Group Theory in Solid State Physics II - ETH Z

Group Theory in Solid StatePhysics II

Preface

This lecture expands on the fundamental group theoretical concepts andmethods introduced in Part I. Again the main aim is to show how to usesymmetry arguments for solving problems in atomic, molecular and solidstate physics. Particular attention will be paid to double groups, as theyare pivotal in understanding problems involving the spin of electrons andrelated topics in other field of Physics such as particle Physics. In partic-ular, we will search for applications of the Clebsch-Gordan coefficients andof the main theorem governing group theoretical methods in Physics – theWigner-Eckart-Koster theorem. Furthermore, we will provide a deep discus-sion of space (double) groups – the true symmetry groups of crystals – andtheir relation to the more common point groups. Finally, we will discuss theapplication of group theoretical methods within the Landau theory of phasetransitions. The relevant literature for the topics presented in this lecturesis:L.D. Landau, E.M. Lifshitz, Lehrbuch der Theor. Pyhsik, Band III, ”Quan-tenmechanik”, Akademie-Verlag Berlin, 1979, Kap. XII and Band V, ”Statis-tische Physik”, Teil 1, Akademie-Verlag 1987, Kap. XIII.

Zurich, October 2003D. Pescia

ii

Contents

Preface ii

1 The Kronecker Product 11.1 Kronecker product of representations . . . . . . . . . . . . . . 1

1.1.1 An example: Product representations of SO(3) . . . . . 31.2 The direct product of groups . . . . . . . . . . . . . . . . . . . 51.3 The Clebsch-Gordan coefficients . . . . . . . . . . . . . . . . . 7

1.3.1 Clebsch-Gordan coefficients for point groups . . . . . . 91.4 The Wigner-Eckart theorem . . . . . . . . . . . . . . . . . . . 18

2 Energy bands in solids 252.1 The translation group . . . . . . . . . . . . . . . . . . . . . . 252.2 Space groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2.1 The irreducible representations of space groups . . . . 342.3 Symmetry adapted plane waves . . . . . . . . . . . . . . . . . 39

2.3.1 The one-dimensional lattice . . . . . . . . . . . . . . . 392.3.2 The square lattice . . . . . . . . . . . . . . . . . . . . . 412.3.3 Free electron energy bands in three-dimensional lattices 47

3 Landau theory of phase transitions 48

iii

Chapter 1

The Kronecker Product

So far we have considered elementary groups. In this chapter we treat twocomplementary subjects. In the first one we consider the Kronecker productof representations, find out which irreducible components it contains and de-velop methods to find the symmetry adapted basis functions. In the secondone we deal with enlarging the symmetry group to account for extra symme-tries. Example: consider a system whith SO(3) as symmetry group. Havingnoticed that the inversion I is also a symmetry element, we ask how the sym-metry group can be expandend and which are the irreducible representationsof the expanded group.

1.1 Kronecker product of representations

The Kronecker (or direct or tensor) product of vector

spaces

Consider the two spaces Ln and Lm. The direct-product space is a space ofdimension p = n ·m defined by the p basis vectors

e1 ⊗ i1, e1 ⊗ i2, ..., e1 ⊗ im, e2 ⊗ i1, ...en ⊗ im

with Lp = Ln ⊗Lm. The scalar product in the spaces Ln and Lm produces ascalar product in Lp according to

(u1 ⊗ v1, u2 ⊗ v2) = (u1, u2) · (v1, v2)

As any vector of Lp can be expressed as u =∑

i,j uijei⊗ ij, the scalar productbetween any vector in Lp can be calculated according to

(u, v) = (∑i,j

uijei ⊗ ij ,∑l,m

vlmel ⊗ im) =∑i,j

u∗ijvij

1

CHAPTER 1. THE KRONECKER PRODUCT 2

(because ei ⊗ ij , el ⊗ ik) = δilδjk). The action of the operator A⊗B on u⊗ vis defined as (A⊗B)u⊗ v = Au⊗Bv. With this definition one can find theaction of A ⊗ B on each vector of Lp as (A⊗ B)u =

∑i,j uijAei ⊗ Bij . The

matrix representation of A⊗B is the direct product of the two matrices. Let[Alm] be a matrix of the order LxM and [Bpq] a matrix of order PxQ. Thedirect (tensor or Kronecker) product of these two matrices is a matrix C ofthe order L · P ×M ·Q, which can be written as

A11B A12B ... A1MBA12B A22B ... A2MB. . . .. . . .

AL1B AL2B ... ALMB

where the ’element’ Alm · B stands for a matrix of order PxQ given by

AlmB11 AlmB12 ... AlmB1Q

AlmB21 AlmB22 ... AlmB2Q

. . . .

. . . .AlmBP1 AlmBP2 ... AlmBPQ

If A1, A2 and B1, B2 are any matrices whose dimensions are such that theordinary matrix product A1A2 and B1B2 are defined, then the direct producthas an important property:

(A1 ⊗B1)(A2 ⊗ B2) = (A1A2) ⊗ (B1B2)

Further, if F is the direct product of a number of matrices A, B,C, ..., then

trF = (trA) · (trB) · .....

Exercise: prove this.The operation of the direct product of matrices is associative, so that

A⊗ (B ⊗ C) = (A⊗ B) ⊗ C = A⊗ B ⊗ C

The operation is distributive with respect to the matrix addition,

A⊗ (B + C) = A⊗B + A⊗ C

Moreover, (AB)⊗k = A⊗k · B⊗k, where A⊗k = A⊗ A⊗ A⊗ A... (k-times).


Consider two matrix representations T1 and T2 (reducible or irreducible)of a group G. Let us take the direct product of the corresponding matricesfor the two representations and denote it by

T (a) = T1(a) ⊗ T2(a)

Theorem: T(a) is also a representation of the group.Proof:

T (a)T (b) = (T1(a) ⊗ T2(a)(T1(b) ⊗ T2(b)

= (T1(a)T1(b)) ⊗ (T2(a)T2(b))

= T1(ab) ⊗ T2(ab)

= T (ab)

Thus, we see that the set of matrices constructed by taking the direct productof two representations also form a representation. An important property ofthe direct product representation is that

χ(a) = χ1(a) · χ2(a)

i.e. the characters of the direct product representations are the product of thecharacters of the individual representations. In general, the direct productrepresentation is reducible, certainly if T1 or T2 are reducible. For example,for the reduction of the direct product of irreducible representations Ti andTj we expect an expansion of the type

Ti ⊗ Tj = ⊕kni,jk Tk

where ni,jk are non-negative integers. A sum of this type is also known as the

Clebsch-Gordan series.Exercise: work out all direct products of all irreducible representations ofC4v.

1.1.1 An example: Product representations of SO(3)

According to the rules worked out in Part I, the reduction of a representationD in irreducible components relies on the knowledge of the characters alone.Let nowG be SO(3) andD with corresponding characters χ(φ) the charactersof the representation D. Then the coefficients cj in D = ⊕cjDj are given bythe general formula

cj =1

2π2

∫do · dφ sin[(2j + 1)φ]χ(φ) sinφ


which is equivalent of expanding the function χ(φ) sinφ in sin(2j + 1)φ, j ∈N . We now apply this equation to the product representation Dm ⊗ Dl oftwo irreducible representations Dm, Dl (without limitation of generality, weconsider m ≥ l. Dm ⊗Dl has dimension (2m+ 1) · (2l+ 1) and its characteris given by χ = χl · χm. Rewriting χl(φ) · χm(φ) sinφ by using the identitysin a cos b = 1/2[sin(a+ b) + sin(a− b)], as

(l |integerhalfint.) = 2(cos 2lφ+ cos(2(l − 1)φ+ .....+ |1/2

cosφ) sin(2m+ 1)φ

= sin[2(m+ l) + 1]φ+ sin[2(m− l) + 1]φ+

+ sin[2(m+ l − 1) + 1]φ+ sin[2(m− l + 1) + 1]φ+

+ .......

+ |sin(2m+1)φsin[2(m+1/2)+1]φ+sin[2(m−1/2)+1]φ

we recognize that the coefficients of the representation Dm+l, ....Dm−l are justequal one, all others are vanishing. Thus

Dm ⊗Dl = Dm+l +Dm+l−1 + ...+Dm−l

This results is the so called Clebsch-Gordan series for the reduction of directproduct representations of SO(3).

A practical method for reducing any representation D of SO(3) withoutthe explicit use of the characters and for finding the invariant subspaces isgiven by Cartan.

First Cartan Algorithmus

This is a practical way to work out the irreducible representations Dj andthe numbers nj in D = ⊕jnjDj

1. Let D be a representation of SO(3) and ID3 the corresponding matrix

representing the generator I3.

2. Among the diagonal matrix elements (which are called Cartan’s weightsor multiplicators) lets pick up the largest one jk. Then the reductionof the representation D contain Djk

at least once.

3. The representation Djkuses the corresponding 2jk +1 diagonal matrix

elements.

4. the remaining eigenvalues are sorted out starting from the largest re-maining one and repeating the above procedure until all eigenvaluesare used


An important application of this algorithm is the reduction of Dm ⊗ Dl.Without limitation of generaliy, we take m ≥ l. To perform the reduction, weconsider only rotation about the z axis, for which the matrix representationsare

e−imϕ 0 . . 0 00 e−i(m−1)ϕ . . 0 00 0 . . 0 00 0 . . 0 00 0 . . e−i(−m+1)ϕ 00 0 . . 0 e−i(−m)ϕ

e−ilϕ 0 . . 0 00 e−i(l−1)ϕ . . 0 00 0 . . 0 00 0 . . 0 00 0 . . e−i(−l+1)ϕ 00 0 . . 0 e−i(−l)ϕ

The matrix representing I3 for Dm ⊗Dl is given by

−1

i

d

dϕ(Dm ⊗Dl) |ϕ=0= I3,m ⊗ E + E ⊗ I3,l

This matrix has the following form:

m+ l 0 . . 0 0 00 m+ l − 1 . . 0 0 00 0 . . . .00 0 . m− l . . 00 0 . . . 00 0 . . . . .0 0 . . 0 0 −m− l

The top diagonal elements m+ l, ..., m−l represent all the largest eigenvaluesand can be used to classify the irreducible representations into which Dm⊗Dl

reduces. The result is the Clebsch-Gordan series that we have worked ourabove using the theory of characters.

1.2 The direct product of groups

Let H = e, h1, h2, ... and K = e, k1, k2, ... be two groups such that all theelements hj commute with kj. If we multiply each element of H with each of


K we obtain a new set of elements.Theorem: This set build a group: the group is called the direct product ofH and K and is denoted by G ≡ H ⊗K.Proof. With G = e, g11, g12, ...g1k, g21, g22, ..., gij, ...., where gij = hikj andhihm = hp and kjkn = kq, then

gijgmn = (hikj)(hmkn) = hihmkjkn = hpkq = gpq

Exercise: Prove that H and K are subgroups of G.Let Th be a representation of H and Tk a representation of K, i.e.

Th(hi)Th(hm) = Th(hp)

Tk(kj)Tk(kn) = Tk(kq)

Theorem: the direct product Th⊗Tk of the representations of two commutinggroups is a representation of the direct product group.Proof:

(Th ⊗ Tk)(gij)(Th ⊗ Tk)(gmn)

= (Th(hi) ⊗ Tk(kj))(Th(hm) ⊗ Tk(kn))

= Th(hp) ⊗ Tk(kq)

= Th ⊗ Tk(gpq)

Theorem: if Th and Tk are irreducible representations of H and K, thenTh ⊗ Tk ≡ Tg is an irreducible representation of G.Proof: Th and Tk are irreducible representations, i.e.

