Invitation to a Contemporary Physics (2004)

Chapter 6. Quantum Computation

455

6.2The nand gate is a combination of the and and not gates. The not gate, which interchanges 0 and 1, can be obtained from the xor ( ) gate with the help of an auxiliary bit 1, because 0 1 = 1 and 1 1 = 0.

6.3d c = d if c = 0, and d c = d if c = 1.

6.41000 = 13 + 47 × 21, hence 1000 mod 21 = 13.

6.5Di erent gates:

1. (1d) → cn → (1, 1 d) = (1, d).

2. (cc 0) → ccn → (c, c , c × c ).

3.(cd) → cn → (c, c d).

The last bit of these three operations give the desired result.

4.(ab) → fanout1 → (aab) → fanout3 → (aabb)

→	and13	→	(a × b, a, b) →	xor23	→ (a × b, a b)
			(a × b) (a b) .
→	xor12	→

The final result is 1 if either a or b is 1, and is 0 if neither of them is 1. Hence, it is a or b.

6.6Hadamard gate:

0

1

( 0

+ 1 )

1

( 0

+ 1 )

e−iθ 0

h

→

√

→

rX

→

e−iθ √

h

→

,

2

| →

2

|

|

|

| →

|

|1 →

1

(|0 − |1 )

1

(|0 − |1 ) → h → eiθ|1 .

h → √

→ rX

→ eiθ √

2

2

6.7 Consider h2(cu)h2 operating on |cd , where c is the control bit and d the data bit. If cu were absent, then nothing changes because h2 h2 = 1. With cu sandwiched in-between, the intermediate state |11 gets a minus sign, thereby

1			1
interchanging the two intermediate states |1	√		(|0 +|1 ) and |1	√		(|0 −|1 ).
		2			2

After the operation by the final h2, this has the e ect of interchanging the initial |10 with |11 , while leaving |00 and |01 untouched. This is therefore

a cn gate.

6.8The state |0 is given by H|Υ , where H is the 4-bit hadamad gate, and

|Υ = 4

(−1)f(x)|x = 4

|x −

|x

= 4

(|0abc −|1abc ) .

1

15

1

7

15

1

x=0

x=0

x=8

a,b,c=0,1

Now pass this state through the 4-bit h gate. Then

1

H|

|

|

− |

|

|

2

Υ

= h1h2h3h4 Υ

= h1

√

( 0000

1000 ) = 1000

= a

.

6.9Finding the greatest common divisor of two numbers:

1. R2 = 19, R3 = 2, R4 = 1, R5 = 0. Hence, gcd(124, 21) = 1.

456	Solutions to Problems

2.R2 = 0. Hence, gcd(126, 21) = 21.

3.R2 = 0. Hence, gcd(21, 7) = 7.

4.R2 = 3, R3 = 0. Hence, gcd(21, 9) = 3.

6.10Factorizing a number:

1.r = 6 because 56 = 15625 = 1 + 744 × 21. Therefore, b = 53 − 1 = 124 and c = 53 + 1 = 126. From Problem 6.9, we know that gcd(124, 21) = 1 and gcd(126, 21) = 21. Thus, a = 5 and r = 6 cannot be used to find the prime factors p and q.

2.r = 6 because 26 = 64 = 1 + 3 × 21. Therefore, b = 23 − 1 = 7 and c = 23 + 1 = 9. From Problem 6.9, we know that gcd(21, 7) = 7 and gcd(21, 9) = 3. Thus, the prime factors of N = 21 are 7 and 3.

Chapter 8. Bright Stars and Black Holes

8.1Wien’s law, λ = 0.0029 m K/T , gives for the earth λ = 10 µm (infrared), the sun λ = 500 nm (yellow) and a class A star λ = 290 nm (blue).

8.2 (a) Use L = 7.14 × 10−7R2T 4 (in SI units) to get for the earth L = 2.37 × 1017 W, and for the sun L = 0.39 × 1027 W.

(b)(T/T )4 = 23/(1.8)2 = 7.1, hence T = 1.63T = 9450 K.

8.3(a) Using d = 1 AU = 1.496 × 1011 m and L = 0.39 × 1027 W, we get the solar apparent luminosity L = 1.4 kW/m2.

(b)Use the formula d2 = L/(4πL) and the data to get d = 21 pc.

