Solution Manual for Modern Condensed Matter Physics, 1st Edition

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

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Chapter 2Ex. 2.1(i)The radiated electric field is~a≈reeikRDRD[ˆn×(ˆn×~Ein)]e−iωte−i~q·~r.(1)Replacee−i~q·~rby〈e−i~q·~r〉=f(~q). The vector (ˆn×~Ein) is perpendicular to both ˆnand to~Einandhas length|Einsinθ|. Hence|ˆn×(ˆn×~Ein)|2=E2insin2θ. Thus|a|2=E2inr2eR2Dsin2θ|f(~q)|2.(2)The total radiated power passing through a sphere of radiusRDisP=cR2D∫dΩ(2a8π×2)(3)=c4π r2eE2in∫dΩ sin2θ|f(~q)|2.(4)Let us normalize the incident electric field to that associated with a single photon in the normal-ization volumeL3E2in4π=ℏωL3=ℏckL3(5)which yieldsP=ℏc2k r2eL3∫dΩ sin2θ|f(~q)|2.(6)(ii)Now compare this to the quantum result using the photon scattering matrix element in Eq. (2.28)M=ref(~q)∧2kˆ~kλ·ˆ~k′λ.(7)2

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3Fermi’s Golden Rule for the transition rate isΓ = 2πℏ∑λ′L3(2π)3∫d3k′r2e∧4k[ˆ~kλ·ˆ~k′λ]2δ(ℏω−ℏck′)|f(~q)|2.(8)Noting that the two polarization vectors ˆ~k′λ′and the vector ˆk′are all mutually perpindicular, wefind∑λ′[ˆ~kλ·ˆ~k′λ]2= 1−[ˆ~kλ·ˆk′]2= 1−cos2θ= sin2θ. The radiated power isP=ℏc2k r2eL3∫dΩ sin2θ|f(~q)|2,(9)in agreement with the result from the semiclassical calculation.Ex. 2.2S(~q)=1N <|W(~q)|2>=1N <N∑i=1ei~q.~riN∑j=1e−i~q.~rj>(10)=1N <N∑i=jei~q.~ri−i~q.~rj>+ 1N <N∑i6=j∫d3~rd3~r′ei~q.~ri−i~q.~rjδ(~r−~ri)δ(~r′−~rj)>(11)=1N N+ 1N∫d3~rd3~r′ei~q·(~r−~r′)<N∑i6=jδ(~r−~ri)δ(~r′−~rj)>(12)Remembering<N∑i6=jδ(~r−~ri)δ(~r′−~rj)>=n(2)(~r′i−~rj),then obviouslyS(~q) = 1 + 1N∫d3~rd3~r′ei~q·(~r−~r′)n(2)(~r′−~r) = 1 +n∫d3~rei~q.~rg(~r)where we usedN/V=nandn(2)(~R) =n2g(~R).P.S. ”<>” indicates thermal average in liquid or amorphous materials. It is unnecessary onlyfor perfect lattices. Generally ”<>” must be in the formula.

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Chapter 3Ex. 3.1•SCFCCBCCradiusa/2√24a√34avolume of one sphereπ6a3√2π24a3√3π16a3number of sites in unit cell142volume fractionπ6√26π√38πEx. 3.2F CC8×18+ 6×12= 4BCC8×18+ 1 = 2Ex. 3.3(i)Suppose the lattice spacing isa. The three primitive vectors are~a1=a(0,12,12 )~a2=a( 12,0,12 )~a3=a( 12,12,0)Thus, the coordinates of the four points of the tetrahedron spanned by the three vectors areO= (0,0,0)A=a(0,12,12 )B=a( 12,0,12 )C=a( 12,12,0)4

