errata cor ridge

8/22/2019 Errata Cor Ridge

1/41

Introduction

We are indebted to our colleague ShangYung Wang of the Tamkang University, Taiwan for a thorough criticalreading of our book, the enlightening correspondence and the very helpful suggestions. His precious work hasmade these errata possible.

1


2/41

2


3/41

Chapter 1

In the first line of the second full paragraph in p. 8, change the words associated to into associated withand do the same across the book.

Change every occurrence of Zender into Zehnder. Moreover, in the first paragraph, p. 15, replace thesentence When T = R = 1/

2, we have a symmetric (or a 50%50%) beam splitter by:

When T = R = 1/

2, we have a 50%50% beam splitter

In the end of the last paragraph of Sec. 1.2 in p. 20, the phrase as for example x and px are is modified into

as for example the position and the momentum are

In the first paragraph, Sec. 1.4, p. 28, change Nt = N0et into Nt = N0et , where > 0 is the decayconstant.

Modify Eq. (1.53), p. 29, intoa | a = |b | a|2 + |b | a|2 = 1 . (1.53)

Replace the first paragraph of Subsec. 1.5.3 (starting with the words At the beginning of and ending withthe words in the partition of energy) with:

At the beginning of the 19th

century it was already experimentally known that the specific heat per mole ofmonatomic, diatomic, and multiatomic ideal gases is given by 32 R,52

R, 3R, respectively, where R = NAkBand NA is the Avogadro number. According to the classical equipartition law, an equilibrium system hasan average energy of kBT /2 for each harmonic degree of freedom (i.e. quadratic term appearing in itsHamiltonian). As a consequence, the specific heat per mole should be equal to R/2, where is here thenumber of the degrees of freedom of the molecule. While this correctly explains the experimental value ofthe specific heat of monoatomic ideal gases, it cannot account for the above experimental values in the caseof diatomic and multiatomic ideal gases. This disagreement between the classical equipartition predictionand the experimental value of the molar heat capacity cannot be explained by using a more complex modelof the molecule, since adding more degrees of freedom can only increase the predicted specific heat, not de-crease it. This discrepancy can be solved by taking into account that energy levels are quantized accordingto the prediction of quantum mechanics [see Sec. 4.4 on the quantum harmonic oscillator].

Change Eq. (1.74), p. 37, intoEn = 2

2me4

h2n2(n = 1, 2, 3, . . .) . (1.74)

Change the title of Subsec. 1.5.6 from Intrinsic magnetic momentum into

Intrinsic magnetic moment

In the first line of the last paragraph, p. 40, change Schrodinger s into Schrodingers. Drop Problem 1.10 (and also the solution in the web page).

3


4/41

4 CHAPTER 1.


5/41

Chapter 2

Replace the text of the footnote 6, p. 46, by:This is a consequence of a theorem which states that a linear transformation O, on a Hilbert space, is

Hermitian if and only if (iff)

| O

is real for all vectors | [Byorn/Fuller 196970, I, 154].

Substitute the paragraph after the box, p. 47 (starting with the word Therefore, and ending with Eq.(2.10)) with the following:

Moreover, let P = | | and consider two arbitrary states |a and| b . Then,aP b = b P a = (b | | a) = a | | b = a P b . (2.10)

Change the phrase in the box 2.1 below Eq. (2.8), p. 47 where | , | are vectors in the Hilbert spacesuch that

O and O are finite. intowhere

O = O | , O is the bra corresponding to O , and | , | are vectors in the Hilbertspace such that

O and O are finite. In the first line below Eq. (2.53) in p.55, caj =n Uj,ncbj should read

caj =

n Uj,ncbn .

In Eq. (2.55) n,k should read nk. In the first line below Eq. (2.82) in p. 61, Ujk = okj should read Ujk = ojk. In Subsec. 2.1.5 and thereafter change noncommutability into noncommutativity. Change the last line of Eq. (2.90), p. 63 into

=

0 sin cos

sin cos 0

= 0 . (2.90)

Replace the last equality of Eq. (2.102) by n OjnOnk.

