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Identical Particles
Bosons and Fermions
In Quantum Mechanics there is no difference between particles and fields. The objects which we
refer to as “fields” in classical physics (electromagnetic field, field of distortions of an elastic medium,
etc.) turn out to consist of quanta which to a certain extent can be considered as particles. The objects
which we refer to as “particles” in the Classical Mechanics, can actually demonstrate a classical-field
behavior. A typical example of the Quantum Mechanical field behavior is the interference phenomenon
associated with the wave function. The wavefunction itself is not yet a classical field, because it is not
directly observable. There is, however, a more deep property that relates quantum particles to fields.
This property—the so-called exchange symmetry—is expressed by the axiom that the wavefunction
of the system of identical particles should remain the same, up to a possible change of sign, with
respect to interchange of the coordinates, ri and rj , of any two identical particles:
Ψ(. . . , ri . . . , rj , . . .) = σ Ψ(. . . , rj . . . , ri , . . .) ,
σ = ±1 .
(1)
Clearly, the exchange symmetry implies the indistinguishability of identical particles, because now we
cannot change the state of the particle i without identically changing states of all the other particles.
For a given sort of particles, σ is always the same. Otherwise, the superposition principle of
Quantum Mechanics would be violated. The particles for which σ = +1 are called bosons (after
Bose), and the particles for which σ = −1 are called fermions (after Fermi). The sign of σ is of
crucial importance for the properties of the particles. The properties of bosons and fermions are
generally speaking radically different.
The exchange symmetry reduces the number of allowed quantum states for identical particles.
Suppose we have two non-interacting particles. Let ψa (r) and ψb (r) be the wavefunctions of two
orthogonal single-particle states. Consider the following two very simple two-particle states:
ΨI (r1 , r2 ) = ψa (r1 ) ψb (r2 ) ,
(2)
ΨII (r1 , r2 ) = ψa (r2 ) ψb (r1 ) .
(3)
The physical meaning of the state ΨI is that the particle 1 is in the single-particle state a, while the
particle 2 is in the single-particle state b. And ΨII is the same up to the permutation 1 ↔ 2. The states
ΨI and ΨII do not feature the interchange symmetry. The only exception is the case a ≡ b, which
is good for bosons. To construct proper bosonic and fermionic states, we need to (anti-)symmetrize
the wave function. This means that if we want to deal with the two different single-particle states a
and b for two identical particles, we have to write
√
ΨBose (r1 , r2 ) = (1/ 2) [ψa (r1 ) ψb (r2 ) + ψa (r2 ) ψb (r1 )] ,
(4)
√
ΨFermi (r1 , r2 ) = (1/ 2) [ψa (r1 ) ψb (r2 ) − ψa (r2 ) ψb (r1 )] .
(5)
This symmetrization procedure straightforwardly generalizes to any number of particles. For bosons
we sum over all permutations of coordinates between different single-particle states. In the fermionic
case, two or more equal single-particle states are not allowed, since this immediately leads, upon the
anti-symmetrization, to zero wavefunction (Pauli exclusion principle). The crucial observation for
both fermions and bosons is that for N identical particles and a given set of N single-particle states
there is only one N -body state with the proper exchange symmetry that can be constructed out of the
given set. Hence, a state of identical particles can be unambiguously specified by listing all the singleparticle states participating in the symmetrized wavefunction. And this is a very good news for the
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Statistical Mechanics, since despite the entangled character of the eigenstate wavefunctions of identical
particles, their nomenclature turns out to be quite simple. This nomenclature deals with the notion
of occupation numbers. For example, if we have a box with periodic boundary conditions (PBC)
so that the single-particle eigenstates are labelled by momenta (or wavevectors, which is basically
one and the same), then each many-particle state of a system of identical particles is specified by
corresponding set of occupation numbers {nk }, where each nk is a non-negative integer equal to the
number of single-particle states ψk = (V )−1/2 eik·r participating in the N -body wavefunction. Though
the set of the occupation numbers is infinite—because of the infinite number of possible wavevectors,
only a finite number of them (≤ N ) is different from zero for a given N -particle state. By definition
of the occupation numbers,
X
nk = N .
(6)
k
For bosons each nk can assume any integer value from zero to infinity, while for fermions nk is either
zero or unity.
In terms of the occupation numbers, the eigenstate spectrum of a system of identical particles in
a PBC box reads
X
E=
ε k nk ,
(7)
k
where εk is the single-particle energy. The form of εk depends on what particles we are dealing with.
