10 Group theory

10.1 What is a group? A group G is a set of elements f, g, h, ... and an operation called multipli- cation such that for all elements f,g, and h in the group G:

1. The product fg is in the group G (closure); 2. f(gh)=(fg)h (associativity); 3. there is an identity element e in the group G such that ge = eg = g; 1 1 1 4. every g in G has an inverse g in G such that gg = g g = e.

Physical transformations naturally form groups. The elements of a group might be all physical transformations on a given set of objects that leave invariant a chosen property of the set of objects. For instance, the objects might be the points (x, y) in a plane. The chosen property could be their distances x2 + y2 from the origin. The physical transformations that leave unchanged these distances are the rotations about the origin p x cos ✓ sin ✓ x 0 = . (10.1) y sin ✓ cos ✓ y ✓ 0◆ ✓ ◆✓ ◆ These rotations form the special orthogonal group in 2 dimensions, SO(2). More generally, suppose the transformations T,T0,T00,... change a set of objects in ways that leave invariant a chosen property property of the objects. Suppose the product T 0 T of the transformations T and T 0 represents the action of T followed by the action of T 0 on the objects. Since both T and T 0 leave the chosen property unchanged, so will their product T 0 T . Thus the closure condition is satisfied. The triple products T 00 (T 0 T ) and (T 00 T 0) T both represent the action of T followed by the action of T 0 followed by action of T 00. Thus the action of T 00 (T 0 T ) is the same as the action of 414 Group theory

(T 00 T 0) T , and so the transformations are associative. The identity element e 1 is the null transformation, the one that does nothing. The inverse T is the transformation that reverses the action of T . Thus physical transformations that leave a chosen property unchanged naturally form a group.

Example 10.1 (Permutations). A permutation of an ordered set of n objects changes the order but leaves the set unchanged.

Example 10.2 (Groups of coordinate transformations). The set of all trans- formations that leave invariant the distance from the origin of every point in n-dimensional space is the group O(n) of rotations and reflections.The rotations in n-space form the special orthogonal group SO(n). Linear transformations x0 = x + a, for di↵erent n-dimensional vectors a, leave invariant the spatial di↵erence x y between every pair of points x and y in n-dimensional space. They form the group of translations. Here, group multiplication is vector addition. The set of all linear transformations that leave invariant the square of the Minkowski distance x2 + x2 + x2 x2 between any 4-vector x and the origin 1 2 3 0 is the Lorentz group (Hermann Minkowski 1864–1909, Hendrik Lorentz 1853–1928). The set of all linear transformations that leave invariant the square of the Minkowski distance (x y )2 +(x y )2 +(x y )2 (x0 y0)2 between 1 1 2 2 3 3 any two 4-vectors x and y is the Poincar´egroup, which includes Lorentz transformations and translations (Henri Poincar´e1854–1912).

In the group of translations, the order of multiplication (which is vector addition) does not matter. A group whose elements all commute

[g, h] gh hg = 0 (10.2) ⌘ is said to be abelian. Except for the group of translations and the group SO(2), the order of the physical transformations in these examples does matter: the transformation T 0 T is not in general the same as TT0.Such groups are nonabelian (Niels Abel 1802–1829).

Matrices naturally form groups with group multiplication defined as ma- trix multiplication. Since matrix multiplication is associative, any set of n n 1 ⇥ non-singular matrices D that includes the inverse D of every matrix in the set as well as the identity matrix I automatically satisfies three of the four properties that characterize a group, leaving only the closure property uncertain. Such a set D of matrices will form a group as long as the prod- { } 10.1 What is a group? 415 uct of any two matrices is in the set. As with physical transformations, one way to ensure closure is to have every matrix leave something unchanged. Example 10.3 (Orthogonal groups). The group of all real matrices that leave unchanged the squared distance x2+ +x2 of a point x =(x ,...,x ) 1 ··· n 1 n from the origin is the group O(n) of all n n orthogonal (1.37) matrices ⇥ (exercises 10.1 & 10.2). The group O(2n) leaves unchanged the anticommu- tation relations (section 10.18) of the real and imaginary parts of n complex fermionic operators ,..., .Then n orthogonal matrices of unit deter- 1 n ⇥ minant form the special orthogonal group SO(n). The group SO(3) describes rotations in 3-space.