Substitute the whole Proof of Corollary 2.1 on p. 67 by the following:

Proof

In fact, if O and O commute, then we have a common eigenbasis {|ok, ok} such thatO |ok, ok = ok |ok, ok , (2.104a)O |ok, ok = ok |ok, ok . (2.104b)

This means that there is a common basis in which both observables are perfectly determined. In otherwords, for each state |ok, ok of the basis, the observable O, if measured, gives with certainty the eigenvalueok as outcome as well as the observable O

, if measured, gives with certainty the eigenvalue ok as outcome.Q.E.D

5


6/41

6 CHAPTER 2.

Change Eq. (2.123), p. 71, intox | x = x x | . (2.123)

Change Eq. (2.127) into:x |x| = x x | , (2.127)

Substitute the whole paragraph on ps. 72-73, starting with words In classical mechanics and ending with

the words up to a constant factor just after Eq. (2.133) by:

In classical mechanics, momentum is defined as the quantity which is conserved under global spatial trans-lations or, alternatively, as the generator of spatial translations. Let us then consider a onedimensionalparticle described by the wave function (x). A rigid translation by a quantity a of the particle will change(x) into

(x + a) = (x) + a

x +

a2

2

2

x2 +

a3

6

3

x3 + . . . +

an

n!

n

xn + . . .

=

j=0

aj

j!

j

xj(x))

= Ua(x) .

(2.130)

The unitary operator Ua [see Eq. (2.34)], translating the wave function of the system by an amount a, whichis called translation operator, can be written as

Ua = eaGT . (2.131)

As a consequence, the generator GT of the spatial displacements can be identified as

GT = x

(2.132)

up to a constant factor and must represent the quantummechanical momentum operator of the particle(see also Subsec. 3.5.4), i.e.

px =

x (2.133)

up to a constant factor.

Add the following footnote after the word into immediately before Eq. (2.130):

In the onedimensional case we could use the total derivative instead of the partial one. However, weprefer to use partial derivatives throughout the book.

Replace px in the line just before Eq. (2.139), p. 73, by px. Replace Eq. (2.140), p. 74, by:

px =h

= kx . (2.140)

Change Eq. (2.142) with:+

dx|p(x)|2 =

0 for px = 0+

|C|2dx = for px = 0 . (2.142)

Replace Eq. (2.143) by+

dxp(x)p (x) = |C|2+

dxe(p

xpx)x = 2|C|2 (px px) , (2.143)


7/41

7

In the line below Eq. (2.144) the factor (2) 12 should be replaced by (2) 12 . Similarly in Eqs. (2.145),(2.146), (2.150a), (2.150b), (2.163), (2.191), (2.194) and (2.215).

In Eq. (2.147) the factor 183

should be replaced by

1833

In Eqs. (2.159) and (2.161) replace d by dd . Replace Eq. (2.166) by

H|E = E|

E , (2.166)

In the line below Eq. (2.170) replace k =

p2/ by k =

p2x/.

At the end of Eq. (2.192) add a last line with the equality:= (px px) .

In the line below E. (2.193) replace |(x)|2 = (2)1 by |(x)|2 = (2)1

Replace Eq. (2.194) by(px) = x0(px) = 1

2e

pxx0 , (2.194)

In Figs. 2.5 and 2.6, pp. 8586, (2)1(pxpx) and (xx0) should read (2)1(pxpx) and 2(xx0).Moreover, the labels of Figs. 2.5(b) and 2.6(b) should read px instead of x. Finally, drop the symbol p

x

below Fig. 2.6(b).

Replace the whole paragraph starting with the words We assume that three lines before Eq. (2.201) andending with Eq. (2.206) with what follows:

We assume that BS1 has (in general, complex) transmission (T) and reflection (R) coefficients that forthe sake of simplicity we impose here to be real. These coefficients may be changed, still satisfying therelation T2 + R2 = 1. The BS1 transformation can be described as

| iBS1

T |1 + R |2 , (2.201)that is, a quantum superposition of the states photon in lower path and photon in upper path. Afterthe two mirrors and the phase shifter the state becomes

T |1 + eR |2 , (2.202)describing the fact that only the upper path acquires a phase factor e. The second beam splitter is assumedto be a 50%50% one. Its action may be described by the transformations

|1 BS2 12

(|3 |4 ) , |2 BS2 12

(|3 + |4) , (2.204)

so that we can write the outgoing state as

T

|1

+ eR

|2

BS2

1

2 T (|3 |4) + eR (

|3

+

|4

) .

