30 Photons and internal motions

Summary

∗ Radiation field is understood as a collection of quantized harmonic oscillators.

∗ The resultant Planck’s radiation formula gives a finite energy density of radia-

tion field.

∗ The internal degrees of freedom have (often well-separated) characteristic en-

ergy scales.

Key words

photon gas, Planck’s radiation formula, ultraviolet catastrophe, Stefan-

Boltzmann law, internal degrees of freedom, rotational partition functions

The reader should be able to:

∗ Derive Planck’s formula.

∗ Recognize clearly the main features of Planck’s formula.

∗ Itemize internal degrees of freedom of a molecule and telling their energy scales

(in K).

∗ Sketch the molecular ideal gas specific heat as a function of T .

30.1 Quantization of harmonic degrees of freedom

Photons and phonons are obtained through quantization of the systems that can

be described as a collection of harmonic oscillators.1 Possible energy levels for the

i-th mode whose angular frequency is ωi2 are (n+ 1/2)~ωi, where n = 0, 1, 2, · · ·

(A.29). The canonical partition function of a system with modes {ωi} is given by

(cf. Section 23)

Z(β) =∏i

( ∞∑ni=0

e−β(ni+1/2)~ωi

), (30.1)

since no modes interact with each other. Here, the product is over all the modes. The

sum in the parentheses gives the canonical partition function for a single harmonic

oscillator, which we have already computed (Section 23). The canonical partition

1 That is, the system whose Hamiltonian is quadratic in canonical coordinates (quantum

mechanically in the corresponding operators).2 What ‘mode’ means was explained in 20.7.

353

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354 Photons and internal motionstfunction (30.1) may be rewritten as:

Z(β) =

[∏i

(e−β~ωi/2

)]∏i

(∑ni

e−βni~ωi

)=

[∏i

(e−β~ωi/2

)]Ξ(β, 0), (30.2)

where

Ξ(β, 0) =∏i

( ∞∑n=0

e−βn~ωi

)=∏i

(1− e−β~ωi)−1, (30.3)

which may be obtained from the definition of the grand partition function (see

(28.16) and (28.17)) by setting εi = ~ωi, and µ = 0. As long as we consider a single

system, the total zero-point energy of the system∑i ~ωi/2 is constant and may be

ignored by shifting the energy origin.3

Therefore, the canonical partition function of the system consisting of harmonic

modes (or equivalently, consisting of photons or phonons) may be written as the

bosonic grand partition function with a zero chemical potential ΞBE(β, 0) (recall

28.6), regarding each mode ~ωi as a one particle state energy. From this observa-

tion, the reader should immediately recognize that T dependence of various ther-

modynamic quantities can be computed easily (or dimensional analytic approaches

allow us to guess many T -dependent behaviors).

30.2 Warning: grand partition function with µ = 0 is only a gimmick4

The thermodynamic potential for the system consisting of photons or phonons is

the Helmholtz free energy A whose independent variables are T and V . The ex-

pected number 〈ni〉 of phonons (photons) of mode i is completely determined by

the temperature T and the volume V . Notice that we do not have any more ‘handle’

like µ to modify the expectation value.

If we set formally µ = 0, then dA = −SdT − PdV , so we have A = −PV(see Section 27 around the Gibbs-Duhem equation 27.4). That is, our observation

logZ(β) = log Ξ(β, 0) is consistent with thermodynamics. Thus, we may consis-

tently describe systems consisting of phonons and photons in terms of the grand

partition function for non-interacting bosons with zero chemical potential (as long

as the zero-point energy is constant).

However, do not understand this relation to indicate that the chemical poten-

tials of photons and phonons are indeed zero; actually they cannot be defined. The

3 〈〈Warning: zero-point energies can shift〉〉 However, if the system is deformed or chemicalreactions occur, the system zero-point energy can change, so we must go back to the original

formula with the total zero-point energy and take into account its contribution. For theelectromagnetic field, the change of the total zero-point energy may be observed as force. This

is the Casimir effect.4 This is emphasized by Akira Shimizu. There is a report on the Einstein condensation of

photons (Klaers, J. et al. (2011). Bose-Einstein condensation of photons in an opticalmicrocavity, Nature 468, 545-548). In this case, the experiment is performed in an artificial

environment in which the number of photons mimics a conserved quantity.

