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Solution Manual for Modern Condensed Matter Physics, 1st Edition

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    Solutions to Exercises in Modern Condensed Matter Physics S. M. Girvin and Kun Yang [Compiled: August 12, 2019]

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    Chapter 2 Ex. 2.1 (i) The radiated electric field is ~  a ≈ r e e ikR D R D [ ˆ n × (ˆ n × ~ E in ) ] e − iωt e − i~ q · ~ r . (1) Replace e − i~ q · ~ r by 〈 e − i~ q · ~ r 〉 = f ( ~ q ). The vector (ˆ n × ~ E in ) is perpendicular to both ˆ n and to ~ E in and has length | E in sin θ | . Hence | ˆ n × (ˆ n × ~ E in ) | 2 = E 2 in sin 2 θ . Thus |  a | 2 = E 2 in r 2 e R 2 D sin 2 θ | f ( ~ q ) | 2 . (2) The total radiated power passing through a sphere of radius R D is P = cR 2 D ∫ d Ω (  2 a 8 π × 2 ) (3) = c 4 π r 2 e E 2 in ∫ d Ω sin 2 θ | f ( ~ q ) | 2 . (4) Let us normalize the incident electric field to that associated with a single photon in the normal- ization volume L 3 E 2 in 4 π = ℏ ω L 3 = ℏ ck L 3 (5) which yields P = ℏ c 2 k r 2 e L 3 ∫ d Ω sin 2 θ | f ( ~ q ) | 2 . (6) (ii) Now compare this to the quantum result using the photon scattering matrix element in Eq. (2.28) M = r e f ( ~ q ) ∧ 2 k ˆ  ~ kλ · ˆ  ~ k ′ λ . (7) 2

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    3 Fermi’s Golden Rule for the transition rate is Γ = 2 π ℏ ∑ λ ′ L 3 (2 π ) 3 ∫ d 3 k ′ r 2 e ∧ 4 k [ˆ  ~ kλ · ˆ  ~ k ′ λ ] 2 δ ( ℏ ω − ℏ ck ′ ) | f ( ~ q ) | 2 . (8) Noting that the two polarization vectors ˆ  ~ k ′ λ ′ and the vector ˆ k ′ are all mutually perpindicular, we find ∑ λ ′ [ˆ  ~ kλ · ˆ  ~ k ′ λ ] 2 = 1 − [ˆ  ~ kλ · ˆ k ′ ] 2 = 1 − cos 2 θ = sin 2 θ . The radiated power is P = ℏ c 2 k r 2 e L 3 ∫ d Ω sin 2 θ | f ( ~ q ) | 2 , (9) in agreement with the result from the semiclassical calculation. Ex. 2.2 S ( ~ q ) = 1 N < | W ( ~ q ) | 2 > = 1 N < N ∑ i =1 e i~ q.~ r i N ∑ j =1 e − i~ q.~ r j > (10) = 1 N < N ∑ i = j e i~ q.~ r i − i~ q.~ r j > + 1 N < N ∑ i 6 = j ∫ d 3 ~ rd 3 ~ r ′ e i~ q.~ r i − i~ q.~ r j δ ( ~ r − ~ r i ) δ ( ~ r ′ − ~ r j ) > (11) = 1 N N + 1 N ∫ d 3 ~ rd 3 ~ r ′ e i~ q · ( ~ r − ~ r ′ ) < N ∑ i 6 = j δ ( ~ r − ~ r i ) δ ( ~ r ′ − ~ r j ) > (12) Remembering < N ∑ i 6 = j δ ( ~ r − ~ r i ) δ ( ~ r ′ − ~ r j ) > = n (2) ( ~ r ′ i − ~ r j ) , then obviously S ( ~ q ) = 1 + 1 N ∫ d 3 ~ rd 3 ~ r ′ e i~ q · ( ~ r − ~ r ′ ) n (2) ( ~ r ′ − ~ r ) = 1 + n ∫ d 3 ~ re i~ q.~ r g ( ~ r ) where we used N/V = n and n (2) ( ~ R ) = n 2 g ( ~ R ). P.S. ” <> ” indicates thermal average in liquid or amorphous materials. It is unnecessary only for perfect lattices. Generally ” <> ” must be in the formula.