Ordering the terms, the final state is

|f = 12

T + eR

|3 T eR |4 . (2.205)It is easy to calculate the final detection probabilities at the two detectors D3 and D4, which are given bythe square moduli of the amplitudes of the states |3 and |4 in the above expression, respectively, i.e.

3 =1

2

T + eR

T + eR

=

1

2+ TR cos ,

4 =1

2

T eR T eR = 1

2 TR cos .

(2.206)


8/41

8 CHAPTER 2.

LASER

BS1M2

M1

BS2

PS

BD'DA

D'A

BD

Substitute Fig. 2.11, p. 94, with the figure above. The solution of Problem 2.10 is 3N/16.

In Problem 2.17 replace + dx(px)2 = 1 by + dpx

(px)2 = 1. In Problem 2.19 replace Eq. (2.215) by

P(x) |px = 12

epxx |x . (2.215)


9/41

Chapter 3

Replace Eq. (3.49) bya

0

dx|n(x)|2 = 1 . (3.49)

In the fourth line of item (iii) in p. 110, which is the energy has to be negativeshould read

that is, for negative values of the energy

In the fifth line of the first paragraph in p. 111, the sentence beginning with the words For instance, Galileanrelativity and ending with the words quantum field theoryshould be replaced by:

For instance, Galilean relativity is the relativity attached to classical mechanics and quantum mechanics,whereas special relativity underlines relativistic quantum mechanics and quantum field theory.

In the last equality of Eq. (3.57) in p. 116, the pre-factor

2

ashould be

2

a.

Replace Eq. (3.68), p. 118, by:

O

OU=j o

j

U |oj oj |

U =

U

O

U . (3.68)

Replace the second sentence of Subsec. 3.5.2 (We wish now to show . . . unitary transformations), by whatfollows:

Generally speaking, the action of a beam splitter can be described by an inputoutput relation of thetype

|out = UBS | in ,

where the matrix U is given by

UBS = T R

R T

,

T and R being the (in general, complex) transmission and reflection coefficients, respectively. By imposingthe requirement of unitarity of the matrix UBS, we must have

1

R = R , T = T ,

which implies |R| = |R| and |T| = |T|. For the common case of a beam splitter that has the same effect ona beam incident through port 1 as on a beam incident through port 2, that is, a symmetric beam splitter,we have R = R and T = T. By imposing this condition together with that of a fiftyfifty beam splitter

1See Holbrow, C. H., Galvez, E., and Parks, M. E., Photon Quantum Mechanics and Beam Splitters, American Journal ofPhysics 70 (2002): 26065.

9


10/41

10 CHAPTER 3.

(|R| = |T|, which together with the previous one implies |R| = |R| = |T| = |T|), it is possible to prove thatthe unitary matrix above turns into

UBS =1

2

1 1

. (3.72)

Relabel Eqs. (3.72)-(3.73)-(3.74) as (3.73)-(3.74)-(3.75) (shift of one digit) and drop the paragraph startingwith the words For this reason and ending with Eq. (3.75).

Replace the line just after Eq. (3.77) together with Problem 3.12 (and its solution) with the following:

We remark that in the derivation of Subsec. 2.3.4 we have made use of two beamsplitter matrices withreal transmission and reflection coefficients, that is, of the matrix

T RR T

,

for BS1 and of the matrix

12

1 11 1

for BS2. These two matrices can derived from the general case by imposing the reality of T and R butcannot account for symmetric beamsplitting. Moreover, the latter describes a 50%50% beam splitter.

Replace Eq. (3.90) by ke H(tt0)j = G(k, t;j, t0) , (3.90)

Replace Eq. (3.93), p. 123, by

RH() =

+0

d e e

H , (3.93)

In the second line below Eq. (3.95) replace eta by .

Replace Eq. (3.96) byPj = 1

2

fj

dRH() , (3.96)

In the second line after Eq. (3.97) before the words In the case insert the following sentence:

In the continuous case the function R may have a cut such that the parts above and below it are notanalytical continuations one of the other through the cut.


