Download Chapter 20 statistical mechanics

Chapter 20. The fluctuation-dissipation theorem
When chemical reactions reach macroscopic equilibrium, they continue microscopically:
individual molecules still exchange between reactant and product, even though there is no
net reaction. One would expect the fluctuations of a single molecule going back and
forth to be related to the one-way reaction of a large number of molecules. This is indeed
the case, as Lars Onsager showed.
1) Hamiltonian, polarization P and susceptibility χ
Consider again a system with Hamiltonian H = H 0 − Pε , P 0 = 0 . H 0 is the
unperturbed Hamiltonian, ε is an external perturbation, and P is the molecular property
that couples the external perturbation to the system to change its energy levels. P is often
called the polarization. For example, if the system is a molecule and ε is an externally
applied electric field, P would be the dipole moment of the molecule.
We saw in the review chapter on statistics that for a small external driving force in
steady-state,
P (ω ) = χ (1) (ω ) ε (ω ) .
χ is the susceptibility. This is simply the first order Taylor expansion of P:
χ (1) (ω ) = ∂P(ω , ε ) / ∂ε |ε = 0 and χ (0) (ω ) = P(ω , ε = 0) = 0 . The latter is usually true for
macroscopic P. For example, if P is the total dipole of an ensemble of molecules in a
liquid or gas, the random orientations ensure that P(ε=0)≈0. Only when the electric field
is turned on is a net dipole moment induced in the ensemble of molecules. Fourier
transform relates P (ω ) and P ( t ) :
∞
1
P(t) =
dω eiω t P(ω )
2π −∞∫
Hence the convolution theorem relates P ( t ) and χ (t)
∞
P(t) = ∫ dt ' χ (t ')ε (t − t ')
0
2) Lars Onsager’s fluctuation-dissipation theorem
As it turns out, in the linear response limit, the relaxation of P is related to the
spontaneous fluctuations δ P of P . Thus fluctuations and dissipations are not
independent.
Theorem:
An perturbation ε is switched off at time t=0. The response of the system is related to the
autocorrelation function of the response by
P(t) = βε i P(0)P(t)
ε =0
+ O(ε 2 ) when
P
0
=0.
When P
0
≠ 0 , we simply subtract the long term average from P to get a more general
version of the theorem. The key is that the behavior of P when ε is turned off depends on
the autocorrelation function in the absence of the field. Spontaneous fluctuations of the
system contain the same information as measuring the kinetics by perturbation, and viceversa.
Fluctuation-dissipation
ε(t)
<P(t)>E ≠0
P(t)
<P(t)>0 =0
<P(t) 2>0 ≠0
t
t=0
P(t)
Fig. 20.1 Fluctuations and response of P(t) to an applied field switched on and off again. The
fluctuations of P become easy to measure as N→1. The decays P(t) become easy to measure as
N→∞.
Proof:
Let P̂ be the Schrödinger operator for P, and let H = H 0 − Pε and t < 0, H = H 0 for
t>0. The perturbation is switched off at t=0. Choose P so P 0 = 0 .
⇒ ρ0 =
1 − β ( H 0 − Pε )
e
Q(ε )
and
1 − β H0
e
.
Q(0)
initially, and after a long time when equilibrium has been reached. At intermediate times,
i
i
− H0t
+ H0t
i


ρ (t) = [ ρ, H 0 ] or ρ(t) = e
ρ(0)e
.

Making use of the trace rotation property,
P(t) = Tr P̂ ρ(t)
lim ρ(t) =
t →∞
{
}
i
+ H0t ⎫
⎧ − i H 0 t

= Tr ⎨ P̂e
ρ(0)e
⎬
⎩
⎭
i
i
+ H0t
− H0t ⎫
⎧
= Tr ⎨ ρ(0)e  Pe  ⎬
⎩
⎭
{
}
= Tr ρ(0)P̂(t)
Now let β Pε  1
⎧ e− β ( H 0 − ρ̂ε )
⎫
= Tr ⎨
P̂(t) ⎬
⎩ Q(ε )
⎭
(linear response regime).