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5By calculating the distance of any two points, we can prove that the edges of the tetrahedronare equal. So it is a regular tetrahedron.(ii)Another lattice site (on the opposite sublattice) locates atP=a4(1,1,1)The distance of P to each corner of the tetrahedron isP O=a|( 14,14,14 )|P A=a|( 14,−14,−14 )|P B=a|(−14,14,−14 )|P C=a|(−14,−14,14 )|We getP O=P A=P B=P C=√3a4.Therefore, P is at the geometrical center of thistetrahedron.Ex. 3.4(i)~R~m±~n= (m1~a1+m2~a2+m3~a3)±(n1~a1+n2~a2+n3~a3) = (m1±n1)~a1+ (m2±n2)~a2+(m3±n3)~a3Indeed 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 isa. Claim:all sites satisfying 1D version of (3.16) can be written asRm=mawithmbeing an integer.Proof: AssumeR′=α ais a lattice site withαbeing a non-integer. [α] represents the integer partofα, and ∆α=α−[α] is its fractional part. Thus 0<∆α <1.From (3.16), we know [α]aand thusR′−[α]a= ∆α ais also a lattice site, but its distanceto the origin is less thana, leading to a contradiction.For 2D, let us look for the closest site to the origin, located at~a1. This immediately gives us alattice line,m~a1, withmbeing an integer. Any lattice site~R6=m~a1, gives us a parallel lattice line,~R+m~a1. Look for~R=~a2such that~a2+m~a1is the lattice line closest to the linem~a1. Thenm~a1+m2~a2are all lattice sites. Now assumeα1~a1+α2~a2is also a lattice site, with ∆α1>0 and∆α2>0. Then we know (m1+ ∆α1)~a1+ [α2]~a2forms a lattice line, which is closer to them~a1line than the~a2+m~a1line! Contradiction again. It is now obvious how to generalize this to 3D.Ex. 3.5For a Bravais lattice, it has a set of primitive vectors,~ai. The locations of all lattice points couldbe expressed as

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6~R~c=∑ici~ai,ci∈Z(integer).The mid-point of any two lattice sites, say~R~mand~R~nis~Rmid= 12(~R~m+~R~n)=∑i12 (mi+ni)~aiwheremiandniare integers.We can always shift the origin of the coordinate of the lattice to this mid-point so that all latticepoints have new coordinates as~R′~c=~R~c−~Rmid=∑i[ci−12 (mi+ni)~ai](13)If~Rmidis an inversion center, given a lattice site~R′~c,−~R′~cmust be a lattice point as well.Namely, there exists a set ofpiwhich are integers such that−~R′~c=~R′~p=∑i[pi−12 (mi+ni)]~ai(14)Combining Eq. (13) and Eq. (14), we getpi= (mi+ni)−ci∈Z.This tells us that~Rmidis indeed an inversion center.Ex. 3.63-fold axis6-fold axisTriangular Lattice

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73-fold axis6-fold axisHoneycomb LatticeIf the atoms on A and B sublattice are different,C6symmetry will be broken. The original 6-foldaxis will become a 3-fold axis. And the original 3-fold axis is still 3-fold.Ex. 3.7Ex. 3.8A diamond structure could be viewed as a FCC lattice with a basis containing two atoms called Aand B. A and B have aa4(1,1,1) shift whereais the lattice constant. If we take the mid-point of Aand B as an inversion center, the positions of the sublattices will exchange applying to all the latticesites. 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. 3.9Suppose the lattice hasn-fold symmetry. Then rotating byα=2πnabout the origin (assumed to bea lattice site) should leave the lattice invariant. Assume~a0is the shortest lattice vector connecting

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8the origin to one of its neighbors. After a rotation ofα=±2π5, it becomes~a1and~a2respectively,which should be lattice vectors themselves (see figure). Then~a1+~a2should also be a lattice vector.But simple trigonometry finds it is shorter than~a0(see figure), which leads to contradiction. Thus5-fold symmetry is not allowed in 2D. Since 3D lattices are made of parallel 2D planes, this impliessuch symmetry is impossible in 3D as well.2Πê5: lattice site: 5-fold axisa1a2a0a0=a1+a2Ex. 3.10The construction of 1D resiprocal lattice{~bj}with~bj·~am= 2πδmj:~b·~a= 2π⇒~b= 2πaˆx.The reciprocal lattice vectors are~G=m~b=m2πaˆx.(m=integer)1st BZ: [−π/a, π/a)2nd BZ: [−3π/a,−π/a),[π/a,3π/a)nth BZ: [(1−2n)π/a,(3−2n)π/a),[(2n−3)π/a,(2n−1)π/a)

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9Ex. 3.111) FCC~a1=a2 (ˆy+ ˆz) = (0, a/2, a/2)~a2=a2 (ˆx+ ˆz) = (a/2,0, a/2)~a3=a2 (ˆx+ ˆy) = (a/2, a/2,0)w=~a1.(~a2×~a3) =a38 +a38=a34~b1= 2πw(~a2×~a3) = 2πa(−1,1,1)~b2= 2πw(~a3×~a1) = 2πa(1,−1,1)~b1= 2πw(~a1×~a2) = 2πa(1,1,−1)This is indicative that (~b1,~b2,~b3) forms a BCC lattice with lattice constant4πa.2) BCC~a1=a2 (ˆy+ ˆz−ˆx) = (−a/2, a/2, a/2)~a2=a2 (ˆx+ ˆz−ˆy) = (a/2,−a/2, a/2)~a3=a2 (ˆx+ ˆy−ˆz) = (a/2, a/2,−a/2)w=~a1.(~a2×~a3) =a32~b1= 2πw(~a2×~a3) = 2πa(0,1,1)~b2= 2πw(~a3×~a1) = 2πa(1,0,1)~b3= 2πw(~a1×~a2) = 2πa(1,1,0)(~b1,~b2,~b3) constructs a FCC lattice with lattice constant4πa.