8.4(a) For stars of the same radius, Lλ4 = constant.

(b)Because the data are in solar units, we must return to the more general formula L R2/λ4. For Arcturus, we have λ = 1.36 λ = 680 nm (red), while for Orionis, λ = 0.447 λ = 223 nm (blue).

8.5(a) Life-time is given by t = E/L. We take E = 0.1 Mc2 and L Ma. Hence, the star’s life-time on the main sequence is t M1−a.

(b)On the main sequence, we assume the relations R Mb, L Ma and L R2T 4. Together, they give T M(a−2b)/4.

8.6(a) We add electrons on both sides of 4 p → 4He + 2 e+ + 2 ν, so that the initial system is just equivalent to four hydrogen atoms, while the final system is just a helium-4 atom, with the two electron–positron pairs canceling out to give gamma rays (energy). Hence, the mass deficit of the reaction is just the mass di erence between four hydrogen atoms and one helium-4 atom: (4 × 1.007825 − 4.0026) × 931.5 = 26.73 MeV.

(b)14 MeV.

(c)0.862 MeV.

8.7Consider a column of matter extending from the center to the surface and ask yourself what is the weight per unit area of material on top. In a rough

Chapter 8. Bright Stars and Black Holes

457

calculation, we take the average gravitational field felt by a particle in the column as g = GM/R2. The mass per unit area is µ = M/R2. Hence,

P = µg = GM2/R4.

8.8Since ∆x n−1/3 and ∆px = h/∆x = hn1/3, we have vx = px/m (nonrelativistic). Therefore, Pe = nvxpx = h2n5/3/m. The electron number density n is n = Zρ/Amp, where ρ is the mass density given by γ(3M/4πR3), the factor γ is a pure number. So that Pe M5/3/R5. Equating this to Pg M2/R4, we obtain MR3 = constant.

8.9As above, except that now v = c. Hence, Pe = hcn4/3 M4/3/R4. The equation Pe = Pg becomes an equation for M, which yields the mass limit.

8.10If N is the number of ions (N = M/(AmN)) then the total internal energy of the white dwarf is U = 3NkT/2. The cooling rate is −dU/dt, which is equal to L. Hence the cooling time is, approximately, τ = U/L, and exactly τ = 2U/5L (the latter by exact integration). If we take N = 1056 (assuming carbon, A = 12) and T = 107 K, then U = 1040 J and L = 1023 W, so that we find τ = 109 yr.

8.11(a) 0.2 × 10−5; 0.2 × 10−3; 0.15; 0.75.

(b) 28gE; 28 × 104gE; 14 × 1010gE; 14 × 1011gE.

8.12(a) Electron gas in metals: εF = 0.31 eV, TF = 4000 K.

(b) Electron gas in white dwarfs: εF = 3.8 × 105 eV, TF = 109 K.

(c) Neutron gas in neutron stars: εF = 2 × 107 eV, TF = 1011 K.

8.13 Since 1 yr = 3 × 107 s, an accretion rate of 10−10 M /yr is equivalent to

0.7 × 1030	J/s, or 1030 W. If all of this goes into black body radiation, then
L ≥ 1030	W. We use the formula (L/L ) = (R/R )2(T/T )4 to obtain

T ≥ 2000T = 107K. Wien’s displacement law then gives λ ≤ 0.3 nm.

8.14The gravitational binding energy of a proton on the surface of a star of mass M and radius R is ε = GMmp/R = γmpc2, where γ is the surface potential. For a solar-mass neutron star, we know that γ = 0.15, so that the rate of conversion is 15% the accreted mass, or 1.5 × 1016 J/kg.

8.15To measure the mass M of a black hole, one would place a satellite into orbit about the hole. Once one has measured the size and period of the satellite’s orbit, one would use Kepler’s law to determine M. To measure the hole’s spin, one would place two satellites in orbit, one circling in the same direction of rotation of the hole, the other in the opposite direction. By comparing the two orbital periods of the satellites, one can determine the hole’s angular momentum. Finally, one would send a spacecraft, carrying test charge and instruments sensitive to electric fields, passing near the black hole, and would then measure its charge.

8.16For M = M , r = 215 km; for M = 100M , r = 103 km; and for M = 109M , r = 2 × 105 km.

8.17 (a) Setting J = M2			in the relation M2	= M2	+ J2/(4M2 ), we get M =
√				irr	irr
	2	Mirr.