(
)
⎧⎪ e− β H 0 1 + β P̂ε + ... P̂(t)
⎫⎪
⇒ P(t) = Tr ⎨
⎬
− β H0
+ Tr e− β H 0 β P̂ε + ... ⎪
⎪⎩ Tr e
⎭
1
=
Tr e− β H 0 P̂(t) + e− β H 0 βε P̂P̂(t) + 
Q(0)
(
{
)
}
{
{
}
}
{
}
The second line follows because Tr e− β H 0 β P̂ε = βεTr e− β H 0 P̂ = 0 by definition of
<P>0=0. For the same reason, the first term in curly brackets vanishes. The ensemble
average of P(t) when averaged over the population distribution with field off also equals
zero. We are left with
⎧ e− β H 0
⎫
P(t) = βεTr ⎨
P̂P̂(t) + ⎬ = βε P̂P̂(t)
+ O(ε 2 )
ε =0
Q(0)
⎩
⎭
because P(t) 0 = 0 also if averaged over all t>0 despite the small transient.
( )
P(t) = βε P̂(0)P̂(t) + O ε 2 . QED.
0
This important theorem states that small nonequilibrium relaxation phenomena are
completely determined by the equilibrium properties of the system.
Example 1: absorption spectrum
N
Let P̂ = ∑ µ̂ be the polarization (total dipole) of an ensemble of molecules, in terms of
i =1
the dipole of the N individual molecules. Then
P(t) = εβ
N
N
i =1
j =1
∑ µ̂(0)i∑ µ̂(t)
= εβ N µ (0)µ̂ (t)
ε =0
.
ε =0
Why just a factor N instead of N 2 on the right, even though there are two summations?
The dipoles are randomly oriented, so P is a sum of random variables, and by the central
limit theorem, the double sum is proportional to N 2 = N . Note that the dot product
sign is gone in the last equality because the random sum over orientations has been
evaluated. This will usually happen: the quantity P is an extensive quantity that depends
linearly on system size, and it is the randomness of fluctuations that makes sure both
sides are linear in N . However, exceptions are possible. In processes such as
“superradiance,” the dipoles are all pre-aligned and the total emitted or absorbed intensity
goes as (Nµ)2 instead of Nµ2.
For an arbitrary small monochromatic electric field ε=ε(ω) to which corresponds a sinewave field ε(t), we can calculate
∞
P(ω )
χ (ω ) =
= β N ∫ dte−iω t < µ (0)µ (t) >
ε
−∞
The susceptibility is given by the Fourier transform of the dipole autocorrelation
function. From basic electromagnetic theory,
n(ω ) = 1 + 4πχ (ω ) ≈ 1 + 2πχ r (ω ) + i2πχ i (ω ) = nr (ω ) + i2πχ i (ω )
where n is the complex refractive index, and nr is the real refractive index. For a wave
propagating through a medium for a distance z, the wavevector equals k=nω/c and
ε (z) = ε (0)eikz . Combining these and taking the absolute value squared to get intensity
I(z) =| ε (z) |2 /2π from the electric field,
I(z) = 21π | ε (0) |2 e−(4 πωχi (ω )/c) z = I(0) e−α (ω )z
This is Beer’s law of absorption, and it shows that the absorption coefficient can be
calculated directly as the imaginary part of the Fourier transform of the dipole
autocorrelation function. Thus, if we wanted to calculate the absorption spectrum of a
substance, we could simulate the particles by molecular dynamics on a computer,
calculate µ(t) of the simulation, and then get from there to α.
This formula was derived earlier from a microscopic quantum mechanical
Hamiltonian. It also works from any macroscopic Hamiltonian H 0 due to the
fluctuation-dissipation theorem: the fluctuations of the dipoles directly yield the
dissipated optical energy.