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10Ex. 3.12For a direct lattice~R=∑ini~aiwhose reciprocal lattice~K=∑iki~bi, we haveei ~R·~K=e∑ni~ai.∑kj~bj= 1.Let us call the reciprocal lattice of~K ~R′, then we haveei ~K·~R′= 1, thus~R′=~R.Ex. 3.13(a)Begin with a 1D array of disks of radiusr0.If the centers of the disks are on the 1D lattice{~Rj=jd(1,0,0);j∈Z}whered= 2r0, then the disks are just touching as shown in Fig. 1a.a)b)dddFigure 1Now consider adding a second line of atoms as shown in Fig. 1b with lattice positions{~Rj=d(j+δ, y,0);j∈Z}. The lowest possible allowed value ofy(and hence the densest lattice) occursfor horizontal displacementδ= 1/2. Atymin=√32each disk in the second row touches two disksin the first row. Extending this to an arbitrary number of rows yields the triangular lattice withlattice vectors~a1=d(1,0,0) and~a2=d(12,√32,0).(b)Now consider the triangular latticeAdefined by points~Rjk=j~a1+k~a2as shown in the left panelof Fig. 2.UDUDFigure 2

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11We see that each unit cell contains an upward facing triangle (labelledU) and a downwardfacing triangle (labelledD). We can choose to place a second identical layerBon top of the firstpositioned either with the lattice sites located in the center of theUtriangles or theDtrianglesof 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 theUtriangles).If the lower left corner of thetriangle labelledUis taken as the origin, the center of the triangle is located at the ‘center of mass’position~rc=13[~0 +~a1+~a2] =d(12,12√3,0).If each lattice point is the center of a hard sphere of radiusr0=d/2, the spheres in a givenlayer are close-packed. If the second layer is raised a distanceλdabove the first, we can define athird vector~b=~rc+λd(0,0,1) connecting this point to the origin. The close-packed hard-sphereconstraint requires|~b−~0|=|~b−~a1|=|~b−~a2|=dwhich impliesλ=√23. This minimum verticaldisplacement 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 patternABABAB...AB to form a hexagonal close-packed (HCP) lattice. The vectors~a1, ~a2defined aboveconstitute two of the lattice vectors for the hcp lattice. The vector~bdefined above isnota latticevector because (for example) there is no lattice site at 2~b.~bsimply defines the location of a secondatom in the unit cell of this non-Bravais lattice.Under the ABABAB...AB repetition, the thirdplane matches the A plane and hence the lattice vector lies in thezdirection and is elevated twiceas high as the B plane.It thus has length 2~b·ˆzin order to satisfy the hard-sphere constraint:~a3=d(0,0,2√23).(d)If the third layer is centered on the other choice (downward facing triangles) we can repeat thisABCABCABC...ABCC stacking to obtain the face-centered-cubic (FCC) lattice. In this case thethird lattice vector is~a3=~b.In this case~b, defined above,isa lattice vector because the ABCstacking places an atom at both~band 2~b, unlike the case of AB stacking. To see that this is indeedthe FCC lattice, notice that~a1·~a2=~a1·~a3=~a2·~a3=12d2indicating that there is a 60 degreeangle between each of the three pairs of vectors just as there is for the standard basis vectors forthe FCC lattice:~A1=∆(0,12,12 )~A2=∆( 12,0,12 )~A3=∆( 12,12,0)~A1·~A2=~A1·~A3=~A2·~A3= 12d2,where ∆≡ √2dis the length of the side of the cube defining the conventional unit cell (i.e., thesecond-neighbor distance).There exists an orthogonal rotation matrix that preserves the anglesbetween the vectors while effecting the coordinate change needed to map~aj→Ajforj= 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 beA, B, or C, subject to the constraint that two adjacent layers cannot be the same. For example:ABACBABCBACABCBABABABCBABCBC.Ex. 3.14For diamond lattice the two point basis can be written as~τ1= 0,~τ2=a4(ˆx+ ˆy+ ˆz).Scattering is possible for momentum transfer~q=~G=K~b1+L~b2+M~b3where~b1= 2πa(−ˆx+ ˆy+ ˆz)~b2= 2πa(ˆx−ˆy+ ˆz)~b1= 2πa(ˆx+ ˆy−ˆz)are primitive reciprocal lattice vectors of FCC. The scattering amplitude is proportional tof(~q)=2∑s=1e−i~q·~τfs(~q)(15)=e−i~q·~τ1f1(~q) +e−i~q·~τ2f2(~q).(16)For identical atoms on the two sublatticesf1(~q) =f2(~q) =fa(~q), wherefais the atomic formfactor. Thenf(~G) = (1 +e−i ~G·~τ2)fa(~G) = (1 +e−iπ2(K+L+M))fa(~G) = 0whenK+L+M= 2(2n+ 1) (twice of odd number).Ex. 3.15P(x) =1√2π σ e−x22σ2(i)∫∞−∞dx P(x) =∫∞−∞dx1√2π σ e−x22σ2(Gausian integral) =1√2π σ· √2πσ2= 1∴P(x) is normalized.

