Invitation to a Contemporary Physics (2004)

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445

the degenerate temperature will be said to be degenerate. From the discussion above, T0 must increase with increasing N/N0.3 As a result, it increases with the number density N/V of these particles and decreases with the particle mass m. For example, a proton (hydrogen nucleus) or a hydrogen atom has a mass of m 2 ×10−24 g. At a normal density of N/V 1022 cm−3, the degeneracy temperature is T0 1 K. A photon has m = 0 so a photon gas is always degenerate, as T0 = ∞ in this case. For electrons in metals, typically N/V 1022–1023 cm−3. With the electron mass being m 10−27 g, the degeneracy temperature is T0 (16–20)×103 K, and hence the electronic distribution in a metal at room temperature is degenerate.

We will now proceed to discuss the distributions when the quantum e ects are important. For bosons, the distribution is known as the Bose–Einstein distribution; for fermions, the distribution is known as the Fermi–Dirac distribution. These two distributions are di erent, but for T T0, both approach the Boltzmann distribution discussed before.

Electrons, protons, and neutrons are examples of fermions. They obey the Fermi–Dirac distribution, and two of these particles are not allowed to occupy the same quantum state. Therefore, at absolute zero temperature, the N particles in the system must simply fill out the N quantum states with the lowest energy. The distribution is therefore given by the solid line of Fig. C.2a; the energy of the last occupied state is called the Fermi energy and is denoted by LF.4 When T rises above absolute zero, the distribution smears out a bit as given by the dashed line in Fig. C.2a, but the amount of smearing is still relatively small as long as T T0. Finally, as T T0, the amount of smearing becomes so large that eventually it approaches the Boltzmann distribution given by Fig. C.1.

Helium atoms (4He) are bosons. They obey the Bose–Einstein distribution and these particles have the tendency to occupy the same quantum state. Thus, at absolute zero temperature, these particles all occupy the lowest possible quantum state, as indicated by the heavy vertical bar in Fig. C.2b. This is the complete opposite of the fermions, no two of which are allowed to be in the same state. As T rises from zero, the available thermal energy pumps more and more of these bosons to quantum states of higher energy, as shown by the dashed line in Fig. C.2b. However, as long as T is lower than some condensation temperature Tc, a finite percentage of the particles will remain in the lowest energy state. This unusual phenomenon, where a macroscopic number of particles occupy a single quantum state and therefore exhibit quantum mechanical properties such as phase coherence, is known as Bose–Einstein condensation. The condensation temperature Tc is of the order of the degenerate temperature T0.5 For liquid helium, this turns out to be 3.1 K, if all the inter-particle interactions are ignored as we have done so up to the present. The actual value for Tc is 2.19 K, because of the presence of these interactions. Below this

3The exact formula is T0 = T (N/N0)2/3 = 2π(N/V )2/3 2/mk.

4F = ( 2/2m)(3π2 N/V )2/3 ≡ kTF = 1.31 kT0.

5Tc = 0.527T0.

446 Thermal Physics and Statistical Mechanics

	dN/d3p					dN/d3p			T = 0
					T = 0
					T> 0				T> 0
(a)						(b)
	0
			F				0

Figure C.2: The number of particles per unit momentum volume (dN/d3p) with an energy at temperature T . (a) Fermi–Dirac distribution, with Fermi energy F; (b) Bose–Einstein distribution, where the heavy vertical line shows Bose condensation.

condensation temperature, all sorts of interesting quantum mechanical behaviors (superfluidity) occur. Another interesting example of Bose–Einstein condensation is the phenomenon of superconductivity. This happens because a pair of electrons also behaves like a boson. See Chapter 3 for detailed discussions of these topics.

Photons are also bosons, but they di er from helium atoms in two respects: photons have mass m = 0, and they can be created and absorbed by the walls of the container or any other material body in the box. As a result of their masslessness, the degenerate temperature T0 is infinite. However, photons of zero energy cannot exist, so there is no Bose–Einstein condensation of the photons — these photons must disappear and be absorbed by the walls of the container or other material bodies.