Example 2: chemical reaction
Consider a reaction A  B , initially at equilibrium. At t = 0, a bias P = pq = α qΔT is
introduced, as shown in the figure below. If p>0 as shown, the equilibrium will shift
from B to A, and a net reaction occurs until a new equilibrium population ρeq has been
reached. We’ll call any chemical to the left of † “A”, and any to the right “B.” Of
course, near † itself this distinction is artificial, but the barrier is located there, so
presumably there won’t be too many molecules to worry about near †.
Chemical reaction as bistable system
V(q)
after t=0
†
E†
ΔE†
before t=0
A
EA
B
qA
q†
(turned on
P = α q at t=0)
q
qB
Fig. 20.2 Reaction on energy surface V(q) along coordinate q between wells A and B over barrier
at †. At t=0, a bias PΔT=αqΔT is suddenly applied to the reaction, where ΔT is the external
perturbation (analogous to ε, e.g. a sudden temperature change), and αq is the sensitivity of the
potential to the external pertrubation (analogous to µ(x))
According to the fluctuation-dissipation theorem, the macroscopic reaction rate will be
related to the microscopic fluctuations of individual molecules going back and forth. We
will apply the theorem below, but first, if the barrier is high enough, we can use a
phenomenological rate theory. Let a be the concentration of A, and b be the
concentration of B, and χ the mole fractions.
da
db
= −kAB a + kBAb;
= +kAB a − kBAb
dt
dt
d(a + b)
⇒
= 0 or a + b = const.
dt
By conservation of mass, as much A is made as B disappears, and vice-versa. At
equilibrium, we can set da/dt = db/dt = 0 and find
k AB beq χ B
.
=
=
k BA aeq χ A
Therefore the forward and backward rate coefficients are related to one another by the
equilibrium concentrations or mole fractions. The solution of the two coupled
differential equations is
a(t) − aeq = a(0) − aeq e−t /τ where τ −1 = kAB + kBA ,
(
)
as you can show easily by inserting the solution into the equations. The key thing to
realize here is that the observed relaxation time for the reaction is not the rate from A to
B, or vice-versa, but the sum of the forward and backward rates. Of course, if beq<<aeq
and the reaction goes backwards, kBA>>kAB and so τ-1≈kBA, but this is not generally true.
Solving in terms of mole fractions,
τ −1 = kAB / χ B = kBA / χ A .
The fluctuation dissipation theorem connects τ or the rate coefficients to fluctuation of
particle number n on each side of the well. Let nA (t) be the dynamical variable such that
nA = χ A , i.e.
⎧1 if q ≤ qT ⎫
nA = ⎨
⎬ for any particle in the ensemble.
⎩0 if q > qT ⎭
By the fluctuation-dissipation theorem (normalizing both P(t) and the fluctuations):
a(t) − aeq
δ nA (0)δ nA (t)
=
= e−t /τ
a(0) − aeq
δ nA 2
where δn are the fluctuations of n about the average value. Thus, we can evaluate τ if
we know the spontaneous fluctuations of nA at equilibrium. The correlation expressions
can be simplified as follows.
a
nA = χ A =
a+b
nA can only take on the values 0 and 1, so
nA 2 = χ A
also because nA 2 (q) = nA (q) for all q. Therefore the normalizing denominator becomes
2
δ nA 2 = nA 2 − nA = χ A − χ A 2 = χ A (1 − χ A ) = χ A χ B ,
a symmetrical expression in terms of the mole fractions. We can similarly evaluate
δ nA (0)δ nA (t) = (nA (0)− < nA >)(nA (t)− < nA >) and finally obtain
nA (0)nA (t) − χ A 2
−t / τ
.
e =
χAχB
To bring this into a form where the rate coefficient can be read off, take the time
derivative on both sides:
1
1
1
− e−t /τ =
nA (0)n A (t) = −
n A (0)nA (t) .