For example,
h̄2 k 2
for atoms,
(8)
εk =
2m
εk = h̄ck
for photons,
(9)
where m is the mass of the atom, and c is the velocity of light.
Looking at Eq. (7), we note that if not for the constraint (6), each single-particle mode could be
considered as an independent subsystem. We also see that in the case of bosons, where nk = 0, 1, 2, . . .
each single-mode subsystem is equivalent to a quantum harmonic oscillator, while in the case of
fermions, where nk = 0, 1 the single-mode subsystems are two-level systems. Later on we will show
that in a macroscopical system the constraint (6) is actually equivalent to shifting the single-mode
energies by some global constant µ (chemical potential). Meanwhile, we consider two particular cases:
(i) photons/phonons and (ii) strongly degenerate fermi gas, in which the constraint (6) is either
irrelevant from the very outset [case (i)], or can be easily taken into account by slightly changing the
parametrization [case (ii)].
Equilibrium Photons and Phonons
The constraint (6) is irrelevant to photons and phonons [phonons are the quanta of vibrations of
liquids and solids; they are bosons] for a very simple reason: their total number is not conserved due
to the processes of creation and annihilation. Correspondingly, the total number of photons/phonons
is not an external parameter. Rather, it is just one of the internal thermodynamic characteristics of
the system which essentially depends on temperature (and vanishes in the limit of T → 0). Hence, to
find the thermodynamic properties of our system we can use the spectrum Eq. (7) and treat all nk ’s
as independent quantum numbers. We then immediately have
Z=
Y
Zk ,
(10)
k
where
Zk =
∞
X
e−εk nk /T =
nk =0
2
1
1 − e−εk /T
(11)
is the partition function of a quantum harmonic oscillator—up to an insignificant global energy shift—
with the energy quantum εk . By the standard thermodynamic procedure (line-by-line repeating the
solution of the quantum harmonic oscillator problem) we get
X
F = T
³
ln 1 − e−εk /T
´
,
(12)
k
S =
X·
k
¸
³
´
εk /T
−εk /T
1
−
e
,
−
ln
eεk /T − 1
X
E =
k
µ
CV =
∂E
∂T
¶
=
εk
ε
/T
k
e
−
1
,
X (εk /T )2 eεk /T
V
(eεk /T − 1)2
k
(13)
(14)
.
(15)
It is instructive to compare Eq. (14) with directly averaged Eq. (7):
hEi =
X
εk n̄k ,
n̄k ≡ hnk i .
(16)
k
By this comparison we conclude that the average occupation number is given by
n̄k =
1
eεk /T
−1
.
(17)
Problem 34. Derive Eq. (17) directly from Gibbs distribution by utilizing the trick
∞
X
ne
λn
n=0
∞
∂ X λn
=
e .
∂λ n=0
(18)
Employing a similar trick, find the statistical dispersion, ∆n, of the occupation number and analyze the relative
dispersion, ∆n/n̄, in the two asymptotic cases: T ¿ ε and T À ε.
Eqs. (10)-(15) are valid for any εk . Talking of photons, we are interested in the linear singleparticle dispersion law, Eq. (9). The same linear dispersion law applies to acoustic branch of phonons
(with c the sound velocity), provided the temperature is low enough so that only the small-k linear
part of the phonon spectrum is relevant. We also need to take into account that there are two
different (transverse) polarizations of photons. Polarization can be treated as an internal state of a
boson. Bosons (and fermions) with different internal quantum numbers behave like different sorts of
particles. To account for the internal degrees of freedom we need only to multiply all the answers
for additive quantities by the factor g equal to the number of different internal states. In the case
of photons g = 2—what about phonons? In low-temperature fluids (like 4 He) there is only one
(longitudinal) phonon branch, and thus g = 1. In a solid there are three branches: one longitudinal
and two transverse. Generally speaking, the sound velocities of the longitudinal and transverse
branches are different. This complication is not important for us here and from now on we assume
that all the three velocities are equal. Replacing the summation over k with the integration,
X
k
(. . .) →
V
(2π)3
3
Z
dk (. . .) ,
(19)
we get
F =
gT V
(2π)3
Z
³
dk ln 1 − e−h̄ck/T
´
where
C0 = −
Z ∞
gT V
2π 2
=
Z ∞
0
0
³
dk k 2 ln 1 − e−h̄ck/T
¡
dx x2 ln 1 − e−x
¢
=
´
= −
π4
.
45
gC0 V T 4
,
2π 2 (h̄c)3
(20)
(21)
We arrive at an amazingly simple expression for the free energy:
F = −
gπ 2 V T 4
.