Example 10.4 (Unitary groups). The set of all n n complex matrices that ⇥ leave invariant the quadratic form z z + z z + + z z forms the unitary 1⇤ 1 2⇤ 2 ··· n⇤ n group U(n) of all n n unitary (1.36) matrices (exercises 10.3 & 10.4). Those ⇥ of unit determinant form the special unitary group SU(n) (exercise 10.5). Like SO(3), the group SU(2) represents rotations. The group SU(3) is the symmetry group of the strong interactions, quantum chromodynamics. Physicists have used the groups SU(5) and SO(10) to unify the electroweak and strong interactions; whether Nature also does so is unclear. Example 10.5 (Symplectic groups). The set of all 2n 2n real matrices R ⇥ that leave invariant the commutation relations [qi,pk]=i~ik,[qi,qk] = 0, and [pi,pk] = 0 of quantum mechanics is the symplectic group Sp(2n, R). The number of elements in a group is the order of the group. A finite group is a group with a finite number of elements, or equivalently a group of finite order.

Example 10.6 (Z2 and Zn). The parity group whose elements are 1 and 1 under ordinary multiplication is the finite group Z . It is abelian and 2 of order 2. The group Zn for any positive integer n is made of the phases exp(i2k⇡/n) for k = 1, 2 . . . n. It is abelian and of order n. A group whose elements g = g( ↵ ) depend continuously upon a set of { } parameters ↵k is a continuous group or a Lie group. Continuous groups are of infinite order. A group G of matrices D is compact if the (squared) norm as given by the trace

Tr D†D M (10.3)  ⇣ ⌘ is bounded for all the D G. 2 416 Group theory Example 10.7 (SO(n), O(n), SU(n), and U(n)). The groups SO(n), O(n), SU(n), and U(n) are continuous Lie groups of infinite order. Since for any matrix D in one of these groups

Tr D†D =TrI = n M (10.4)  ⇣ ⌘ these groups also are compact.

Example 10.8 (Noncompact groups). The set of all real n n matrices ⇥ forms the general linear group GL(n, R); those of unit determinant form the special linear group SL(n, R). The corresponding groups of matrices with complex entries are GL(n, C) and SL(n, C). These four groups and the symplectic groups Sp(2n, R) and Sp(2n, C) have matrix elements that are unbounded; they are noncompact. They are continuous Lie groups of infinite order like the orthogonal and unitary groups. The group SL(2, C)represents Lorentz transformations.

10.2 Representations of Groups If one can associate with every element g of a group G a square matrix D(g) and have matrix multiplication imitate group multiplication

D(f) D(g)=D(fg) (10.5) for all elements f and g of the group G, then the set of matrices D(g) is said to form a representation of the group G. If the matrices of the representation are n n,thenn is the dimension of the representation. ⇥ The dimension of a representation also is the dimension of the vector space 1 on which the matrices act. If the matrices D(g) are unitary D†(g)=D (g), then they form a unitary representation of the group.

Example 10.9 (Representations of the groups SU(2) and SO(3)). The defining representations of SU(2) and SO(3) are the 2 2 complex matrix ⇥ cos 1 ✓ + i✓ˆ sin 1 ✓i(✓ˆ i✓ˆ )sin1 ✓ D(✓)= 2 3 2 1 2 2 (10.6) i(✓ˆ + i✓ˆ )sin1 ✓ cos 1 ✓ i✓ˆ sin ✓ 1 2 2 2 3 2 ! 10.2 Representations of Groups 417 and the 3 3 real matrix ⇥ ˆ2 ˆ ˆ ˆ ˆ ˆ ˆ c + ✓1(1 c) ✓1✓2(1 c) ✓3s ✓1✓3(1 c)+✓2s ˆ ˆ ˆ ˆ2 ˆ ˆ ˆ D(✓)= 0✓2✓1(1 c)+✓3sc+ ✓2(1 c) ✓2✓3(1 c) ✓1s1 (10.7) 2 ✓ˆ3✓ˆ1(1 c) ✓ˆ2s ✓ˆ3✓ˆ2(1 c)+✓ˆ1sc+ ✓ˆ (1 c) B 3 C @ A in which ✓ = ✓ , ✓ˆ = ✓ /✓, c = cos ✓, and s =sin✓. | | i i Compact groups possess finite-dimensional unitary representa- tions; noncompact groups do not. A group of bounded (10.3) matrices is compact. An abstract group of elements g( ↵ ) is compact if its space of { } parameters ↵ is closed and bounded.(Asetisclosed if the limit of { } every convergent sequence of its points lies in the set. A set is open if each of its elements lies in a neighborhood that lies in the set. For example, the interval [a, b] x a x b is closed, while (a, b) x asimilarity transformation