11/41

Chapter 4

Delete the words are > 0 in the first line of p. 142. Replace Eq. (4.7) with

= arctank

kand = ka arctan k

k+ n . (4.7)

Replace the phrase where n is . . . and the following sentence (In order to address . . . ) after Eq. (4.7)by the phrase

where arctan(k/k) with range in the interval [/2, /2] is the principal value of the multivalued inversetangent function and n is an integer

Substitute the first sentence after Eq. (4.10) with:

The root k0 = 0 of Eq. (4.8) is excluded because the corresponding eigenfunction vanishes identicallyfor all values of x. The roots kn with n 1 determine the energy eigenvalues En = 2k2n/2m.

no change. Replace the words the parameters NI,NII,NIII and in Eqs. (4.3) by the words:

the parameters NI,NII, and NIII in Eqs. (4.3) Replace Eq. (4.14), p. 144, by:

(x) (b) =xb

dx(x) . (4.14)

Substitute any V and dV in Sec. 4.2 by V and dV, respectively. Replace k21 in Eq. (4.31) by k1. Moreover, substitute k1 in the line after Eq. (4.31) by k2. After Eq. (4.38) insert the sentence:

In order to find the first derivative of the wave function across the point x = 0, we take advantage of

II(0) I(0) = lim0+

dx(x) .

Substitute Fig. 4.9c by the figure in the next page. Change the label of Eq. (4.55a) into (4.55). In the final inequality in Eq. (4.63) change the symbol into >. The second line of Eq. (4.72) should be replaced by:

=1

2m

2x2n,n+1 + 2x2n,n1 +

2x2n,n+1 + 2x2n,n1

11


12/41

12 CHAPTER 4.

r

E

V(r)

0

r1 r2

(c)

In the line immediately below Eq. (4.61), p. 156, substitute the sentence Eq. (4.61) teaches us that the

energy levels of a harmonic oscillator are equally spaced (see Fig. 4.11) by the words:

Eq. (4.61), together with the fact that the energy levels of a one-dimensional harmonic oscillator arenot degenerate, teaches us that the energy levels En are equally spaced (see Fig. 4.11)

In the following sentence, the phrase for any value of j > 0 must be deleted. The sentence immediately below Eq. (4.67), p. 157, is an antisymmetric matrix since should be modi-

fied to:

nm is antisymmetric on its two indices since

Replace Eqs. (4.73), p. 158, by:

a =

m2

x + pxm

, (4.73a)

a =

m2

x pxm

, (4.73b)

Eq. (4.78) must be modified into:

aa = m2

x pxm

x + pxm

= m2

2x2 + m xpx m pxx + p

2x

m2

= m2

2x2 + m [x, px] +

p2xm2

= 1

12

m2x2 +p2x2m

12

,

(4.78)

Substitute x in Prob. 4.10 into px. Replace Eq. (4.97) by:

n() =

14 e

2

2

2n2

n!

Hn() , (4.97)

The text of Prob. 4.14 should be changed into

Verify that the groundstate wave function of the harmonic oscillator is normalized.

Drop the solution of Prob. 4.14.


13/41

13

Replace Eq. (4.112) byV(x) = F x + C . (4.112)

Replace Eq. (4.114) by(x) =

x +

E

F

2m|F|

2

13

, (4.114)

Replace Eq. (4.115) by () + sgn(F) () = 0 , (4.115)

Replace Eq. (4.119) by(x) =

x +

E

F

2m|F|

2

13

. (4.119)

Replace Eq. (4.121) by

=

x +

EnF

2m|F|

2

13

(4.121)

Replace Eq. (4.122) by

() () = 0 , (4.122)

Replace Eq. (4.123) by () = NA () (4.123)

In the line just after Eq. (4.123) replace N by N. After the words Lorentz force just before Eq. (4.127) insert the following footnote:

Here and throughout this book we make use of the SI system of units.

Eq. (4.127), p. 168, must be substituted by

dpdt

= eE + p

mB , (4.127)


L =1

2mr2 + e (r A U) . (4.129)

Eq. (4.130) on page 168 should read:

H =j

qjL

qj L = r L

r L , (4.130)

Eq. (4.131) on page 168 should read:

H =p

m (p + eA) p22m

e

mp A + eU

=1

2mp2 + eU

=1

2m(P eA)2 + eU ,

(4.131)


P =L

r= p + eA (4.132)


14/41

14 CHAPTER 4.

After Eq. (4.132), p. 169, the words we replace the generalized momentum by the operator p = , sothat must be replaced by the words:

we impose the canonical commutation relation [rj , Pk] = jk (see Eq. (2.174)) and replace the gener-

alized momentum P by the operator , so that

Drop the footnote 11, p. 169.

