins, N++ no. of ++ neighbors. Draw one line from all + sites to all their
neighbors:
Number of lines = γN+ = 2N++ + N+−
same for N−−
γN− = 2N−− + N+− ,
=⇒
X
hi,ji
N+ + N− = N
σi σj = N++ + N−− − N+− = 4N++ − 2γN+ + 21 γN
X
σi = N+ − N− = 2N+ − N
i
EIsing = −4JN++ + 2(Jγ − µB)N+ − 21 γJ − µB N
X
X
1
βN
γJ−µB
2
Z=e
g (N+ , N++ ) e4βJN++
e−2β(Jγ−µB)N+
N++
N+
Since configuration counting is the same as in lattice gas we get the same function g(Na , Naa ),
hence the two problems are equivalent as in the following table:
Ising
Lattice gas
N+
Na
4J
0
e−2β(Jγ−µB)
−
1
F
N Ising
ζ
+ 12 γJ − µB
M = N+ − N− , NN+ =
1
2
P
M
N
+1
1
v
=
hNa i
N
order parameter
2. Binary alloy to lattice gas
N11 , N22 , N12 no. of pairs of each type
EA = 1 N11 + 2 N22 + 12 N12
As above
γN1 = 2N11 + N12
γN2 = 2N22 + N12
=⇒
N12 = γN1 − 2N11
N22 =
1
γN
2
47
+ N11 − γN1
N1 + N2 = N
EA = (1 + 2 − 212 ) N11 + γ (12 − 2 ) N1 + 21 γ2 N ,
Z (N1 , T ) =
X
g (N11 ) e−βEA
N11
Lattice gas Binary alloy
Na
N1
−0
1 + 2 − 212
F
F + γ (12 − 2 ) N1 + 21 γ2 N
Ising Model in 1D: Exact Solution
N
X
X
1
βµB
βJ
(σk + σk+1 )
σ
σ
+
k k+1
X X
X
2
k
Z=
...
e k=1
;
σ1=± σ2=±
σ1 = σN +1
σN =±
Consider the bond (k, k + 1) with the elements
e
βJσk σk+1 + 12 βµB(σk +σk+1 )
This element can have 4 values for σk = ± and σk+1 = ±.
Call these elements
P1,1 , P1,−1 , P−1.1 , P−1,−1 and define a matrix

P̂ = 
P1,1
P1,−1
P−1,1 P−1,−1


=
e
β(J+µB)
−βJ
e
e
−βJ
β(J−µB)
e


Note that P̂ is symmetric by the choice of σk + σk+1 for the B term in the Hamiltonian.
P̂ is defined in a spinor space |±i, e.g. h+|P̂ |−i = P1,−1 .
E.g. for N = 3 the partition Z has 23 terms, one of them is
↑↑↓↑ =⇒ P1,1 P1,−1 P−1,1 = h+|P̂ |+ih+|P̂ |−ih−|P̂ |+i
In general we need
Z=
X
all |ii=|±i
h1|P̂ |2i . . . hk|P̂ |k + 1ihk + 1|P̂ |k + 2i . . . hN |P̂ |N + 1i
=
X
N
(P N )σ1 ,σN = T r(P N ) = λN
1 + λ2
σ1 =σN =±
48
λ1 , λ2 are the eigenvalues of P̂ : λ2 − 2λeβJ cosh (βµB) + 2 sinh (2βJ) = 0
1
=⇒ λ1,2 = eβJ cosh (βµB) ± [e−2βJ + e2βJ sinh2 (βµB)] 2
λ2 < λ1 , N → ∞, λ1 dominates, hence
1
N
ln Z → ln λ1 and is an analytic function (except
at T = 0). For the magnetization we have
M=
1 ∂
β ∂B
ln Z = N µ
sinh (βµB)
1
[e−4βJ +sinh2 (βµJ)] 2
=⇒ M (B = 0, T > 0) = 0,
M (T = 0) = N µsign(B), hence an ordered phase is only at
T = 0. The susceptibilty:
χ=
∂M
|
∂B B=0
=
N µ2 2J/kB T
e
|T →0
kB T
→ ∞ indicates ordering at T = 0.
No phase tradition in 1-dimension – general argument:
Low energy excitations are walls ↑↑↑|↓↓↓, energy is 2J (relative to the ground state), while
entropy is S = kB ln N since the wall can be positioned at any bond.
=⇒ F = E − T S = 2J − kB T ln N < 0 ⇒ no long range order in 1-dimension at any finite
temperature.
2D Ising model: Onsager’s solution (1944)
KTc = 2.269J
M ∼ (Tc − T )1/8 near Tc
Cv ∼ − ln (1 −
T
)
Tc
3d. Landau’s Theory for second order phase transitions
Partition sum is done first locally to define a ”‘coarse grained”’ order parameter, slowly
varying. F {M (r)} is determined by symmetry for small M (r), i.e near a 2nd order phase
transition. Coarse graining involves a finite cluster ⇒ F {M (r)} has an analytic expansion.
For ferromagnet, M → −M , symmetry+analiticity ⇒
1
1
1 ~
F {M (r)} = a(T )M 2 (T ) + bM 4 (T ) + c(∇M
)2
2
4
2
where a(T ) = a0 (T − Tc ) so that in Mean Field the transition occur at Tc .
49
Mean Field: M (r) = M is homogenous, no fluctuations.
q
a0
∂F
=
0
⇒
M
=
(Tc − T )
∂M
b

 − 1 a02 (T − T )2 T < T
c
c
4 b
F =
0
T > Tc
continuous (S < 0 at T < Tc , needs correction by fluctuations, see below.)
S = − ∂F
∂T
∂S
cV = T ∂T
has a jump, and is ∼ (Tc − T )0 at T < Tc .
Consider next B 6= 0 and add −M · B to F :
T > Tc ,
a0 (T − Tc )M − B = 0 ⇒ χ =
T = Tc ,
bM 3 − B = 0 ⇒ M ∼ B 1/3
M
|
B 0
∼
1
T −Tc
Consider fluctuations at T > Tc : neglect M 4 , M (r) = V −1/2
R
F {M (~r)}d3 r =
1
2
P
k
(a + ck 2 )|Mk |2 since
R
P
k
Mk eik·r
0
eik·r+ik ·r d3 r = V δ−k,k0 . Note that M−k = Mk∗
for a real M (r).