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    Chapter 3 Ex. 3.1 • SC FCC BCC radius a /2 √ 2 4 a √ 3 4 a volume of one sphere π 6 a 3 √ 2 π 24 a 3 √ 3 π 16 a 3 number of sites in unit cell 1 4 2 volume fraction π 6 √ 2 6 π √ 3 8 π Ex. 3.2 F CC 8 × 1 8 + 6 × 1 2 = 4 BCC 8 × 1 8 + 1 = 2 Ex. 3.3 (i) Suppose the lattice spacing is a . The three primitive vectors are ~ a 1 = a (0 , 1 2 , 1 2 ) ~ a 2 = a ( 1 2 , 0 , 1 2 ) ~ a 3 = a ( 1 2 , 1 2 , 0) Thus, the coordinates of the four points of the tetrahedron spanned by the three vectors are O = (0 , 0 , 0) A = a (0 , 1 2 , 1 2 ) B = a ( 1 2 , 0 , 1 2 ) C = a ( 1 2 , 1 2 , 0) 4

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    5 By calculating the distance of any two points, we can prove that the edges of the tetrahedron are equal. So it is a regular tetrahedron. (ii) Another lattice site (on the opposite sublattice) locates at P = a 4 (1 , 1 , 1) The distance of P to each corner of the tetrahedron is P O = a | ( 1 4 , 1 4 , 1 4 ) | P A = a | ( 1 4 , − 1 4 , − 1 4 ) | P B = a | ( − 1 4 , 1 4 , − 1 4 ) | P C = a | ( − 1 4 , − 1 4 , 1 4 ) | We get P O = P A = P B = P C = √ 3 a 4 . Therefore, P is at the geometrical center of this tetrahedron. Ex. 3.4 (i) ~ R ~ m ± ~ n = ( m 1 ~ a 1 + m 2 ~ a 2 + m 3 ~ a 3 ) ± ( n 1 ~ a 1 + n 2 ~ a 2 + n 3 ~ a 3 ) = ( m 1 ± n 1 ) ~ a 1 + ( m 2 ± n 2 ) ~ a 2 + ( m 3 ± n 3 ) ~ a 3 Indeed have the form of (3.15) and are lattice vectors characterized by ~ m ± ~ n . (ii) Start in 1D. Pick the lattice site closest to the origin, whose distance from the origin is a . Claim: all sites satisfying 1D version of (3.16) can be written as R m = ma with m being an integer. Proof: Assume R ′ = α a is a lattice site with α being a non-integer. [ α ] represents the integer part of α , and ∆ α = α − [ α ] is its fractional part. Thus 0 < ∆ α < 1. From (3.16), we know [ α ] a and thus R ′ − [ α ] a = ∆ α a is also a lattice site, but its distance to the origin is less than a , leading to a contradiction. For 2D, let us look for the closest site to the origin, located at ~ a 1 . This immediately gives us a lattice line, m~ a 1 , with m being an integer. Any lattice site ~ R 6 = m~ a 1 , gives us a parallel lattice line, ~ R + m~ a 1 . Look for ~ R = ~ a 2 such that ~ a 2 + m~ a 1 is the lattice line closest to the line m~ a 1 . Then m~ a 1 + m 2 ~ a 2 are all lattice sites. Now assume α 1 ~ a 1 + α 2 ~ a 2 is also a lattice site, with ∆ α 1 > 0 and ∆ α 2 > 0. Then we know ( m 1 + ∆ α 1 ) ~ a 1 + [ α 2 ] ~ a 2 forms a lattice line, which is closer to the m~ a 1 line than the ~ a 2 + m~ a 1 line! Contradiction again. It is now obvious how to generalize this to 3D. Ex. 3.5 For a Bravais lattice, it has a set of primitive vectors, ~ a i . The locations of all lattice points could be expressed as