Eq. (4.133), p. 169, must be replaced by

H =1

2m( eA)2 + eU . (4.133)


t(r, t) =

1

2m[ eA(r, t)]2 (r, t) + eU(r, t)(r, t) , (4.134)


t (r, t) = P2

2m e2mP A + A P+ e22mA2 + eU(r, t)

=

2

2m +

e

2m A + e

mA + e

2

2mA2 + eU

(r, t) .

(4.135)

The last sentence of Sec. 4.5.3 and the following Eq. (4.136), p. 169 must be replaced by:

In the last expression we have taken into account the fact that P and A do not commute, and their(scalar) commutator is given by (see Prob. 2.26)

P,A

= P AA P = A . (4.136)

In Prob. 4.5 substitute the exponential eEnt

with eEnt

.

In Prob. 4.7 change the phrase in any of the energy eigenstates into in any energy eigenstate. In Prob. 4.8, change the phrase the eigenstates of the discrete spectrum into the energy eigenstate |n

of the discrete spectrum.

Change Prob. 4.13 into

Prove the relation |n =

an

n!

|0 , where |n is the energy eigenstate of the harmonic oscillator withenergy eigenvalue En = (n + 1/2).


15/41

Chapter 5

In the first line after Eq. (5.20), p. 178, replace the word where by and if. In Eq. (5.32b) substitute on the left side the term SVN with S. In the second line of p. 182 replace the words the direct sum H1 H2 by the words:

the direct product

H1

H2

In the third line from the bottom of p. 186 change the expression HS+S = HSHS into HS+S = HSHS . Replace the sentence containing Eq. (5.35), p. 182 ( starting with the words A complete basis for the total

system of the two photons will then be given by the direct product and ending with the words each basiselement ofH2) by:

A complete basis of the direct product Hilbert space H1 H2 for the total system of the two photonswill then be given by the direct product basis

{|h1 |h2 , |h 1 |v2 , |v 1 |h2 , |v 1 |v2} , (5.35)

which is obtained by performing the direct product of each basis vector of H1 and each basis vector of H2.

After Eq. (5.40), p. 183, insert the words:In Eq. (5.40) and in the following repeated indices are dropped.

Just after Eq. (5.43) insert the words:

For the sake of notational simplicity, the direct product sign will be omitted here and henceforth ex-cept where confusion may arise.

Replace the last line of Eq. (5.46), p. 184, by:

=j,l,n

cjncln |j l | , (5.46)

15


16/41

16 CHAPTER 5.


17/41

Chapter 6

Eq. (6.1), p. 193 must be replaced byL = r p , (6.1)

Replace the words where R = n l is called the generator of the rotation. after Eq. (6.10) on p. 196, withthe words:

where R = n l is called the generator of the rotationabout the direction given by the vector n. Replace R2 with R in the third line after Eq. (6.12) on page 197. Transform the text of footnote 1 from See [Byron/Fuller 196970, 11]. to

We follow for the Euler angles the z y zconvention [Byron/Fuller 196970, 11]. In the first line of Subsec. 6.1.4, p. 200, substitute the words angular momentum operator. 4 with

angular momentum operator4.

In the third line of Subsec. 6.1.4, substitute the words only halfinteger values of l are suitable with thewords

only integer or halfinteger values of l are suitable

Substitute the words generates the rotations about z in the first line after Eq. (6.34), p. 201, with thewords

generates the rotations about the z axis

Replace Eq. (6.48), p. 203 withl2 |0, 0 = 0 . (6.48)

P. 206, second line after Eq. (6.70), replace the words around xand yaxes with the words

around x and yaxes

Replace Eq. (6.71) with what follows:

xy

r2,

yz

r2,

zx

r2,

3z2 r2r2

,x2 y2

r2, (6.71)

corresponding to dxy, dyz , dxz, dz2 , and dx2y2 , respectively.