The weight of ±k modes is e−β(a+ck
< M (r)M (0) >= V −1
=V
−1
X
R
ikr
e
k
P
ikr
= V −1
k < Mk M−k > e
2
X
k
Z=
YZ
k
pc
a
. The correlation function is then
ikr
ke
P
2
Q
R
k0
02
Z
∞
dxe−β(a+ck
0
eik·r
kB T −r/ξ
=
e
2
β(a + ck )
4πrc
1
∼ (T − Tc )− 2 is the correlation length. Consider next the free energy
d|Mk |2 e−β(a+ck
2 )|M
k|
2
∼
Y
[β(a + ck 2 )]−1
k
50
0 2
d(ReMk0 )d(ImMk0 )|Mk |2 e−β(a+ck )|Mk |
R
02
2
d(ReMk0 )d(ImMk0 )e−β(a+ck )|Mk0 |
X
d(|Mk |2 )|Mk |2 e−β(a+ck )|Mk |
∂
−1
R
=
V
eikr
ln
2
2
2
−β(a+ck
)|M
|
k
∂(β(a + ck 2 ))
d(|Mk | )e
k
< M (r)M (0) >= V −1
where ξ =
2 )|M |2
k
2 )x
⇒ F = kB T [
X
ln β(a + ck 2 ) + const.]
k
Therefore
∂ 2F
CV = −T
∂T 2
∼
1/a0
Z
dd k
+ less singular terms.
(a + ck 2 )2
a0 is a lattice constant, d=dimensions. Using k 0 = ξk
1
CV ∼ 2
a
Z
ξ
a0
d−4
dd k 0
ξ −d ∼ ξ 4−d ∼ (T − Tc ) 2
02
2
(1 + k )
Validity of mean field:
hM 2 (0)i
hM 2 (0)i ∼
<<
R
hM i2 ; evaluate at T > T c (similar result at T < Tc ) :
dd k
β(a+ck2 )
∼
R
kd−1 dk
(a+ck2 )
√ d−2
∼ ( a)
∼ |T − Tc |(d−2)/2
Comparing with hM i2 ∼ (Tc − T ) shows that mean field is valid at T → Tc if d > 4.
At d < 4
mean field breaks down at T → Tc
Critical exponents:
t =
T −Tc
,
Tc
T > T c; for exponents
βα0 γ 0 ν 0
t =
Tc −T
,
Tc
T < Tc
Exponent
Definition
Mean Field
Fe
α,α0
Specific heat cV ∼ t−α
2 − d/2
−0.12
1/2
0.34
1
1.33
3
4.2 (Ni)
β
γ, γ 0
δ
ν, ν 0
η
Order parameter ∼ tβ
Susceptibility x ∼ t−γ
Order parameters at Tc ∼ B 1/δ
correlation length ξ ∼ t−ν
1/2
at Tc h M (r)M (0)i ∼ 1/rd−2+η
0
0.07
Universality: exponents depend only on symmetry of order parameter and dimensionality.
51
Scaling assumption:
ξ is the single length responsible for singular behavior.
⇒ obtain relations between exponents.
1
e.g. hM (r)M (0)i ∼ rd−2+η
g(r/ξ)
at Tc need g → g(0) 6= 0
R
In terms of M , integration on all configurations M (r), and the Hamiltonian H we have for
the susceptibility
χ=
∂hM i
|
∂B B→0
=
∂
∂B
R
d r)
M (0)e−β(H− M (r)Bd
R
R
−β(H− M (r)Bdd r)
R
M
M
e
|B=0 =
R
R
M
M (0)M (r)dd re−βH
R
−βH
M e
=
R
hM (0)M (r)i dd r
since at T > Tc the hM iB=0 = 0 term vanishes (from ∂/∂B of denominator.)
R
χ ∼
⇒
r−d+2−η g(r/ξ)dd r
∼ ξ −d+2−η · ξ d
∼ (t−ν )2−η
γ = ν(2 − η)
Renormalization Group:
Consider the 1-dimensional Ising model Z =
P
σk =±
eβJ
P
k
σk σk+1
Sum on all even sites to obtain a new form for Z
P
0
0
β(Jσ1 σ2 + Jσ2 σ3 )
= eβ(a + J σ1 σ3 )
σ2 e
0
0
0
Identify a , J by two cases:
σ1 σ3 = − case:
J0
⇒
Z(k) ∼
X
e
X
(a0 −J 0 )β
2 = e
σk σk+2
odd k
odd i σi =±
with the renormalized coupling J 0 given by e
At Tc
0
σ1 σ3 = + case: e2Jβ + e−2Jβ = e(a +J )β
−2βJ 0
ξ → ∞ H is scale invariant, i.e. J 0 = J ⇒
52
= e
2
e
2βJ +
−2βJ
Tc = 0.
4. NON- EQUILIBRIUM
4a. Kinetic theory and Boltzmann’s equation:
f (~r, ~v , t)
~r, ~v
is particle density in phase space d3 rd3 v.
→ ~r + ~v δt ,
~v +
In absence of collisions
~
F
δt
m
so that volume A moves to an infinitesimally close volume B with f changed only by collisions,
f ( ~r + ~v δt, ~v +
~
F
δt
m
, t + δt) − f ( ~r, ~v , t)
≡ ( ∂f
) δt
∂t coll
Effect of collisions: If one initial particle is in A the result is certainly outside A (up to the
differential volume of A) – this is a loss term; conversely with one final particle in B,
Rδtd3 rd3 v = no. of collisions with one initial particle in A
R̄δtd3 rd3 v = no. of collisions with one final particle in B [= A + O(δt)]
( ∂f
) δt = (R̄ − R)δt
∂t coll
Assume binary collisions (dilute gas)
v~1 + v~2 =
v~1 , v~2 → v~1 0 , v~2 0
v~1 0 + v~2 0
v~1 2 + v~2 2 = v~1 0
2
+ v~2 0
2
Define V~ = 21 (v~1 + v~2 ) , ~u = v~2 − v~1 ;
⇒ V~ = V~ 0 , |~u| = |~u0 |
V~ 0 =
1
(v~ 0
2 1
+ v~2 0 ) , ~u0 = v~2 0 − v~1 0
A collision is defined by an angle Ω in the center of mass, i.e. between ~u, ~u0 . V~ , ~u, Ω are
independent parameters which determine V~ 0 , ~u0 .
Consider ~u → ~u + d~u , as Ω is fixed (i.e. dΩ is independent of d3 u).
For ~u0 → ~u0 + d~u0 the triangle (~u, ~u + d~u) is similar to (~u0 , ~u0 + d~u0 ), hence |du0 | = |du|
Choosing an orthogonal basis set d~u1 , d~u2 , d~u3 it transforms to a rotated orthogonal basis
with axes equal in magnitude, hence d3 u = d3 u0
Since d3 V = d3 V 0
⇒
d3 v1 d3 v2 = d3 v10 d3 v20 .
Assumption of molecular chaos: This is the central assumption of the Boltzman equation.