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    6 ~ R ~ c = ∑ i c i ~ a i , c i ∈ Z (integer) . The mid-point of any two lattice sites, say ~ R ~ m and ~ R ~ n is ~ R mid = 1 2 ( ~ R ~ m + ~ R ~ n ) = ∑ i 1 2 ( m i + n i ) ~ a i where m i and n i are integers. We can always shift the origin of the coordinate of the lattice to this mid-point so that all lattice points have new coordinates as ~ R ′ ~ c = ~ R ~ c − ~ R mid = ∑ i [ c i − 1 2 ( m i + n i ) ~ a i ] (13) If ~ R mid is an inversion center, given a lattice site ~ R ′ ~ c , − ~ R ′ ~ c must be a lattice point as well. Namely, there exists a set of p i which are integers such that − ~ R ′ ~ c = ~ R ′ ~ p = ∑ i [ p i − 1 2 ( m i + n i ) ] ~ a i (14) Combining Eq. (13) and Eq. (14), we get p i = ( m i + n i ) − c i ∈ Z. This tells us that ~ R mid is indeed an inversion center. Ex. 3.6 3 - fold axis 6 - fold axis Triangular Lattice

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    7 3 - fold axis 6 - fold axis Honeycomb Lattice If the atoms on A and B sublattice are different, C 6 symmetry will be broken. The original 6-fold axis will become a 3-fold axis. And the original 3-fold axis is still 3-fold. Ex. 3.7 Ex. 3.8 A diamond structure could be viewed as a FCC lattice with a basis containing two atoms called A and B. A and B have a a 4 (1 , 1 , 1) shift where a is the lattice constant. If we take the mid-point of A and B as an inversion center, the positions of the sublattices will exchange applying to all the lattice sites. As a result, the lattice is unchanged if the atoms on different sublattice sites are the same; otherwise, the lattice is not centrosymmetric. Therefore, diamond structures are centrosymmetric, but Zincblende lattices are not. Ex. 3.9 Suppose the lattice has n -fold symmetry. Then rotating by α = 2 π n about the origin (assumed to be a lattice site) should leave the lattice invariant. Assume ~ a 0 is the shortest lattice vector connecting

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    8 the origin to one of its neighbors. After a rotation of α = ± 2 π 5 , it becomes ~ a 1 and ~ a 2 respectively, which should be lattice vectors themselves (see figure). Then ~ a 1 + ~ a 2 should also be a lattice vector. But simple trigonometry finds it is shorter than ~ a 0 (see figure), which leads to contradiction. Thus 5-fold symmetry is not allowed in 2D. Since 3D lattices are made of parallel 2D planes, this implies such symmetry is impossible in 3D as well. 2 Π ê 5 : lattice site : 5 - fold axis a 1 a 2 a 0 a 0 = a 1 + a 2 Ex. 3.10 The construction of 1D resiprocal lattice { ~ b j } with ~ b j · ~ a m = 2 πδ mj : ~ b · ~ a = 2 π ⇒ ~ b = 2 π a ˆ x. The reciprocal lattice vectors are ~ G = m ~ b = m 2 π a ˆ x. ( m =integer) 1st BZ: [ − π/a, π/a ) 2nd BZ: [ − 3 π/a, − π/a ) , [ π/a, 3 π/a ) n th BZ: [(1 − 2 n ) π/a, (3 − 2 n ) π/a ) , [(2 n − 3) π/a, (2 n − 1) π/a )

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    9 Ex. 3.11 1) FCC ~ a 1 = a 2 (ˆ y + ˆ z ) = (0 , a/ 2 , a/ 2) ~ a 2 = a 2 (ˆ x + ˆ z ) = ( a/ 2 , 0 , a/ 2) ~ a 3 = a 2 (ˆ x + ˆ y ) = ( a/ 2 , a/ 2 , 0) w = ~ a 1 . ( ~ a 2 × ~ a 3 ) = a 3 8 + a 3 8 = a 3 4 ~ b 1 = 2 π w ( ~ a 2 × ~ a 3 ) = 2 π a ( − 1 , 1 , 1) ~ b 2 = 2 π w ( ~ a 3 × ~ a 1 ) = 2 π a (1 , − 1 , 1) ~ b 1 = 2 π w ( ~ a 1 × ~ a 2 ) = 2 π a (1 , 1 , − 1) This is indicative that ( ~ b 1 , ~ b 2 , ~ b 3 ) forms a BCC lattice with lattice constant 4 π a . 2) BCC ~ a 1 = a 2 (ˆ y + ˆ z − ˆ x ) = ( − a/ 2 , a/ 2 , a/ 2) ~ a 2 = a 2 (ˆ x + ˆ z − ˆ y ) = ( a/ 2 , − a/ 2 , a/ 2) ~ a 3 = a 2 (ˆ x + ˆ y − ˆ z ) = ( a/ 2 , a/ 2 , − a/ 2) w = ~ a 1 . ( ~ a 2 × ~ a 3 ) = a 3 2 ~ b 1 = 2 π w ( ~ a 2 × ~ a 3 ) = 2 π a (0 , 1 , 1) ~ b 2 = 2 π w ( ~ a 3 × ~ a 1 ) = 2 π a (1 , 0 , 1) ~ b 3 = 2 π w ( ~ a 1 × ~ a 2 ) = 2 π a (1 , 1 , 0) ( ~ b 1 , ~ b 2 , ~ b 3 ) constructs a FCC lattice with lattice constant 4 π a .