Mhi∈Hχ∗h(hi)χh(hi) = 1

Mki∈Kχ∗k(ki)χk(ki) = 1

Taking the product of both sides of these equations leads to

1 = Mhi,kjχ∗

h(hi)χh(hi)χ∗k(kj)χk(kj) = Mgij

χ∗g(gij)χg(gij)

which proves that Tg is indeed an irreducible representation of the productgroup.Theorem: All irreducible representations of G are the direct product of anirreducible representation of H and one of K.Proof (for finite groups): Let the number of irreducible representationsof H be nh, their dimensions be lhi , the number of irreducible representations


of K be nk and their dimensions lkj :

nh∑i

(lhi )2 = h

ng∑j

(lkj )2 = k

Taking the product of both sides gives

h · k = g =∑i,j

(lhi )2(lkj )2

The irreducible representations of G obtained by direct product will havethe dimensions lhi · lkj = lgij. The sum over all such irreducible representationsgives ∑

i,j

(lgij)2 =

∑i,j

(lhi )2(lkj )2 = g

so that the direct product of all irreducible representations exhausts theall irreducible representations of G. The number of such representations isng = nh · nk. This is a very important result as it helps in constructing allirreducible representations of a bigger group from those of smaller groups.Example of group product. The most famous example of group product isSO(3) ⊗ SU(2). This is the full symmetry group of an electron with spinembedded in a spherically symmetric potential. If spin-orbit coupling is ne-glected, then the symmetry group consists of separate rotations in real andspin space, and the energy levels can be classified according to the irreduciblerepresentations Dl⊗D1/2. The dimension of these representations is (2l+1)·2.If spin-orbit coupling terms are allowed, then the symmetry group consistsof simoultaneous rotations in real and spin space. This group is thus essen-tially SU(2), and the Kronecker product representations split into irreduciblerepresentations of SU(2), according to the Cartan algorithm.

1.3 The Clebsch-Gordan coefficients

Once the irreducible components of the product representations are workedout (if any exists), one can ask about the invariant subspaces belonging tothe irreducible components. In other words, we would like to construct out ofthe product functions the symmetrized linear combinations. The coefficientsof the linear combinations with proper symmetry are called Clebsch-Gordancoefficients.


In the case of SO(3) one proceeds, typically, by means of the secondCartan Algorithm.

1. find all basis vectors transforming according to the same Cartan weightjk. This space Vjk

has, in general, more than one dimension.

2. construct the operator I+ for the representation D and solve the equa-tion I+x = 0 within the space Vjk

. The vector x transforms, by con-struction, according to the top weight of the representation Djk

, i.e.when I3(Djk

) is applied to x, x get multiplied by jk.

3. The remaining vectors transforming according to Djkare found by ap-

plying the lowering operator I− to x.

Example. The Clebsch-Gordan coefficients. Let us calculate, as an exampleof this second algorithmus, the invariant subspaces of Dl ⊗D1/2. This repre-sentations is important for describing the phenomenon of spin-orbit couplingin a single electron atom with spherically symmetric potential. We recallbriefly the important matrix elements for the operators I±. The irreduciblerepresentation Dj has dimension 2j + 1 and the non-vanishing matrix ele-ments of the generators are given by (m = j, j − 1, ...− j) (Condon-Shortleyconvention)

(ψj,m, Izψj.m) = m

(ψj,m±1, Ixψj,m) =1

2

√(j ∓m)(j ±m+ 1)

(ψj,m±1, Iyψj,m) = ∓ i

2

√(j ∓m)(j ±m+ 1)

For I± = 12[Ix ± i · Iy] we have

I±Xj,m =1√2·√j(j + 1) −m(m± 1)Xj,m±1

Because of the Clebsch-Gordan series,

Dl ⊗D1/2 = Dl+1/2 ⊕Dl−1/2

Let the basis functions, carrying the irreducible representation Dl, be Xm,with m = l, ...− l and let call the spin functions Yz, z = ±1/2. The only basisfunction transforming according to the Cartan weight l + 1/2 is Xl ⊗ Y1/2.This is also the only solution of the equation L+x = 0 within the space trans-forming according to this Cartan weight. Applying the operator L− to this


function we can find all functions transforming according to Dl+1/2.Exercise: find the function within the invariant space of Dl+1/2 which trans-forms according to the l − 1/2 Cartan weight.

We now consider the space which carries the Cartan weight l−1/2. There aretwo functions in this space, namely Xl ⊗Y−1/2 and Xl−1 ⊗Y1/2. One possiblesolution of the equation

I+(αXl ⊗ Y−1/2 + βXl−1 ⊗ Y1/2) =

α[(I+Xl) ⊗ Y−1/2 +Xl ⊗ (I+Y−1/2)]

+β[(I+Xl−1) ⊗ Y1/2 +Xl−1 ⊗ (I+Y1/2)]

= [1√2α + β

√l]Xl ⊗ Y1/2 = 0

is α =√

2l2l+1

and β = −√

12l+1

. This means that the function

√2l

2l + 1Xl ⊗ Y−1/2 −

√1

2l + 1Xl−1 ⊗ Y1/2

belongs to the invariant subspace of Dl−1/2.Exercise: check that the two functions transforming according to the sameCartan weight l − 1/2 but belonging to different irreducible representationsare orthogonal.Exercise: find all symmetry adapted functions in the reduction of D1⊗D1/2

(these are the functions belonging to the spin orbit split levels 2P3/2 and2P1/2.)The coefficients determining the symmetry adapted product functions, i.e.those basis functions that transform according to a given weigth of a givenirreducible representation in the reduction ofDl⊗Dk are called the Clebsch-Gordan coefficients:

ψ(j,mj) =∑

ml,mk

c(l, k, j,ml, mk, mj)ψ(l,ml) ⊗ ψ(k,mk)

The Clebsch-Gordan coefficent c(l, k, j,ml, mk, mj) is often written as the 3j

symbol

(l k jml mk mj

).

Exercise: show thatml+mk = mj, otherwise the CC-coefficient is vanishing.

1.3.1 Clebsch-Gordan coefficients for point groups

For some physical problems, the full rotational group is not a symmetrygroup of the system. Let us consider a single electron atom embedded – say–


in a cubic cell. The potential experienced by the electron has a symmetrygroup with much less elements than SO(3). In fact, the symmetry group ofthe Hamiltonian is the cubic group Oh, which transforms a cube into itself.Oh is a subgroup of SO(3) with discrete number of elements. Its elements aresummarized in the follwing table as the result of operating on a coordinatesystem placed in the center of the cube. With the aid of the figure, we canillustrate all 48 symmetry elements of Oh.

Figure 1.1: The 48 operations of Oh: the identity e, 3 rotations by π aboutthe axes x, y, z (3C2

4), 6 rotations by ±π/2 about the axes x, y, z (6C4), 6rotations by π about the bisectrices in the planes xy, yz, xz (6C2), 8 rotationsby ±2π/3 about the diagonals of the cube (8C3), the combination of theinversion I with the listed 24 proper rotations

The question is now: what happens to the level scheme of the electronwhen such a symmetry breaking potential is switched on? From the point ofview of group theory, the answer to this question is quite straightforward. Anirreducible representation of SO(3) might no longer be irreduciblewhen it is limited to the elements of a subgroup of SO(3). This leads,in general, to a level splitting, i.e. a reduction of the degeneracy, which is of-ten encountered in the band structures of crystals.Consider for instance the point Γ in the Brillouin zone of the GaAs crystal.The point group symmetry of a GaAs crystal at the Γ-point is Td, which is asubgroup of Oh. The top of the valence band is described (we initially neglectspin) by the single group irreducible representation Γ15 of the group Td. TheΓ15 irreducible representation is a three dimensional one, and it coincideswith D1. Thus, the top of the valence band behaves exactly as if the atomwould have the full rotational symmetry. This is why we can use the CGcoefficients derived from the full rotational group when calculating the spinpolarization of electrons excited in GaAs. This is, however, a coincidence:


Type Operation Coordinate Type Operation CoordinateC1 C1 xyz I I xyzC2

4 C2z xyz IC24 IC2z xyz

C2x xyz IC2x xyzC2y xyz IC2y xyz

C4 C−14z yxz IC4 IC−1

4z yxzC4z yxz IC4z yxzC−1

4x xzy IC−14x xzy

C4x xzy IC4x xzyC−1

4y zyx IC−14y zyx

C4y zyx IC4y zyxC2 C2xy yxz IC2 IC2xy yxz

C2xz zyx IC2xz zyxC2yz xzy IC2yz xzyC2xy yxz IC2xy yxzC2xz zyx IC2xz zyxC2yz xzy IC2yz xzy

C3 C−13xyz zxy IC3 IC3xyz zxy

C3xyz yzx IC3xyz zxyC−1

3xyz zxy IC−13xyz zxy

C3xyz yzx IC3xyz yzxC−1

3xyz zxy IC−13xyz zxy

C3xyz yzx IC3xyz yzxC−1

3xyz zxy IC3xyz zxyC3xyz yzx IC3xyz yzx

Table 1.1: Coordinate transformations under the action of the elements ofthe group Oh


for instance, the D2 irreducible representation of SO(3), which carries thed-electrons, does indeed split at Γ into two irreducible representations of Td.In addition, as soon as we move away from the Γ-point things change dra-matically. We consider now, as an example, the Λ direction in the Brillouinzone. The symmetry group of this direction is C3v.

The single group C3v

The single group C3v contains six elements:

E,C3xyz, C−13xyz, IC2xy, IC2xz, IC2yz, all transforming the k-vector along the

Λ-direction into itself (kΛ = π/a(λ, λ, λ) with 0 < λ < 1). The symbols C3xyz

and C−13xyz, for instance, means rotations by an angle 2π/3 around an axis

with director cosines on the ratio 1 : 1 : 1, see the table and figure for Oh. Toconstruct the character table of the group one must find first the classes ofconjugate elements, because we then know that, for finite groups, the numberof irreducible representations exactly equals the number of classes. There arethree classes: the identity element, the two there-fold rotations and the threereflections. We summarize now some rules for constructing the character table of a finitegroup.

1. It is convenient to display in table form the characters of the irreducible represen-tations. Such a table gives less information than a complete set of matrices, but itis sufficient for classifying the electronic states

2. The number of irreducible representations is equal to the number of classes in thegroup

3. The sum of the squares of the dimensions lα of the irreducible representations isequal to the number of elements g of the group∑

α

l2α = g

4. The characters of the irreducible representation must be mutually orthogonal andnormalized to the order of the group:∑

a

χα(a)∗χβ(a) = g · δαβ

5. Every group admits the one-dimensional identical representation in which eachelement of the group is represented by the number 1. The orthogonality relationbetween characters then shows that for any irreducible representation other thanthe identity representation ∑

a

χ(a) = 0

With the help of these rules we can easily construt the character table ofthe single point group C3v. We are now able to investigate the fate of the top


Irr.Rep. E 2C23 3σv

Λ1 1 1 1Λ2 1 1 -1Λ3 2 -1 0

C3v E 2C23 3σv

Γ15 3 0 1

valence band eigenvalue at Γ. Γ15, restricted to the elements of C3v, is also arepresentation of C3v. It is, however, not irreducible, and its characters aregiven in the Table. Using the theorem for finding the direct sum expansionof reducible representations, we obtain Γ15 = Λ1 ⊕ Λ3. This means that weexpect the theree fold degenerate eigenvalue Γ15 to split into a once degener-ate Λ1-band and a doubly degenerate Λ3 band along the Λ direction. Usingthe projection operator method we can construct symmetrized linear combi-nations transfroming according to each particular irreducible representation.

The double group CD3v

This group has twice the number of elements and twice the number of classesas C3v. The matrices representing the 2C3 group elements in the D1/2 repre-

sentation can be constructed by using l = m = n =√

13

and a rotation angle

±ϕ = 2π3

:

C3xyz =

(1−i2

−1+i2

1−i2

+1+i2

)

Similarly, one can obtain the matrices IC2xy, IC2xz, IC2yz by considering thatthe inversion is followed by the rotation of 2π/2 around an axis with directorcosines in the ration 1 : 1 : 0, i.e. ϕ = π and l = /

√2, m = −1/

√2 n = 0.

Thus, for instance,

IC2xy = I ·(

0 1−i√2

−1+i√2

0

)

The remaining matrices can be constructed in a similar way.