The Bose–Einstein distribution of photons is known as Planck’s distribution. This distribution law is also called the black-body radiation law. It was discovered in 1900 by Max Planck a long time before the general Bose–Einstein distribution was worked out; it is through this law that Planck’s constant was first introduced, and the seeds for quantum mechanics sowed.

Planck’s distribution is sketched in Fig. C.3 for two temperatures. The peaks of the curves at various temperatures are connected by a dashed line.

Note that this di ers from Fig. C.2(b), because we are now plotting the energy density of photons per unit wavelength (which we denote by u) against the wavelength λ of the photon, rather than the photon number per unit momentum volume.6 Since the distribution is fixed for a given temperature, a measure of this distribution can determine the temperature of a body even when this body is unreachable — as long as its light reaches us. This is the case if we want to know the temperature inside a steel furnace, or the temperature of a distant star (see Chapter 8). It is useful to learn more about two special aspects of this distribution, expressed in Wien’s law and Stefan–Boltzmann’s law.

6Momentum p is related to wavelength λ by p = 2π /λ, and frequency ν is related to the wavelength by ν = c/λ. The energy of a photon is = 2π ν.

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447

dρ/d

T2

T2 > T1

T1

Figure C.3: Planck distribution, showing the photon energy density per unit wavelength (dρ/dλ) as a function of the wavelength λ at a given temperature T . The dotted line joins the peaks at di erent temperatures and is the curve for Wien’s law.

Wien’s law states that the wavelength λmax at the peak of Fig. C.3 is inversely proportional to the temperature: λmaxT = 0.2898 cm K. This is so because, quantum mechanically, the energy carried by a photon of wavelength λ is inversely proportional to λ (energy = 2π c/λ). Thermodynamically, this energy is derived from heat and so is proportional to T . Consequently, λmax T −1 as required by Wien’s law. The proportionality constant ( 0.3), however, cannot be obtained without a detailed calculation.

To illustrate this relation, let us just approximate the number 0.2898 by 0.3, and consider the radiation at a room temperature of T = 300 K. The peak wavelength works out to be λmax = 10−3 cm = 10 micron (1 micron = one millionth of a meter). The wavelengths of the visible light being from 0.4 to 0.7 microns, this radiation at the room temperature is too long to be visible. Nevertheless, it is there in the form of the infrared. To emit visible light the body would have to have a higher temperature. The surface of the sun, for example, has a temperature of 6000 K. Its peak wavelength of 0.5 micron is right in the middle of the visible light region. That is why sunlight is visible, and why it contains all these beautiful colors of the rainbow visible at sunrise and sunset.

The Stefan–Boltzmann law states that the total amount of power density, or the radiation energy of all wavelengths emanating from a body per unit surface area and per unit time, is given by f = σT 4, where σ = 5.67 × 10−5 erg/(cm2 s K4) is called the Stefan–Boltzmann constant. Wien’s law tells us what ‘color’ the radiation has; this law tells us how bright the radiation is. The fact that it is proportional to the fourth power of the absolute temperature is the reason why hot objects are so bright.

A related and a more useful quantity for our purpose is the energy density of the radiation. If one imagines putting the heated body and the radiation it emits in a big box, then the radiation and the body will eventually reach an equilibrium state. The energy density ργ of the radiation in the box is then given by ργ = aT 4, where a = 7.56 × 10−15 erg/(cm3 K4) = 4.73 × 10−3 eV/(cm3 K4).

448	Thermal Physics and Statistical Mechanics

Another version of the Stefan–Boltzmann law concerns the number of pho-

tons per unit volume, nγ. It is proportional to T 3: nγ = bT 3, with b = 3.2 × 1013T 3/eV3 cm3 = 20.4T 3/K3 cm3.

C.3 Dimensional Analysis

To understand the various versions of the Stefan–Boltzmann law, let us resort to a very useful tool known as dimensional analysis.

Every quantity in physics carries a dimension (i.e., unit) made up of three fundamental units: second (s) for time, centimeter (cm) for length, and gram (g) for mass (in the cgs system of units). Dimensional analysis is simply the statement that whatever quantity we want to consider must have the right units (dimensions). This by itself is of course trivial, and is not going to get us anywhere. However, if we also know on physical grounds what that quantity may depend on, then quite often this will yield very valuable information as we shall illustrate below with the ‘derivation’ of the Stefan–Boltzmann law.