The number of pairs with velocities (v~1 , v~2 ) about to collide is the product
53
f (~r, v~1 , t)d3 rd3 v1 · f (~r, v~2 , t)d3 rd3 v2
i.e. both particles ”do not know” about each other until they collide, hence no correlation
before the collision.
The scattered flux of v~2 hitting v~1 is
Z
Z
3
I = d v2 dΩσ(Ω) |v~1 − v~2 | f (~r, v~2 , t)
where σ(Ω) is the scattering cross section.
f (r, v~1 , t)d3 v1 is the number of target particles, therefore Rd3 v1 = f (~r, v~1 , t)d3 v1 · I
For R̄ define a collision by variables v~0 , v~0 → v~1 , v~2 so that v1 ∈ A.
1
2
The scattered flux of v~20 on v~10 is
0
I =
Z
d3 v20
Z
Z
Z
~0
3 0
0
0
~
~
dΩσ (Ω) v1 − v2 f (r, v2 , t) = d v2 dΩσ(Ω) |v~1 − v~2 | f (~r, v~20 , t)
0
using microscopic time reversal σ 0 (Ω) = σ(Ω) (spin and internal quantum states are neglected).
f (~r, v~10 , t)d3 v10 is the number of target particles (no integral since (v~1 , v~20 , Ω) determine v~10 ).
Therefore, Rd3 v1 = f (~r, v~0 , t)d3 v 0 · I 0 .
1
1
Since d3 v1 d3 v2 = d3 v10 d3 v20 the probability for the reverse collision process is
Z
R=
3
d v2
Z
dΩσ(Ω) |v~1 − v~2 | f (~r, v~10 , t)f (~r, v~20 , t)
Hence Boltzman’s transport equation:
∂
~ r + 1 F~ · ∇
~ v1 )f (~r, v~1 , t) =
( + v~1 · ∇
∂t
m
Z
Z
σ(Ω)dΩ d3 v2 |v~1 − v~2 | [f (~r, v~20 , t)f (~r, v~10 , t) − f (~r, v~2 , t)f (~r, v~1 , t)]
Time reversal would imply that f (~r, −v~1 , −t) also solves this equation. However, changing
t → −t, and all ~v → −~v in this equation, the left side changes sign (acting on f (~r, −v~1 , −t))
while the right hand side does not change sign (acting on f (~r, −~v , −t) with the various v).
Hence time reversal, although valid on the microscopic level, is broken on the macrospcopic
54
level. [If t → −t and also v~i → −v~i0 the equation is invariant, i. e. reversing the order of
colliding particles; assumption of ”molecular chaos” looses this symmetry.]
Consider the case with no external force F = 0 and no ~r dependence (e.g. no density waves).
In equilibrium
∂f
∂t
= 0, therefore the right side is equal to 0 for all v1 , which requires
f (v2 )f (v1 ) = f (v20 )f (v10 )
hence ln f (~v ) is conserved in all collisions. The conserved quantities are momentum and
energy, therefore
~ · ~v + c · (~v )2 = −b(~v − v~0 )2 + ln a
ln f (~v ) = A + B
2
⇒ f (~v ) = ae−b(~v−v~0 )
which is the Maxwell distribution (up to a center of mass velocity).
Boltzman’s H theorem
As a candidate for entropy, Boltzman defined the function H(t):
Z
H(t) ≡ f (~v , t) ln[f (~v , t)]d3 v
where f (~v , t) is a solution of Boltzmann’s equation.
We show now that
dH(t)
dt
≤ 0 so that H(t) decreases with time. Define as a shorthand
f1 = f (v~1 , t)
f2 = f (v~2 , t)
f10 = f (v~10 , t)
f20 = f (v~20 , t)
(2)
Thus we get:
∂f1
=
∂t
dH
=
dt
Z
Z
d v2
Z
3
d v2
Z
3
d v1
by changing v1 ↔ v2 , using eq. for
dH
1
=
dt
2
Z
3
d v1
Z
Z
3
3
d v2
Z
dΩσ(Ω) |v~1 − v~2 | (f20 f10 − f2 f1 )
dΩσ(Ω) |v~1 − v~2 | (f20 f10 − f2 f1 )(1 + ln f1 )
∂f2
∂t
and taking
1
2
of both, we get
dΩσ(Ω) |v~2 − v~1 | (f20 f10 − f2 f1 )[2 + ln(f1 · f2 )]
55
Now, using equations for
dH
1
=
dt
2
Z
3
d v1
∂f10
∂t
Z
∂f20
,
∂t
,
3
i.e. fi0 ↔ fi we get
Z
dΩσ 0 (Ω) v~20 − v~10 (f2 f1 − f20 f10 )[2 + ln(f10 · f20 )]
d v2
And finally, adding the last two equations
Z
Z
Z
1
dH
3
3
=
d v1 d v2 dΩσ(Ω) |v~2 − v~1 | (f20 f10 − f2 f1 )[ln(f1 · f2 ) − ln(f10 · f20 )]
dt
4
Since (x − y) ln xy > 0 for all x, y we have
dH
≤0
dt
In equilibrium f20 f10 = f2 f1 and
dH
dt
(3)
= 0.
Therefore, a candidate for entropy is (−H) which increases with time and is maximal in
equilibrium. [Note that f (r, v, t) is coarse grained by the assumption of molecular chaos,
therefore initial velocities vi and final velocities vi0 are distinguished].
4b. Brownian motion
Brown (1828) - the random motion of pollen particles in a solution.
Einstein (1905) - random walk problem.
Consider one dimension: Pn (m) is the probability of arriving at coordinate m after n steps,
i.e. one needs 12 (n + m) steps to the right and 21 (n − m) steps to the left,
Pn (m) =
1 n
2
[ 21 (n
n!
+ m)]![ 21 (n − m)]!
P
is the equal probability of going right or left. Normalization: nm=−n Pn (m) = 1
p
√
Averages: m = 0, m2 = n. For m n (since m2 = n n)) and using Stirling’s limit
where
1
2
we get:
2
2
e−m /2n
2πn
Defining x = ml (l is a step size in space) and t = nτ (τ is the time for each step)), we get
Pn (m) ≈ √
56
x2 =
so that D =
l2
2τ
l2
t = 2Dt
τ
is the diffusion coefficient.
In this continuoum form
1
2
e−x /4Dt
4πDt
In general, D is defined by the linear response:
P (x) = √
~ r, t)
j(~r, t) = −D∇n(~
If there is a gradient, then there is a current which responds to reduce this gradient.