Replace Eq. (6.88), p. 210, with

pr = 1r

rr =

r+

1

r

, (6.88)

17


18/41

18 CHAPTER 6.

Replace Eq. (6.98), p. 211, withlimr0

r2V(r) = 0 , (6.98)

Replace Eq. (6.102), p. 212, withA = B

2y +

B

2x + 0k , (6.102)


H =1

2m

px +

eB

2y

+

py eB

2x

+ pzk

2=

p2z2m

+1

2m

p2x + p

2y

eB2m

lz +e2B2

8m

x2 + y2

.

(6.103)

Eq. (6.104), p. 212 must be replaced by

Hr =1

2m

p2x + p

2y

eB2m

lz +e2B2

8m

x2 + y2

, (6.104)

Eq. (6.110), p. 213, must be replaced by(2n + 1) eB

2m(6.110)

Eq. (6.111) must be replaced byEn =

2k2z

2m+ (2n + 1)

eB

2m. (6.111)

Substitute ekr2 with er2/2 in the first line after Eq. (6.131) as well as in Eq. (6.132), p. 217. Replace Eq. (6.136), p. 219, with

p 11000

e , (6.136)

Replace Eq. (6.143), p. 220, with

(r, s) =

(r)(r)

= (r)

10

+ (r)

01

. (6.143)

Replace Eq. (6.146), p. 221 with

(r, s) = r | = r | | z + r | | z . (6.146)

Replace Eq. (6.148), p. 221 withsz =

1

2

1 00 1

. (6.148)

Replace the expressions electric dipole momentum, magnetic dipole momentum and magnetic mo-mentum used in Sec. 6.3 (and in the caption of Fig. 6.10) by electric dipole moment, magnetic dipole

moment and magnetic moment, respectively.

In the line immediately below Eq. (6.160), replace the phrase and, ifO is real, also and must be real by

and O is Hermitian if and only if and are real

In the sentence just before Eq. (6.161) replace the reference (see Eq. (6.145)) by (see Eqs. (6.143) and(6.145)).

After Eq. (6.162) add the words:

where the transmission and reflection coefficients T and R (see Subsecs. 2.3.4 and 3.5.2) are absorbedinto the amplitudes 1 and 2.


19/41

19

Replace Eqs. (6.165) by

sx =1

2

0 1 01 0 1

0 1 0

, sy = 1

2

0 0 0

0 0

, sz = 1

2

1 0 00 0 0

0 0 1

. (6.165)


l

=e

2ml (6.166)

Eq. (6.167), p. 224, must be replaced byB =

e

2m(6.167)

Eq. (6.169), p. 225, must be replaced bys = g

e

2ms (6.169)


H =

1

2m (px + eBy)2

+

p2y2m +

p2z2m szB , (6.172)

Eq. (6.173), p. 225, must be replaced by = g

e

2m. (6.173)

Eq. (6.174), p. 226, must be replaced by1

2m(px + eBy)

2+

p2y2m

+p2z

2m szB

(r) = E(r) . (6.174)

Eqs. (6.177), p. 226, must be replaced by

B

= eBm

and y0 = pxeB

. (6.177)

P. 228, Eq. (6.182): replace all occurrences of l1,2 by l as well as of l1,2 by l. Do the same in the whole

subsection (for instance the whole paragraph after Eq. (6.184).

Replace Eqs. (6.191) by:

Jx =

2

0 1 01 0 1

0 1 0

, Jy =

2

0 0 0

0 0

, Jz =

1 0 00 0 0

0 0 1

, J2 = 22

1 0 00 1 0

0 0 1

.

(6.191)

In Subsec. 6.4.1 all the occurrences ofl1,2 should be replaced by l as well as all the occurrences of l1,2 shouldbe replaced by l.

Replace |+1, 1 1 |2, 1 2 in caption of Fig. 6.13 by |1, 11 |2, 12. Insert a right parenthesis at the end of the second line of Prob. 6.10, p. 242. Replace Eq. of Prob. 6.25, p. 243 by:

( f) ( f) = (f f) I+ (f f) ,


20/41

20 CHAPTER 6.


21/41

Chapter 7

In the caption of Tab. 7.1, p. 250, replace the expressions the particle and particles by the lepton and leptons, respectively.