To begin the analysis for the photon number density nγ, we must first decide what it can depend on. It will clearly depend on the temperature T . It must also depend on the dynamics and the kinematics governing the emission and the absorption of photons, the balance of which generates the eventual photon density in a volume. The details of these do not concern us, as long as we realize that they are controlled by two fundamental physical constants: the speed of light c = 3 × 1010 cm/s, and the Planck’s constant = 1.054 × 10−27 erg s. It depends on c because anything that has to do with electromagnetic radiation involves c, and it depends on because we are in a degenerate regime where quantum e ects are important. The numerical magnitudes of these two constants do not even matter for our purpose, but their units do.

So the photon number density nγ is a function of T , c, and : nγ = f(T, c, ), where f is some appropriate function which we seek to determine. The unit for nγ is number per unit volume, say cm−3. We must now choose the function f above so that f(T, c, ) comes out to have this unit cm−3. The unit of T is energy (with units chosen so that k = 1), say erg = g (cm/s)2; the unit of c is cm/s, and the unit of is erg s. The only combination of these three variables, T, c, , to yield something of a unit cm−3 is clearly c−3 −3T 3, which gives a unit (cm/s)−3(erg s)−3 erg3 = cm−3. Thus, we must have nγ = b0T 3/( c)3, and the proportionality constant b0 is dimensionless (i.e., without any unit, or is a pure number). To calculate this number, we must know the technical details of quantum electrodynamics and statistical mechanics. Nevertheless, as we have just seen, the very fact that nγ is proportional to T 3 can be deduced simply from dimensional analysis.

The fact that the energy density ργ is given by ργ = a0T 4/( c)3 can be obtained in exactly the same way through dimensional analysis, once we note that the dimension of energy density is erg/cm3.

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The beauty of this argument is not only that it is simple, but that it is also quite general. We used the argument above to obtain nγ and ργ, but nothing in it says that we are dealing with photons γ rather than some other relativistic particles. The appearance of the speed of light c, which is the speed of travel of a massless particle, however suggests that this argument is not valid for massive particles unless they are at such a high temperature to render the mass m negligible (mc2 kT ) and their velocity approaches that of light. Indeed, it is easy to see that if we had to assume f to be a function of m as well as T , c, and , then the dimensional argument would fail because there are infinitely many combinations of these four variables to yield the right dimension. No useful conclusion can then be drawn. However, as indicated before, the argument for photons is equally valid for any relativistic particle (meaning m kT/c2), in which case its number density is still given by n = b0T 3/( c)3, and its energy density is given by ρ = a0T 4/( c)3.

The proportionality constants a0 and b0 cannot be obtained without detailed dynamical inputs. With them, they are calculable and we shall quote the results. For bosons, the values are a0B = (π2/30)g = 0.33g, and b0B = (1.202/π2)g = 0.12g, where g is the spin degeneracy factor, or the number of spin orientations the particle can have. This number is g = 1 for pions and g = 2 for photons. For fermions, the two proportionality constants are a0F = (7/8)a0B and b0F = (3/4)b0B. That in both a0 and b0 the values for fermions are smaller than the values for bosons is understandable. It is due to the fact that fermions tend to avoid one another while bosons tend to stick with their own clans. As a result, there must be more bosons per unit volume than fermions, everything else being equal. Hence, b0B > b0F, and

similarly	a0B > a0F.
If we	write the energy density for the photon as ργ	= aT 4 = a0BT 4/( c)3,
and substitute into it the known values of a0B, , and		c, we get a = 4.73 ×

10−3 eV/cm3/K4. Similarly we can calculate the constant b defined by nγ = bT 3 from the value of b0B and get b = 20.4/cm3/K3.