1 D0(g)=S D(g)S (10.8) which preserves the law of multiplication

1 1 D0(f) D0(g)=S D(f)SS D(g)S 1 1 = S D(f) D(g)S = S D(fg)S = D0(fg). (10.9)

A proper subspace W of a vector space V is a subspace of lower (but not zero) dimension. A proper subspace W is invariant under the action of a representation D(g)ifD(g) maps every vector v W to a vector 2 D(g) v = v W . A representation that has a proper invariant subspace 0 2 is reducible. A representation that is not reducible is irreducible. A representation D(g)iscompletely reducible if it is equivalent to a representation whose matrices are in block-diagonal form

D1(g)0... 1 S D(g) S = 0 D2(g) ... (10.10) 0 . . . 1 . . . B C @ A in which each representation Di(g) irreducible. A representation in block- diagonal form is said to be a direct sum of the irreducible representations 418 Group theory

Di 1 S DS = D D ... . (10.11) 1 2

10.3 Representations Acting in Hilbert Space A symmetry transformation g is a map (1.186) of states that pre- ! 0 serves probabilities 2 2 0 0 = . (10.12) |h | i| |h | i| The action of a group G of symmetry transformations g on the Hilbert space of a quantum theory can be represented either by operators U(g) that are linear and unitary (the usual case) or by ones K(g) that are antilinear (1.184) and antiunitary (1.185), as in the case of time reversal. Wigner proved this theorem in the 1930’s, and Weinberg improved it in his 1995 classic (Wein- berg, 1995, p. 51) (Eugene Wigner, 1902–1995; Steven Weinberg, 1933–).

Two operators F1 and F2 that commute F1 F2 = F2 F1 are compatible (1.373). A set of compatible operators F1, F2,...iscomplete if to every set of eigenvalues there belongs only a single eigenvector (section 1.31).

Example 10.10 (Rotation Operators). Suppose that the hamiltonian H, 2 the square of the angular momentum J , and its z-component Jz form a complete set of compatible observables, so that the identity operator can be expressed as a sum over the eigenvectors of these operators

I = E,j,m E,j,m . (10.13) | ih | E,j,mX Then the matrix element of a unitary operator U(g) between two states | i and is | i

U(g) = E0,j0,m0 E0,j0,m0 U(g) E,j,m E,j,m . h | | i h | | ih | | ih | i E0X,j0,m0 E,j,mX (10.14) 2 Let H and J be invariant under the action of U(g) so that U †(g)HU(g)=H 2 2 2 2 and U †(g)J U(g)=J .ThenHU(g)=U(g)H and J U(g)=U(g)J , and so if H E,j,m = E E,j,m and J 2 E,j,m = j(j + 1) E,j,m ,wehave | i | i | i | i HU(g) E,j,m = U(g)H E,j,m = EU(g) E,j,m | i | i | i (10.15) J 2U(g) E,j,m = U(g)J 2 E,j,m = j(j + 1)U(g) E,j,m . | i | i | i 10.4 Subgroups 419 Thus U(g) cannot change E or j, and so

(j) E0,j0,m0 U(g) E,j,m = E Ej j j, m0 U(g) j, m = E Ej jD (g). h | | i 0 0 h | | i 0 0 m0m (10.16) The matrix element (10.14) is a single sum over E and j in which the irreducible representations D(j) (g) of the rotation group SU(2) appear m0m (j) U(g) = E,j,m0 D (g) E,j,m . (10.17) h | | i h | i m0m h | i E,j,mX0,m This is how the block-diagonal form (10.10) usually appears in calculations. The matrices D(j) (g) inherit the unitarity of the operator U(g). m0m

10.4 Subgroups If all the elements of a group S also are elements of a group G,thenS is a subgroup of G. Every group G has two trivial subgroups—the identity element e and the whole group G itself. Many groups have more interest- ing subgroups. For example, the rotations about a fixed axis is an abelian subgroup of the group of all rotations in 3-dimensional space. A subgroup S G is an invariant subgroup if every element s of the ⇢ subgroup S is left inside the subgroup under the action of every element g of the whole group G, that is, if