Replace the expression En1 En2 in the second line of p. 251 by:En1 and En2

Replace the expression where l1 , l2 , . . . , lN in the line immediately below Eq. (7.17), p. 251, by:

where l1 , l2 , . . . , lN

Replace the formula k!/(k n)! in the fifth line below Principle 7.2, p. 251, by:k!/(k N)!

Replace the term disposition in the 5th, 7th and 8th lines below Principle 7.2 and in Footnote 11, p. 251,by the term permutation.

Replace Eq (7.21), p. 253, by:

(1, 2, . . . , N) = (l1 , l2 , . . . , lN) , (7.21)

Replace Eq. (7.24), p. 254, by:Ylm(, ) Ylm( , + ) = (1)lYlm(, ) . (7.24)

Replace Eq. (7.26), p. 255, by:

T =

h2

2mkBT, (7.26)

In the 8th line of Subsec. 7.5.2, replace formula V(r) 1/r byV(r) 1/|r|

21


22/41

22 CHAPTER 7.


23/41

Chapter 8

Replace Eq. (8.4), p. 260, by

| UBS | =

=1

2(|1 + |2) . (8.4)

Replace Ee. (8.5), p. 261, by:1 = 12

(|1 |2) ,2 = 1

2(|1 + |2 ) . (8.5)

In the line immediately below Eq. (8.20), p. 264, for an arbitrary eigenstate ketx should readfor an arbitrary eigenstate |x

Replace Eq. (8.22), p. 264, by:Ux(a)xU

x(a) = x a , (8.22)

Replace Eq. (8.27), p. 265, byUp(v) = e

vr (8.27a)

orUp(v)pU

p(v) = p v . (8.27a)

Replace Eq. (8.30), p. 266, by:UR() = e

nJ , (8.1)

Change the title of Subsec. 8.3.1 from Space reflection into Space reflection or parity. In the last line of the last paragraph of Subsec. 8.3.1, p. 267, the sentence time reversal operator commutes

with the generator of rotations. should be replace by

time reversal operator anticommutes with the generator of rotations.

Replace the sentence A noticeable example of isomorphism is provided by the groups C. and D. of theprevious section by

A noticeable example of isomorphism is provided by the groups 3. and 4. of the previous subsection

Replace the words check the correctness if this statement immediately above Eq. (8.39), p. 269, by:check the correctness of this statement

Replace Eq. (8.43), p. 270, by:3

b=1

RTabRbc = ac. (8.43)

23


24/41

24 CHAPTER 8.

Replace Eq. (8.44), p. 270, by(R t)abc =

3d,e,f=1

RadRbeRcftdef . (8.44)

Replace Eq. (8.45), p. 270, by:3

a=1taab =

3a=1

taba =

3a=1

tbaa = 0 . (8.45)

Replace Eq. (8.48), p. 271, by:r (t + v) =

db=1

rab(tb + vb) . (8.48)

Replace Eq. (8.52), p. 273, by:d

a,b=1

RTaagabRbb = gab , (8.52)

which for gab = ab and d = 3 reduces to Eq. (8.43).

Replace Eq. (8.53), p. 273, by:gab = abfa , (8.53)

Replace the sentence This group is usually called O(d1, d2), where d1 + d2 = 1. by:

This group is called the indefinite orthogonal group O(d1, d2), where d1 + d2 = d.

In the second last line of the third last paragraph in p. 273, g(0) = 1 should be replaced by g(0) = 1. Replace Eq. (8.55), p. 274, by:

m =

3i=1

Rivi , (8.55)

Eq. (8.56), p. 274, should be replaced by:

[Ti, Tk] =l

CiklTl , (8.56)

Replace Eq. (8.57), p. 274, by:C r = Ar , (8.57)

Replace Eq. (8.58), p. 274, by:C = l2x + l

2y + l

2z (8.58)

Replace the displayed equation in Problem 8.1, p. 275, by:

|1

=

1

2 1 + 2 ,

In Further Reading, p. 276, replace Eisberg, R./Resnick, R. by Eisberg, R. and Resnick, R..


25/41

Chapter 9

The sentence immediately above Eq. (9.9) (In general, there will be a unitary operator describing theevolution during the interaction time , i.e. ) should be replaced by:

If the interaction Hamiltonian dominates during the interaction time over the free terms of theHamiltonian, the unitary operator describing the evolution during the interaction time may bewritten as

Three lines above Eq. (9.26), p. 284, replace U

SAby USA .