The continuity equation is ∇ · j +
∂
n(r, t)
∂t
= 0. Taking a gradient yields
∇2 n(~r, t) −
1 ∂n(~r, t)
=0
D ∂t
This is the diffusion equation. The solution with an initial value n(r, t = 0) = N δ 3 (r) is
n(~r, t) =
with
R∞
0
N
(4πDt)
3/2
e−r
2 /4Dt
~ >= 0 and
n(~r, t)4πr2 dr = N . This solution shows < r(t)
Z ∞
1
2
n(~r, t)r4 dr = 6Dt
< (~r(t)) >= 4π
N
0
Langevin’s equation
A particle of mass M is immersed in a medium which produces a random fluctuating force
D
E
~
~
~
~
M A(t). A(t) has time correlation A(t + τ )A(t) which decays fast with τ .
The correlation time τcol is the time between collisions of medium particles and the particle
of mass M . The medium leads also to irreversibility, i.e. friction.
.
~
~v = −γ~v + A(t)
1/γ is the time scale for approaching equilibrium (see below).
D
E
~
We assume τcol << 1/γ so that the average on the distribution of random forces is A(t)
=
~ + τ ) · A(t)i
~
0, while hAi (t + τ )Aj (t)i = 13 C · δ(τ )δij , i.e. hA(t
= Cδ(τ ).
D
E
D
E
~ · ~r = 0, while A
~ · ~v 6= 0 (exercise).
Note that A
Evaluate now hr2 i assuming that the particle is in equilibrium at temperature T, 21 M hv 2 i =
3
kT :
2
57
~r · ~v =
1d 2
r
2 dt
1 d2 2
~
~r· ~v =
r − v 2 = ~r · (−γ~v + A)
2
2 dt
2
d 2
d2 2 +
γ
=
2
v = 6kT /M
⇒ average :
r
r
dt2
dt
6kT
⇒ r2 =
t + c1 + c2 e−γt
γM
.
Initial value ~r(t = 0) = 0 ⇒
d 2
r
dt
= 2~r · ~v = 0 at t = 0.
6kT
1
[t − (1 − e−γt )]
⇒ r2 =
γM
γ
2 2 2
r = v t ballistic
2
6kT
r →
t diffusive
γM
t << 1/γ
t >> 1/γ
Friction γ is related to diffusion D, D =
kT
.
γM
γ is also related to mobility µ - responce to
constant force , e.g electric field E for particle with charge e.
~
.
~ + eE
~v = −γ~v + A(t)
M
.
In steady state ~v = 0 ⇒ hvi = eE/γM ≡ µE
D = kT µ/e
Einstein’s relation
~ was not explicit. Consider next the velocity fluctuations. We
So far the random force A
~
assume now E = 0; if E 6= 0 one needs just to shift ~v → ~v + eE/M
γ.
~v (t) = ~v (0)e
−γt
−γt
Z
+e
t
~
eγu A(u)du
0
2 v (t) = v 2 (0)e−2γt + e−2γt
Z tZ
0
t → ∞ should have equilibrium
⇒
t
h~v (t)i = ~v (0)e−γt
eγ(u1 +u2 ) hA(u1 )A(u2 )idu1 du2 = v 2 (0)e−2γt +
0
1
M
2
hv 2 i = 32 kT
C
(1 − e−2γt )
2γ
⇒ C = 6kT γ/M .
Strength of fluctuating force ∼ γ, both fluctuations and friction are due to the medium.
58
3kB T
v 2 (t) = v 2 (0) + (
− v 2 (0))(1 − e−2γt )
M
General solution for hr2 i (v 2 (0) 6= 3kB T /M )
:
2
d 2
d 2
+
γ
=
2
v
r
r
dt2
dt
1
3kB T
6kB T
⇒ r2 = 2 v 2 (0)(1 − e−γt )2 −
t=
(1 − e−γt )(3 − e−γt ) +
2
γ
Mγ
Mγ
=

v 2 (0)t2 + O(t3 )

6kB T
t
Mγ
t << 1/γ
t >> 1/γ
Velocity correlations: Kv (τ ) ≡ hv(t)v(t + τ )i
R t R t+τ
0
= v 2 (0)e−γ(2t+τ ) + e−γ(2t+τ ) 0 0 eγ(u+u ) Cδ(u − u0 )dudu0 =
=


v 2 (0)e−γ(2t+τ ) +
C −γ(2t+τ ) 2γt
e
(e
2γ

v 2 (0)e−γ(2t+τ ) +
C −γ(2t+τ ) 2γ(t+τ )
e
(e
2γ
⇒ Kv (τ ) = v 2 (0)e−γ|τ | + (
=
3kB T −γ|τ |
e
M
− 1)
− 1)
τ >0
τ <0
3kT
− v 2 (0))(e−γ|τ | − e−γ(2t+τ ) )
M
at long times
t + τ, t >> 1/γ .
Note that replacing v 2 (0) by its equilibrium average yields Kv (τ ) = (3kB T /M )e−γ|τ | at all
times t.
Reaching equilibrium at large times Kv (τ ) becomes t independent ⇒ stationary medium.
Note that for correlations of just one component, e.g. vx , we have
Kvx (τ ) = hvx (0)vx (τ )i =
kB T −γ|τ |
e
.
M
Below the Fourier transform is needed:
Z ∞
γ
2kB T
Φvx (ω) =
Kvx (τ )eiωτ dτ =
2
M ω + γ2
−∞
59
4c. Fluctuation Dissipation Theorems (FDT)
General properties of averages of stationary medium
Consider a variable x(t) such that Kx (τ ) = hx(t) · x(t + τ )i is independent of the time t.
Kx (0) > 0; h[x(t) ± x(t + τ )]2 i = 2[Kx (0) ± Kx (τ )] ≥ 0
τ →∞
it follows that |Kx (τ )| ≤ Kx (0) is a decreasing function, and usually Kx (τ ) → 0. Also
Kx (τ ) = hx(t − τ )x(t)i = Kx (−τ ), with shift t → t − τ .
Consider now the Fourier transform x(ω) :
x(t) =
R∞
−∞
x(ω)e−iωt dω
2π
For a precise derivation one needs x(t) to be finite only in the interval [− T2 , T2 ]. The spacing
∆ω between distinct ω values with cos(ω T2 ) = 0 or sin(ω T2 ) = 0 is ∆ω =
2π
.
T
(See Wannier
p. 481). For T → ∞ consider
R
R
0
Kx (τ ) = dω dω 0 e−iωt−iω (t+τ ) hx(ω)x(ω 0 )i /(2π)2
R
Since this is t independent, perform the integral ... dt
T
Kx (τ ) =
R
dω
R
dω
0
2π
δ(ω
T
0
0
+ ω )e−iω τ
hx(ω)x(ω‘ )i
(2π)2
= ∆ω
R
dω |x(ω)|2 eiωτ /(2π)2
This is the Wiener-Khinchin theorem:
Φx (ω) =
R
|x(ω)|2
⇒ |v(ω)|2 =
|A(ω)|2
(ω 2 +γ 2 )
hx(t)x(t + τ )i e−iωτ dτ =
∆ω
2π
|x(ω)|2 is called the intensity, or the power spectrum of x.