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    10 Ex. 3.12 For a direct lattice ~ R = ∑ i n i ~ a i whose reciprocal lattice ~ K = ∑ i k i ~ b i , we have e i ~ R · ~ K = e ∑ n i ~ a i . ∑ k j ~ b j = 1 . Let us call the reciprocal lattice of ~ K ~ R ′ , then we have e i ~ K · ~ R ′ = 1, thus ~ R ′ = ~ R . Ex. 3.13 (a) Begin with a 1D array of disks of radius r 0 . If the centers of the disks are on the 1D lattice { ~ R j = jd (1 , 0 , 0); j ∈ Z } where d = 2 r 0 , then the disks are just touching as shown in Fig. 1a. a) b) d d d Figure 1 Now consider adding a second line of atoms as shown in Fig. 1b with lattice positions { ~ R j = d ( j + δ, y, 0); j ∈ Z } . The lowest possible allowed value of y (and hence the densest lattice) occurs for horizontal displacement δ = 1 / 2. At y min = √ 3 2 each disk in the second row touches two disks in the first row. Extending this to an arbitrary number of rows yields the triangular lattice with lattice vectors ~ a 1 = d (1 , 0 , 0) and ~ a 2 = d ( 1 2 , √ 3 2 , 0). (b) Now consider the triangular lattice A defined by points ~ R jk = j~ a 1 + k~ a 2 as shown in the left panel of Fig. 2. U D U D Figure 2

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    11 We see that each unit cell contains an upward facing triangle (labelled U ) and a downward facing triangle (labelled D ). We can choose to place a second identical layer B on top of the first positioned either with the lattice sites located in the center of the U triangles or the D triangles of the first layer. Both choices result in a honeycomb lattice as shown in the right panel of Fig. 2 (where we have arbitrarily chosen to center on the U triangles). If the lower left corner of the triangle labelled U is taken as the origin, the center of the triangle is located at the ‘center of mass’ position ~ r c = 1 3 [ ~ 0 + ~ a 1 + ~ a 2 ] = d ( 1 2 , 1 2 √ 3 , 0). If each lattice point is the center of a hard sphere of radius r 0 = d/ 2, the spheres in a given layer are close-packed. If the second layer is raised a distance λd above the first, we can define a third vector ~ b = ~ r c + λd (0 , 0 , 1) connecting this point to the origin. The close-packed hard-sphere constraint requires | ~ b − ~ 0 | = | ~ b − ~ a 1 | = | ~ b − ~ a 2 | = d which implies λ = √ 2 3 . This minimum vertical displacement required to satisfy the hard-sphere constraint maximizes the packing density. (c) Calling the lower lattice A and the upper lattice B, we can repeat the stacking in the pattern ABABAB...AB to form a hexagonal close-packed (HCP) lattice. The vectors ~ a 1 , ~ a 2 defined above constitute two of the lattice vectors for the hcp lattice. The vector ~ b defined above is not a lattice vector because (for example) there is no lattice site at 2 ~ b . ~ b simply defines the location of a second atom in the unit cell of this non-Bravais lattice. Under the ABABAB...AB repetition, the third plane matches the A plane and hence the lattice vector lies in the z direction and is elevated twice as high as the B plane. It thus has length 2 ~ b · ˆ z in order to satisfy the hard-sphere constraint: ~ a 3 = d (0 , 0 , 2 √ 2 3 ). (d) If the third layer is centered on the other choice (downward facing triangles) we can repeat this ABCABCABC...ABCC stacking to obtain the face-centered-cubic (FCC) lattice. In this case the third lattice vector is ~ a 3 = ~ b . In this case ~ b , defined above, is a lattice vector because the ABC stacking places an atom at both ~ b and 2 ~ b , unlike the case of AB stacking. To see that this is indeed the FCC lattice, notice that ~ a 1 · ~ a 2 = ~ a 1 · ~ a 3 = ~ a 2 · ~ a 3 = 1 2 d 2 indicating that there is a 60 degree angle between each of the three pairs of vectors just as there is for the standard basis vectors for the FCC lattice: ~ A 1 = ∆(0 , 1 2 , 1 2 ) ~ A 2 = ∆( 1 2 , 0 , 1 2 ) ~ A 3 = ∆( 1 2 , 1 2 , 0) ~ A 1 · ~ A 2 = ~ A 1 · ~ A 3 = ~ A 2 · ~ A 3 = 1 2 d 2 , where ∆ ≡ √ 2 d is the length of the side of the cube defining the conventional unit cell (i.e., the second-neighbor distance). There exists an orthogonal rotation matrix that preserves the angles between the vectors while effecting the coordinate change needed to map ~ a j → A j for j = 1 , 2 , 3. Hence the lattice is FCC.