90(h̄c)3
(22)
How does it turn out that the expression for the bunch of harmonic oscillators is much simpler than the
expression for just one of them? It is a consequence of the so-called scale invariance of the problem.
In the case of one oscillator of the frequency ω0 , there is a characteristic scale of energy, h̄ω0 , so that
the answers can be written as some non-trivial functions of dimensionless variable T /h̄ω0 . In the case
of the bunch of oscillators with the linear (or any other power law) dependence of εk on k, there is no
any special energy scale. All the energy scales are similar to each other. The only possible function
of temperature thus is a power law function. The physics of a single oscillator is totally absorbed by
the dimensionless integral (21).
In view of the simplicity of the expression (22) for the free energy, there is no reason in using
generic relations (10)-(15). We readily obtain all the thermodynamic relations directly from (22):
µ
∂F
S=−
∂T
¶
CV = T
∂S
∂T
¶
µ
=
V
µ
P =−
∂F
∂V
(23)
gπ 2 V T 4
,
30(h̄c)3
(24)
V
E = F + TS =
µ
2gπ 2 V T 3
,
45(h̄c)3
=
∂E
∂T
¶
=
T
¶
=
V
2gπ 2 V T 3
,
15(h̄c)3
gπ 2 T 4
.
90(h̄c)3
(25)
(26)
Problem 35. Make sure that Eqs. (23)-(25) agree with the generic relations (13)-(15). Hint: The dimensionless
coefficients can be related to each other by doing corresponding integrals by parts.
Spectral density. It is easy to note that the generic answers (12)-(15) for the thermodynamic
quantities are actually depend on one function of one variable—the so-called spectral density, w(ε),
which, up to a normalizing coefficient, is the “distribution function” for the energies of the harmonic
oscillators: the quantity w(ε) dε is proportional to the number of oscillators (per unit volume) with
the energies within the interval [ε, ε + dε].
To reveal this fact, we replace summation with integration, in accordance with (19), and then
write—below F is an arbitrary function:
Z
dk
F(εk /T ) ≡
(2π)3
Z
dk
(2π)3
Z
dε δ(ε − εk ) F(εk /T ) .
(27)
That is we utilize the identity for the δ-function:
Z ∞
−∞
dx δ(x − x0 ) = 1 .
4
(28)
This trick allows us to make the replacement F(εk /T ) → F(ε/T ) and then change the order of
integrations. We thus get
Z
Z
dk
F(ε
/T
)
=
dε w(ε) F(ε/T ) ,
(29)
k
(2π)3
where
Z
dk
δ(ε − εk ) .
(2π)3
w(ε) =
(30)
Eq. (30) is the formal definition of the spectral density. If there are more than one branches of the
bosons, we just sum up all the spectral densities. If the branches have one and the same dispersion
law, this summation reduces to multiplying w(ε) by the factor g.
For example,
ε2
for photons .
(31)
w(ε) = 2
π (h̄c)3
Note the scale-invariance of this expression.
Now we rewrite (12)-(15) in terms of w(ε).
Z
F = VT
Z
S = V
³
dε w(ε) ln 1 − e−ε/T
´
,
·
(32)
¸
³
´
ε/T
dε w(ε) ε/T
− ln 1 − e−ε/T
,
e
−1
Z
dε w(ε) ε
,
eε/T − 1
E = V
Z
CV = (V /T 2 )
dε w(ε)
ε2 eε/T
.
(eε/T − 1)2
(33)
(34)
(35)
We can also use (17) to find the total number of bosons by summing up all the occupation numbers:
Z
dε w(ε)
.
eε/T − 1
N = V
(36)
The form of Eqs. (34) and (36) is quite transparent:
Z
N ≡ V
dε w(ε) nε ,
(37)
dε w(ε) ε nε ,
(38)
Z
E ≡ V
where
nε =
1
eε/T
−1
(39)
is the average occupation number of the mode with energy ε. Clearly, the functions Wparticls (ε) =
w(ε) nε and Wenergy (ε) = w(ε) ε nε play the roles of the distribution functions (over the single-particles
energies, or frequencies—which is one and the same up to the Planck’s constant) for the particles
and energy, respectively. This distribution functions should not be confused with the spectral density
w(ε). The latter tells us only how the oscillators are “distributed,” saying nothing on whether they
are excited or not.
If w(ε) is a monotonically increasing function (normally this is the case up to some qualitatively
irrelevant details), then the maxima of the distributions of energy and particles correspond to ε ∼ T ,
which is a very useful relation for order of magnitude estimates.