As an example, consider Langevin‘s equation:
(−iω + γ)v(ω) = A(ω)
from the theorem above Φv (ω) =
γ
2kB T
M γ 2 +ω 2
and ΦA (ω) =
2kB T γ
,
M
ΦA (ω)
γ 2 +ω 2
which is consistent with previous result: Φv (ω) =
the latter corresponds to white noise.
Note also that −iωx(ω) = v(ω), hence
|x(ω)|2 =
|v(ω)|2
ω2
⇒ Φx (ω) =
60
2kB T
γ
M (γ 2 + ω 2 )ω 2
Dissipation and Response functions
An external force adds the Hamiltonian term −F (t)x. The Hamiltonian becomes:
H=
p2
2m
+ V (x; env) − F (t)x
where ”env” stands for the coordinates and momenta of the environment. For an explicit
system + environment see the Appendix.
The equations of motion are:
ẋ =
∂H
∂p
=
p
,
m
ṗ = − ∂H
= − ∂V
+ F = mẍ
∂x
∂x
hence F (t) is indeed an external force.
The rate at which the system’s energy changes is
=0
dH
dt
z
}|
{
∂H
∂H
=
ṗ +
ẋ +... +
∂p
∂x
∂H
∂t
= −x dF
dt
where the sum of the first two terms on the right hand side vanishes by Hamilton’s equations
for x, p; the ... terms are similar derivatives with respect to the coordinates and momenta
of the environment which also vanish due to their corresponding Hamilton’s equations.
The dissipation rate is the rate of energy absorption from the external source, averaged
on time. This includes an average on the enviorment hx(t)i (e.g. on the random force in
Langevin’s equation) and on the explicit time dependence in F (t), hence
dF
dE
= −hxi
dt
dt
The dissipation rate can be expressed in terms of a response function αx (ω) where the linear
response at small forces F is defined by
hx(ω)i = αx (ω)F (ω)
It is more compact to consider a single frequency F (t) = 21 f0 e−iωt + 12 f0∗ eiωt (although a
hx(t)i = 12 αx (ω)f0 e−iωt + 12 αx∗ (ω)f0 eiωt
with αx∗ (ω) = αx (−ω)
The dissipation is then:
61
iω
dE
∂F
iω
=− x
= [αx (ω)f0 e−iωt + αx∗ (ω)f0∗ eiωt ][f0 e−iωt − f0∗ eiωt ] = − |f0 |2 [αx (ω)−αx∗ (ω)]
dt
∂t
4
4
⇒
dE
1
= ω|f0 |2 Imαx (ω)
dt
2
Consider now the specific case of Langevin’s equation, where the environment leads to
friction and random force terms. Averaging on the random force,
M (−ω 2 − iωγ) hx(ω)i = F (ω), hence the response has the same frequency as the source and
the response function is
⇒
αx (ω) =
−1
hx(ω)i
=
2
F (ω)
M (ω + iωγ)
The dissipation rate is therefore proportional to Imαx (ω) =
γω
M (ω 2 +γ 2 )ω 2
Comparing with Φx (ω) we get the ”Fluctuation - Dissipation theorem”:
Φx (ω) =
2kB T
Imαx (ω)
ω
Note: same result holds for Φv (ω), αv (ω) with an external source term in the Hamiltonian
−F (t)p/M (exercise).
In conclusion, Φx (ω) → are fluctuations in absence(!) of external force.
αx (ω) → linear response and energy dissipation due to external force.
Applications
1. Kapplers experiment (1931) : fluctuations in angle θ of a mirror suspended on a fine wire
with a restoring force 21 Cθ2 = 12 kB T → determine kB . Note : Variance is independent of
gas density. Equilibrium is achieved after time τcol , 1/γ.
62
Low density: rare strong deflections
High density: frequent weak deflections
2. A square vane of area 1cm2 , painted white
on one side and black on the other, is attached to a vertical axis and can rotate freely
about it as shown. Assume that the vane can
support a temperature difference between its
black and white sides. Suppose the arrangements is placed in He gas at room temperature and sunlight is allowed to shine on the
vane. Explain qualitatively why : (a) At extremely small densities the vane rotates. (b)
At some intermediate (very low) density the
vane rotates in a sense opposite to that in (a);
estimate this intermediate density and the corresponding pressure. (c) At a higher (but still
low) density the vane stops.
(a) Radiation pressure on white > black.
(b) Black heats up and nearby hotter atoms exert higher pressure. Gas is in local equilibrium
with density n 1cm−3 .
(c) Viscosity is effective at τcol << 1/γ and global equilibrium is achieved.
63
3. Electrical circuit:
= −RI + V (t)
L dI
dt
V (t) - voltage fluctuations
Langevin analogy:
dvx
dt
= −γvx + Ax (t) => γ → R/L
M →L 1
L hI 2 i
2
Correlation of A is determined by 21 M hvx2 i = 12 kB T =
Φvx (w) =
2kB T
M
ΦAx (ω) =
2kB T γ
M
·
γ
ω 2 +γ 2
→ ΦI (ω) =
2kB T R
L2
→ ΦV /L (ω) =
KI (τ = 0) = hI 2 (t)i =
R
2kB T
L
·
ω<<R/L 2k T
R/L
B
=
R
ω 2 +(R/L)2
=> ΦV (ω) = 2kB T R
ΦI (ω) dω
2π
Fluctuations measured in interval ω1 < |ω| < ω2 : h∆I 2 iω1 ↔ω2 = 2 ·
Rω2
ω1
. The factor
ΦI (ω) dω
2π
2 accounts for ω < 0.
h∆I 2 iω1 ↔ω2 '
2kB T
πR
(ω2 − ω1 )
[ω1,2 << R/L]
hδV 2 iω1 ↔ω2 = π2 kB T R (ω2 − ω1 )
4. Brownian particle : τ0 response time of eye, i.e. only frequencies |ω| <
hvx2 iobs = [Kv (τ = 0)]obs = 2
1
τ0 γ
−6
' 10
⇒
hvx2 iobs
'
4kB T
M
2π/τ
R 0
0
·
1
τ0 γ
Φv (ω) dω
=
2π
<<
2kB T
πM
tan−1
2π
τ0
2π
τ0 γ
are observed.
kB T
M
p
i.e instead of kB T /M ' 10−1 cm/ sec a small fraction of the fluctuations is observed
p
hvx2 iobs ' 10−4 cm/ sec.
