Download Gibbs Distribution in Quantum Statistics

Gibbs Distribution in Quantum Statistics
Quantum Mechanics is much more complicated than the Classical one. To fully characterize a state
of one particle in Classical Mechanics we just need to specify its radius vector and velocity, while
in Quantum Mechanics the state of the same system corresponds to a complex-valued function of
coordinates. One may expect thus that the Quantum Statistics is much more sophisticated than the
Classical one. And this is indeed the case if we mean the complexity of statistical calculations for some
strongly interacting systems. It is amazing, however, that the general statistical considerations leading
to Gibbs distribution—and subsequently to the derivation of all generic thermodynamic relations—are
much simpler in the quantum case.
The derivation of the quantum Gibbs distribution is not compulsory in this course. Below I
present it for those who really like mathematical rigor of theoretical physics. The others may safely
skip the proof and go directly to the final result, Eq. (33).
Derivation of quantum Gibbs distribution. We start with considering the simplest quantummechanical system—two-level system—and then generalize the analysis.
Two-level system. By definition, a two-level system is a system which wavefunction is a two-component
complex vector and, correspondingly, the Hamiltonian is a hermitian matrix 2×2. Each wavefunction
|ψi can be represented as a linear combination of two orthogonal eigenstates of the Hamiltonian
|ψi = C1 |1i + C2 |2i ,
H|1i = E1 |1i ,
H|2i = E2 |2i ,
(1)
(2)
where E1 and E2 are the energies of the two eigenstates. If the system is isolated and |ψi is its initial
state, then the time evolution is given by the formula
|ψ(t)i = C1 e−iE1 t/h̄ |1i + C2 e−iE2 t/h̄ |2i .
(3)
Suppose now that our system interacts with an equilibrium heat bath. Equilibrium heat bath is a
crucial notion of Statistical Mechanics. Basically, it is a sort of intuitively clear axiom saying that a
macroscopic system with an extremely large excited degrees of freedom admits statistical description
in the sense that in the limit of t → ∞—in practice, when t is much larger than the relaxation
time—the observables of the system can be treated as random numbers with some distributions.
By equilibrium heat bath we understand such a macroscopic system interacting with a particular
system of interest (two-level system in our case). We assume that the interaction with a heat bath is
infinitesimally weak. Otherwise the properties of our system would be significantly distorted by this
interaction. The effect of a weak interaction with the heat bath can be represented as some weak
time-dependent noise in the parameters C1 , C2 , E1 , and E2 . Let us trace the consequences of this
noise. To this end we consider some observable, A, which in our case is a hermitian matrix 2 × 2.
The quantum-mechanical expectation value of the observable A is given by
hψ(t)|A|ψ(t)i = |C1 |2 A11 + |C2 |2 A22 + C1 C2∗ ei(E2 −E1 )t/h̄ A21 + C2 C1∗ ei(E1 −E2 )t/h̄ A12 ,
(4)
A11 = h1|A|1i , A22 = h2|A|2i , A12 = h1|A|2i , A21 = h2|A|1i .
(5)
where
1
Whatever weak, the interaction with heat bath qualitatively changes the problem at large times: the
system forgets about its initial state. Hence, the only meaningful quantity is the average of the
expectation value (4) over large times. Using symbol hh. . .ii to denote this average we write
hhAii = hh|C1 |2 ii A11 + hh|C2 |2 ii A22 + hhC1 C2∗ ei(E2 −E1 )t/h̄ ii A12 + hhC2 C1∗ ei(E1 −E2 )t/h̄ ii A21 .