1 g sg = s0 S for all g G. (10.18) 2 2 This condition often is written as g 1Sg = S for all g G or as 2 Sg= gS for all g G. (10.19) 2 Invariant subgroups also are called normal subgroups. AsetC G is called a conjugacy class if it’s invariant under the action ⇢ of the whole group G, that is, if Cg = gC or

1 g Cg= C for all g G. (10.20) 2 A subgroup that is the union of a set of conjugacy classes is invariant. The center C of a group G is the set of all elements c G that commute 2 with every element g of the group, that is, their commutators [c, g] cg gc = 0 (10.21) ⌘ vanish for all g G. 2 420 Group theory Example 10.11 (Centers Are Abelian Subgroups). Does the center C al- ways form an abelian subgroup of its group G?Theproductc1c2 of any two elements c1 and c2 of the center commutes with every element g of G since c1c2g = c1gc2 = gc1c2. So the center is closed under multiplication. The identity element e commutes with every g G,soe C.Ifc C,then 2 2 0 2 c g = gc for all g G, and so multiplication of this equation from the left 0 0 2 and the right by c 1 gives gc 1 = c 1g, which shows that c 1 C.The 0 0 0 0 2 subgroup C is abelian because each of its elements commutes with all the elements of G including those of C itself.

So the center of any group always is one of its abelian invariant subgroups. The center may be trivial, however, consisting either of the identity or of the whole group. But a group with a nontrivial center can not be simple or semisimple (section 10.27).

10.5 Cosets If H is a subgroup of a group G, then for every element g G the set of 2 elements Hg hg h H is a right coset of the subgroup H G. ⌘{ | 2 } ⇢ (Here means is a subset of or equivalently is contained in.) ⇢ If H is a subgroup of a group G, then for every element g G the set of 2 elements gH is a left coset of the subgroup H G. ⇢ The number of elements in a coset is the same as the number of elements of H, which is the order of H. An element g of a group G is in one and only one right coset (and in one and only one left coset) of the subgroup H G. For suppose instead that ⇢ g were in two right cosets g Hg and g Hg , so that g = h g = h g 2 1 2 2 1 1 2 2 for suitable h1,h2 H and g1,g2 G.ThensinceH is a (sub)group, we 1 2 2 have g = h h g = h g , which says that g Hg . But this means that 2 2 1 1 3 1 2 2 1 every element hg Hg is of the form hg = hh g = h g Hg .Soevery 2 2 2 2 3 1 4 1 2 1 element hg Hg is in Hg : the two right cosets are identical, Hg = Hg . 2 2 2 1 1 2 The right (or left) cosets are the points of the quotient coset space G/H. If H is an invariant subgroup of G, then by definition (10.19) Hg = gH for all g G, and so the left cosets are the same sets as the right cosets. In 2 this case, the coset space G/H is itself a group with multiplication defined 10.6 Morphisms 421 by

(Hg1)(Hg2)= hig1hjg2 hi,hj H { |1 2 } = h g h g g g h ,h H i 1 j 1 1 2| i j 2 = h h g g h ,h H { i k 1 2| i k 2 } = h g g h H = Hg g (10.22) { ` 1 2| ` 2 } 1 2 which is the multiplication rule of the group G. This group G/H is called the quotient or factor group of G by H

10.6 Morphisms An isomorphism is a one-to-one map between groups that respects their multiplication laws. For example, a similarity transformation (10.8) relates two equivalent representations 1 D0(g)=S D(g)S (10.23) and is an isomorphism (exercise 10.8). An automorphism is an isomor- 1 phism between a group and itself. The map gi ggi g is one to one 1 1 ! because gg1 g = gg2 g implies that gg1 = gg2, and so that g1 = g2. 1 1 This map also preserves the law of multiplication since gg1 g gg2 g = 1 gg1 g2 g . So the map 1 G gGg (10.24) ! is an automorphism. It is called an inner automorphism because g is an element of G. An automorphism not of this form (10.24) is an outer automorphism.