64
Fluctuation Dissipation Theorem (FDT): quantum version
Fluctuations
Consider a system in equilibrium, no external force. For x̂ (t) (Heisenberg representation)
define a symmetrized correlation
K (τ ) =
1
hx̂ (t) x̂ (t + τ ) + x̂ (t + τ ) x̂ (t)i .
2
where a subscript x is omitted here, for convenience. The symmetrized correlation involves
a hermitian operator and has an obvious classical limit. note, however, that there are FDT’s
for non-symmetrized correlations.
Note that K (τ ) = K (−τ ) and K(τ ) is real. The average is defined by
h...i =
X
n
hn| ... |ni e−βEn /Z.
e−βEn is the partition sum.
P
Introduce a unit operator
|mi hm| = 1 and define (time independent) matrix elements
where Z =
P
n
m
xmn by
hn| x̂ (t) |mi = ei(En −Em )t/~ xnm .
Therefore
P
1
i(En −Em )t/~
K (τ ) = 2Z
xnm xmn ei(Em −En )(t+τ )/~ e−βEn + (t ↔ t + τ ) =
n,m e
P
1
−iωnm τ
m
= 2Z
where
ωnm = En −E
|xnm |2 e−βEn + c.c.
n,m e
~
Φ (ω) =
R
K (τ ) eiωτ dτ =
π
Z
P
n,m
e−βEn |xnm |2 [δ (ω − ωnm ) + δ (ω + ωnm )]
Interchange n ↔ m in the first δ (ω − ωnm ),
e−βEm + e−βEn |xnm |2 δ (ω + ωnm ) =
n,m
P −βEn −β(Em −En )
e
e
+ 1 |xnm |2 δ (ω + ωnm ) =
n,m
P −βE
n
1 + e−β~ω
e
|xnm |2 δ (ω + ωnm )
Φ (ω) =
=
π
Z
=
π
Z
π
Z
P
n,m
Dissipation
Add coupling to an external force, F (t):
Ĥ = Ĥ0 + V̂ (t)
where
65
V̂ (t) = −F (t)x̂ .
We use here Shrdinger’s representation to avoid the form HH (t) 6= H(t). The response
function α(τ ) gives hx̂(t)i in terms of F (t − τ ) at all previous times, i.e. τ > 0:
hx̂(t)i =
Z
∞
0
α(τ )F (t − τ )dτ.
Assume, without loss of generality,
1
1
F (t) = f0 e−iωt + f0∗ eiωt .
2
2
Therefore
1
hx̂(t)i =
2
Z
∞
α(τ ) f0 e−iω(t−τ ) + f0∗ eiω(t−τ ) dτ.
0
The Fourier transform is
Z
α(ω) =
∞
α(τ )eiωτ dτ
(α(τ ) = 0 for τ < 0),
0
therefore
1
1
hx̂(t)i = α(ω)f0 e−iωt + α(−ω)f0∗ eiωt ,
2
2
since α(−ω) = α∗ (ω) [α(τ ) is real]. In terms of the density matrix ρ̂(t) corresponding to the
full Hamiltonian H, the dissipation rate is
dE
d
i
∂H
= Tr{ (ρ̂H)} = Tr{[ρ̂, H]H} + Tr{ρ̂
}
dt
dt
~
∂t
∂F
∂F
= −Tr{ρ̂x̂
} = −hx̂i
∂t
∂t
(4)
(5)
where the equation of motion of ρ̂ is used, as well as the cyclic property of the trace. The
result is formally the same as in the classical case, except that hx̂i is a quantum expectation
value.
Using the response function and averaging on time,
dE
1
iω
= (α(ω)f0 e−iωt + α(−ω)f0∗ eiωt )
(f0 e−iωt − f0∗ eiωt ),
dt
2
2
⇒
dE
iω
ω
= |f0 |2 [−α(ω) + α∗ (ω)] = |f0 |2 Im[α(ω)].
dt
4
2
66
Therefore
ωIm[α(ω)]
measures
the
dissipation
rate.
Since
this
Im[α(ω)] > 0 for ω > 0.
Golden Rule for the Dissipation
The rate of transition between states at time T is
Z
2
1
1 =
Wn→m = 2 hm|V̂ |nidt
~
T
1
= 2
~
Z
1
i(Em −En )t/~
2
−iωt
∗ iωt 1
xmn (f0 e
+ f0 e ) 2 dt e
T
We note that
Z
2
Z
Z T /2
1 1 T /2 −i(ωmn −ω)t
0
i(ωmn −ω)t e
dt =
e
dt
ei(ωmn −ω)t dt0 =
T
T −T /2
−T /2
1
=
T
Z
T /2
−T /2
e−i(ωmn −ω)t [2πδ(ωmn − ω)]dt = 2πδ(ωmn − ω).
⇒ Wn→m =
π
|f0 |2 |xmn |2 [δ(ωnm − ω) + δ(ωnm + ω)].
2~2
The energy dissipation rate can now be calculated
Q=
=
1X
(Em − En )Wn→m e−βEn =
Z n,m
X
π
2 1
ω|f
|
e−βEn |xnm |2 [δ(ω + ωnm ) − δ(ω − ωnm )] =
0
2
2~
Z n,m
=
X
π
2 1
ω|f
|
(e−βEn − e−βEm )|xnm |2 δ(ω + ωnm ).
0
2
2~
Z n,m
Noting that
e−βEn − e−βEm = e−βEn 1 − e−β(Em −En ) = e−βEn 1 − e−βωmn ,
and using the δ function with ωmn = ω, we get
Q=
1 X −βEn
π
ω|f0 |2 (1 − e−β~ω )
e
|xnm |2 δ(ωnm + ω)
2~
Z n,m
67
is
positive
⇒ Im[α(ω)] =
X1
π
(1 − e−β~ω )
e−βEn |xnm |2 δ(ωnm + ω).
~
Z
n,m
⇒ Φ(ω) = ~ coth
The coth
1
β~ω
2
1
β~ω
2
Im[α(ω)]
can be interpreted as the mean fluctuation of an oscillator coordinate:
x2osc
~ 2
=
(a + a† )2 ∼ β~ω
+ 1 = coth
2mω
e
−1
1
β~ω
2
.
Note that in the limit ~ → 0 (i.e. for kB T >> ~ω) the quantum FDT becomes
Φ(ω) →
2kB T
Im[α(ω)].
ω
which is our previous classical FDT.
4d. Onsager’s Relations
Onsager’s relations (1931) consider the response to deviations from equilibrium. The responses are measured by a flow processes ẋi in coordinates xi (e.g heat flow, electric current,
mass transfer, etc.), while the deviation from equilibrium corresponds to ”forces” fi , e.g
gradients in temperature, potential, pressure, etc. The forces fi can be external ones, or
they can form spontaneously as a fluctuation.