(6)
There is a crucial difference between the first two terms in (6) and the last two ones. The first two
are averages of non-negative numbers and at least one of them is nonzero. Indeed, if
w1 = hh|C1 |2 ii ,
w2 = hh|C2 |2 ii ,
(7)
then
w1 + w2 = 1 ,
(8)
since |C1 |2 + |C2 |2 ≡ 1 (wavefunction is always normalized to unity). The two last terms in (6) are
complex numbers which phases, ϕ, can take on any value in the interval [0, 2π] with equal probability
densities: The only possible reason for one phase to be more probable than another would be the
interaction with the heat bath, in which case the difference of probability densities for two phases,
W (ϕ1 )−W (ϕ2 ), would depend on the strength of the interaction. But we assume that the interaction
is arbitrarily weak and thus the distribution of the phases should be flat: W (ϕ) ≡ 1/2π. The averaging
over time will thus include averaging over all possible and equally probable phases. With this averaging
we have
Z
Z
heiϕ i =
2π
0
eiϕ W (ϕ) dϕ = (1/2π)
2π
0
eiϕ dϕ = 0 ,
(9)
and conclude that
hhC1 C2∗ ei(E2 −E1 )t/h̄ ii = 0 ,
hhC2 C1∗ ei(E1 −E2 )t/h̄ ii = 0 .
(10)
That is only diagonal terms survive time averaging:
hhAii = w1 A11 + w2 A22 .
(11)
In effect, the system lives only in its eigenstates, with probability wj (j = 1, 2) of finding it in the
j-th eigenstate. The question now is: What are these probabilities? We get the clue to the answer by
analyzing the degenerate case: E1 = E2 . In a degenerate two-level system, each state is an eigenstate.
This means that instead of the basis { |1i, |2i } we can use, say, { |ai, |bi }, where
√
√
|ai = (|1i + |2i)/ 2 ,
|bi = (|1i − |2i)/ 2 .
(12)
In terms of this new basis, Eq. (11) reads
hhAii = wa Aaa + wb Abb .
(13)
It is supposed to yield exactly the same value for hhAii, and we are going to check how it can be the
case. With (12) we find
Aaa ≡ ha|A|ai = (1/2)A11 + (1/2)A22 + (1/2)A12 + (1/2)A21 ,
(14)
Abb ≡ ha|A|ai = (1/2)A11 + (1/2)A22 − (1/2)A12 − (1/2)A21 .
(15)
Substituting it into (13) leads to
hhAii = (1/2)(wa + wb ) A11 + (1/2)(wa + wb ) A22 + (1/2)(wa − wb ) (A12 + A21 ) .
2
(16)
And this has to be identically equal to (11) for any matrix A. The only way to satisfy this requirement
is to have
wa = wb = w1 = w2 ,
(17)
and we arrive at the fundamental statement that the probabilities for the system to be found in
each of its eigenstates of the same energy are equal. This statement as well as the whole analysis
presented above apply to any system weakly interacting with a heat bath. One just need to expand
the wavefunction in terms of the eigenstates of the Hamiltonian and repeat all the derivation, from
(3) to (10). Eq. (10) now reads
∗ i(Em −En )t/h̄
hhCn Cm
e
ii = 0 ,
Eqs. (11) and (8) generalize to
hhAii =
X
X
if n 6= m .
wn Ann ,
(18)
(19)
n
wn = 1 ,
(20)
n
where the sum is over all the eigenstates. To prove that the probabilities of two states with equal
energies are equal we can actually utilize the derivation for two-level system, because for the two
given states of interest it is enough to consider operators A of a special form—having non-zero matrix
elements with only these two states.
Now we are ready to derive the explicit form of wn . We already know that wn can depend only
on the energy: wn ≡ f (En ), where f is some function.—This is nothing else than a mathematical
expression of the requirement that wn = wm , if En = Em . Next, we take into account the following
consideration. If we have two independent (non-interacting with each other, but interacting with the
heat bath) systems, A and B, we can formally treat them as a united system AB. The Hamiltonian
of this system is a sum of Hamiltonians:
HAB = HA + HB ,
(21)
and, correspondingly, the eigenstates of such systems are direct products of the eigenstates of HA
and HB :
|AB : n, mi = |A : ni |B : mi,
(22)
HA |A : ni = En(A) |A : ni ,
(23)
(B)
HB |B : mi = Em
|B : mi ,
(24)
(AB)
HAB |AB : n, mi = Enm
|AB : n, mi ,
(25)
(AB)
(B)
Enm
= En(A) + Em
.