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Problem 36. Give order of magnitude estimates for the following quantities:
(a) Wavelength of the equilibrium EM radiation corresponding to the maximum of the energy distribution, at
room temperature,
(b) The same at the temperature of the Sun,
(c) Temperature at which the maximum of the energy distribution of the equilibrium EM radiation corresponds
to the red monochromatic light,
(d) Pressure of the EM radiation corresponding to the case (c).—Compare this pressure to the atmospheric
pressure.
Degenerate Fermi Gas
What is the distribution of the occupation numbers for fermions at T = 0? At T = 0 the system
is supposed to be in its ground state. So, we need to minimize the energy (7) under the constraint
(6). This constraint requires that N of the two-level systems be in their upper states anyway. Clearly,
the minimum of energy corresponds to the case when the excited modes are just the first N minimal
energies εp (in this section we label single-particle eigenstates by momentum rather than wave vector).
We thus have:
(
1,
if p ≤ pF
np =
(T = 0) ,
(40)
0,
if p > pF
where pF is called Fermi momentum. It is related to the total number of particles by the formula
X
1=N .
(41)
|p|<pF
In practice, this formula needs to be slightly corrected. The point is that fermions always have
internal degrees of freedom because they feature half-odd-integer spin, and at least two internal
states are guaranteed. Different internal states behave like different sorts of particles, which means
that Eq. (41) should be written for each sort of the particles, with the value of N understood as the
number of particles of the given sort. We will assume that all the internal states of our fermions
are equivalent (basically, this means that there is no magnetic field), and that corresponding total
numbers of particles equal to each others. If we want to reserve symbol N for the total number of
particles, we need to replace N with N/g in Eq. (41). Using integration instead of summation, we
get
Z pF
X
V
V p3
1= 2 3
dp p2 = 2 F3 .
(42)
N/g =
2π h̄ 0
6π h̄
p<pF
That is
Ã
pF =
6π 2
g
!1
3
h̄ n1/3 ,
(43)
where n = N/V is the number-density of the gas. The single-particle energy corresponding to the
Fermi-momentum is called Fermi-energy,
εF =
p2F
.
2m
(44)
The total energy of the gas at T = 0 is
E0 = g
X
p<pF
p2
gV
= 2 3
2m
2π h̄
Z pF
0
6
dp p2 p2 /2m =
3
εF N .
5
(45)
With (43) taken into account, we have
E0 =
3 (6π 2 /g)2/3 h̄2
10
m
µ
N
V
¶2
3
N.
(46)
Differentiating with respect to volume, we find the equation of state
µ
∂E0
P =−
∂V
¶
N
(6π 2 /g)2/3 h̄2
=
5
m
µ
N
V
¶5
3
(T = 0) .
(47)
Hence, the pressure of a Fermi gas is finite even at zero temperature and scales like n5/3 with density.
Particles and holes. Consider now the case of finite, but low temperature, T ¿ εF . How can we
excite our system? We can take some particle from an occupied state p1 < pF and place it into a free
state p2 > pF . Actually, this way we produce two elementary excitations: a hole at the momentum
p1 and a particle at the momentum p2 . The language of quasi-particles—particles with p > pF and
holes with p < pF —is very constructive. It will allow us to easily describe the thermodynamics of a
Fermi gas at T ¿ εF .
First, we need to elaborate on the particle-hole language. We introduce the notion of Fermi
sphere—the sphere in the single-particle momentum space defined by the condition p = pF . We treat
the genuine particles which fill the ball under the Fermi sphere as vacuum, and our quasi-particles
are the elementary excitations above this vacuum. From now on we use the word “particle” only for
the states with p > pF . Particles and holes behave like anti-particles with respect to each other.—A
particle-hole pair can annihilate. We introduce the occupation numbers for the quasi-particles:
(
ñp =
The constraint (6) then reads
1 − np ,
np ,
X
if
if
ñp =
X
p ≤ pF ,
p > pF .
ñp .
(48)
(49)
p>pF
p≤pF
That is the numbers of particles and holes should coincide. We will see that in the limit T ¿ εF this
constraint will be satisfied automatically, the total number of particles being fixed by the position
of the Fermi momentum. This is the main advantage of using the quasi-particle language. We then
introduce effective quasiparticle energy:
(
ε̃p =
εF − εp ,
εp − εF ,
if
if
p ≤ pF ,
p > pF ,
(50)
which takes into account the fact that if we want to excite a single hole/particle we have to simultaneously put/take a genuine particle onto/from the Fermi surface. Now we can identically rewrite
Eq. (7) in the form
X
ε̃p ñp .