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355 Photons and internal motionstrelation is only a mathematical formal relation that is useful sometimes.5

30.3 Expectation number of photons

The µ = 0 boson analogy tells us that the average number of photons of a harmonic

mode with angular frequency ω is given by (as we know, cf. (23.13) and a footnote

1* there)

〈n〉 =1

eβ~ω − 1. (30.4)

30.4 Internal energy of photon systems

The photon contribution to the internal energy of a system may be computed just

as we did for the Debye model (Section 23). We need the density of states (i.e.,

photon spectrum = the distribution of the frequencies of the modes) D(ω). The

internal energy of all the photons is given by

E =∑

modes

〈n(ω)〉~ω =

∫ ∞0

dωD(ω)~ω

eβ~ω − 1. (30.5)

This is the internal energy without the contribution of zero-point energy.

A standard way to obtain the density of statesD(ω) is to study the wave equation

governing the electromagnetic waves, but here we use our usual shortcut (28.13).

The dispersion relation for photons is ε = c|p| = ~ω, so∫ ω

0

D(ω′)dω′ =V

h3

∫|p|≤~ω/c

d3p. (30.6)

Differentiating the above equality, we obtain exactly as in 23.10

D(ω) =4πV

h3

(~ωc

)2 ~c

=V ω2

2π2c3. (30.7)

Photons have two polarization directions, so the actual density of the modes is

this formula × 2.

30.5 Planck’s distribution, or radiation formula

The internal energy dEω and the number dNω of photons in [ω, ω + dω) in a box

of volume V are given by

dEω = 2D(ω)~ω

eβ~ω − 1dω, (30.8)

dNω = 2D(ω)1

eβ~ω − 1dω. (30.9)

The factor 2 comes from the polarization states (i.e., D here is given by (30.7)).

5 〈〈When chemical potential is definable〉〉 Intuitively speaking, chemical potential may be

defined only for particles we can ‘pick up.’ More precisely speaking, if no (conserved) chargeof some kind (say, electric charge, baryon number) is associated with the particle, its chemical

potential is a dubious concept.

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356 Photons and internal motionstTherefore, the energy density u(T, ω) at temperature T due to the photons with

the angular frequencies in [ω, ω + dω) reads

u(T, ω) =ω2

π2c3~ω

eβ~ω − 1. (30.10)

This is Planck’s radiation formula (19006) (Fig. 30.1).

tFig. 30.1 Classical electrodynamics gives the Rayleigh-Jeans formula (30.12) (gray); this is the result ofequipartition of energy. The density is not integrable due to the contributions from the UV modes(the total energy diverges). Wien reached (30.13) empirically (dotted curve). Planck initially guessedhis formula (black) by interpolation of Wien’s formula and another empirically known result:u(T, ω) ∝ T for small frequencies. Although the latter agrees with the Rayleigh-Jeans formula, it wastheoretically derived only after Planck’s formula.

30.6 Salient features of Planck’s law

It is important to know some qualitative features of Planck’s formula (30.10):

(i) Planck’s law can explain why the spectrum blue-shifts as temperature increases:

The peak position is 2.82kBT corresponding to the wavelength λ = 2.9 × 106/T

nm.7

(ii) The total energy density u(T ) = E/V of a radiation field at temperature T is

finite. u(T ) is obtained by integration:

u(T ) =

∫ ∞0

dω u(T, ω). (30.11)

With Planck’s law (30.10) this is always finite (we will study this in the next entry).

(iii) In the classical limit ~→ 0, we get

u(T, ω) =kBTω

2

π2c3( = 2D(ω)kBT ) , (30.12)

which is the formula obtained by classical physics (i.e., the equipartition of energy;

recall the last portion of 20.6) called the Rayleigh-Jeans formula. Upon integra-

tion, the classical limit gives an infinite u(T ). This divergence is obviously due to

the contribution from the high frequency modes. Thus, this difficulty is called the

ultraviolet catastrophe, which destroyed classical physics.

(iv) In the high frequency limit ~ω � kBT Planck’s law (30.10) goes to

u(T, ω) ' kBT

π2c3ω2e−β~ω, (30.13)

which was empirically proposed by Wien (1896) (indicating clearly the existence of

6 [1900: The Paris World Exhibition; Boxer Rebellion; de Vries rediscovers Mendel’s Laws;Sibelius Finlandia premiered; Evans begins to unearth Knossos.]

7 〈〈The Sun〉〉 The surface temperature of the Sun is about 5800 K, and its radiation roughly

follows Planck’s law of T = 5500 K with the peak wavelength about 525 nm (green).

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357 Photons and internal motionstthe energy gap corresponding to the quantum ~ω).