(26)
where
The distribution function for the system AB should have the form
(AB)
(B)
wnm
= fAB (En(A) + Em
).
(AB)
On the other hand, the two systems are independent, and wnm
individual distributions for the two systems:
(27)
should be equal to the product of
(AB)
(B)
wnm
= wn(A) wm
.
(28)
Recollecting that
wn(A) = fA (En(A) ) ,
(B)
(B)
wm
= fB (Em
),
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(29)
we obtain the requirement
(B)
(B)
fAB (En(A) + Em
) = fA (En(A) ) fB (Em
)
(30)
that should be valid for any two systems weakly coupled to one and the same heat bath, but not to
each other. Mathematically, there is only one function that satisfies Eq. (30). It is the exponential:
fA (E) ∝ fB (E) ∝ fAB (E) ∝ e−βE .
(31)
The only free parameter here is β, since the proportionality coefficients are fixed by the normalization
conditions.
We arrive at the following formulation of quantum Gibbs distribution. As we have demonstrated,
without loss of generality we can believe that any quantum mechanical system in equilibrium with a
heat bath is in one of the eigenstates of its Hamiltonian. If we label these state by n and introduce
corresponding energies En :
H |ni = En |ni ,
(32)
then the probability of finding the system in the n-th eigenstate is
wn = e−βEn /Z ,
where
Z=
X
e−βEn
(33)
(34)
n
is a normalization constant; it is called partition function. The particular value of parameter β is
related to (and characteristic of) the heat bath. The inverse of β is called temperature, T = 1/β.
The physical meaning of the parameter T becomes clear from further analysis of generic and specific
properties of Gibbs distributions. The statistical expectation value of any observable represented by
an operator A—all the observables in Quantum Mechanics are operators—is given by
hAi =
X
Ann wn ,
(35)
n
where
Ann = hn|A|ni
(36)
is the quantum mechanical expectation value of A in the state |ni. Hence, we have a two-fold averaging of each observable: (i) with respect to each energy eigenstate, and (ii) with respect to Gibbs
distribution of these eigenstates.
Simplest example: Two-level system. There are only two eigenvalues of energy: E1 and E2 . Physically,
only the energy difference ε = E2 − E1 is relevant, and without loss of generality, we can set E1 = 0,
E2 = ε. Then the partition function is
Z = 1 + e−ε/T ,
and the probabilities are
(37)
1
1
,
w2 =
.
(38)
−ε/T
1+e
1 + eε/T
We introduce a simple observable—polarization, P . Its operator has the following matrix elements:
P11 = 1, P22 = −1, P12 = P21 = 0. The expectation value of P —average polarization, or simply
w1 =
4
polarization (for briefness)—yields the difference between probabilities of finding the system in the
state 1 and 2, respectively. If we are dealing with an ensemble of many identical two-level systems,
then polarization times the total number of systems yields the difference between the number of
systems in the state 1 and the number of systems in the state 2. Calculating P̄ ≡ hP i, we find
P̄ =
sinh(ε/T )
.
1 + cosh(ε/T )
(39)
Polarization of the two-level system provides us with a simplest model of thermometer. Given P̄ from
experiment, we restore the temperature T = T (P̄ ) from Eq. (39).
Problem 16. Draw the plot of the function T (P̄ ). Note that for the systems with bounded spectrum a
negative temperature is also meaningful.
One of the most important observables is the energy. Its operator is the Hamiltonian of the
system, Hnn = En . For the mean value of energy, E ≡ hHi, we have
E =
X
En wn = Z −1
n
X
En e−βEn .
(40)
n
Another useful quantity is the heat capacity
dE
.
dT
C =
(41)
Heat capacity shows how fast the mean energy of the system changes per unit variation of temperature.
By directly differentiating Gibbs distribution, we can show that
C =
1 X
(En − E)2 wn .
T2 n
(42)
Hence, the heat capacity is always equal to the square of the dispersion of energy divided by T 2 . Note
that C ≥ 0.
Problem 17. Derive Eq. (42). [Do not forget to differentiate Z.]