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    12 (e) There are an infinite number of ways to do the stacking by randomly choosing each layer to be A, B, or C, subject to the constraint that two adjacent layers cannot be the same. For example: ABACBABCBACABCBABABABCBABCBC. Ex. 3.14 For diamond lattice the two point basis can be written as ~ τ 1 = 0, ~ τ 2 = a 4 (ˆ x + ˆ y + ˆ z ). Scattering is possible for momentum transfer ~ q = ~ G = K ~ b 1 + L ~ b 2 + M ~ b 3 where ~ b 1 = 2 π a ( − ˆ x + ˆ y + ˆ z ) ~ b 2 = 2 π a (ˆ x − ˆ y + ˆ z ) ~ b 1 = 2 π a (ˆ x + ˆ y − ˆ z ) are primitive reciprocal lattice vectors of FCC. The scattering amplitude is proportional to f ( ~ q ) = 2 ∑ s =1 e − i~ q · ~ τ f s ( ~ q ) (15) = e − i~ q · ~ τ 1 f 1 ( ~ q ) + e − i~ q · ~ τ 2 f 2 ( ~ q ) . (16) For identical atoms on the two sublattices f 1 ( ~ q ) = f 2 ( ~ q ) = f a ( ~ q ), where f a is the atomic form factor. Then f ( ~ G ) = (1 + e − i ~ G · ~ τ 2 ) f a ( ~ G ) = (1 + e − i π 2 ( K + L + M ) ) f a ( ~ G ) = 0 when K + L + M = 2(2 n + 1) (twice of odd number). Ex. 3.15 P ( x ) = 1 √ 2 π σ e − x 2 2 σ 2 (i) ∫ ∞ −∞ dx P ( x ) = ∫ ∞ −∞ dx 1 √ 2 π σ e − x 2 2 σ 2 (Gausian integral) = 1 √ 2 π σ · √ 2 πσ 2 = 1 ∴ P ( x ) is normalized.

