Report CopyRight/DMCA Form For : Solutions To Exercises In Modern Condensed Matter Physics
Solutions to Exercises in Modern Condensed Matter Physics S M Girvin and Kun Yang c 2019 Compiled August 12 2019 Note to Instructors For a few of the more di cult problems we include notes to the instructor suggesting simpli cations specializations and hints that the instructor may wish to give the students when assigning those problems Note also that there are some useful exercises
Note to Instructors, For a few of the more difficult problems we include notes to the instructor suggesting simplifications . specializations and hints that the instructor may wish to give the students when assigning those. problems , Note also that there are some useful exercises within the appendices of the textbook . 1, Chapter 2,Ex 2 1, i ,The radiated electric field is. eikRD h i, in e i t e i q r , a re n n E 1 , RD, in is perpendicular to both n and to E. Replace e i q r by he i q r i f q The vector n E in and. 2 2 2,has length Ein sin Hence n n Ein Ein sin Thus. re2 2, a 2 Ein, 2, q 2 , 2 sin f 2 , RD, The total radiated power passing through a sphere of radius RD is. Z 2 , 2 a, P cRD d 2 3 , 8 , Z, c 2 2, r E d sin2 f q 2 4 . 4 e in, Let us normalize the incident electric field to that associated with a single photon in the normal . ization volume L3, 2, Ein ck, 3 3 5 , 4 L L,which yields. r2, Z, P c k e32, d sin2 f q 2 6 , L, ii , Now compare this to the quantum result using the photon scattering matrix element in Eq 2 28 . M re f q 2k k k0 7 , 2, 3,Fermi s Golden Rule for the transition rate is. 2 X L3, Z, d3 k 0 re2 4k , k k0 2 ck 0 f q 2 8 , 0 2 3. , Noting that the two polarization vectors k0 0 and the vector k 0 are all mutually perpindicular we. k k 0 2 1 cos2 sin2 The radiated power is, k k0 2 1 . P,find 0 , re2, Z, P c2 k d sin2 f q 2 9 , L3, in agreement with the result from the semiclassical calculation . Ex 2 2, N N, 1 1 X X, S q W q 2 ei q ri e i q rj 10 . N N i 1 j 1, N N Z, 1 X 1 X, ei q ri i q rj d3 rd3 r 0 ei q ri i q rj r ri r0 rj 11 . N i j, N, i6 j, Z N, 1 1 0 X, N d3 rd3 r 0 ei q r r r ri r0 rj 12 . N N, i6 j, Remembering, N, X, r ri r0 rj n 2 ri0 rj . i6 j, then obviously, Z Z, 1 0, S q 1 d3 rd3 r 0 ei q r r n 2 r0 r 1 n d3 rei q r g r . N, where we used N V n and n 2 R n2 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 . Chapter 3,Ex 3 1, SC FCC, , BCC, , 2 3, radius a 2 4. a 4, a, 3 2 3 3 3, volume of one sphere 6a 24 a 16 a. number of sites in unit cell 1 , 4 , 2, 2 3, volume fraction 6 6 8 . Ex 3 2,F CC 8 18 6 12 4, BCC 8 18 1 2,Ex 3 3, i , Suppose the lattice spacing is a The three primitive vectors are. 1 1, a1 a 0 , 2 2, 1 1, a2 a 0 , 2 2, 1 1, a3 a 0 . 2 2, Thus the coordinates of the four points of the tetrahedron spanned by the three vectors are. O 0 0 0 , 1 1, A a 0 , 2 2, 1 1, B a 0 , 2 2, 1 1. C a 0 , 2 2, 4, 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 , a, Another lattice site on the opposite sublattice locates at P 4 1 1 1 . The distance of P to each corner of the tetrahedron is. 1 1 1, P O a , 4 4 4, 1 1 1, P A a , 4 4 4, 1 1 1. P B a , 4 4 4, 1 1 1, P C a , 4 4 4, , 3a, We get P O P A P B P C 4 Therefore P is at the geometrical center of this. tetrahedron ,Ex 3 4, m , i R a1 m2 a2 m3 a3 n1 a1 n2 a2 n3 a3 m1 n1 a1 m2 n2 a2 . n m1 , m3 n3 a3, n , Indeed have the form of 3 15 and are lattice vectors characterized by m. 