If the system is close to equilibrium, we assume a linear relation between the forces and the
responding currents:
ẋi = γij fj ,
where summation convention is used (i.e. repeated indices are summed). γij are known as
kinetic coefficients. [In general, however, the relation is non-local in time and can be written
Rt
as ẋi (t) ∼ −∞ γij (t − t0 )fj (t0 )dt0 .]
We consider xi as coordinates of the entropy S so that the forces are fi = ∂S/∂xi . For small
deviations from equilibrium values x̄i we expand near the entropy maximum
1
S = S(x̄i ) − βij (xi − x̄i )(xj − x̄j )
2
⇒ fi =
∂S
= −βij (xj − x̄j )
∂xi
68
ẋi = −γij βjk (xk − x̄k ) .
Consider
R∞
hxi fj i = kB
−∞
xi ∂x∂ j e−βij (xi −x̄i )(xj −x̄j )/2kB
R∞
−∞
e−βij (xi −x̄i )(xj −x̄j )/2kB
where the ±∞ limits are allowed due to fast convergence.
By partial integration hxi fj i = −kB δij
⇒ hẋi xj i = γik hfk xj i = −kB γij .
This is a type of FDT – the fluctuations hẋi xj i are related to the dissipation ∼ γij .
Consider hxi (τ )xj (0)i with microscopic time reversal (dissipation ∼ γij and irreversibility
arise after average on the environment, or on the ensemble.)
hxi (τ )xj (0)i = hxi (−τ )xj (0)i = hxi (0)xj (τ )i
where in the second form both times are shifted by τ , allowed in equilibrium.
∂
|0 ⇒ hẋi (0)xj (0)i = hxi (0)ẋj (0)i ⇒
∂τ
γij = γji
These are Onsager’s relations.
Define
F̃ = 12 γij fi fj
Ṡ =
⇒
ẋi =
∂ F̃
∂fi
∂S
∂ F̃
ẋi = fi ẋi = fi
= 2F̃
∂xi
∂fi
Ṡ > 0, hence F̃ > 0 for all choices of fi , therefore the matrix γij is positive definite.
A few more properties of γij :
In presence of a magnetic field B time reversal involves B → −B, hence γij (B) = γji (−B);
e.g. for the Hall conductance σxy (−B) = σyx (B).
In presence of angular velocity Ω, similarly, γij (Ω) = γji (−Ω).
Furthermore, if for some coordinate xi → −xi , xj → +xj under time-reversal, then
γij = −γji
69
Alternative derivation:
Consider fi as external forces with a Legendre transformed entropy S → S̃ = S − (xi − x̄i )fi .
Since fi =
∂S
∂xi
⇒
dS̃ = −(xi − x̄i )dfi , i.e. S̃ = S̃(fi ).
Weight of a configuration is now
eS̃/kB = eS/kB −(xi −x̄i )fi /kB
here S = S{xi } has all orders of xi .
hẋi if =0 = 0, i.e. flow is only in response to fj 6= 0.
R
R
ẋi eS/kB [1 − (xj − x̄j )fj /kB ]
ẋi eS/kB −(xj −x̄j )fj /kB
+· · · = − hẋi xj i0 fj /kB +O(f 2 )
hẋi i = R S/k −(xj −x̄j )fj /k = R S/k
B
B
B
e
[1
−
(x
−
x̄
)f
/k
]
e
j
j j
B
Here linear response is explicit ⇒ γij = − hẋi xj i0 /kB as above.
hẋi xj i0 is evaluated with fi = 0 and microscopic time reversal can be used as above
⇒
γij = γji
Application: Bi-metal junctions
Using E1 + E2 =const, and Q1 + Q2 =const we have
dE1→2 = −dE1 = dE2 , dQ1→2 = −dQ1 = dQ2
and since dE = T dS + V dQ we obtain
dE1 dE2 V1
V2
1
1
V1 V2
dS =
+
− dQ1 − dQ2 = dE1→2 − +
+ dQ1→2
−
| {z }
| {z } T1 T2
T1
T2
T1
T2
T1 T2
{z
}
{z
}
|
|
x1
x2
f1
Ė1→2
f2
1
1
V1 V2
= γ11 − +
+ γ12
−
T1 T2
T1 T2
Q̇1→2 = γ21
1
1
− +
T1 T2
70
+ γ22
V1 V2
−
T1 T2
Seebeck coefficient Ψ is defined by ∆V = Ψ∆T in an open circuit, i.e. Q̇1→2 = 0:
V1 − V2 =
γ21 T2 − T1
⇒
γ22 T
Ψ=−
γ21
γ22 T
Peltier coefficient Π, corresponds to T1 = T2 and is defined by
Ė = ΠQ̇ ⇒
γ12 = γ21 ⇒
Π = −T Ψ
71
Π=
γ12
γ22
This is Kelvin’s relation.
Appendix: Langevin’s equation from a Hamiltonian
This appendix considers a microscopic model for an environment that produces the dissipation γ and the random forces ξ(t) in Langevin’s equation
∂V
= ξ(t) .
∂q
mq̈ + γ q̇ +
The derivation is based on Caldeira and Legget, Ann. Phys. 149, 374(1983) and provides a
basis for Brownian motion.
Consider a linear coupling of the particle coordinate q to a set of Harmonic oscillators, with
coordinates Qi and momenta Pi ; the linearity is valid when the particle’s trajectories are
sufficiently confined.
N
X
P2
+ V (q) +
H=
2m
λi qQi
|i=1 {z
}
Coupling of oscillators to a heat bath
X P2
1
i
2 2
+
+ Mi Ωi Qi +∆V (q)
2M
2
i
|i
{z
}
Bath0 s Energy
Where ∆V (q) is chosen to retain the original minimum of V (q). Minimizing H with respect
P λ2i 2
1
i
to Qi gives Qi = M−λ
2 q, hence q dependent potential terms appear
i Mi Ωi q (−1 + 2 ) which
Ω
i
i
we wish to cancel by
∆V (q) =
λ2i
q2 .