E =

(1 00 1

); E =

(1 00 1

)

C3xyz = 1/2

(1 − i −1 − i1 − i 1 + i

); C3xyz = 1/2

( −1 + i 1 + i−1 + i −1 − i

)


Irr.Rep. E E 2C3 2c3 3σv 3σv

Λ1 1 1 1 1 1 1Λ2 1 1 1 1 -1 -1Λ3 2 2 -1 -1 0 0Λ4 1 -1 -1 1 i -iΛ5 1 -1 -1 1 -i iΛ6 2 -2 1 -1 0 0

C−13xyz = 1/2

(1 + i 1 + i−1 + i 1 − i

); C−1

3xyz = 1/2

( −1 − i −1 − i1 − i −1 + i

)

IC2xy =√

1/2I

(0 1 − i

−1 − i 0

); IC2xy =

√1/2I

(0 −1 + i

1 + i 0

)

IC2xz =√

1/2I

( −i ii i

); IC2xz =

√1/2I

(i −i−i −i

)

IC2yz =√

1/2I

(i −11 −i

); IC2yz =

√1/2I

( −i 1−1 i

)

The 12 elements are thus divided into 6 classes, and the equation∑

α(lα)2 =12 has the solution 12 + 12 + 22 + 12 + 12 + 22 = 12. Thus, there are atotal of four one dimensional representations and two two-dimensional rep-resentations. The three irreducible representations of C3v can be extendedas irreducible representations of CD

3v by representing E with E and the Celements as the non-C elements. There are three extra additional irreduciblerepresentations to be found. We notice that the two dimensional D1/2 repre-sentation is irreducible: in fact with the characters

χ(E) = 2, χ(E) = −2, χ(C3) = 1, χEC3 = −1, χ(σv) = 0

we have∑

A χ(A)2 = 12, which is a necessary and sufficient condition foran irreducible representation. The remaining two-extra one dimensional rep-resentations are found by applying the unitarity conditions. The charactertable is given above. (Λ6 is the name of D1/2 when restricted to this pointgroup).

We now would like to know what happens to the crystal split bands Λ3

and Λ1 when spin-orbit coupling is switched on. In the absence of spin-orbitcoupling, the Hamiltonian is invariant with respect to separate rotations inorbit and spin space, i.e. the symmetry group is the direct product of twogroups, namely G × SU(2). We consider a subspace spanned by the eigen-functions f1, ..., flα which act as basis for representation Dα a the single


Irr.Rep. E E 2C23 2C2

3 3σv 3σv

Λ1 +1 +1 +1 +1 +1 +1Λ2 +1 +1 +1 +1 -1 -1Λ3 2 2 -1 -1 0 0Λ4 +1 -1 -1 +1 i -iΛ5 +1 -1 -1 +1 -i iΛ6 2 -2 +1 -1 0 0

D1/2 2 -2 +1 -1 0 0

Λ1 ⊗D1/2 2 -2 1 -1 0 0Λ2 ⊗D1/2 2 -2 1 -1 0 0Λ3 ⊗D1/2 4 -4 -1 1 0 0

group of the Hamiltonian, i.e, when the spin is neglected. The product of thefunctions fi with u+, u− which are the basis functions for the representationD1/2, constitutes a basis for the representation Dα ⊗D1/2 of the direct prod-uct group G× SU(2), with characters χα(R) · χ1/2(R). If the the symmetrygroup of the Hamiltonian consists of independent operations in coordinateand spin space, then Dα⊗D1/2 is irreducible. Should the Hamiltonian containspin-orbit coupling, then the symmetry group of the Hamiltonian containsonly simultaneous operations in coordinate and spin space, i.e. it is GD: asa consequence, the direct product representation, when limited to simoul-taneous operations, is, in general, reducible. Using known theorems, we canreduce the product representation to a sum of irreducible representations ofthe double group, and we can also find the linear combinations of productfunctions building the basis for the irreducible representations of the doublegroup (Clebsch-Gordan coefficients). The number of irreducible representa-tions contained in the sum and the dimension of the representation givesthe splitting of the eigenvalue and the degeneracy of the sublevels when spinterms – like spin-orbit interaction – are included in the Hamiltonian. Con-sider now the Λ direction in k space again. The character table of CD

3v isextended to include the characters of Λi ⊗D1/2. From the characters of theadditional representations one sees that

Λ1 ⊗D1/2 = Λ6

Λ2 ⊗D1/2 = Λ6

Λ3 ⊗D1/2 = Λ4 + Λ5 + Λ6

Thus, the Λ3 band split into three subbands, each carrying one irreduciblerepresentation of the double group. The band Λ1 remains degenerate and


carries the irreducible representation Λ6 of the double group.

The symmetry adapted basis functions

We might ask now what are the symmetry adapted wave functions belongingto these bands. The answer to this question is a nice piece of application ofthe CGC for point groups. As the groups involved are no longer the continuosgroup SO(3), the method of Cartan can no longer be applied. We have toresort to a different method, which is called the projection operator method.This method is based on the following theorem:Theorem (without proof): The operator

P (p) =lpg

∑T

χ∗(p)(T )OT

projects out of any normalizable wave function the sum of all basis functionstransforming according to the columns of the p-th irreducible representationof any group G.Having determined φ(p), we can then apply to φ(p) the operations of the sym-metry group to obtain exactly lp-linear independent functions which can bemade orthonormal and thus can be used as a basis set for the irreduciblerepresentation Dp (the application of g operation of the group to a functionbelonging to an invariant subspace carrying the irreducible representationDp cannot bring us outside this subspace). Alternatively, one can apply theprojector operator to a different function with the hope that the resultingfunction is a further linearly independent basis function transforming as Dp.Notice that, if the irreducible representation is one dimensional, this operatorproduce the only required symmetry adapted basis function.The important aspect of this theorem is that it is only necessary to knowthe characters of a given representation in order to obtain the whole setof symmetry adapted basis functions transforming according to it. Noticethat, after having generated the basis functions transforming according tothe irreducible representation Dp, it is possible to construct the matrix rep-resentation of Dp itself (in practice, this is the procedure most adopted toconstruct matrix representations of the irreducible representations), using

OTφ(p)n =

∑m

Dpnmφ

pm

Thus, in is possible to generate a complete matrix representation knowingonly the characters.


Example 1.

We would like to find, within the subspace Y00, Y11, Y10, Y1−1, the symmetryadapted wave functions transforming according to Λ1 and Λ3. We start forinstance from Y00 and we apply the projection operator:

PΛ1Y00 =1

6

∑1 · Y00 = Y00

This means that Y00 transforms according to Λ1. The lowest conduction bandalong the Λ direction of GaAs or Ge is indeed a s-state with Λ1 symmetry.Applying the projector operator to the same eigenfunctions in oder to projectout the wave functions transforming according to Λ3 will inevitably leads tozero, as Y00 does not contain parts transforming like Λ3. Starting from Y10

we obtain (for convenience, we set the z axis along the Λ direction)

PΛ1Y10 =1

6

∑1 · Y10 = Y10

i.e. Y10 transforms according to Λ1 as well. The deep valence Λ1 band alongthe Λ- direction in GaAs and Ge has mostly z-character. In order to computethe transformation properties of x, y we use

Oϕf(x, y) = f(x cosϕ+ y sinϕ,−x sinϕ+ y cosϕ)

The projector applied to x respectively y gives x respectively y: thus theyare basis functions carrying the representation Λ3. The highest valence bandalong Λ has this symmetry. Thus, the p-states, which are degenerate at theΓ-point, split into two components, the one containing Y10 (Λ1) and the onecontaining Y1±1 (Λ3).

Example 2.

So far we have neglected spin-orbit coupling. This interaction leads to furthersplitting of bands, as we shall illustrate now. We would like to find the linearcombinations of

X1Y1/2, X1Y−1/2, X13Y1/2, X

13Y−1/2, X

−13 Y1/2, X

−13 Y−1/2

transforming according to the irreducible representations Λ6, Λ4 and Λ5 ap-pearing in Λ1 ⊗ D1/2 and Λ3 ⊗ D1/2. We will use as basis functions for Λ3

Y±1: the matrices representing rotations by an angle ϕ aroung the z axis inthis basis are diagonal, with diagonal elements exp(∓iϕ).Exercise: prove this last statement.


The matrices representing such rotations in the D1/2 representation are alsodiagonal, their matrix elements being exp(∓iϕ

2).

Exercise: prove this.The symmetry adapted basis functions build up from orbital wave functionof the type X1 and spin fuctions Y±1/2 can be found from

PΛ6X1Y±1/2 ∝ X1Y±1/2

Thus, the Λ6 band originating from the coupling of a Λ1 band with spin isdoubly degenerate and its basis functions are simpy X1Y±1/2. The symmetryadapted basis functions transforming according to the Λ6 irreducible repre-sentation within the subspace containing the four functions X±1

3 Y±1/2 arefound again by the projector method.

PΛ6X13Y1/2 ∝ 4 ·X1

3Y1/2 + 4 cos3

2ϕX1

3Y1/2 = 0 = PΛ6X−13 Y−1/2

so that the wave functions X±13 Y±1/2 do not transform as Λ6. In contrast,

PΛ6X±13 Y∓1/2 ∝ 4 ·X±1

3 Y∓1/2 + 4 cosϕ

2X±1

3 Y∓1/2 ∝ X±13 Y∓1/2

i.e. the two wave functions X±13 Y∓1/2 belong to the Λ6 representation. Thus,

the Λ6 band derived from the coupling of Λ3 orbital wave functions with thespin is doubly degenerate and contains the functions X1

3Y−1/2 and X−13 Y1/2.

The remaining two functions must be in the subspace transforming accordingto Λ4 and Λ5. We notice that the time reversal operation is also a symmetryoperation of the Hamiltonian, and states which are transformed into eachother by the time reversal operator must belong to the same energy eigen-value. One can show that the two states X±1

3 Y±1/2 are exactly such states,i.e. states related by time reversal symmetry, so that the bands belonging toΛ4 and Λ5 symmetry are degenerate when it comes to their energy (althoughthey behave differently under the operation of the point group). In the bandstructure of GaAs and Ge the Λ3 band is found to generate only two spin-orbit split bands, namely Λ6 and Λ4+5. The level scheme, used for instancein band structure theory or to study optical transitions along the Λ directionin GaAs or Ge is given in the following figure.

1.4 The Wigner-Eckart theorem

The construction of the symmetry adapted wave functions is one of the mainapplications of of group theory to quantum mechanical problems. The second


Figure 1.2: Level scheme Λ-direction. The arrows show allowed optical tran-sition (see later in this chapter)

one relies on the Wigner-Eckart-Koster theorem. The WE (Koster) theoremis the consequence of the following Lemma:Lemma: Let G be a unitary matrix group with measure and φ(k,mk, r) be abasis function transforming according to the mk-th column of the irreduciblerepresentation Dk of the group G.The index r must be introduced because functions having the same symmetry under theoperations of G might differ on aspects other then their symmetry. In the case of SO(3), forexample, functions which have a certain symmetry under rotations might have a differentradial dependence (i.e. they might have a different quantum number n).Let also φ(l,ml, s) be a basis function transforming according to the ml-thcolumn of the irreducible representation Dl. Then(

φ(k,mk, r), φ(l,ml, s))

= δklδmkmlC(k)

rs

i.e. the scalar product of two basis functions is vanishing if the basis functionsbelong to different irreducible representations or to different columns of thesame irreducible representation. Else is the scalar product a constant whichdoes not depend on mk but only on k, r, s.Proof: Since the scalar product of two functions is invariant under unitarytransformations, for any operation T ∈ G(φ(k,mk, r), φ(l,ml, s)

)=

(OTφ(k,mk, r), OTφ(l,ml, s)

)=⇒

(φ(k,mk, r), φ(l,ml, s)

)= (

lk∑mp

Dkmpmk

(T )φ(k,mp, r),ll∑

mq

Dlmqml

(T )φ(l,mq, s))

=∑

mp,mq

D∗kmpmk

(T )Dlmqml

(T )(φ(k,mp, r), φ(l,mq, s))

Taking the measure over both sides we obtain(φ(k,mk, r), φ(l,ml, s)

)=

∑mp,mq

(φ(k,mp, r), φ(l,mq, s)) ·(MT∈GD

∗kmpmk

(T )Dlmqml

(T ))


=1

lkδklδmkml

∑mpmq

δmpmq(φ(k,mp, r), φ(l,mq, s))

= δklδmkml

1

lk

∑mp

(φ(k,mp, r), φ(l,mp, s))

This proves the orthogonality of the basis functions. As the right hand side isindependent of mk and ml, provided mk and ml are equal, the scalar productdoes not depend on the column index but only on k, r and s. The remainingconstant i.e. 1

lk· ∑mp

(φ(k,mp, r), φ(l,mp, s)) must, in general, be explicitlycomputed, as symmetry arguments no longer help.