Problem 18. For the two-level system, calculate E by Eq. (40) and C by both (41) and (42). Draw the plots
E(T ) and C(T ). [Do not forget about negative temperatures.]
Entropy. One of the central thermodynamic variables is entropy. In the probability theory, the
entropy of a distribution is defined as
S=−
X
wn ln wn .
(43)
n
[The entropy defined by Eq. (43) is dimensionless. If one works in units kB 6= 0, then for thermodynamical purposes it proves convenient to include the factor kB in the definition of entropy.] What
does the entropy of a distribution tell us about? It characterize the complexity of the distribution. If
a system consists of a number of independent parts—typical situation in Statistical Physics, then the
entropy of the distribution is equal to the sum of entropies of the distributions of each independent
subsystem. To prove this statement, consider a system AB which is a combination of two independent
systems, A and B. With the distribution (28) we have
S (AB) = −
X
n,m
(AB)
(AB)
wnm
ln wnm
=−
X
(B)
(B)
wn(A) wm
(ln wn(A) + ln wm
) = S (A) + S (B) .
n,m
5
(44)
Problem 19. Calculate the entropy of the two-level system using Eq. (43). Draw the plot S(T ). What
happens when T → +0 and T → −0 (note that these are different limits) ?
Simplest Thermodynamic Relations
Once the partition function Z ≡ Z(T ) is found for a given system from a direct calculation, the rest
of the thermodynamic quantities can be obtained from generic relations which we establish below.
We start with introducing one more quantity—Helmholtz free energy
F = −T ln Z .
(45)
wn = e(F −En )/T .
(46)
S = (E − F )/T ,
(47)
E = F + TS .
(48)
dZ
d X −En /T X
=
e
=
(En /T 2 ) e−En /T ,
dT
dT n
n
(49)
In terms of F , Gibbs distribution reads
Plugging this into (43), we see that
or
Consider now the derivative
and note that
T 2 dZ
=E.
Z dT
Expressing Z through F as Z = exp(−F/T ), we rewrite this relation as
F −T
dF
=E.
dT
(50)
(51)
Comparing this to (48), we see that
dF
= −S .
(52)
dT
Finally, by differentiating (48) with respect to temperature and taking into account (52), we get
C = T
dS
.
dT
(53)
Conclusion: To establish all basic thermodynamic characteristics of a system it is enough to directly
evaluate from Gibbs distribution only its partition function. The rest is done by the following algorithm:
• Find F in accordance with (45).
• Find S in accordance with (52).
• Find E in accordance with (48).
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• Find C in accordance with either (53), or the definition (41) [or both, to make sure that the two
answers coincide, and thus your algebra is free of errors which, unfortunately, happen too often.]
Problem 20. Apply this algorithm to the two-level system.
Macroscopic System: Heat Bath for Itself
By macroscopic system we mean a system which can be viewed as a set of a very large number of
weakly coupled (and thus almost independent) subsystems. Typical example: Consider a liquid, gas,
or solid containing a large number of molecules. Divide the whole thing into cells, each cell much
smaller than the whole system, but still large enough to contain many particles. Statistically, the
cells are almost independent, because they interact with each other only through the surface.—The
number of close-to-surface particles is much smaller than the number of the particles in the bulk.
The key idea behind the notion of macroscopic system is that for each its small quasi-independent
cell the rest of the system behaves like a heat bath. A macroscopic system thus forms a heat bath
for itself, and one may apply statistical description even if the system is isolated from any other heat
baths and its energy is exactly conserved. This conclusion is known as the equivalence of the canonical
and microcanonical distributions. Here by canonical distribution one means Gibbs distribution, while
microcanonical distribution corresponds to the distribution over the states within limitingly small
energy interval—each state having the same probability—and models an isolated macroscopic system
where the energy is conserved. The above conclusion is also clear from the Central Limiting Theorem,
because the energy of a macroscopic system can be represented as a sum a large number of statistically
independent energies of its cells. Hence, the relative fluctuations of energy in a macroscopic system
are negligibly small and cannot be relevant to any observable quantity.