458			Solutions to Problems
			√
	(b) The initial mass is Mirr, the final mass is			2	Mirr. Hence, the mass increase
	√		− 1)Mirr. The rotational energy is δM = 0.29M.
	is δM = (	2
8.18	(a) With E = 3NkT/2, where N = 1057, we have S = 2 × 1057k.
	(b) Since LP2 = 2.6 × 10−70 m2 and A = 4πrg2 = 1.13 × 108 m2, we have the
	entropy SBH = kA/4LP2 = 1077k, which is 1020 times larger than the entropy

of the sun.

Chapter 9. Elementary Particles and Forces

9.1 List of particles:

	Name		mc2 in MeV	spin (J)			charge (e)
photon γ			0	1			0
gluon g			0	1			0
electron e−			0.51	1			−1
				2
muon µ−			105.66		1		−1
				2
electron-neutrino νe			<3 × 10−6	1			0
				2
muon-neutrino νµ			<0.19	1			0
				2
tau-neutrino ντ			<18.2	1			0
				2
u-quark u			1.5 to 4.5	1			+	2
				2				3
d-quark d			5 to 8.5	1			−	1
				2				3
s-quark s			80 to 155	1			−	1
				2				3
pion π±			139.57	0			±1
pion π0			134.98	0			0
η meson			547.30	0			0
f0(600) meson			400 to 1200	0			0
ρ meson			771.1	1			±1, 0
ω meson			782.57	1			0
η mesons			957.78	0			0
f0(980) meson			980	0			0
a0 meson			984.7	0			0
K± meson			493.68	0			±1
K0,	K0	mesons	497.67	0			0
K (892) mesons			892	1			±1, 0, 0
proton p			938.27		1		+1
				2
neutron n			939.57	1			0
				2

Chapter 10. Cosmology

459

9.2Using c = 197 MeV fm, and the rest energy mc2 from the table in Problem 1, we obtain the ranges for the exchange of ρ, ω, η, η to be respectively 0.26 fm,

0.25 fm, 0.36 fm, and 0.21 fm.

9.4From the table in Problem 1, the energy needed to create a π+π− pair is 2 ×139.57 = 279.14 MeV. This translates into a distance r = 279.14 MeV/σ =c/573 MeV = 0.34 fm.

9.5Verification of relation Q = I3 + q/6.

	I3			q	Q			I3 + q/6
u	+	1		+1	+	2		+	2
		2				3			3
d	−	1		+1	−	1		−	1
		2				3			3
u¯	−	1		−1	−	2		−	2
		2				3			3
¯		1		−1		1			1
d	+	2			+	3		+	3

9.5Since the central value may not be the correct value, the fit cannot be exact, so the parameters λ, A, ρ, η cannot be determined uniquely. From Footnote 3, the magnitude of the matrix elements of V are given by (we have assumed λ to be small and positive):

1

1 −

λ2

λ

|A|λ3 ρ2 + η2

2

V

=

λ

1

−

1

λ2

A λ2

.

2

| |

| |

A λ3

(1 ρ)2 + η2

A λ2

1

| |

−

|

|

The central values on the webpage are

0.9749 0.223 0.0037

|V | =	0.223	0.974 0.041
	0.009	0.041 0.9992

λ is fixed from |Vus| to be 0.223. With that, |A| can be fixed from |Vcb| to be 0.041/(0.223)2 = 0.824. The remaining parameters can be determined from |Vub| and |Vtd| to satisfy ρ2 +η2 = (0.0037/|A|λ3)2 = 0.164, and (1−ρ)2 +η2 = (0.009/|A|λ3)2 = 0.969. This gives |ρ| = 0.086 and |η| = 0.396.

Chapter 10. Cosmology

10.1 Measurements of astrophysical parameters:

460	Solutions to Problems

1.v = zc = 3 × 104 km/s.

2.s = v/H0 = (3 × 104/72) Mpc = 420 Mpc = 4.2 × 108 pc = 1.5 × 109 light-years.

3.(a) Recall that a di erence of 5 magnitudes corresponds to a brightness ratio of 100, and that

(b)The apparent brightness is (10 pc/s)2 times the intrinsic brightness.

(c)If y is the di erence between the apparent magnitude and the absolute magnitude, then the apparent brightness is (100)−y/5 times the

intrinsic brightness.