This orthogonality Lemma for basis functions can be applied for calculat-ing the matrix elements of operators (this application goes often under thename of Wigner-Eckart theorem).Definition: The set of ll operators Ql

1(r), Ql2(r), ...., Q

lll(r) transform accord-

ing to the columns of the ll-dimensional irreducible representation Dl of thegroup G if for every T of G

OTQli(r)O

−1T =

∑p

DlpiQ

lp(r)

This equation has to be understood as an operator equation. That is, bothsides must produce the same result when acting on any scalar function. Wealso assume that in the left-hand side every operator acts on everything toits right. We consider now matrix elements of one of these operators be-tween scalar functions transforming according to the columns of irreduciblerepresentations:

OTφ(k,mk) =∑mq

Dkmqmk

φ(k,mq)

OTφ(j,mj) =∑mr

Djmrmj

φ(j,mr)

and (φ(j,mj), Q

liφ(k,mk)

)As both Ql

i and φ(k,mk) transform according to representations of the sym-metry group, the wave function Ql

iφ(k,mk) will be in the subspace trans-forming according to the direct product representation Dl ⊗Dk. The repre-sentation constructed over the set of basis function Ql

iφ(k,mk), i = 1, ..., ll,mk = 1, ..., lk will be in general reducible, according to Dl ⊗Dk = ⊕snsD

s,and the function Ql

iφ(k,mk) will occur in the subspace of one or more ofthese irreducible representations. Thus, according to our previous theorem,


this function will have non-vanishing matrix element only with wave func-tions transforming according to the same irreducible representations. Thismeans that a matrix element involving a final state wave function transform-ing according to a representation Dj which is not contained in the directsum decomposition of the representations Dl ⊗ Dk, is certainly vanishing(selection rule theorem). Notice that, to apply this selection rule the-orem to a concrete example, only the Clebsch-Gordan series is needed, sothat the vanishing of a matrix element can be determined with certaintyand using symmetry arguments only. Thus, group theory forbid, with ab-solute certainty, certain matrix elements. Those which are not forbidden bysymmetry can still be vanishing by some reason other than symmetry. Thisselection rule theorem is, however, the weakest form of a more general ma-trix element theorem, known as the Wigner-Eckart-Koster theorem (G.F.Koster, Phys. Rev. 109, 227 (1958)). In fact, if we consider more closely theprocedure adopted to analyze the matrix element, we will notice that thewave function Ql

iφ(k,mk) enters the subspace transforming according to Dj

exactly nj times, nj being specified by the general formulas for the decom-position of product representations. The coefficients of the linear combina-tion determining the various basis functions, for which the representationDl ⊗ Dk is in block form, are the Clebsch-Gordan coefficients and can bedetermined by group theoretical arguments (for SO(3) see Cartan’s algorith-mus). Clearly, the basis function Ql

iφ(k,mk) will enter these basis functionswith given Clebsch-Gordan coefficients. Thus, the matrix element will be alinear combination of nj Clebsch-Gordan coefficients C(l, k, j,ml, mk, mj , r).Notice that for SO(3) nj = 1 and only one coefficient is needed to describematrix elements. In summary: if the irreducible representation Dj occurs nj

times in the direct sum decomposition of Dl ⊗Dk, then(φ(j,mj), Q

liφ(k,mk)

)= a1 · C(l, k, j,ml, mk, mj, 1)

+ a2 · C(l, k, j,ml, mk, mj, 2)

+ ......

+ anj· C(l, k, j,ml, mk, mj, nj)

The coefficients ai must be calculated explicitely.

Example 1: Spin polarized excitation in a spherically symmetric poten-tial (or at the Γ-point of a cubic lattice) by circularly polarized light.Consider the optical transition between p and s states. The operator produc-ing a transition in an atom is proportional to ( A · r). If we choose lineralypolarized light along z and circularly polarized light in the (x− y)-plane, weobtain the operators A0 ≈ z ≈ Y10 and A± ≈ x± iy ≈ Y1±1. These operators


transform according to the representation D1 of SO(3). As a consequence,p → s, p → p and p → d transitions are optically allowed. The p → p tran-sition is forbidden if one consider that D1-wave functions are uneven withrespect to the parity operation, so that an optical transition is only allowedbetween states of different parity (if the inversion is a symmetry element).The sought for matrix elements are

< 0, 0 | Y1,i | Y1,j >

i = 1(−1) for right(left)-hand circularly polarized light and j = 1, 0,−1.The states on the right-hand side are all possible initial states and the stateY0,0 is the final state of the p → s transition. To find the value of thesematrix elements we search for the linear combinations of product functions| 1, m >| 1, m > which transfom according to the D0. There is only onelinear combination, which can be read out from the table for Clebsch-Gordancoefficients given e.g. in the book by E.U. Condon and G.H. Shortley, ”TheTheory of Atomic Spectra”, Cambridge Univeristy Press, 1935. From Table2 on p.76 we obtain

| 0, 0 >=

√1

3| 1,−1 >| 1, 1 > −

√1

3| 1, 0 >| 1, 0 > +

√1

3| 1, 1 >| 1,−1 >

From this last equation we deduce that the only non-vanishing matrix ele-ments are

< 0, 0 | Y1,−1 | Y1,1 >=< 0, 0 | Y1,1 | Y1,−1 >=

√1

3

In the spirit of the Wigner-Eckart theorem, these non-vanishing matrix ele-ments are determined up to a common constant, which includes the detailsof the radial part of the wave function. If we consider now the spin-orbitsplit states p3/2 and p1/2 constructed by means of the linear combination ofp = 1 wave functions multiplied by spin functions, we obtain the schemeshown in the figure. Accordingly, for e.g. left hand circularly polarized light(Y1,−1) there are two possible transitions from the p3/2 states and one fromthe p1/2 state. The relative strength of the transition from the p3/2 leading tothe wave function with ”up” spin and of the one leading to ”down” spins is3. Accordingly, the spin polarization of the electrons excited into the s- stateis

< σz >=(3 − 1)

(3 + 1)= 50%

If these electrons are in some way extracted, we will have a spin polarizedelectron beam. The spin polarization is an ideal quantity for applying the


Figure 1.3: Level scheme for an optical transition at the Gamma point ofGaAs.

WE-theorem, because all details other than symmetry cancel out completely.Example 2: spin polarization of the optical transitions along the Λ-direction.We refer to the figure above describing the spin-split bands along the Λ-direction. The opical transitions indicated for circularly polarized light pro-duces polarized electrons, as one can easily derive using the WEK-theorem.There is, however, something very unphysical in this level scheme: the spinpolarization achieved is 100%, albeit with opposite sign when starting fromthe Λ6 and Λ4+5 band. Upon approaching the Γ-point, both bands comeclose, so that one must assume that both are excited. As the matrix elementfor the excitation is the same, we obtain the unphysical result that the netspin polariztion goes to zero when the Γ point is approached, although ourcalculation at the Γ-point gives a spin polarization of 50%. Clearly, there issomething wrong in the level scheme. The solution to this contradiction isthat bands with the same symmetry – in this case Λ6 symmetry, are allowedto hybridize, in particular when they are close in energy. This means thatthey are allowed to contain orbital components coming from different singlegroup symmetries, provided these orbital components end up to have thesame double group symmetry when the spin is introduced. As we approachthe Γ-point, the two Λ6 bands originating from the single group irreduciblerepresentation Λ1 and Λ3 come also close together, so that we might hybridizethem. The recipe for determining the degree of hybridization is given by therequirement that the net spin polarization achievd when both Λ6 and Λ4+5

are simoultaneouly excited is 50%. This is obtained by modifying the levelscheme as in the following figure.


Figure 1.4:

Figure 1.5: Level scheme for optical transitions including hybridized bands.

Chapter 2

Energy bands in solids

So far we have assumed that the discrete symmetry observed in crystals canbe dealt with using point groups, i.e. symmetry groups which leave one pointinvariant. We will soon see that this approach is a useful one in many cases.However, strictly speaking, a crystal consists of an infinite regular array ofidentical unit cells, and translations – which move all points in space – mightbe symmetry operations as well. Thus, we must consider so called spacegroups.

2.1 The translation group

A crystal has the fundamental property of remaining unchanged under thetranslation tn = n1a1 + nea2 + n3a3, where the subscript n on t indicatesa collection of three integers n1, n2, n3 and ai are the primitive translationvectors. The translation operation can be indicated by a symbol introducedby Seitz, (E | tn), where R = E indicate the identity operation in the generalmapping r → rt = Rr+tn. In order for the functions defined over the crystalto be normalizable (square integrable), we consider the crystal to have finitedimensions and to terminate at the planes r ·a1 = N1, r ·a2 = N2, r ·a3 = N3,Ni being large but finite integer numbers. In order to preserve translationalsymmetry, we impose the cyclic (Born-von Karman) boundary conditions

(E | t(N1,0,0)) = (E | t(0,N2,0)) = (E | t(0,0,N3)) = (E | 0)

The collection of elements with this boundary condition constitutes a finitegroup T with as many elements as the number of unit cells of the crystal,i.e. g = N1 · N2 · N3: in fact (E | tn) + (E | tm) = (E | tn + tm), which isagain an element of the set, even if the sum of the two vectors lie outside thecrystal, in virtue of the boundary condition. The group is Abelian, because

25

CHAPTER 2. ENERGY BANDS IN SOLIDS 26

all transformations consisting of pure translations commute, so that each ele-ment is a class on its own and the group has only one-dimensional irreduciblerepresentations. To find the irreducible representations, notice that the fulltranslation group T is the direct product of the three translation groups T1,T2 and T3. This groups have orderNi and elements n1ai, n2ai, ......nsai, ...Niai,with nsai = (ai)

ns, i.e. they form a cyclic groups. The matrix element of theirreducible representation Dpi of the element nsai is,

e−i·2π

Nipi·ns

with pi = 0, ..., Ni−1 labeling the irreducible representations. As a conse-quence, the translation (E | tn) is represented by

e−i( 2π

N1p1·n1+

2πN2

p2·n2+2πN3

p3·n3)

the set of integers p1, p2, p3 labeling an irreducible representation. We cansimplify this formula by introducing the following notation. Define the basicvectors of a fictional reciprocal lattice b1, b2, b3 by ai

bj = 2πδij, i, j = 1, 2, 3,so that explicitly

b1 = 2πa2 × a3

a1 · (a2 × a3)

b2 = 2πa3 × a1

a1 · (a2 × a3)

b3 = 2πa1 × a2

a1 · (a2 × a3)

then define the so called allowed k-vectors

k = k1b1 + k2

b2 + k3b3

with kj = pj/Nj. Because of the special definition of the basis vectors of thereciprocal lattice, we have the important equation

k · tn =2π

N1p1 · n1 +

2π

N2p2 · n2

2π

N3p3 · n3

so that the element (E | tn) is represented in the k-irreducible representationby

D(�k)(E | tn) = e−i·�k·�tn

where now the N irreducible representations are labeled by the N allowed kvectors. These representations are clearly unitary. The N irreducible repre-sentations of T are described by the allowed k vectors. These k vectors can


Figure 2.1:

be imaged as laying on a very fine lattice within and upon three faces of theparallelepiped having edges b1,b2,b3 that is shown in the figure. With thebasis vectors of the reciprocal space one can define a set of lattice vector inthe reciprocal lattice by Km = m1

b1 + m2b2 + m3

b3 where mi are integers.Notice that

e±i· �Km·�tn = 1

so that two k-vectors differing by a reciprocal lattice vector label the sameirreducible representation. Using this property, it is possible to redefine theallowed k-region into a more symmetric polyhedron in reciprocal space, whichis called the first Brillouin zone, and consists of all points of k space thatlie closer to k = 0 than any other reciprocal lattice points. Its boundariesare therefore the planes that are perpendicular bisectors of the lines joiningthe point k = 0 to the nearer reciprocal lattice points. Thus, only the firstBrillouin zone is required to label completely all irreducible representationof T (and ultimately all electronic states of a crystal). In the following, wegive the first Brillouin zones of important lattices and indicate the label usedto indicate special point in the Brillouin zone, which we will discover laterto have special extra symmetries (the present labeling was introduced byBouckaert et al., Phys. Rev. 50, 58 (1936).Body centered cubic lattice.For this lattice,

b1 =2π

a(0, 1, 1)

b2 =2π

a(1, 0, 1)

b3 =2π

a(1, 1, 0)