30.7 Statistical thermodynamics of black-body radiation

The total energy density u(T ) of the radiation field at T is given by

u(T ) =

∫ ∞0

ω2

π2c3~ω

eβ~ω − 1dω = β−4

∫ ∞0

(βω)2

π2c3~βω

eβ~ω − 1d(βω). (30.14)

We see (as seen above)

u(T ) ∝ T 4, (30.15)

which is called the Stefan-Boltzmann law.8

Since we know the T 3-law of the phonon low temperature specific heat (the

Debye theory; 23.10), this should be expected. This is understandable by counting

the number of degrees of freedom (as explained in Fig. 29.1). Clearly recognize that

the radiation field problem and the low temperature specific heat of crystals are

the same problem.

30.8 Black-body equation of state

The proportionality (30.15) was obtained purely thermodynamically by Boltzmann

before the advent of quantum mechanics (as shown below). The proportionality

constant contains ~, so it was impossible to obtain the constant theoretically before

Planck.

Photons may be treated as ideal bosons with µ = 0, so the equation of state

is immediately obtained as (‘2’ in front of the following integral is due to the

polarization states)

PV

kBT= log Ξ = −2

∫dεD(ε) log(1− e−βε). (30.16)

For 3D superrelativistic particles, D(ε) ∝ ε2, so∫ ε

0

dεD(ε) =1

3εD(ε). (30.17)

This gives us (review what we did to derive PV = 2E/3 for the ordinary particles;

recall 28.13 and 28.14)

PV =1

3E. (30.18)

Just as PV = 2E/3 is a result of pure mechanics, (30.18) is a result of pureelectrodynamics, so this was known before quantum mechanics. Boltzmann startedwith (30.18) to obtain the Stefan-Boltzmann law9 as follows.

Since we know generally

E = TS − PV = TS − 1

3E, (30.19)

8 The proportionality constant can be computed as k4Bπ

2/15~3c3. Cf. 23.11.9 Stefan deduced the law from Tindal’s data in 1879. [1879: Edison’s electric light bulb; Ibsen’s

A Doll’s House; Frege’s Begriffsschrift; Einstein was born.]

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358 Photons and internal motionstST =

4

3E or S =

4

3

E

T. (30.20)

Differentiating S wrt E under constant V and noting (∂S/∂E)V = 1/T , we obtain

1

T= − 4

3T 2

(∂T

∂E

)V

E +4

3T(30.21)

or

1

3T=

4

3T 2

(∂T

∂E

)V

E, (30.22)

that is, under constant V

dE

E= 4

dT

T. (30.23)

This implies the Stefan-Boltzmann law E ∝ T 4. Compare this with (23.30) in 23.11.

Let us look at the quantum effect on internal degrees of freedom in gas molecules.

30.9 Internal degrees of freedom of classical ideal gas

If noninteracting particles are sufficiently dilute (µ � 0), we know classical ideal

gas approximation is admissible. However, the internal degrees of freedom may not

be handled classically, because energy gaps may be huge. We have already glimpsed

at this when we discussed the gas specific heat (Section 23).

Let us itemize internal degrees of freedom of a molecule:

i) Each atom has a nucleus, and its ground state could have a nonzero nuclear spin.

This interacts with electronic angular momentum in the atom to produce the ultra-

fine structure. The splitting due to this effect is very small, so for the temperature

range relevant to the gas phase we may assume all the levels are energetically equal.

Thus, (usually) we can simply assume that the partition function is multiplied by

a constant g = degeneracy of the nuclear ground state.10

ii) Electronic degrees of freedom has a large excitation energy (of order of ioniza-

tion potential ∼a few eV), so unless the ground state of the orbital electrons is

degenerate, we may ignore it.11

iii) If a molecule contains more than one atom, it can exhibit rotational motion.

The quantum of rotational energy (ΘR below) is usually of order 10 K.12

10 〈〈Nuclear spin-rotation interference in light homonuclear diatomic molecules〉〉 Inthe case of light homonuclear diatomic molecules (e.g., H2, D2, T2), nuclear spins could

interfere with rotational degrees of freedom through quantum statistics, so we cannot take thenuclear effect simply by g. This effect is historically important in showing protons arefermions, but is not discussed in this book [See Kubo’s problem book]. For not so light

homonuclear diatomic molecules, usually we may ignore this effect.11 〈〈What if the ground state is degenerate?〉〉 If the ground state is degenerate, then it

could have a fine structure with an energy splitting of order a few hundred K due to the

spin-orbit coupling. For ground state oxygen (3P2) the splitting energy is about 200 K, so we

cannot simply assume that all the states are equally probable nor that only the ground slateis relevant.