(d)Equating (b) and (c), and taking the logarithm to base 10 on both

sides, we get −2y/5 = 2 ×log(10 pc/s). Hence, y = 5×log(s/10 pc) = 5 × log(4.2 × 107) = 38.1, so the apparent magnitude of the galaxy is 38.1 − 20 = 18.1.

10.2Let ρ0 = ρ01 + ρ02 be the present energy density of universe, where ρ01, ρ02 are, respectively, the matter and dark energy components. Similar notations without the subscript 0 will be used to denote the densities at any time. The equations of state for these two components are, respectively, P1 = 0 and P2 = −0.8ρ2.

Recall from Sec. 10.3.5 that the equation of state P = −wρ leads to the

dependence ρ a(t)3(w−1). Hence, ρ1 = ρ01[a(t0)/a(t)]3 = ρ01a(t)−3, and ρ2 = ρ02[a(t0)/a(t)]0.6 = ρ02a(t)−0.6. The ratio of the two densities at any time is therefore ρ2/ρ1 = (ρ02/ρ01)a(t)2.4 = 2a(t)2.4.

1. The e ective force on a galaxy is proportional to ρ + 3P = ρ1 + ρ2 + 3(P1 + P2) = ρ1 + (1 − 3w)ρ2 = ρ1 − 1.4ρ2. Deceleration turns into acceleration when this force is zero. This occurs when ρ2/ρ1 = 1/1.4

= 2a(t)2.4, or a(t) = (1/2.8)1/2.4 = 0.65. 2. z = 1/a(t) − 1 = 0.54

10.3Recall from Footnote 13 that entropy conservation implies g(aT )3 to be a constant. Using a subscript 0 to represent the present time, we have a(t0) = 1, T0 = 2.728 K, and g0 = 1.68 (see Footnote 14). At the beginning of

the Big Bang, g	=	gU , T = 10	24		28	K. Hence, a =
			1/3eV/kB = 1.16 × 10
(T0/T )(g0/gU )1/3	=	2.8 × 10−28/gU		. Let a denote the scale factor at the

beginning to inflation. Hence, a = 10−25a = 2.8 × 10−53. Taking the present size of the universe to be r = 14×109 light-years = 14×109 ×0.946×1016 m = 1.3×1026 m, the size before inflation is therefore s = a r = 3.6×10−27/gU1/3 m.

10.4Evolution of the universe:

1.a(t) = a(t0)(T0/T ) = 3/(0.2 × 106 × 1.16 × 104) = 1.29 × 10−9.

2.z = a(t0)/a(t) − 1 = 7.7 × 108.

3.Using Eq. (10.6) with the coe cient equal to 1.32, the time it takes is t 1.32 × (1/0.2)2 = 33 s. This estimate is accurate to approximately 1 s because we have used coe cient 1.32 throughout. The time it takes to

Chapter 10. Cosmology

461

reach a temperature of 100 MeV is quite negligible. Assuming e+e− annihilation to occur around 1 MeV, the time it takes to reach that can be calculated from Eq. (10.6) using the coe cient 0.74, which is 0.74 s. Since this is short compared to 33 s, we can ignore the correction in this epoch.

10.5Wien’s law:

1.The photon density at T0 = 2.728 K is (nγ)0 = 414/(cm)3 (see Sec. 10.5.2). At T = 27◦C = 300 K, the photon density is nγ = (nγ)0(T/T0)3 = 414 × (300/2.728)3 = 5.5 × 108/(cm)3.

2.According to Wien’s law, the peak wavelength occurs at λm = (2π/4.967) × ( c/kBT ) = 1.265( c/kBT ). Since kB = 1 eV/(1.16 × 104 K) and c = 1.97 × 10−7 eV m, c/kB = 2.29 × 10−3 m K. With T = 300 K, we get λm = 1.265 × 2.29 × 10−3/300 = 0.96 × 10−5 m = 9.6 µm.

3.It is dark because the photons are in the infrared, beyond the visible range.

10.6Elements in the universe:

1.Hydrogen.

2.4He and D are produced in Big Bang nucleosynthesis.

3.4He.