The Brillouin zone is shown in the figure. The position vectors of the high


Figure 2.2:

symmetry points (see later) marked are as follows:

k = (0, 0, 0) → Γ

k =π

a(2, 0, 0) → H

k =π

a(1, 1, 0) → N

k =π

a(1, 1, 1) → P

Face centered cubic lattice.For this lattice,

b1 =2π

a(−1, 1, 1)

b2 =2π

a(1,−1, 1)

b3 =2π

a(1, 1,−1)

The Brillouin zone is shown in the figure. The position vectors of the highsymmetry points (see later) marked are as follows:

k = (0, 0, 0) → Γ

k =π

a(2, 0, 0) → X

k =π

a(3/2, 3/2, 0) → K

k =π

a(1, 1, 1) → L


Figure 2.3:

Basis functions

Let φ(r) be a scalar function in the Hilbert space of the crystal and (E |t) be a transformation mapping any vector r into rt = r + t. Under thistransformation, the scalar function φ will be transformed into the scalarfunction φt

.= O(E|�t)φ, defined by

O(E|�t)φ(r) = φ(r − t)

If φ is a basis function of a irreducible representations of T , φ will obey thefollowing equation:

O(E|�tn)φ(r) = φ(r − tn) = e−i�k·�tnφ(r)

By multiplying on the both sides with e−i�k(�r−�tn) we obtain

e−i�k(�r−�tn)φ(r − tn) = e−i�k·�rφ(r)

which shows that e−i�k·�rφ(r).= u(r) is a function invariant under translations

(a periodic function). Thus, the basis functions of the irreducible representa-tion of the translation group can be written in the familiar form first obtainedby Bloch

φ(r) = ei�k·�ru(r)

where u(r) = u(r − tn). This equation is known as Bloch theorem, whichin this case has been derived from symmetry properties alone. The wavefunctions are called Bloch functions. As the Hamiltonian H of a crystal com-mute with the operator O(E|�tn), its eigenfunctions must be Bloch functions. In


general, there is an infinite set of energy eigenfunctions and eigenvalues corre-sponding to each irreducible representation of the group of the Schroedingerequation and hence to each allowed k vector. Eigenfunctions and eigenvaluescorresponding to the vector k can be indicated by φn(k,r) and En(k), wherethe index n labels the members of the set of energies corresponding to thewave vector k. The set of energy eigenvalues En(k) corresponding to a par-ticular n are said to form the n−th-energy band, and the set of energy bandsis said to constitute the band structure. To visualize the band structure, itis convenient to consider one at a time the axes of the Brillouin zone thatjoin the high symmetry points and for every allowed k vector on each axisto plot the energy levels En(k). A typical example of such a plot is shownin the figure, which gives the energy bands along ∆ and Λ in fcc Cu. Theoccurrence of degenerate values and the apparent opening of gaps within aband of a given index is a consequence of the rotational symmetry which hasbeen so fare neglected.In the one electron approximation on which the whole theory is based, thePauli exclusion principle implies that no two electrons can ’occupy’ the sameelectron state, here specified by an allowed k vector, a band index n anda spin quantum number that can only take two possible values. It followsthat each energy level En(k) can ’hold’ two electrons and hence each energyband can hold 2N electrons. If there are V valence (or conduction) electronsper atoms, and A atoms per unit cell, there will be NV A electrons in thelarge basis block of the crystal which will therefore require 1

2V A bands to

hold them. In the ground state of the system, all energy levels En(k) will bedoubly occupied up to a certain energy EF , the Fermi energy and all levelsabove this energy will be unoccupied. The surface in k space defined by

EF = En(k)

is called the Fermi surface. If one and only one band contains the Fermienergy, and all other are entirely above or below it, then the Fermi surfacemerely consist of one sheet. If no band contains the Fermi surface, as hap-pened for insulators and semiconductors, there is no Fermi surface. In allother cases, the Fermi surface consists of several sheets, and it is the dis-tribution of energy levels near the Fermi energy that largely determines theelectronic properties of a solid.

2.2 Space groups

A crystal structure is fully specified by the primitive vectors a1,a2,a3 of thetranslation group and by the basis of vectors d1, d2, ..., dj which determine the


Figure 2.4:

position of the atoms in the unit cell. There are symmetry operations whichleaves the crystal unchanged. Besides the translation symmetry operations,proper or improper rotations, followed by an appropriate translation, whichis not necessarily a primitive translation, can also transform a crystal intoitself. The symmetry elements can be denoted by (R | a), where R is a realorthogonal matrix indicating a proper or improper rotational part, and a isan appropriate translation. All possible translation vectors associated with Rhave the form a = tn + fR, where tn is the translation defined in the previoussection and fR is a fractional translation required for some rotations. Themultiplication of two elements is defined as

(R | a)(S | b) = (RS | Rb+ a)


(R | a)−1 = (R−1 | −R−1a)

Provided the result of the multiplication also leaves the crystal unchanged,the set of symmetry transformation form a group, which is called the spacegroup. There is a restriction on the possible rotations R that can occur inthe operations of the space group, namely that if (R | tn) is a memberof the space group and tn is a primitive translation, then Rtn must alsobe a primitive translation. This follows because if (R | tn) is a member ofthe group, so is (R | tn)(E | tn)(R | tn)−1 = (E | Rtn) which is a puretranslation and therefore must be a primitive one. Let us find out whichrotational symmetries are compatible with the translational symmetry. LetA (see the Figure) be a point of a crystal lattice, and let us assume that it

Figure 2.5:

has a n-fold rotational axis passing though it. In B there is another latticepoint, and A and B can be joined by a primitive translation vector. Let, forsimplicity, the rotational axis be perpendicular to the plane formed by A andB. We perform now a rotation by ϕ = 2π/n about A. This bring B into B′.Because B is equivalent to A there is also a n fold axis passing though B,and a rotation by −ϕ brings A into A′. As the rotation is a symmetry ofthe lattice, A′ and B′ are two lattice point, and their distance must be pa, pbeing some integer and a the lattice constant. From the figure we determinethe following trigonometric equation:

a+ 2a sin(ϕ− π

2) = a− 2a cosϕ = pa

i.e.

cosϕ =1 − p

2

As | cosϕ |≤ 1, we can only allow p to be 3, 2, 1, 0. These values lead to thepossible n values being 2, 3, 4, 6. A crystal can only have two-three-four-or six fold rotational axes. Notice that the rotational part of a space groupconstitute a group by itself: it is therefore one of the point groups. Because ofthe restriction on the rotational symmetry, only 32 point groups are allowed


as rotational part of a space group. Space groups G having the same pointgroup G0 belong to the same crystal class, and there are 32 different crystalclasses. In the classification of Schonflies, a space group is denoted by theSchonflies symbol for its point group together with a superscript. Thus, forexample the space group of the face centered cubic structure with point groupOh is denoted O5

h. The diamond structure belong to the same class and isdenoted O7

h (the superscript is rather arbitrary). Notice that the restrictionson rotations impose also corresponding restrictions on possible lattice vectors,and, in fact only 14 different lattices are allowed. They are known as theBravais lattices. A more detailed description of the crystallography of spacegroup is given in the book by Landau Lifshitz on ’Statistical Physics’, PartI.Example: The operations of the space group O7

h.The diamond lattice, see figure, is invariant for translations (E | tn) with

Figure 2.6:

tn =∑

i niai and

a1 =a

2(0, 1, 1)

a2 =a

2(1, 0, 1)

a3 =a

2(1, 1, 0)

where a is the length of the cube edge. The first Brillouin zone is the one ofthe face centered cubic lattice. There are two equal atoms in the unit cell inthe positions

d1 = (0, 0, 0)

d2 =a

4(1, 1, 1)


The diamond lattice can be thought as consisting of two interpenetratingface-centered cubic sublattices displaced with respect to each other by thevector d2 ≡ f . The point group of the diamond lattice is the cubic group Oh.Notice that some of the symmetry operations of the cubic group appear inthe space group associated with the fractional translation

f =a

4(1, 1, 1) =

1

4(a1 + a2 + a3)

By choosing a lattice point as the origin of a cubic coordinate system we ob-tain for the symmetry operations the explicit expressions given in the table.

2.2.1 The irreducible representations of space groups

We now set up to construct all irreducible representations of a space group.The problem is complicated by two facts: first, the presence of fractionaltranslations associated with some rotation. If all fractional translations arezero, the space group is said to by symmorphic, otherwise the space groupis non-symmorphic. We will see that constructing the irreducible representa-tions of symmorphic space groups is somewhat easier. The second is that evenin the case of symmorphic space groups, the operations of the point groupG0 do not necessarily commute with the primitive translations, so that ingeneral G = T ⊗G0.

We now look at an invariant subspace carrying an irreducible representa-tion of the space group. If G coincides with T , the basis functions are Bloch

functions, and can be written as φ(k,r) = u(k, n)ei�k�r. u(k, n) are invariantunder primitive translations. The index n signifies the possibility of the exis-tence of linearly independent periodic functions having the same symmetryunder translations k (n is the band index). In the case of G = T all irre-ducible representations are one-dimensional. If we introduce extra symmetryelements, the one dimensional spaces might no longer be invariant under theoperation of these new elements, i.e. the various Bloch functions belonging todifferent k and n are mixed by these extra symmetry operations, and essentialdegeneracies occur. These enlarged spaces are the new invariant subspacessought for. For instance, the operation (R | a) transform the Bloch function

of vector k into a Bloch function of vector Rk. In fact,

O(R|�k)φ(k,r) = ei�k·(R−1�r−R−1�a)u(R−1r − R−1a)

= eiR�k·�re−i�kR−1�au(R−1(r − a))

= eiR�k·�ru1(r)


Type Operation Coordinate Type Operation Coordinate

C1 (E | 0) xyz I (I | f) x+ a4y + a

4z + a

4

C24 C2z xyz IC2

4 (IC2z | f) x+ a4y + a

4z + a

4

C2x xyz (IC2x | f x+ a4y + a

4z + a

4

C2y xyz (IC2y | f x+ a4y + a

4z + a

4

C4 (C−14z | f) y + a

4x+ a

4z + a

4IC4 IC−1

4z yxz

(C4z | f) y + a4x+ a

4z + a

4IC4z yxz

(C−14x | f) x+ a

4z + a

4y + a

4IC−1

4x xzy

(C4x | f) x+ a4z + a

4y + a

4IC4x xzy

(C−14y | f) z + a

4y + a

4x+ a

4IC−1

4y zyx

(C4y | f) z + a4y + a

4x+ a

4IC4y zyx

C2 (C2xy | f) y + a4x+ a

4z + a

4IC2 IC2xy yxz

(C2xz | f) z + a4y + a

4x+ a

4IC2xz zyx

(C2yz | f) x+ a4z + a

4y + a

4IC2yz xzy

(C2xy | f) y + a4x+ a

4z + a

4IC2xy yxz

(C2xz | f) z + a4y + a

4x+ a

4IC2xz zyx

(C2yz | f) x+ a4z + a

4y + a

4IC2yz xzy

C3 C−13xyz zxy IC3 (IC−1

3xyz | f) z + a4x+ a

4y + a

4

C3xyz yzx (IC3xyz | f) y + a4z + a

4x+ a

4

C−13xyz zxy (IC−1


4y + a

4


4x+ a

4

C−13xyz zxy (IC−1


4y + a

4


4x+ a

4

C−13xyz zxy (IC3xyz | f) z + a

4x+ a

4y + a

4


4x+ a

4


where u1(r) is some periodic function, in general different from u(r).The general strategy to find the invariant subspaces starts with picking out ofthe point group those elements which rotate k into itself or into an equivalentone

R�kk = k + h

h being a reciprocal lattice vector (including 0). This group is a subgroup of

G0 and is called the small point group of the wave vector k and is indicated byG0(k). All the symmetry operations in the space group of the form (R�k | a)constitute a group G�k called the little group of k or simply the group of k.Notice that at a general point of the Brillouin zone only the identity satisfiesthe above condition. However, there are special points, lines and planes ofthe Brillouin zone where G0(k) is larger than the identity operation. Theselocations are so called high-symmetry locations (or simply symmetry loca-tions). For example, the Brillouin zone of the body centered cubic lattice hassymmetry points Γ, H , N and P , symmetry axes ∆, Σ, Λ, and G and thesymmetry planes are those containing two symmetry axes. Among the basisfunctions for an irreducible representation of the space group, we chose thesubset of functions with a given k vector. As a consequence of the transfor-mation property under the action of (R | a), this subset form a basis of a

irreducible representation of the group of k. The irreducible representationsof the group of the k vector (whose basis functions are Bloch functions of

vector k ) allow the classification of the eigenvalues at any given k vector.

case 1: k inside the first Brillouin zone.In this case the small point group of k is selected from those operations ofG0 for which Rk = k.Theorem: From any irreducible representation D(p)(R�k) of the small point

group G0(k) an irreducible representation D�k,p of the little group of k is ob-

tained by associating to every element (R�k | a) of the group G�k the matrix

e−i�k·�aDp(R�k).Proof: we have to show that these matrices follow the same multiplicationrule as the elements of the group G�k. The matrix product corresponding tothe left-hand side of

(R�k | a)(S�k | b) = (R�kS�k | R�kb+ a)

ise−i�k·�aDp(R�k)e

−i�k·�bDp(S�k) = e−i�k·(�a+�b)Dp(R�kS�k)

The matrix corresponding the right-hand side is

e−i�k·(R�k�b+�a)Dp(R�kS�k) = e

−i(R−1�k

�k�b+�k·�a)Dp(R�kS�k)


The phase factors of the above expressions are identical because R−1�kk = k.