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    13 (ii) Consider the integrals << x 2 n >> = 1 √ 2 π σ ∫ ∞ −∞ dx x 2 n · e − α x 2 = 1 √ 2 π σ ( − ∂ ∂α ) n ∫ ∞ −∞ dx e − α x 2 = 1 √ 2 π σ ( − 1) n ( ∂ ∂α ) n √ π α − 1 2 = ( 2 σ 2 ) n √ π Γ ( n + 1 2 ) = σ 2 n (2 n )! 2 n n ! where we set α = 1 2 σ 2 at the last steps. We thus have << x 2 n >> = ∫ ∞ −∞ dx x 2 n P ( x ) as desired. (iii) << e − iαx >> = ∫ ∞ −∞ dx e − iαx P ( x ) = 1 √ 2 π σ ∫ ∞ −∞ dx e − iαx · e − x 2 2 σ 2 = 1 √ 2 π σ ∫ ∞ −∞ dx e − 1 2 σ 2 ( x + iασ 2 ) 2 · e − α 2 σ 2 2 = e − α 2 σ 2 2 . According to (ii), << x 2 >> = σ 2 . Hence, we get << e − iαx >> = e − α 2 <<x 2 >> 2 . Note here α is a fixed (but arbitrary) parameter unrelated to the auxiliary variable we introduced in (ii). (iv) << e − iαx >> = ∫ ∞ −∞ dx e − iαx P ( x ) = 1 √ 2 π σ ∫ ∞ −∞ dx e − iαx · e − x 2 2 σ 2 = 1 √ 2 π σ ∫ ∞ −∞ dx ( ∞ ∑ n =0 1 (2 n )! ( α x ) 2 n + ∞ ∑ n =0 ( − i ) 2 n +1 (2 n + 1)! ( α x ) 2 n +1 ) e − x 2 2 σ 2 The second sum vanishes because odd functions have no contribution to this integral.

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    14 ∴ << e − iαx >> = 1 √ 2 π σ ∫ ∞ −∞ dx ( ∞ ∑ n =0 ( − 1) n (2 n )! ( α x ) 2 n ) e − x 2 2 σ 2 = 1 √ 2 π σ ∞ ∑ n =0 ( − 1) n (2 n )! α 2 n ∫ ∞ −∞ dx x 2 n e − x 2 2 σ 2 = ∞ ∑ n =0 ( − 1) n (2 n )! α 2 n << x 2 n >> = ∞ ∑ n =0 ( − 1) n n ! ( α 2 2 ) n σ 2 n = e − α 2 σ 2 2 , in agreement with (iii). Ex. 3.16 Let us introduce M ( α ) = << e αA >> = 1 + ∞ ∑ n =1 << A n >> n ! α n = e ∞ ∑ n =1 cn n ! · α n = e C ( α ) where M ( α ) is the moment generating function and C ( α ) = ∞ ∑ n =1 c n n ! α n is cumulant generating function. We thus have c n = d n dα n ln M ( α ) | α =0 , << A n >> = d n dα n e C ( α ) | α =0 . By applying the relations above, we can get cumulant coefficients c n .

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    15 << A >> = d dα ( e ∞ ∑ n =1 cn n ! α n ) = ∞ ∑ n =1 c n · 1 ( n − 1)! · α n − 1 e C ( α ) | α =0 = c 1 ∴ c 1 = << A >> = 0 << A 2 >> = d 2 dα 2 ( e ∞ ∑ n =1 cn n ! α n ) | α =0 = ∞ ∑ n =2 c n · 1 ( n − 2)! α n − 2 e C ( α ) | α =0 + ( ∞ ∑ n =1 c n · 1 ( n − 1)! α n − 1 ) 2 e C ( α ) | α =0 = c 2 + c 2 1 ∴ c 2 = << A 2 >> − << A >> 2 = << A 2 >> << A 3 >> = d 3 dα 3 ( e ∞ ∑ n =1 cn n ! α n ) | α =0 = ∞ ∑ n =3 c n · 1 ( n − 3)! α n − 3 e C ( α ) | α =0 + 3 · ( ∞ ∑ n =2 c n · 1 ( n − 2)! α n − 2 e C ( α ) | α =0 ) · ( ∞ ∑ n =1 c n · 1 ( n − 1)! α n − 1 e C ( α ) | α =0 ) + ( ∞ ∑ n =1 c n · 1 ( n − 1)! α n − 1 ) 3 e C ( α ) = c 3 + 3 · c 2 c 1 + c 3 1 ∴ c 3 = << A 3 >> << A 4 >> = d 4 dα 4 ( e ∞ ∑ n =1 cn n ! α n ) | α =0 = e C ( α ) [ C (4) + 4 · C (1) C (3) + 3 · ( C (2) ) 2 + 6 · ( C (1) ) 2 C (2) ] | α =0 = c 4 + 4 · c 1 c 3 + 3 · c 2 2 + 6 · c 2 1 c 2 = c 4 + 3 · c 2 2 ∴ c 4 = << A 4 >> − 3 << A 2 >> 2 where C ( n ) ≡ d n d α n C ( α ) .

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