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13(ii)Consider the integrals<< x2n>>=1√2π σ∫∞−∞dx x2n·e−α x2=1√2π σ(−∂∂α)n∫∞−∞dx e−α x2=1√2π σ(−1)n(∂∂α)n√π α−12=(2σ2)n√πΓ(n+ 12)=σ2n(2n)!2nn!where we setα=12σ2at the last steps. We thus have<< x2n>>=∫∞−∞dx x2nP(x) as desired.(iii)<< e−iαx>>=∫∞−∞dx e−iαxP(x)=1√2π σ∫∞−∞dx e−iαx·e−x22σ2=1√2π σ∫∞−∞dx e−12σ2(x+iασ2)2·e−α2σ22=e−α2σ22.According to (ii),<< x2>>=σ2. Hence, we get<< e−iαx>>=e−α2<<x2>>2. Note hereαis afixed (but arbitrary) parameter unrelated to the auxiliary variable we introduced in (ii).(iv)<< e−iαx>>=∫∞−∞dx e−iαxP(x)=1√2π σ∫∞−∞dx e−iαx·e−x22σ2=1√2π σ∫∞−∞dx(∞∑n=01(2n)! (α x)2n+∞∑n=0(−i)2n+1(2n+ 1)! (α x)2n+1)e−x22σ2The second sum vanishes because odd functions have no contribution to this integral.

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14∴<< e−iαx>>=1√2π σ∫∞−∞dx(∞∑n=0(−1)n(2n)! (α x)2n)e−x22σ2=1√2π σ∞∑n=0(−1)n(2n)!α2n∫∞−∞dx x2ne−x22σ2=∞∑n=0(−1)n(2n)!α2n<< x2n>>=∞∑n=0(−1)nn!(α22 )nσ2n=e−α2σ22,in agreement with (iii).Ex. 3.16Let us introduceM(α) =<< eαA>>= 1 +∞∑n=1<< An>>n!αn=e∞∑n=1cnn!·αn=eC(α)whereM(α) is the moment generating function andC(α) =∞∑n=1cnn!αnis cumulant generatingfunction. We thus havecn=dndαnlnM(α)|α=0,<< An>>=dndαneC(α)|α=0.By applying the relations above, we can get cumulant coefficientscn.

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15<< A >>=ddα(e∞∑n=1cnn!αn)=∞∑n=1cn·1(n−1)!·αn−1eC(α)|α=0=c1∴c1=<< A >>= 0<< A2>>=d2dα2(e∞∑n=1cnn!αn)|α=0=∞∑n=2cn·1(n−2)!αn−2eC(α)|α=0+(∞∑n=1cn·1(n−1)!αn−1)2eC(α)|α=0=c2+c21∴c2=<< A2>>−<< A >>2=<< A2>><< A3>>=d3dα3(e∞∑n=1cnn!αn)|α=0=∞∑n=3cn·1(n−3)!αn−3eC(α)|α=0+ 3·(∞∑n=2cn·1(n−2)!αn−2eC(α)|α=0)·(∞∑n=1cn·1(n−1)!αn−1eC(α)|α=0)+(∞∑n=1cn·1(n−1)!αn−1)3eC(α)=c3+ 3·c2c1+c31∴c3=<< A3>><< A4>>=d4dα4(e∞∑n=1cnn!αn)|α=0=eC(α)[C(4)+ 4·C(1)C(3)+ 3·(C(2))2+ 6·(C(1))2C(2)]|α=0=c4+ 4·c1c3+ 3·c22+ 6·c21c2=c4+ 3·c22∴c4=<< A4>>−3<< A2>>2whereC(n)≡dnd αnC(α).

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