Case 2: k at the surface of the Brillouin zone. The same procedure can beapplied if all f vectors are zero, i.e. for symmorphic space groups. However,for non-symmorphic space groups, it might happens that some points at

the surface are transformed to k + h. e−i�h·�b is one if b is some primitive

translation vector, but, in general e−i�h·�f = 1. Then the phase factors in theexpressions above are different and the above matrix is not a representation.The procedure of finding the irreducible representations in this special casesis outside the scope of this lecture.

In both cases, considering the group of the wave vector, we are able toconstruct a number of irreducible representation of this group which obeythe equation ∑

p

(lp)2 = gk

with gk being the order of the small group of k, i.e. G0(k). So far, we have

considered only those symmetry operations which transform k into itself.In general, the effect of a point group symmetry is to send k into a starof vectors k1, k2, ..ks. The eigenvalues at all points of the star are the sameand the eigenfunctions at all points of the star are obtained by applying thesymmetry operations of the space group to the eigenfunctions of the vectork. It is convenient to define the restricted volume inside the Brillouin zoneas a volume containing all points that do not belong to the same star. Thevolume of this restricted region is the volume of the Brillouin zone divided byg0, the order of G0. The basis functions φ(k, (1)), φ(k, (2), ..., φ(k, (lp) of the

irreducible representation D�k,p of G�k and the corresponding eigenfunctions at

the vectors of the star of k constitute a basis for the irreducible representation

D�k,pspace of the space group G. Its dimension is s�k · lp, where s�k is the number

of points in the star of k. We now show that the so constructed irreduciblerepresentations represent all possible irreducible representations of the spacegroup. From ∑

�k∈res.vol.,p

(s�k · lp)2 =∑

�k∈res.vol.

(s�k)2 · gk

and skgk = g0 we obtain

∑�k∈res.vol.

sk · g0 = g0 ·∑

�k∈Brill.zone

= N · g0

which is exactly the order of the space group.Summarizing: This construction the invariant subspace shows that the irre-ducible representations of G are specified by an allowed k-vector and a label


p, specifying an irreducible representation of G�k. To classify the energy levels

along one specific k, it is enough to know the irreducible representation ofG�k. This group consists of element of the type (R�k | fR�k

) and the multi-plication of these elements with the pure primitive translations. The lp × lprepresentation matrices of the elements (R�k | fR�k

) within the Brillouin zone

can be easily calculated following our discussion. Those for k on the surfaceof the Brillouin zone require more work. Once these matrices are known,

their characters χ(p,�k)(R�k | fR�k) can be obtained taking the trace of the di-

agonal matrix element. The matrices for the elements (R�k | fR�k+ tn) obey

the following equation:

Dp,�k(R�k | fR�k+ tn) = e−i�k·�tnDp,�k(R�k | fR�k

)

Correspondingly, their characters can be obtained by multiplying the char-

acters χ(p,�k)(R�k | fR�k) with e−i�k·�tn. Thus, in order to specify the character

table of the group G�k is enough to give the character table of the h elements

(R�k | fR�k). For symmorphic space groups, this is the character table of the

point group G0(k).

Exercise. Show that the representations Dp,�k of G�k (Case 1) are irreducible.Example: some irreducible representations of O7

h.Point Γ. This is the point at the origin of the first Brillouin zone. The smallpoint group of k is the entire point group Oh. Since we are inside the Bril-

louin zone and e−i�k·�f = 1, the characters of the irreducible representationsat Γ are simply given by the characters of the point group Oh, without anymodification.Line Λ. k = π

a(λ, λ, λ), λ < 1. This is a point on the line joining the center

of the Brillouin zone to the midpoint of an hexagonal face. The small pointgroup of k is made up by the operations whose rotational parts interchange(x, y, z) among themselves, because these are the only operations which leavek invariant. The small point group of k is thus C3v. Since no fractional trans-lation is associated with the elements of C3v, the characters of the irreduciblerepresentations at Λ are the characters of the group C3v without any modi-fication.Line ∆ (k = 2π

a(δ, 0, 0), with δ < 1). This is a point on the line joining the

center of the Brillouin zone to the midpoint of the square face. the smallpoint group of k is given by the operations whose rotational part leave xunchanged. The small point group of k is thus C4v. The irreducible repre-sentations at ∆ are obtained by multiplying the characters of the irreducible

representations of the group C4v by the phase factor e(−i�k·�f) = e−i(π2·δ).


Irr.Rep. (E | 0) (C2x | 0) (C−14x | f) (IC2x | f) (IC2xy | 0)

(C4x | f) (IC2y | f) (IC2yz | 0)

∆1 1 1 e−i(π2·δ) e−i(π

2·δ) 1

∆′1 1 1 e−i(π

2·δ) −e−i(π

2·δ) -1

∆2 1 1 −e−i(π2·δ) e−i(π

2·δ) -1

∆2′ 1 1 −e−i(π2·δ) −e−i(π

2·δ) 1

∆5 2 -2 0 0 0

2.3 Symmetry adapted plane waves

We would like now to address the problem of calculating, at least in a qual-itative way, the band structure of real crystals. A possible approach is tostart with symmetry adapted plane waves and introduce a crystal potential.This will not be very precise in general, but will give at least a feeling for thedifficulties and tasks lying behind a real band structure calculation.

2.3.1 The one-dimensional lattice

We consider a chain of N -atoms with lattice constant a. The fundamentaltranslation group has N elements n ·a, with n = 0, ..., N − 1. The group hasN irreducible representations. In the representation with label p the n-th

element has the character e−i·2π·p·n

N , with p = 0, 1, ..., N−1. This establishes areciprocal lattice vector of length 2π

ahosting the allowed k-vectors k = 2π

Na·p.

The first Brillouin zone of the 1-dimensional lattice is the intervall [−πa, π

a].

The point group G0 of the system is Ci, while the small point group of anyvector inside the intervall is just E. This means that – at general k insidethe intervall – k is the only quantum number labeling the energy levels andthe symmetry adapted wave functions are of the form eik·xu(x), where u(x)is a periodic function. As possible basis functions we might decide to choose

√1

Naeik·x · eiG·x

with G = n · 2πa

, n ∈ Z being a reciprocal lattice vector. For point insidethe intervall, the inversion operation tranfsorms k into −k, which togetherwith k forms the star of k in the one dimensional lattice. As the energy levelsbelonging to the star of k are degenerate, we obtain that energy levels in theone dimensional lattice are at least double degenerate.


The representation theory of the k point at the end of the intervall (includ-ing k = 0) must be modified because the small group of k contains bothoperations E and I. For this reason, the energy levels at the edge of the onedimensional Brillouin zone are classified according to the two one-dimensionalirreducible representations of Ci, namely the one carrying symmetric and theone carrying anti-simmetric wave functions.

We are now able to calculate the empty lattice band structure (alsofree electron band structure), which sets the crystal potential to zero andonly consider the kinetic energy of the electrons. Depending of the choice ofG we can generate a set of empty lattice energy bands with energy

E =h2

2m(k +G)2

with G = 0,±2πa, ... The first band is the one characterized by G = 0, see

Figure 2.7:

the figure. The G = 0 band starts at Γ and ends at the point πa

with a dou-


ble degenerate energy level with wave functions√

2Na

cos πax and

√2

Nasin π

ax.

Taking −2πa

as the next G-vector we produce a further band taking off fromthe Brillouin zone edge, see the figure. This band ends at k = 0 with a dou-

ble degenerate band with wave functions√

2Na

cos 2πax and

√2

Nasin 2π

ax. The

next band is generated by G = 2πa

, and so on. The general expression for the

wave functions at k = 0 is√

2Na

cos n·2πa

·x and√

2Na

sin n·2πa

·x, n = 0, 1, 2, ....

From this expression it is clear that the deepest lying level (k = G = 0) isnot degenerate, while the degeneracy of the higher lying levels it two. The

general expression for the wave function at k = πa

is√

2Na

cos[πa

+ n2πa

]x and√2

Nasin[π

a+ n2π

a]x, n = 0, 1, .....

The band structure calculated for the empty lattice using symmetryadapted wave function is exact. However, the same functions can be used,within in a perturbational approach, to find the first order correction causedby a small crystal potential V (x). The most remarkable feature of switchinga small crystal potential is the lifting of the accidental double degeneracy atthe center and at the edge of the Brillouin zone. In fact, from general sym-metry considerations, we expect that the double degenerate level splits intotwo single degenerate energy levels, thus opening a gap between the variousbands at k = 0 and at k = π

a. The energy of the levels at these two singular

points of the BZ can be calculated, in first order perturbation theory, to be

E±(k = 0) =h2

2m(2π

a)2 · n2 +

1

4Na

∫ Na

0dx · V (x)

± 1

4Na

∫ Na

0dx · V (x) · cos(2n · 2π

a)x

(the - sign-level is suppressed for n = 0) and

E±(k =π

a) =

h2

2m(2π

a)2 · (2n+ 1

2)2 +

1

4Na

∫ Na

0dx · V (x)

± 1

4Na

∫ Na

0dx · V (x) · cos[(2n+ 1) · 2π

a]x

(n = 0, 1, 2, ....). Thus, the gap is twice the Fourier-transform of the crystalpotential perturbing the empty lattice band structure.

2.3.2 The square lattice

The Figure shows the Brillouin zone of a plane square lattice of lattice con-stant a. The special points and lines of symmetry and the correspondingsmall point group are indicated in the table.


Figure 2.8:

k Symbol G0(k)(kx, ky) general C1

kx, kx) Σ C1,h12, kx) Z C1,h

kx, 0 ∆ C1,h12, 0 X C2,v

12, 1

2M C4,v

(0, 0) Γ C4,v

Table 2.1: The special point and the lines of symmetry in the Brillouin zoneof the planar square lattice, in units of 2π

a.