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 Rm ma with m being an integer . Proof Assume R 0 a is a lattice site with being a non integer represents the integer part. of and is its fractional part Thus 0 1 , From 3 16 we know a and thus R 0 a 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 a1 This immediately gives us a. 6 m a1 gives us a parallel lattice line , lattice line m a1 with m being an integer Any lattice site R. , R m a1 Look for R a2 such that a2 m a1 is the lattice line closest to the line m a1 Then. m a1 m2 a2 are all lattice sites Now assume 1 a1 2 a2 is also a lattice site with 1 0 and. 2 0 Then we know m1 1 a1 2 a2 forms a lattice line which is closer to the m a1. line than the a2 m a1 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 ai The locations of all lattice points could. be expressed as, 6, c P ci ai , R ci Z integer , i. m, The mid point of any two lattice sites say R and R . n is, mid 1 R, , R m R , n, 2, X 1, mi ni ai, i, 2. where mi and ni 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, X 1, , R c R, 0 R mid ci mi ni ai 13 . c, i, 2, mid is an inversion center given a lattice site R. If R 0 R, 0 must be a lattice point as well , , c . c, Namely there exists a set of pi which are integers such that. X 1, , 0 R, R 0 pi mi ni ai 14 , c p, , i, 2,Combining Eq 13 and Eq 14 we get. pi mi ni ci Z , mid is indeed an inversion center . This tells us that R,Ex 3 6, Triangular Lattice, 6 fold axis. 3 fold axis, 7, Honeycomb Lattice, 6 fold axis, 3 fold axis. If the atoms on A and B sublattice are different C6 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 a4 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 2 n about the origin assumed to be. a lattice site should leave the lattice invariant Assume a0 is the shortest lattice vector connecting. 8, the origin to one of its neighbors After a rotation of 2 5 it becomes . a1 and a2 respectively , which should be lattice vectors themselves see figure Then a1 a2 should also be a lattice vector . But simple trigonometry finds it is shorter than a0 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 , a1, 2 5. a0 a0 a1 a2, a2 lattice site, 5 fold axis,Ex 3 10. The construction of 1D resiprocal lattice bj with bj am 2 mj . b a 2 b 2 x , a, The reciprocal lattice vectors are. m b m 2 x , G m integer , a, 1st BZ a a , 2nd BZ 3 a a a 3 a . nth BZ 1 2n a 3 2n a 2n 3 a 2n 1 a , 9,Ex 3 11,1 FCC. a, a1 y z 0 a 2 a 2 , 2, a, a2 x z a 2 0 a 2 , 2, a. a3 x y a 2 a 2 0 , 2, a3 a3 a3, w a1 a2 a3 , 8 8 4. b1 2 a2 a3 2 1 1 1 , w a, b2 2 2 , a3 a1 1 1 1 , w a. b1 2 a1 a2 2 1 1 1 , w a, This is indicative that b1 b2 b3 forms a BCC lattice with lattice constant 4 . a , 2 BCC, a, a1 y z x a 2 a 2 a 2 , 2, a, a2 x z y a 2 a 2 a 2 . 2, a, a3 x y z a 2 a 2 a 2 , 2, a3, w a1 a2 a3 , 2. b1 2 a2 a3 2 , 0 1 1 , w a, b2 2 2 , a3 a1 1 0 1 . w a, b3 2 a1 a2 2 , 1 1 0 , w a, b1 b2 b3 constructs a FCC lattice with lattice constant 4 . a , 10,Ex 3 12, P ki bi we have, P ni ai whose reciprocal lattice K. For a direct lattice R, i i, K kj bj, P P, iR ni , ai . e e 1 , R R, 0 then we have eiK 0 0 R , , Let us call the reciprocal lattice of K 1 thus R. Ex 3 13, a , Begin with a 1D array of disks of radius r0 If the centers of the disks are on the 1D lattice. j jd 1 0 0 j Z where d 2r0 then the disks are just touching as shown in Fig 1a . R, a , d, d, b , d, Figure 