Note that the photon density is fixed at a given temperature. This could not possibly be the case for the electron density at low temperatures because the electron density depends on how many electrons we put in the volume, so it is a number which could in principle be anything. This is not so for photons, because photons can be created and absorbed. If we force more photons in the volume at a given temperature, they will simply be absorbed by the walls. If we take out photons, then the walls will radiate enough photons so that the Stefan–Boltzmann law remains valid.

We also mention that Stefan–Boltzmann’s law is valid for any relativistic particle, and that the proportionality constants a0 and b0 are completely fixed as well. That this is so is again because all these particles can also be created and absorbed if the temperature is high enough. For electrons, if the temperature is considerably larger than 1 MeV (million eV), then there is su cient heat energy present to create electron–positron pairs. At low temperatures, such creations are

450	Thermal Physics and Statistical Mechanics

energetically forbidden (electrons and positrons both have a mass of 0.5 MeV), and such automatic adjustments of their numbers cannot occur.

It should also be mentioned that the pressure P these particles exert on one another and on the walls of the box is given by P = ρ/3, both for bosons and for fermions. The fact that P is proportional to ρ can be obtained from dimensional analysis, but the factor three in the denominator can be obtained only through detailed calculations.

Appendix: Solutions to	D
Problems

Chapter 2. Lasers and Physics

2.1We know that E/ν, λν and λE are universal constants, which can be obtained from the stated equivalences. So that, with λ given, we can calculate the

corresponding ν and E by the relations ν = (λ1ν1)/λ and E = (λ1E1)/λ, where λ1ν1 = (1240 nm)(2.4 × 1014 Hz) = 3 × 1017 nm Hz and λ1E1 = (1240 nm)(1 eV) = 1240 nm eV.

So 1 nm is equivalent to 3 × 1017 Hz or 1.24 keV; 500 nm is equivalent to 6 × 1014 Hz or 2.48 eV; and 104 nm is equivalent to 3 × 1013 Hz or 0.12 eV.

2.2(a) 10.2 eV. (b) An energy of 1.9 eV is released. (c) No, because absorption would excite the atom to an energy of 3.6 eV, where no atomic state exists.

2.3Since τ = 1/γ, we have τ2/τ3 = γ3/γ2 = 1/100. Level 3 lives 100 times longer than level 2.

2.4We know that ∆ν/ν = vx/c. Now vx2/c2 = kT/Mc2. We have Mc2 = 109 eV (mass of the hydrogen atom) and kT = 25 × 10−3 eV (room temperature). It follows that ∆ν/ν = 5 × 10−6.

2.5E/T = k is a universal constant, so that E = T (E1/T1), where E1/T1 = 1/12000 (eV/K). So 300 K is equivalent to 25 meV, and 6500 K to 550 meV.

2.6In thermal equilibrium, excited states have very small atomic populations. The stimulated emission rate depends on the population of the upper transition level, and so must be much smaller than the absorption rate from the ground state because of the large population of the latter.

2.7When the first spontaneous radiations occur, they run into all directions, and so the stimulated photons that follow also go into all directions, since no cavity is there to guide them in a particular direction. No amplification is possible without a cavity. So directionality and the high intensity of laser light are the qualities determined by the resonant cavity.

2.8(a) The irradiance is 10−3 W/(π(10−3 m)2/4) = 1273 W/m2. A photon at a wavelength of 700 nm has energy 1.8 eV or 2.88 × 10−19 J. So a power of

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452	Solutions to Problems

1 mW or 10−3 J/s corresponds to a flux of 3.5 × 1015 photons per second. Given that the speed of light is 3 ×108 m/s, a cavity 1 m long should contain 107 photons at all times to sustain such a flux.

(b)NOVA produces 100 kJ of infrared light in 3 ns pulse lengths. This is equivalent to a power of 100 kJ/3 ns, or 34 × 1012 W. Assuming a diameter of 10 mm for the beam, the irradiance is 430 × 1012 W/m2.

2.9(a) With the laser wavelength λR = 500 nm, we must have L = 250 nm to guarantee a monochromatic beam at the resonant wavelength λ1 = 2L = 500 nm. If L = 5 cm, the resonant mode corresponds to n = 2 ×105, so there will be at least 4 × 105 modes. The most likely are λR = 500 nm and a few more adjacent ones separated from it (roughly) by multiples of 0.0025 nm.