The next figure gives the stars of the various k vectors for a plane squarelattice. Every dashed vector is related to one solid vector by a reciprocallattice vector and is therfore not counted separately. The number of distinctvectors in the star is given by s. Notice that the full point group G0 is C4v

and that the space group is symmorphic.We now proceed toward mapping the empty lattice energy bands. The energyversus k relation for a two-dimensional lattice may be represented by sur-faces in the three-dimensional space kx, ky, E. Typically, one shows the crosssections of these surfaces along variuos lines of symmetry of the Brillouinzone. The first task is to construct symmmetry adapted wave functions fromthe plane waves

φ�k, �G =1√Na2

ei(�k+ �G)·�r


Figure 2.9:

G is some reciprocal lattice vector with coordinates 2πa

(m,n), where m,n ∈Z. We consider for instance, the ∆-direction. Ths small group of k consistof [E, σx] and thus has two irreducible representations, which we call ∆1 and∆2. Upon reflection with respect to the x axis in the BZ, states belonging to∆1 remains invariant, while states belonging to ∆2 change sign. The enegybands generated by the numbers (m,n) have energy

E(m,n) =h2

2m(2π

a)2[(κ +m)2 + n2]

and the corresponding plane waves are

ei·2π

a[(κ+m)·x+n·y]

The lowest lying band is the one with m = n = 0 and has ∆1 symmetry. Thenext band is charactereized by m = −1, n = 0 and has, again ∆1 symmetry.The degeneracy of these bands is 1. These two bands are followed by a doubledegenerate band corresponding to the indices (m = 0, n = 1) and (m = 0, n =−1). This band has an accidental degeneracy involving a wave function with∆1 symmetry and a wave function with ∆2 symmetry. Them = 1, n = 0 bandhas ∆1 symmetry and has higher energy than the (0, 1)-band, except at theΓ-point where the bands meet. The next reciprocal lattice vectors are the four(1, 1) vectors. The (−1, 1) and (−1− 1) bands are degenerate along ∆, theirenergy being equal to the energy of the 0, 1, 0,−1 band at the X-point. Theresulting free electron band structure, compared with a calculation includinga realistic crystal potential, is given in the Figure. We now proceed analyzing


Figure 2.10:

the Γ and X-points. Both have a larger symmetry group and a larger numberof irreducible representations. The plane wave corresponding to the lowestlying energy level E = 0 is (0, 0)Γ. It is a constant and transforms accordingto the irreducible representation Γ1. The next fours waves (1, 0)Γ, (1, 0)Γ,(0, 1)Γ and (0, 1)Γ are degenearte with energy 1. They carry some irreduciblerepresenations of C4v, which can be found by decomposing the representationconstructed over these four plane waves. We find Γ1 ⊕ Γ3 ⊕ Γ5. Using theprojector operator technique one can easily find the symmetryized linearcombination corrsponding to these components. For instance, the function


Irr.Rep. e C4, C34 C2 2m 2σ

∆1,Γ1 1 1 1 1 1∆′

1,Γ2 1 1 1 -1 -1∆2,Γ3 1 -1 1 1 -1∆′

2,Γ4 1 -1 1 -1 1∆5,Γ5 2 0 -2 0 0

Γ1 Γ2 Γ3 Γ4 Γ5

∆1 ∆2 ∆1 ∆2 ∆1∆2

X1 X2 X3 X4

Table 2.2: Compatibility table for the square lattice along the ∆-direction

with ∆1 symmetry is clearly

1√4Na2

[ei 2πa

(x) + ei 2πa

(y) + ei 2πa

(−x) + ei 2πa

(−y)]

The next four plane waves (1, 1)Γ, (1, 1)Γ, (1, 1)Γ and (1, 1)Γ belong to the fourfold degenerate energy level E = 2 and transform according to Γ1 ⊕ Γ4 ⊕ Γ5.

The symmetry group at X is C2v = [E,mx, my, C24 ] and has four irre-

ducible representations which we denote X1, X2, X3, X4 for the square lattice.We obtain the following set of degenerate plane waves at X:

1. the set (0, 0)X and (0, 1)X generating the represenations X1 ⊕X3 withE = 0.25

2. the set (0, 1)X , (0, 1)X , (1, 1)X , (1, 1)X transforming according to X1 ⊕X2 ⊕X3 ⊕X4 with E = 1.25

3. and so on

The connection between the band structure along ∆ with the end-pointsof the BZ can be directly read out in general terms from the compatibilityrelations given in the Table.

We turn now to explore the aspects of the band strucure when a crystalpotential is switched on. The key to solve the full problem including crystalpotential is to use symmetrized linear combinations. Even few symmetrizedplane waves are quite successful in predicting the real band structure, al-though in some materials the real wave functions require many of them, seethe figure. In fact, it turns out that the energy bands of many simple metals,


particularly the one-electron metals such as the alkalis, are not significantlydifferent from the empty lattice band structure, except for some general fea-tures wich can be easily worked out from symmetry arguments alone.In fact, as we learned from the one-dimensional band strucure, accidentaldegeneracies are removed when a potentiat is switched on. This mechanismintroduces typical gaps in the band structure. In addition, group theory alsoallows to treat special crossing points (such as the one around E = 1.5 inthe empty lattice band structure). When the crystal potential is taken intoaccount, there will be, in general, non-vanishing matrix elements betweensymmetrized wave function belonging to the same representation. Thismeans that the proper eigenfunctions must be constructed by taking linearcombinations of as many functions having the same symmetry as practicallypossible. This phenomenon is known as hybridization and has the obviouseffect of lifting the degeneracy of band having the same symmetry. A hy-bridization gap arises. The ”birth” of an hybridization gap is illustrated inthe figure. Close to the crossing point the bands strongly hybridize, i.e. thewave function close to the crossing point will contain a sizeable amount ofthe waves belonging to the crossing bands. Away from the crossing point, ac-cording to perturbation theory, the hybridization decreases and the originalwave funtion will be restored. The consequence of the lifting of the crossingis that, moving along the ”new” bands will produce a change of wave func-tion from one type to the other. As the rule that bands having the samesymnmetry do not cross is a quite general one, we expect such hybridizationgaps to be introduced upon switching on the spin-orbit interaction as well,when two bands belonging to two different single group representations endup into the same extra-representation of the double group.

Figure 2.11:


2.3.3 Free electron energy bands in three-dimensional

lattices

Using the same methods we can produce and label the empty lattice bands inthree-dimensional lattices, as illustrated in the following figure. More resultscan be found in the monumental work by J.C. Slater, Quantum Theory ofMolecules and Solids, Volume 2, Mc Graw-Hill book company, New York,1965.

Figure 2.12: Empty-lattice band structure for a bcc-lattice (right) and forNa (left).

Chapter 3

Landau theory of phasetransitions

The Van der Waals theory of real gases contains a most important feature:liquid and vapour can transform into each other along a well defined phasetransition line in the P − T -phase diagram. It is also well known experimen-tally that the solidification establishes a phase transition line as well, andthat there are important phenomena which are also realized by a phase tran-sition, such as supercoductivity and magnetism. Landau established in 1937(L. Landau, Phys.Z.Sowjet.11, 26 (1937) and ibidem, 545) that in some casesthe phase transitions proceeds by a continuous change of the thermodynamicpotential when the phase transition line is crossed. In this case the phase tran-sition line is called the Curie line and the transition is a continuous one (alsocalled 2nd order phase transition, in contrast to first order phase transitionswhere there is a discontinuous change of the thermodynamic potential andof other relevant quantities). The main idea of Landau is to characterize thesystem at the Curie point by mean of a symmetry group G0 containing thosesymmetry operations which leave the system invariant. Landau showed thatG0 determines quite univocally the structure of the thermodynamic potentialduring the phase transition. Thus, in this respect, Landau theory of phasetransition is a non-trivial application of group theory. In this third chapterwe will discuss the essential elements of this theory.

General arguments

To be concrete, we describe the system by a spatially dependent densityρ0(x, y, z) (in the case of magnetic system the quantity describing the systemis the effective current distribution j0(x, y, z)). Let G0 be the symmetry group

48

CHAPTER 3. LANDAU THEORY OF PHASE TRANSITIONS 49

at the transition point (G0 can be one of the 230 space groups, but also somecontinuous group like the full group of translations in a liquid or the full groupof rotationsO3 in an Heisenberg exchange ferromagnet). We will consider onlythose transitions where the state of the system changes continuously, i.e. ρchanges continuously. A clear example of transition of this kind is representedin Fig.1.

A small decrease of the maxima, which can occur in a continuous wayupon varying some parameter like the temperature, produces a drastic changeof symmetry, in this case a doubling of the translation period of the lattice.Thus, while ρ changes continuolsly, the symmetry of the system can changeabruptly. A general feature of any function ρ(x, y, z) is that it can be decom-posed into a linear combination of invariant subspaces transforming accordingto an irreducible representation of G0:

ρ(x, y, z) =∑n,i

ηi,nϕi,n

The index n runs over all irreducible representations Γn, the index i labels thevariuous partners transforming according to the n-dimensional representationΓn. Without loosing generality, we require that the functions ϕi,n are realfunctions and thus all matrix elements of Γn are real. Among the functionsϕi,n there is one – which we call ρ0 – transforming according to the ′1′-representation of G0. In other words: ρ0 is invariant with respect to anyoperation of G0. Thus, we can write

ρ(x, y, z) = ρ0 + (∑n,i

)′ηi,nϕi,n

where δρ.=∑′

n,i ηi,nϕi,n contains all irreducible representations with the ex-ception of the ′1′ one. Notice that the function δρ has a lower symmetry thanG0, because not all transformations of G0 leaves δρ invariant. We call thesymmetry group of δρ G. G is some subgroup of G0. The thermodynamicpotential Φ of the body is some functional of ρ, Φ = Φ[ρ] and depends onT and P . When P and T are given, Landau suggests that the form of the


function ρ is determined from the condition that Φ[ρ] should have a mini-mum. This also determines the symmery G of the crystal at any moment,because once the coefficients of the linear combination are established thefunctions appearing in the equilibrium ρ are established and thus G. At thetransition point we require that the symmetry group is G0: this means thatexactly at the transition point ρ = ρ0 and δρ = 0. It the transition has toproceed in a continuous way, δρ – which is exactly zero at the transition point– must assume very small values close to the transition point. Therefore thethermodynamic potential can be expanded in powers of δρ close to the tran-sition point. Individual terms of this expansion are integral operators of δρ.To determine which terms appear in the expansion, we notice that Φ, as aqunatity characterizing any physical property of the body, should remainsinvariant under all possible coordinate transformations, in particular thosebelonging to G0:

Φ[ρ0 + δρ] = Φ[ρ′0 + δρ′]

ρ0 is left invariant by G0 but δρ is changed. During this transformation,the functions belonging to the same irreducible representation trasform intothemselves, i.e. the space remains invariant. Thus, we can consider a trans-fomation of G0 as acting on the coefficients of the linear combination inthe same way as it acts on the functions themselves. As a consequence, theinvariance property of Φ with respect to G0 means that the expansion ofΦ in powers of ηn,i should contain invariants of the relevant degrees con-structed over the coefficients ηn,i. As one cannot construct linear invariantsfrom quantities transforming according to a given irreducible representations(other than the ′1′ representation) (Exercise 1: prove this statement),the expansion of Φ will start with a scalar and with invariant polynomialsof the second degree. We will prove later that the only polynomial of sec-ond degree build up from the coefficients transforming according to a givenirreducible representation is the sum of the squares of the coefficients. Thus,

Φ = Φ0 +∑n

An(P, T )∑

i

η2n,i

the coefficients An depending on T and P . At the transition point the coef-ficienst ηn,i are all vanishing. For the minimum of Φ to be at ηn,i = 0 all An

must be non negatives. If at the transition point all coefficients are positivethey would remain positive in the vicinity of the transition point and nosymmetry change occurs. Thus, at least one of the coefficients must changesign at the transition point and vanish at the transition point. Notice thatit is very improbable that more than one coefficients vanishes along a linein the P − T plane. Such a simoultaneoous vanishing of two coefficients can


only occur in an isolated point of the plane which represents the intersec-tion of two line of continuous phase transitions. The change of sign of thecoefficients means that at one side of the transition point all coefficients arepositive and the value minimizing Φ is ηn,i = 0. At this side, the symmetrygroup G0 is realized. On the other side, instead, a set of coefficients assumefinite – albeit small in the vicinity of the transition point – values, and thesubgroup G is realized. In the following, we will set all positive coefficients tozero and retain only the irreducible representation which is realized on thelow symmetry side of the transition point. By introducing the quantities