12 However, for H2 it is 85.4 K. For other molecules, the rotational quantum is rather small: N2:2.9 K; HCl: 15.1 K.

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359 Photons and internal motionstiv) Also such a molecule can vibrate. The vibrational quantum (ΘV below) is of

order 1000 K.13

30.10 Rotation and vibration

Notice that there is a wide temperature range, including room temperature, where

we can ignore vibrational excitations and can treat rotation classically (Fig. 30.2).

Thus, equipartition of energy applied to translational and rotational degrees of

freedom can explain the specific heat of many gases.

vibrational

rotational

translational

vC /Nk BT

TΘ Θ

R V0

3/2

5/2

7/2

RTtFig. 30.2 The constant volume specific heat. RT means room temperature. ΘR is the rotational energyquantum (in K) and ΘV the vibrational energy quantum (in K). The hump in the rotationalcontribution can only be found by actual computation (no simple physical reason for it).

The Hamiltonian for the internal degrees of freedom for a diatomic molecule

reads

H =1

2IJ

2+ ~ω

(n+

1

2

), (30.24)

where I is the moment of inertia, J the total angular momentum operator and

n the phonon number operator. Therefore, the partition function for the internal

degrees of freedom reads zi = zrzv: the rotational contribution is14

zr =∞∑J=0

(2J + 1)e−(ΘR/T )J(J+1), (30.25)

with ΘR = ~2/2kBI and the vibrational contribution is

zv =∞∑n=0

e−(ΘV /T )(n+1/2). (30.26)

with ΘV = ~ω/kB .

If the temperature is sufficiently low, then

zr ' 1 + 3e−2ΘR/T . (30.27)

13 N2 3340 K; O2: 2260 K; H2: 6100 K.14 A rigid rotor is a symmetric top that cannot rotate around its symmetry axis. This restricted

eigenspace is characterized by the total angular momentum quantum number J with therotational energy ~2J(J + 1)/2I which is 2J + 1-tuple degenerate.

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360 Photons and internal motionstThe contribution of rotation to specific heat is

Crot ' 3NkB

(ΘR

T

)2

e−2ΘR/T . (30.28)

For T � ΘR, we may approximate the summation by integration (Large Js

contribute, so we may approximate J ' J + 1):

zr ' 2

∫ ∞0

dJJe−J2(ΘR/T ) =

T

ΘR. (30.29)

This gives the rotational specific heat, but it is more easily obtained by the equipar-

tition of energy, because the rotational energy is a quadratic form. Thus, Crot = kB(per molecule) in the high temperature limit for a diatomic molecule (because two

orthogonal rotations are allowed; each degree of freedom classically contributes

kB/2. Recall (20.25)).

The vibrational contribution has already been studied in Section 23.

Q30.1 [Einstein’s A and B]

Let us consider a system with two energy levels with energy 0 (ground state) and ε (> 0)

(excited state). If many such systems are maintained at temperature T , we know the

occupation ratio of these states must be given by a Boltzmann factor e−βε. Now, let us

assume that the system is also interacting with the electromagnetic field (radiation field) of

frequency ν = ε/h with radiation energy density ρ (= the number density of the photons

of energy hν; no other field interacts with the system because the energy exchange is

through photons of energy hν).15

Interacting with the radiation field, the system may be excited at the rate ρB↑n0,

where n0 is the number of the systems in their ground state and B↑ a positive constant

(imagine a collisional reaction between the atom and the photon). It is also de-excited at

the rate ρB↓nε, where nε is the number of the systems in their excited state and B↓ a

positive constant. It is also known that there is the so-called ‘spontaneous radiation’: even

if there exists no radiation field, a photon is emitted spontaneously from the excited state

with the rate Anε, where A is a positive constant.

If T is very large and the intensity of the radiation field is very strong, then nε ' n0,

and the spontaneous radiation may be ignored relative to the induced transitions by

radiation, so we must assume B↑ = B↓ = B, a common constant. In equilibrium there

must be a transition balance between going up and going down. From this balance derive

Planck’s radiation formula ρ ∝ 1/(eβhν − 1).Soln.The equilibrium condition must be (up rate = down rate)

Bρn0 = (A+Bρ)ne. (30.30)

This implies

ρ =A/B

eβhν − 1. (30.31)

15 We follow the argument by Einstein in Einstein, A., (1917). Zur Quantentheorie derStrahlung, Phys. Z. 18, 121-128 [To the quantum theory of radiation]. [1917: The RussianRevolution; the Balfour declaration]

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PART III

ELEMENTS OF PHASE

TRANSITION

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