τ
χAχB
χAχB
The second equality follows if we invoke time-shift invariance, that is
nA (0)nA (t) = nA (t 0 )nA (t 0 + t) = nA (−t)nA (0)
if t0 = -t. We can rewrite the time derivative of nA as
dn dq
n A = A
= vδ (q − q† ) .
dq dt
The delta function appears because nA is a step function at q=q†. We can now write the
derivative of the fluctuation-dissipation theorem as
1
τ −1e−t /τ =
v(0)δ ( q(0) − q† ) nA (t)
χAχB
1
kAB e− kAB t / χ B =
v(0)δ ( q(0) − q† ) nA (t)
χA
In the second line we made use of the relationship between τ, mole fraction, and forward
rate. As t → 0 , kAB cannot be time-independent because reactants that already happen to
be near the transition state ballistically cross the line q = q† without recrossing, so
kBA (0) ≥ kBA (t) . However, let us evaluate the rate coefficient at t=0:
1
kAB =
v(0)δ ( q(0) − q† ) nA
χA
1
kAB =
v0 δ ( q(0) − q† )
2χA
In the second line, we made use of the fact that nA=1 for all particles near the transition
state that will react forward, but only half those particles have the velocity in the right
direction to react and turn into B. This equation says that the rate coefficient is
proportional to the product of particle velocity at the transition state, and the expectation
value that the particle is located at the transition state.
In transition state theory, the correct function nA (t) is replaced by the approximation
⎧1 if v(0) goes toward a ⎫
H A (t) = ⎨
⎬,
⎩0 if v(0) goes toward b ⎭
evaluated at q = q† where flux of particles is smallest (the bottleneck for reaction).
This is not a perfect criterion for reactions because particles could switch velocity v after
passing † and not react after all. Not every particle that crosses † actually makes it down
to B. The resulting expression for the rate in transition state theory is
1
TST
kAB
=
v(0)δ (q(0) − q† )H A (t)
χA
1
v(0) δ (q(0) − q† )
2χA
The second line follows because 1/2 of the particles move in the wrong direction; it must
be 1/2 because at t < 0 before the perturbation was switched on, we were at equilibrium.
Thus if we define † to be the location of minimal particle flux on the energy surface,
transition state theory gives the same result as the exact expression evaluated at t=0.
TST
kBA
= kBA (0) ≥ kBA (t).
The real rate coefficient will always be smaller than the transition state theory rate
coefficient because particles going through † could be bumped back and return to A
before they ever make it to B. After a time τ mol  τ , kAB (t) reaches a plateau at the
actual rate constant (smaller because of recrossing). τmol is the time it takes particles to
diffuse across the transition state, allowing a steady-state flux of particles to be
established that is smaller than the initial flux.
The above equation for the transition state rate coefficient may look somewhat
unusual, but consider the following: v has units of m/s, and δ(q) has units of 1/m, so kAB
has units of 1/s, as expected for the rate coefficient. Finally, consider what the
expectation value δ (q(0) − q† ) actually is: the probability of finding the particle per
=
unit length around position q† in figure 20.2, which is at an energy ΔE† = E †-EA higher
than state A. If we make the approximation that all particles on the “A” side are near the
bottom and thus at energy 0, the canonical probability ratio at constant temperature is
p† W† − ΔE † / kB T
.
=
e
pA WA
Realizing that χA = pA,
1
TST
kAB
=
v(0) p†
2χA
.
W† − E † / kB T
1
=
v(0)
e
2
WA
The microcanonical partition function ratio equals exp(ΔS†/kBT), and defining <v(0)>/2 =
ν†, we can write
TST
k AB
= v†e− ΔG / kB T ,
the familiar expression for the transition state rate. Remember that this expression
requires a number of approximations beyond the exact fluctuation-dissipation expression.
In some cases, these approximations are quite severe and transition state theory breaks
down.
†