(b)Wavelength spacing ∆λn = |λn+1 −λn| = 2L/[n(n+1)]; frequency spacing ∆νn = |νn+1 − νn| = c/(2L). The latter is independent of n, whereas the former decreases rapidly with increasing n.

2.10(a) 30 µm. (b) 300 m. (c) 2.5 cm.

2.11(a) 3.1 × 10−4 rad. (b) 3.5 × 10−3 rad.

2.12D = 2R = 2θx = 2(6.94 × 10−7)(400 × 106) ≈ 550 m.

2.13(a) When a slit is covered, the photons pass through the uncovered slit and go on to strike the screen: we will see a bright stripe on the screen opposite the uncovered slit. There might be faint dark lines near the edges of the stripe due to the bending of light around the edge of the slit (di raction).

(b)When a detector is placed behind each slit, we will hear a clicking sound from the detector through which a photon passes, and not from both detectors at the same time. So a photon must pass through either slit, not both.

(c)When both slits are left uncovered, we will not see two separate bright bands, as we would expect from observations in (b), but rather a pattern of overlapping bright and dark stripes (with a white ripple at the center) opposite the two slits. This indicates that the light waves have constructively and destructively interfered.

2.14The condition T 1/∆ν also says that ∆ν 1/T . (a) To have T = 100 fs, we must have ∆ν 1/(100 fs) = 1013 Hz. For a 1 m long cavity, adjacent modes are separated by ∆ν0 = c/2L = (3 × 108 m)/(2 m · s) = 1.5 × 108 Hz. So, to have T = 100 fs, we must lock in at least ∆ν/∆ν0, or 6.7 × 104 modes.

Chapter 5. Nanostructures

5.1If the film thickness tends to zero, the particle position uncertainty also tends to zero, which implies that the conjugate momentum uncertainty and the momentum itself tend to infinity: no confinement is possible.

5.2∆E21 = 3h2/8m d2 = 1.185(me/m )/d2 in eV if d is in nm. For d = 10 nm and m = 0.07me one has ∆E21 = 0.170 eV. Since ∆E32 = 5h2/8m d2 is 5∆E21/3, we have ∆E32 = 0.283 eV. Converting to wavelength λ = hc/∆E,

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453

we have respectively λ21 = 7290 nm and λ32 = 4380 nm in the infrared. When d is reduced by a factor of 2, the energy is increased by a factor of 4 and the wavelength reduced by the same factor.

5.3Electrons in a grain scatter photons whose energy is less than a minimum determined by the size of the grain, and absorb those with higher energy. The color we see is determined by the scattered light. The larger crystallites can absorb lower energy photons and so appear red, whereas the smaller grains absorb higher-energy quanta and so appear yellow.

5.4The condition ∆E21 > kT translates into d2 < 1.2me/m kT , with d in nm and kT in eV. At room temperature, kT = 0.026 eV, so that d < 6.8 nm for m = me and d < 25 nm for m = 0.07me.

5.5The Pauli exclusion principle dictates that only two electrons can occupy any wave-guide mode. The detection of a current (I = e/∆t) means an electron has scattered inelastically inside the output reservoir (so that the energy changes by ∆E = eV ), thereby vacating a waveguide mode and enabling an incoming electron to occupy it. The uncertainty relation ∆E ∆t h then becomes e2V/I h, hence I e2V/h for each electron. With two electrons filling a mode, we must have I 2e2V/h. So the conductance for each mode in the wire is I/V = 2e2/h.

5.6The condition Ec > kT is equivalent to R < e2/εkT . Hence, the limiting value for R is 4.6 nm, which corresponds to a diameter of 9.2 nm.

5.7As the atomic size changes, there are two opposing e ects. First, the Coulomb energy arising from repulsion between electrons orbiting around the nucleus decreases when the average distance between electrons increases. But there is another energy scale in the problem: the energy spacings of the orbits of electrons. As the atomic size increases, the di erences in the orbital energies decrease faster than the Coulomb energy. It follows that the e ects of the electron–electron interaction are relatively more important in artificial atoms than in natural atoms.