Sometimes instead of canonical and microcanonical distributions one speaks of canonical and
microcanonical ensembles representing these distribution. Speaking the ensemble language, an equilibrium macroscopic system can be viewed as an ensemble of its smaller subsystems (cells).
The natural thermodynamic variable for an isolated system is the energy. It is of crucial importance, however, that the equivalence of the microcanonical and canonical descriptions allows one to
use the notion of temperature, entropy, and other statistical characteristics for isolated macroscopic
systems. By definition, these characteristics are the same as if the system is in contact with some
heat bath such that the mean energy of the Gibbs distribution equals to the energy of the isolated
system of interest. Moreover, with a macroscopic system even the microcanonical distribution is not
necessary to speak of temperature, entropy, etc. Just one typical state of the system will do, because
dividing the system into cells we still have the statistical description in terms of the ensemble of cells.
Energy as a thermodynamic parameter. Now we derive thermodynamic relations in which the
(average) energy (rather than temperature) is the main thermodynamic parameter. These relations
are especially useful for macroscopic systems, while being also valid for microscopic systems. For a
macroscopic system, mean energy is a very good thermodynamic variable because the fluctuations of
energy from its mean value are negligibly small. Suppose we have expressed entropy as a function of
(mean) energy: S = S(E). What about the first derivative of this function?
dS
≡
dE
µ
dS
dT
With (53) we get the famous result
¶µ
dT
dE
¶
µ
≡
dS
dT
¶µ
dE
dT
dS
1
= =β.
dE
T
7
¶−1
≡ C −1
dS
.
dT
(54)
(55)
What about the second derivative? Differentiating (55) with respect to E, we find
d2 S
1 dT
1
=− 2
=− 2
2
dE
T dE
T
µ
dE
dT
¶−1
,
(56)
and see that
d2 S
C −1
=
−
≤ 0,
(57)
dE 2
T2
since, in accordance with (42), we have C ≥ 0. Below we will use this inequality to prove the increase
of entropy.
Problem 21. A macroscopic system is formed by N À 1 two-level systems with interlevel separation ε.
(Interactions between the systems are negligible.) Plot the function S(E), where E is the total energy and
S(E) is the entropy of corresponding equilibrium state.
Increase of Entropy
Heat exchange. Suppose we have two isolated macroscopic systems, A and B, each of which is
in equilibrium and is characterized by its own temperature. We then allow the systems to weakly
interact (exchange energy). As a result of this energy exchange, the systems will relax to some
common temperature. In such a process, the entropy always increases. Let us prove this fundamental
fact. It is convenient to characterize the thermodynamic states of our systems with energy rather
than temperature, because the total energy, E, is conserved and we can consider the energy EA of
the system A as the only free parameter describing the state of the two systems. For the entropy,
which is an additive function, we then have
SAB (EA ) = SA (EA ) + SB (EB = E − EA ) .
(58)
We want to understand the properties of the total entropy as a function of EA . To this end we
calculate the first and the second derivatives:
dSA
dSB
dSAB
=
−
,
EB = E − EA ,
(59)
dEA
dEA dEB
d2 SAB
d2 SA d2 SB
=
EB = E − EA .
(60)
2
2 + dE 2 ,
dEA
dEA
B
First derivative tells us about the extremum points, and the second derivative allows us to distinguish
the maxima from minima. Now we note that all the derivatives in the right-hand sides of (59) and
(60) correspond to equilibrium states, and we can use Eqs. (55), (57). From (55) we see that
dSAB
= βA − βB ,
(61)
dEA
that is the extremal point of the total entropy corresponds to the case when two temperatures are
equal. From (57) we have
d2 SAB
(62)
2 ≤0,
dEA
which means that our extremal point is the point of maximum. Hence, while the system evolves
towards its equilibrium, βA = βB , the entropy always increases. And the energy always goes from
the body with the smallest β to the body with the largest β. Indeed, from (61) we have
∆SAB = (βA − βB ) ∆EA .
(63)
The increase of entropy means that the l.h.s. is positive. Hence, the sign of ∆EA should be equal to
the sign of (βA − βB ).
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