(51)
E = E0 +
p
[We have also utilized Eq. (49).]
At T ¿ εF the excited holes and particles will typically have momenta |p − pF | ¿ pF —we will
also see it explicitly a posteriori—which allows us to use the approximate relation
ε̃p ≈ vF |p − pF | ,
7
(52)
where vF = pF /m is the Fermi velocity.
If we assume that all occupation numbers are independent parameters, we can easily find their
average values, because these immediately follow from the statistics of the two level system. To find
hñp i we need to consider a two-level system which first energy level is ε̃p · 0 = 0, and the second level
is ε̃p · 1 = ε̃p . We then write
hñp i = 0 · w0 + 1 · w1 = w1 ,
(53)
where w0 and w1 are the probabilities to find our system in the lower and upper energy states,
respectively. Hence,
1
hñp i = ε̃ /T
.
(54)
p
e
+1
We are ready to calculate the numbers of particles and holes, Npart and Nhole .
X
Npart = g
ñp =
p>pF
gV
(2πh̄)3
Z
p>pF
dp ñp =
gV
2π 2 h̄3
Z ∞
dp p2
pF
evF (p−pF )/T + 1
.
(55)
Note that at T ¿ εF the exponential in the denominator of (55) will cut off the integration at
p − pF ∼ T /vF ¿ pF . This means that we can replace p2 in the numerator with p2F . Then,
introducing dimensionless variable x = vF (p − pF )/T , we get.
Npart =
gV pF mT
2π 2 h̄3
Z ∞
0
dx
3 ln 2 T
=
N.
+1
2 εF
(56)
ex
Note that Npart /N ¿ 1.
Analogously,
Nhole = g
X
gV
ñp =
(2πh̄)3
Z
gV
dp ñp = 2 3
2π h̄
p≤pF
Z pF
dp p2
gV p2
≈ 2 F3
+1
2π h̄
Z pF
dp
.
+1
(57)
With the same accuracy with which we replace p2 with p2F , we can formally set the lower limit of
integration to −∞, since the actual cutoff of the integral corresponds to pF − P ∼ T /vF ¿ pF .
Finally, introducing the new integration variable, x = vF (pF − p)/T , we get absolutely the same
expression as Eq. (56). Hence, the constraint is satisfied.
We could further proceed by standard thermodynamic algorithm: calculate partition function,
free energy, and so forth. However, if we are interested only in the energy and heat capacity, we can
skip these procedure by directly calculating the energy. Indeed, the average energy is related to the
average occupation numbers by
X
E = E0 + g
εp hñp i .
(58)
p≤pF
0
evF (pF −p)/T
evF (pF −p)/T
0
p
Utilizing, (54) and making the same approximations as when calculating the numbers of quasiparticles,
we readily find
X
ε̃p hñp i =
p
V
(2πh̄)3
Z
dp ñp ε̃p ≈
V
2π 2 h̄3
It is known that
Z ∞
dp p2 vF |p − pF |
evF |p−pF |/T + 1
0
Z ∞
dx x
0
ex + 1
=
≈
π2
,
12
V pF mT 2
π 2 h̄3
Z ∞
dx x
0
ex + 1
.
(59)
(60)
which leads to the final answer
E = E0 +
π2T 2
N
4εF
(T ¿ εF ) .
8
(61)
The heat capacity then is
CV =
π2 T
N
2 εF
(T ¿ εF ) .
(62)
Note that heat capacity per particle is much smaller than unity: As compared to the gas of distinguishable particles, Fermi gas at T ¿ εF has, in effect, much less excited degrees of freedom. In view
of this fact, Fermi gas in this regime is called degenerate.
Problem 37. Give an order of magnitude estimate for the Fermi energy of valence electrons in copper (in
Kelvins). Comment: To a reasonably good approximation, valence electrons in metals can be treated as
non-interacting Fermi gas [on the uniform background of ions—the so-called jelly model]. Give an order of
magnitude estimate of the temperature T∗ (in Kelvins), at which the contributions to CV of copper from the
phonon and electron subsystems are of the same order. Comment: In this temperature region, only the acoustic
phonons are relevant.
Problem 38. Calculate free energy of the degenerate Fermi gas (T ¿ εF ) and use it to calculate the finite temperature correction to the pressure. At what temperature this correction becomes comparable to the
zero-temperature pressure?
9