η · γi = ηi

with∑

i γ2i = 1 we define a unit vector with ”amplitude” η within a n-

dimensional space. This vector, carrying the Γn-irreducible representation ofthe group G0, appears below the transition point and is called the orderparameter of the system and the corresponding space is the order parameterspace. Depending on the dimension of Γn, the OP can have more than onecomponent. The appearance of an OP below the transition point establishes anew symmetry group G. Which group G is realized below the transition pointdepends on the dierction of the OP. The expansion of the thermodynamicpotential can be continued with the third, fourth,...,order terms and can bewritten in general as

Φ = Φ0 + A(T, P )η2 + η3∑α

Cα(P, T )f (3)α (γi) + η4

∑α

Bα(P, T )f (4)α (γi) + .....

where f (3)α , f (4)

α , .... are homogeneous polynomials of degree 3,4,..., which areinvariant under the operations of G0. The sum over α takes into accountthat there might be more than one invariant polynomial of a given degree.However, terms of order 3 can be vanishing in some cases. At least one termof order 4 do always exist, and is given be (

∑i γ

2i )

2.Before we address more technical problems related to group theoretical

aspects we would like to investigate the role of the third and fourth orderterms. Consider a thermodynamic potential that includes third order terms.At Tc, the minimum corresponds to ηi = 0. Thus, for any value of ηi closeto ηi = 0 Φ − Φ0 must be positive. However, a third order term cannot bealways positive for any set of infinitesimal ηi: if f3 is positive for some setcorresponding to a given direction in the OP space, it will be negative forthe opposite direction. Thus, in the presence of a third order term, the valueηi = 0 does not correspond to a minumum but to a saddle point of the po-tential. Hence, in order to ensure that the equilibrium state of the system atTc is indeed a minimum with values ηi = 0, there must be no third-degree


term in the expansion of the potential. The presence of a third order invari-ant leads to the mimimum at Tc being at a finite value for ηi and thus to adiscontinuous (first order) phase transition. Thus, the Landau condition fora phase transition being continuous is the absence of any third order invari-ant in the expansion of the thermodynamic potential. The absence of thirdorder invariants can be due to the impossibility to find 3rd degree invariantpolynomials for the group G0 . This absence, due to symmetry reasons, willhold at any temperature and pressure. It might also be possible at some iso-lated point of the P − T -plane, that the coefficient of the third order termvanishes ”accidentally”. In this case the Curie point is the intersection of thelines A(T, P ) = 0 and C(T, P ) = 0 and is therfore an isolated point in theP − T -plane.In contrast to the third order terms, a fourth order term can be positive inany direction and therefore can sustain a second order phase transition. Thefourth degree terms are essential in determining the actual direction of theOP in the OP space: as the 2nd order term is independent of γi, these quan-tities must be determined by the requirement that the fourth oder terms beminimized with respect to the γi. The amplitude η of the OP is determinedby minimizing globally Φ.

Invariant polynomials

Invariant polynomials play a central role in the Landau theory of phase tran-sitions. The procedure for costructing invariant polynomials starts by con-sidering the f -dimensional irreducible representation Γ of G0 which carry theinvariant subspace Λ defined by φ1, φ2, ..., φf . We discuss in depth the con-struction of the 2nd degree polynomials, the costruction of the higher degreepolynomials proceed in a similar way. The first step is the realization of theKronecker product representation Γ ⊗ Γ by the n2 basis functions φiφk. Itscharacter is

χ(Γ ⊗ Γ)(g) = (χ(g))2

The Kronecker product representation can be reduced to the direct sum oftwo representations constructed over the f(f + 1)/2 functions φiφj + φjφi

and the f(f − 1)/2 functions φiφk − φkφi (i = k). Both sets of functionsare invariant with respect to the operations of the group and thus carry arepresentation of the group. The first representation is called the symmetrized2nd power representation of Γ and is denoted by [Γ2]. The correspondingspace is denoted by [Λ2] and the character of the representation is [χ2](g).


[χ2](g) can be determined from the following equations:

g(φiφk + φkφi) =∑l,m

gligmk(φlφm + φmφl)

=1

2

∑l,m

(gligmk + gmiglk)(φlφm + φmφl)

The character is

[χ2(g)] =∑i,k

< (φiφk + φkφi), g(φiφk + φkφi >=1

2

∑i,k

(giigkk + gkigik)

Because

∑i

gii = χ(g)

∑i,k

gikgki = χ(g2)

we arrive at the important formula

[χ2(g)] =1

2(χ(g)2 + χ(g2))

which allows to express the characters of the symmetrized 2nd power rep-resentation of Γ by means of the characters of Γ. A similar expression canbe found for [χn](g), i.e. for the character of the symmetrized n-th powerrepresentation: for instance

[χ3(g)] =1

3χ(g3) +

1

2χ(g2) · χ(g) +

1

6χ3(g)

[χ4(g)] =1

4χ(g4) +

1

3χ(g3) · χ(g) +

1

8χ2(g2) +

1

4χ(g2)χ2(g) +

1

24χ4(g)

These group theoretical results can be used to calculate the invariant poly-nomials – and in particular to determine the occurrence of third power termsin the expansion of the theormodynamic potential – because the monomi-als of a given degree costructed over the components ηi transform as thecorresponding symmetric product functions. Thus, to find the presence ofinvariant polynomials of the n-th degree one has to check for the occurrenceof the ′1′-representation in the reduction of [Γn]. If the ′1′-representation isabsent, the corresponding invariant polynomial does not exist. Should the′1′ representation occur nα times, then there are nα different invariant poly-nomials of the given degree. To find them, one can use the usual projection


operator technique.Exercise 2: Show that there is only one second degree invariant polynomialand that it takes the form

∑i η

2i .

Exercise 3: Show that if the OP corresponds to a one dimensional non-totally symmetric IR, then the Landau condition is necessarily verified.Exercise 4: For the group C3,v and C4,v find the invariant polynomials ofthe third degree constructed with the basis functions belonging to the 2-dimensional IR Λ3.

Magnetic phase transitions

The most famous case for the application of the Landau theory of phasetransitions is the classical Ising magnet on a lattice (specified by an index i)described by the Hamilton function

H = −J∑i,j

σi · σj

where the variable σ takes the value 1 or −1. The symmetry group of thissystem consists of two operations: the identity operation E and the inversionI which transforms simoultaneously all σi into −σi. This group – called oftenZ2 – has two irreducible representations, the non-trivial one establishing aone-dimensional OP space. Third order invariants are absent. The only fourthorder invariant is η4. The Landau functional Φ[η] is

Φ[T, P, η] = A(T, P )η2 +B(T, P )η4

At one side of the Curie point (which we take to be the high temperatureside) A > 0: provided B > 0 the functional has a minimum at η = 0, i.e.thesystem has the full Z2 symmetry. At the Curie point A = 0 and below itA < 0. From the minimization of Φ, i.e. from the equation ∂Φ/∂η = 0 wefind

Aη + 2Bη3 = 0

with two solutions: η = 0 and η = ±√− A

2B. The solution η = 0 represents,

below the Curie point, a maximum of Φ, see the figure. The other two so-lutions are minima of the Landau functional and both realize a symmetrybroken state below the phase transition. If the system assume one of theminima, its symmetry group will consist of only the identity operation. Theremaining operation of the full symmetry group transforms one minimuminto the other. Thus, the existence of a larger symmetry group above thephase transitiuon manifests itself below the phase transition by the existence


of different states which are related by operations of the full symmetry groupZ2.

The application of Landau Theory to the Heisenberg exchange ferromag-net follows the same arguments. The Heisenberg classical ferromagnet is de-fined by the Hamiltonian

H = −J∑i,j

ni · nj

where the variables n are unit (pseudo) vectors in a three dimensional Eu-clidean space. The symmetry group of this Hamiltonian is the full rotationgroup O3, which includes improper rotations. We search for a phase transi-tion where the IR D+

1 is realized below the Curie point. Third order degreepolynomials – which contain odd power of the basis functions of D+

1 – cannotbe excluded from the O3 symmetry alone. However, the time reversal sym-metry represents a further symmetry element of the Hamiltonian, and oddorder polynomials of the order parameter η ·(nx, ny, nz) are not invariant withrespet to time reversal symmetry and cannot appear in Φ. We have at leastone invariant of the fourth degree – η4, which is independent on the directionof the OP. In fact, as the group O3 contains rotations by arbitrary angles, wecome to the conclusion that this is the only fourth degree invariant polyno-mial. Thus, below the phase transition we will have a continuous degeneracyof the system corresponding to the surface of a sphere of radius 1, as anydirection minimize the fourth order invariant. This continuous degeneracyreflects again the continuous symmetry of the system above the transitionpoint. The expansion of Φ contains only the amplitude of the OP and itsminimization proceeds along the same lines encountered for the Ising model.Once the system has assumed one of the many degenerate states correspond-ing to some direction of the OP, the symmetry of the system is lowered fromO3 to O2, because the state of the system below the transition point will onlybe invariant with respect to rotations around the direction of the OP.

Landau’ s model of the liquid-solid phase tran-

sition

The symmetry group of the liquid is primarily the full group T of continuoustranslations in space. Since the liquid is isotropic, ρ0 = const. At the pointof transition ρ becomes ρ0 + δρ, where δρ will, in general, assume the formof a Fourier integral

δρ(r) =∫η(k)e−i�k·�rdk


(to ensure reality of δρ we impose η(−k) = η∗(k)). In the case of a continuous

group of translations, the integral extends over all k-vectors in the three-dimensional space, while for a discrete translations the integral is restricted toa fine mesh within the first Brillouin zone. The fully symmetric representationof T is the one dimensional one with k = 0. We proceed now to expandingΦ. Near the tansition point Φ will be expanded in different powers of thecoefficients η�k. Different terms of this expansion will contain the integrand

η�k1· η�k2

· η�k3......

The condition that the polynomials are invariant with respect to translationsis

k1 + k2 + ..... = 0

Terms linear in η�k do not exist. The quadratic contribution to Φ writes

∫A(T, P,k)dk | η�k |2

where we have used the symmetry between η�k and η−�k. Because of the sym-metry of the liquid with respect to the full rotation group we can takeA = A(T, P, | k |) and perform the integral over spherical surfaces of varying

radius. As usual in the Landau theory, we select that isolated vector k0 forwhich the coefficient A vanishes at the transition point. In general, this isan isolated point as it is very difficult that the coefficients of different k van-ishes simultaneously. Thus, we restrict the expansion to an integral over thesurface of a sphere of radius | k0 | and call, the coefficient A(k0) = A. Belowthe transition point the second order coefficient is negative and the stabilityof the solution is determined by the high-degree terms in the Landau expan-sion. As the third degree terms are not required by symmetry to be absent,we will consider them with the requirements | ki |= k0 and

k1 + k2 + k3 = 0

i.e. the three vectors build an equilateral triangle. In all third order termsthese triangles have equal size and differ only by their orientation in space.Again, because of the rotational symmetry, the coefficients of the third orderterms depend only on the size, not on the orientation of the triangles. Ther-fore, all coefficints in the third order terms are equal: their common valuewill be denoted by B(T, P, k0). Thus, the third order contribution writes

f3 = B(T, P, k0) · η3 · [ ∑|�kli

|=k0

γ(kl1) · γ(kl2) · γ(kl3)]


with the usual condition∑

i | γ(kli) |2= 1. the integral has been changed to

a sum due to the discrete nature of the mesh in k-space in a crystal. Thestable phase obtained upon solidification corresponds to the minimum of thethird order term with respect to the γ(kli). Notice that if f3 is positive fora certain direction in OP space, it is negative for the opposite one. As bothdirections have the same invariance group, the symmetry below Tc does notdepend on the sign of B. Thus, the stable phase corresponds to a set of γ(kli)associated with the maximum value of the expression

ϕ3 = [∑

|�kli|=k0

γ(kl1) · γ(kl2) · γ(kl3)]

The question about which restriction is imposed on the solid by the geo-metrical requirement of equilateral triangels was answered by Alexander andMcTague , Phys.Rev.Lett. 41, 702 (1978). They pointed out that such vec-tors cab be essentially seen as terminating at the edges of three-dimensionalpolyhedra with faces consisting of identical triangles, see the figure. The octa-hedron on the left generates a fcc Brillouin zone, which corresponds to a bcclattice. The one on the right is a regular icoshedron and generates icosahedralquasi-crystals. One can show that ϕ3(bcc) > ϕ3(icosahedron).

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