In the line immediately below Eq. (9.42c), p. 298, replace {|ej} by {|ej}. Replace the text of Footnote 49, p. 311, by:

See Box 13.1: p. 494.

In the line immediately below Eq. (9.81), p. 313, replace |1102 |0112 by |1102, |0112. In the third line from the end of the first paragraph, p. 315, photosensory should be replaced by photo-

sensor.

Drop the Phase shifter PS in Fig. 9.12, p. 316. Eq. (9.91), p. 316, should be replaced by:

|i BS1 12

(|1 + |2) M1,M2 12

(|1 + |2)BS2 1

2(|3 |4) + 1

2(|3 + |4) = |3 , (9.91)

Drop the words As usual, upon reflection the state vector acquires an imaginary factor. three lines afterEq. (9.91).

Replace Eq. (9.92), p. 316, by:

|i BS1 12

(|1 + |2) O 12

(|1 + |a)M1,M2

1

2(

|1

+

|a

)

BS2

1

2

(

|3

|4

) +

1

2 |a

, (9.92)

Replace the formula in Footnote 66, p. 324, by:

limn

1 +

x

n

n= ex

In the line immediately below Eq. (9.104), p. 325, replace o|SS () = (o) by o|SS () 2= (o). In the line immediately below Eq. (9.112), p. 328, substitute = Sup {Tr [T()]} with

T = Sup {Tr [T()]} Replace Elzevier by Elsevier in the second item of QND measurement, p. 354.

25


26/41

26 CHAPTER 9.


27/41

Chapter 10

27


28/41

28 CHAPTER 10.


29/41

Chapter 11

Eq. (11.22), p. 407, must be replaced byB =

e

2m. (11.22)

In Figure 11.4, p. 412, the r axis is misplaced. It should be moved down so that it become the asymptoteof the solid line.


Bp = E v = Zer3r v = Ze

me

1

r3L = Bpk , (11.57)

Eq. (11.58), p. 415, must be replaced bys =

1

2

e

meS , (11.58)


V = s Bp = Ze2

2m2er3L S = 1

2

Zc

m2e

L Sr3

, (11.59)


(0)nlm |V| (0)nlm

=

Ze2

2m2e

L S

ang,spin

1

r3

rad

, (11.63)


E(1)nlm =

Z4ce6

25n3l

l + 12

(l + 1)

L S

ang,spin

= l sang,spin

, (11.65)


=Z4ce6me

23n3l

l + 12

(l + 1)=

Z44mec4

2n3l

l + 12

(l + 1), (11.66)


H = HA e2me

l + 2s

Bext f(r) l s , (11.68)

Eq. (11.69), p. 416, must be replaced byf(r) =

Ze22

2m2er3

. (11.69)

29


30/41

30 CHAPTER 11.


H = HA e2me

Bext

lz + 2 sz

f(r) l s = H0 + HI , (11.71)

Eq. (11.72) must be replaced by

H0 = HA e2me

Bext

lz + 2sz

, (11.72)


H0nlmlms =

HA e

2meBext

lz + 2 sz

nlm

lms

=

E(0)nl BBext (ml + 2ms)

nlm

lms , (11.75)


HI =

e

2meBext lz + 2 sz . (11.80)


j = l cos + s cos =e

2me(l cos + 2s cos ) , (11.82)


j =e

2me

j2 + l2 s2

2j+ 2

j2 + s2 l22j

=

e

2megLj , (11.86)

Replace the words Gauss, law in the immediately immediately above Eq. (11.97), p. 424, by:

the Gauss law


31/41

Chapter 12

31


32/41

32 CHAPTER 12.


33/41

Chapter 13

33


34/41

34 CHAPTER 13.


35/41

Chapter 14

35


36/41

36 CHAPTER 14.


37/41

Chapter 15

37


38/41

38 CHAPTER 15.


39/41

Chapter 16

39


40/41

40 CHAPTER 16.


41/41

Chapter 17

At the end of Footnote 2 add the words:

The only difference with formulation (5.33) is that here we have taken the Boltzmann constantkB = 1. The reason is that we deal here with pure informational aspects and not with thermody-namical or statisticmechanics considerations.

errata cor ridge

Documents