10.7Doppler shift:

1.If δλ/λ = (λ − λ)/λ is the percentage shift of wavelength, then according to the Doppler shift formula (Sec. 10.1.1), δλ/λ = v/c = 371/(3 × 105) = 1.24 × 10−3. In particular, this is true for the peak wavelength λ = λm of the black-body radiation.

2.According to Wien’s formula, λm 1/T , hence −δT/T = δλm/λm = 1.24 × 10−3.

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Glossary

absolute zero The lowest possible temperature. It is by definition the zero of temperature on the Kelvin scale, or about −273.15 degrees on the Celsius scale. At this temperature, thermal random motions completely disappear, but not quantum fluctuations.

asymptotic freedom A principle in quantum chromodynamics which says that the gluon-exchange forces acting between quarks become weaker as the quarks move in very close, as they often do in very high-energy collisions, so that the quarks become free of the forces at asymptotically high energies.

attractor A stable end-state toward which a system with appropriate dynamical properties will ultimately approach.

beta-decay (β-decay) Transmutation of a neutron into a proton, accompanied by the production of an electron and an electron-antineutrino; also, similar transmutation of a neutron-rich (radioactive) atomic nucleus. It is the most familiar manifestation of the weak interaction.

baryons Heavy particles of half-integral spins made up of three quarks, such as the proton and other, higher-mass particles. They can interact through any of the four known fundamental forces.

beauty One of the quark flavors; same as bottom.

bifurcation Splitting of an equilibrium state in two, progressively, as a control parameter is varied; hence, the bifurcation diagram.

Big Bang The singularity at the beginning of the universe.

black hole A region of space-time, centered around a gravitational singularity, that cannot be seen or otherwise observed by distant observers, because everything, light or matter, once inside, cannot escape to the outside. Physicists think that black holes exist in three forms: stellar, supermassive and primordial.

Bose–Einstein condensate (BEC) Below a critical temperature that depends on the number density, a system with fixed number of indistinguishable Bose particles has a finite fraction of it occupying the lowest-energy single-particle state. This macroscopically occupied state is called the BEC. Thus, we have

463

464

Glossary

the BEC of 4He, 87Rb, etc. The BEC shows quantum coherence e ects like interference, and is associated with superfluidity — zero viscosity.

bosonic stimulation It refers to the phenomenon of a Bose particle being scattered preferentially to a state which already has a higher population of the bosons of the same kind. This follows from Bose statistics. It is involved in the kinetics of Bose–Einstein condensation.

bosons Particles having an integral-valued intrinsic spin (measured in units of ). Photons, gravitons, gluons, W and Z — all particles associated with the transmission of forces — are bosons. Composite objects made up of even number of fermions (particles possessing half-integral spin) — e.g., mesons, 4He, 16O — also behave as bosons. Bosons obey Bose–Einstein statistics, which favors the occupation of the same state by many particles. They also exhibit a phenomenon called Bose condensation observed at a temperature near the degeneracy temperature.

bottom-up nanotechnology Attempts to create structures by connecting molecules. See chemical reduction, electrochemical processes, thermal decomposition and self-assembly.

Cepheid variable A type of star with variable luminosity due to pulsations (regular oscillating changes in size). The luminosity can be deduced from the period of pulsation, making these stars celestial standard candles, useful for the determination of distance.

chaos Aperiodic and apparently random behavior of a dynamical system despite governing laws being deterministic; hence, deterministic chaos.

charm One of the quark flavors.

chemical reduction Method by which crystals of pure metal or metal alloys can be produced by dissolving an inorganic salt of the desired metal in an appropriate solvent and allowing the solution to react with a reducing agent.

color A quantum number for quarks and gluons.

conduction band Set of closely spaced single-electron energy levels in a metal that are partially occupied by current-carrying electrons. In an insulator or a semiconductor, it is the lowest normally empty band to which electrons may be excited from lower levels to carry electric current.

conservation Principle according to which the total amount of some physical quantity in an isolated system always remains the same, whatever internal changes may occur in the system.

Cooper pair Stable complex of two fermions having equal and opposite momenta and opposite spins in a degenerate-fermion system; such a pair behaves as a condensed boson if an e ective attractive force exists between the two fermions in the pair.

cosmic rays High-energy particles (mostly protons, electrons and helium nuclei) present throughout our galaxy. Their origin is not well known. They can be detected above the earth’s atmosphere. Cosmic rays are the object of study in cosmic-ray astronomy.