1, Now consider adding a second line of atoms as shown in Fig 1b with lattice positions R j . d j y 0 j Z The lowest possible allowed value of y and hence the densest lattice occurs. for horizontal displacement 1 2 At ymin 23 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 a1 d 1 0 0 and a2 d 12 23 0 , b , jk j a1 k a2 as shown in the left panel. Now consider the triangular lattice A defined by points R. of Fig 2 , D D, U U, Figure 2, 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 rc 13 0 a1 a2 d 12 2 , 1, 3, 0 , If each lattice point is the center of a hard sphere of radius r0 d 2 the spheres in a given. layer are close packed If the second layer is raised a distance d above the first we can define a. third vector b rc d 0 0 1 connecting this point to the origin Theq close packed hard sphere. , constraint requires b 0 b a1 b a2 d which implies 2 This minimum vertical. 3, 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 a1 a2 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. , q B plane It thus has length 2b z in order to satisfy the hard sphere constraint . as high as the, 2, a3 d 0 0 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 a3 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 a1 a2 a1 a3 a2 a3 21 d2 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 , 1 0 1 1 , A, 2 2, 2 0 1 , A, 1, 2 2. 1, 3 0 1, A, 2 2, A 1 A, 2 A 1 A, 3 A 3 1 d2 , 2 A. 2, , where 2d 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 aj Aj for j 1 2 3 . Hence the lattice is FCC , 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 1 0 2 a4 x y z . K b1 L b2 M b3 where, Scattering is possible for momentum transfer q G. b1 2 x y z , a, b2 2 x y z , a, b1 2 , x y z , a, are primitive reciprocal lattice vectors of FCC The scattering amplitude is proportional to. 2, X, f q e i q fs q 15 , s 1, i , q , e 1, f1 q e i q 2 f2 q 16 . For identical atoms on the two sublattices f1 q f2 q fa q where fa is the atomic form. factor Then, 2, 1 e iG 1 e i 2 K L M fa G , 0, f G fa G . when K L M 2 2n 1 twice of odd number ,Ex 3 15, 1 x2. P x e 2 2, 2 , i , Z Z , 1 x2, dx P x dx e 2 2, 2 . 1 , Gausian integral 2 2, 2 , 1, P x is normalized . 13, ii ,Consider the integrals, Z , 1 2, x2n dx x2n e x. 2 , n Z , 1 2, dx e x, 2 , n, , , 1 n 1, 1 2, 2 , n . 2 2, , 1, n , 2, 2n , 2n, 2n n , 1, R , where we set 2 2 at the last steps We thus have x2n . dx x2n P x as desired , iii , Z , e i x dx e i x P x . , Z , 1 x2, dx e i x e 2 2, 2 , Z , 1 1, x i 2 2 2 2. dx e 2 2 e 2, 2 , 2 2, e 2 , 2 x2 , According to ii x2 2 Hence we get e i x e 2 Note here is a. fixed but arbitrary parameter unrelated to the auxiliary variable we introduced in ii . iv , Z , i x, e dx e i x P x , , Z , 1 x2, dx e i x e 2 2. 2 , , , , i 2n 1, Z, 1 X 1 X x2, dx x 2n x 2n 1 e 2 2. 2 n 0, 2n n 0, 2n 1 , The second sum vanishes because odd functions have no contribution to this integral . 14, , , , 1 n, Z, i x 1 X x2, e dx x 2n e 2 2, 2 n 0. 2n , , 1 n 2n , Z, 1 X x2, dx x2n e 2 2, 2 n 0 2n . , X 1 n 2n, x2n , n 0, 2n , , X 1 n 2 n 2n, , n 0. n 2, 2 2, e 2 ,in agreement with iii ,Ex 3 16,Let us introduce. , An n cn, n, P, X n , A, M e 1 en 1 eC , n 1, n . , cn, n is cumulant generating, P, where M is the moment generating function and C n . n 1,function We thus have, dn, cn ln M 0 , d n, dn C . An e 0 , d n, By applying the relations above we can get cumulant coefficients cn .