2Mi Ω2i
X
i
The equations of motion for Qi are
Q̈i + Ω2i Q = −
λi
q(t)eηt
Mi
where retarded response is defined by η → 0+ , i.e. at t → −∞ the coupling to the environment vanishes. The solution is
Qi (t) =
Q0i (t)
λi
+
Mi
Z
0
0
dt q(t )
Z
0
0
dω eiω(t−t )+ηt
2π ω 2 − Ω2i
where Q0i is the solution of Q̈i + Ω2i Q = 0. Let us shift ω → ω − iη and rewrite
Z
Z
0
λi
dω
eiω(t−t )
0
0
0
Qi (t) = Qi (t) +
dt q(t )
eηt
Mi
2π (ω − Ωi − iη) (ω + Ωi − iη)
0
To solve this integral we need to take the upper contour for t > t0 (see figure, so that eiω(t−t )
vanishes on the upper half circle). The closed contour integration has two poles (see figure)
and therefore
Qi (t) =
Q0i (t)
λi
+
Mi
0
0 eiΩi (t−t ) e−iΩi (t−t ) ηt0
dt q(t )i
−
e
2Ωi
2Ωi
−∞
Z
t
0
0
72
Hence,
Z t
λi
0
−
dt0 q(t0 ) sin Ωi (t − t0 )eηt
Qi (t) =
Mi Ωi −∞
Z t
λi
λi
0
= Qi (t) −
q(t) +
dt0 q̇(t0 ) cos Ωi (t − t0 )
Mi Ω2i
Mi Ω2i −∞
Q0i (t)
(*)
0
where partial integration is done, and at t0 → −∞ we use eηt → 0; in the final form we set
η = 0.
Equation of motion for q(t):
X
∂∆V
∂V
+
=−
λi Qi
∂q
∂q
i
X
X λ2 Z t
X λ2
i
i
0
=−
λi Qi +
q(t) −
dt0 q̇(t0 )cosΩi (t − t0 )
2
2
Mi Ωi
Mi Ωi −∞
i
i
mq̈ +
Note the cancelation of
∂∆V
∂q
. Define the spectral density as
J(ω) =
so that
∂V
2
mq̈ +
+
∂q
π
Z
0
∞
π X λ2i
δ(ω − Ωi )
2 i Mi Ωi
J(ω)
dω
ω
Z
t
dt0 q̇(t0 ) cos ω(t − t0 ) = −
X
λi Q0i .
i
To obtain linear dissipation, as in Langevin’s equation, we assume an ”ohmic” bath
J(ω) = γω. Using
Z
0
∞
0
dω cos ω(t − t ) =
1
2 Re
Z
∞
−∞
0
eiω(t−t ) dω = πδ(t − t0 )
we get
mq̈ + γ q̇ +
X
∂V
=−
λi Q0i
∂q
i
We identify now the random force ξ(t) = −
Q0i (t) =
P
i
λi Q0i (t). Free oscillators evolve as
1 0
Q̇ (0) sin Ωi t + Q0i (0) cos Ωi t
Ωi i
73
With random initial conditions we have hQ0i (0)i = hQ̇0i (0)i = 0 so that hQ0i (t)i = 0.
For a thermal distribution (classical oscillators)
kB T
Q0i (t)Q0i (0) = (Q0i (0))2 cos Ωi t =
cos Ωi t
Mi Ω2i
Z ∞
X λ2
J(ω)
2
i
dω
hξ(t)ξ(0)i = kB T
cos Ωi t = kB T
cos ωt .
2
M
Ω
π
ω
i
0
i
i
⇒
With J(ω) = γω we obtain
hξ(t)ξ(0)i = 2γkB T δ(t)
(∗∗)
i.e. the noise correlation relates to the dissipation γ.
To check FDT, we need to evaluate the response of ξ(t) to a source external to the oscillators
qex (t); this source may or may not be the particle’s position q(t). Adding a term −qex (t)ξ(t)
to the Hamiltonian, where qex (t) = qex (ω)e−iωt+ηt , the response is then defined by ξ(ω) =
α(ω)qex (ω). Since qex (t) affects the equation of motion for Qi (t) in the same way as q(t) in
Eq. (*) above,
X λ2
X λ2 Z t
i
i
ξ(t) =
q (t) −
dt0 q̇ex (t0 )cosΩi (t − t0 )
2 ex
2
M
Ω
M
Ω
i i
i i −∞
i
i
Z ∞
Z
Z
0
2
2 ∞ 0 J(ω 0 ) t
0 J(ω )
−iωt+ηt
0 −iωt0 +ηt0
dω
dω
=
q
(ω)e
+
dt
e
cos[ω 0 (t − t0 )]iωqex (ω)
ex
π 0
ω0
π 0
ω0
−∞
Z ∞
Z ∞
0
0
ω
1
1
2
0 J(ω )
0 J(ω )
dω
−
dω
(
+
) qex (ω)e−iωt+ηt .
=
0
0
0
0
π 0
ω
2 0
ω
ω + ω + iη ω − ω + iη
Therefore, for a general bath (not necessarily an ohmic one)
Imα(ω) = J(ω)
For an Ohmic bath J(ω) = γω with Eq. (**) we recover the classical FDT. This relates the
fluctuations of the bath to the response α(ω) of the bath to an external force.
Note that there is a separate FDT relating the fluctuations of the particle hq(t)q(0)i and its
response αq (ω) to an external force, i.e. adding to the Hamiltonian a term −q(t)ξex (t). For
V (q) = 0 this was checked in a previous section on Langevin’s equation. For V (q) 6= 0 it is
nontrivial to find either the fluctuations or the response, yet FDT is guaranteed.
We proceed to check the quantum FDT, allowing for a general J(ω). Consider the harmonic
oscillator’s creation a†i and annihilation ai operators,
74
r
Q̂0i (t)
where Ĥ0 =
P
i
=
~
eiĤ0 t ai + a†i e−iĤ0 t =
2Mi Ωi
r
~ −iΩi t
e
ai + eiΩi t a†i
2Mi Ωi
Ωi (a†i ai + 12 ) is the free Hamiltonian for the oscillators. Hence the correlations
become
=
~
4Mi Ωi
E
1D 0
Q̂i (t)Q̂0i (0) + Q̂0i (0)Q̂0i (t) =
2
D
E
=
e−iΩi t ai + eiΩi t a†i ai + a†i + ai + a†i e−iΩi t ai + eiΩi t a†i
=
D
E
Boson occupation ni = a†i ai =
~
(2ni + 1) cos Ωi t
2Mi Ωi
1
,
eβ~Ωi −1
hence
1
Kξ (t) = hξ(t)ξ(0) + ξ(0)ξ(t)i =
2
Z
X ~λ2
~ ∞
1
1
i
=
β~Ωi cosΩi t =
dωJ(ω) coth
β~ω cosωt
coth
2
2M
Ω
2
π
2
i
0
i
i
Using J(−ω) = −J(ω) we identify the Fourier transform
Φξ (ω) = ~J(ω) coth
1
2 β~ω
.
Since the response was found as Imα(ω) = J(ω), we obtain the full quantum FDT
Φξ (ω) = ~ coth( 12 β~ω)Imα(ω) .
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