5.8(a) E(N, δ) = −N(N + 2δ)/2C; ∆+ = (1 − 2δ)/2C; ∆− = (1 + 2δ)/2C; and ∆+ +∆− = 1/C. We note that the first three depend on the capacitance and the voltage, but the last depends only on the capacitance. The last three do not depend on N. (b) The results follow directly from the cancelation of ∆+ for δ = 1/2 at N = 0, 1, N, and ∆− for δ = −1/2 at N = 1, 2, N + 1.

5.9Just one kind, since all carbons in C60 are exactly equivalent. There are 12 pentagons and 20 hexagons. A simple way is to calculate first the total

surface area of 12 pentagons and 20 hexagons, all of equal sides a, which is 72.6 a2. Equating this to 4πR2 gives R = 2.4 a. For a = 0.142 nm, we have R = 0.34 nm, i.e., a diameter of 0.68 nm.

5.10The width is equal to the diameter of C60, that is 0.68 nm, and the length is the same diameter increased by 0.142 nm, that is 0.82 nm. One could also use the method of the preceding exercise; the answer is the same.

454	Solutions to Problems

5.11By the same reasoning as given in the text, but now including also heptagons and octagons, one has the relation 2e = 3v = 5p+6h+7s+8g, v−e+f −2 = 0, f = p + h + s + g, where s is the number of heptagons, and g the number of octagons. It follows that p = s + 2g + 12.

5.12This is a purely geometrical problem. One should draw a hexagon of edges

√			2 2 2	2 2		|	1|	|	2	|	= 2a cos 30◦ =
with length a, and apply trigonometry to obtain						a		= a
	3	a. Next, C	= n a1	+ m a2	+ 2nma1 · a2. Given 60◦ as the angle between

a1 and a2, one obtains the answer. Similarly, with C · a1 = |C| |a1| cos θ = na21 + ma1 · a2, one obtains θ, as stated.

5.13Since C60 has a radius R = 0.34 nm, the circumference is 2.14 nm. Then, C2 = 4.58 nm2, which is also equal to 0.058(n2 + nm + m2). It follows that

the possible values of (n, m) are (9, 0), (8, 1), (8, 2), (7, 3), (6, 4) and (5, 5). Of these, (9, 0), (8, 2) and (5, 5) are metallic.

5.14In the copper atom, there is a single electron in the external orbital, which happens to be an s-state (which allows only two spin states). When the atoms get together in a bulk volume, another electron of the same kind will fill up the only empty spin state to complete the orbital, corresponding to an anti-aligned spin configuration. Since there is a single possibility, there is no exchange interaction and no splitting.

5.15Using the conversion 1 in2 = 6.45 × 1014 nm2, we see that 1 bit occupies an area of 6.45×105 nm2/x. One bit also occupies an area equal to NL2. Hence, x = 6.45 × 105 nm2/NL2(nm2). Now assume L = 15 nm and x = 1, we have N = 2900.

5.16With L = 10 nm, we have x = 6450/N, which amounts to 2.15 for N = 3000, and 6450 for N = 1. A CD, 12.8 square inches in area, can hold 27.50 Gb in

the first case, and 82 560 Gb (or 82.50 Tb) in the second. Now, 1 Pb is 106 Gb, hence storing 1 Pb requires 36 400 CDs in the first case, and 13 CDs in the second. If L = 3.5 nm, the grain area size is ten times smaller. Hence, the density is correspondingly larger, x = 64 500 for N = 1. In this case, one CD can hold 825 Tb, and so two CDs are enough to store 1 Pb, with plenty of space to spare.

Chapter 6. Quantum Computation

6.1Two theorems and one game:

1.10000000, 1001100.

2.(a) When we decrease any binary number sj, some of its bits must change. The sum da at that bit must become odd.

(b)Suppose among the n sums, da, db, . . . are odd, and a is the leftmost bit among these odd sums. To make the sums all even, all we have to do is to pick a pile whose ath bit is 1, and change its ath, bth, . . . bits.