10.7 Schur’s Lemma

Part 1:IfD1 and D2 are inequivalent, irreducible representations of a group G, and if D (g)A = AD (g) for some matrix A and for all g G,thenthe 1 2 2 matrix A must vanish, A = 0. Proof: First suppose that A annihilates some vector x , that is, A x = 0. | i | i Let P be the projection operator into the subspace that A annihilates, which is of at least one dimension. This subspace, incidentally, is called the null space (A) or the kernel of the matrix A. The representation D must N 2 leave this null space (A) invariant since N AD2(g)P = D1(g)AP =0. (10.25) 422 Group theory If (A) were a proper subspace, then it would be a proper invariant subspace N of the representation D2, and so D2 would be reducible, which is contrary to our assumption that D and D are irreducible. So the null space (A) 1 2 N must be the whole space upon which A acts, that is, A = 0. A similar argument shows that if y A = 0 for some bra y ,thenA = 0. h | h | So either A is zero or it annihilates no ket and no bra. In the latter case, A 1 must be square and invertible, which would imply that D2(g)=A D1(g)A, that is, that D1 and D2 are equivalent representations, which is contrary to our assumption that they are inequivalent. The only way out is that A vanishes. Part 2: If for a finite-dimensional, irreducible representation D(g) of a group G,wehaveD(g)A = AD(g) for some matrix A and for all g G,then 2 A = cI. That is, any matrix that commutes with every element of a finite- dimensional, irreducible representation must be a multiple of the identity matrix. Proof: Every square matrix A has at least one eigenvector x and eigen- | i value c so that A x = c x because its characteristic equation det(A cI)=0 | i | i always has at least one root by the fundamental theorem of algebra (5.76). So the null space (A cI) has dimension greater than zero. The assumption N D(g)A = AD(g) for all g G implies that D(g)(A cI)=(A cI)D(g) for 2 all g G. Let P be the projection operator onto the null space (A cI). 2 N Then we have (A cI)D(g)P = D(g)(A cI)P = 0 for all g G which 2 implies that D(g)P maps vectors into the null space (A cI). This null N space therefore is a subspace that is invariant under D(g), which means that D is reducible unless the null space (A cI) is the whole space. Since by N assumption D is irreducible, it follows that (A cI) is the whole space, N that is, that A = cI. (Issai Schur, 1875–1941)

Example 10.12 (Schur, Wigner, and Eckart). Suppose an arbitrary observ- able O is invariant under the action of the rotation group SU(2) represented by unitary operators U(g) for g SU(2) 2

U †(g)OU(g)=O or [O, U(g)] = 0. (10.26)

These unitary rotation operators commute with the square J 2 of the angular momentum [J 2,U] = 0. Suppose that they also leave the hamiltonian H unchanged [H, U] = 0. Then as shown in example 10.10, the state U E,j,m | i 10.8 Characters 423 is a sum of states all with the same values of j and E. It follows that

E,j,m O E0,j0,m0 E0,j0,m0 U(g) E0,j0,m00 h | | ih | | i Xm0 (10.27) = E,j,m U(g) E,j,m0 E,j,m0 O E0,j0,m00 h | | ih | | i Xm0 or in the notation of (10.16)

(j ) (j) E,j,m O E0,j0,m0 D 0 (g) = D (g) E,j,m0 O E0,j0,m00 . h | | i m0m00 mm0 h | | i m0 m0 X X (10.28) Now Part 1 of Schur’s lemma tells us that the matrix E,j,m O E ,j ,m h | | 0 0 0i must vanish unless the representations are equivalent, which is to say unless j = j0.Sowehave (j) (j) E,j,m O E0,j,m0 D (g) = D (g) E,j,m0 O E0,j,m00 . h | | i m0m00 mm0 h | | i m0 m0 X X (10.29) Now Part 2 of Schur’s lemma tells us that the matrix E,j,m O E ,j,m h | | 0 0i must be a multiple of the identity. Thus the symmetry of O under rotations simplifies the matrix element to

E,j,m O E0,j0,m0 = O (E,E0). (10.30) h | | i jj0 mm0 j This result is a special case of the Wigner-Eckart theorem (Eugene Wigner 1902–1995, Carl Eckart 1902–1973).

10.8 Characters Suppose the n n matrices D (g) form a representation of a group G g. ⇥ ij 3 The character D(g) of the matrix D(g) is the trace n D(g)=TrD(g)= Dii(g). (10.31) Xi=1 Traces are cyclic, that is, TrABC =TrBCA =TrCAB.Soiftworepresen- 1 tations D and D0 are equivalent, so that D0(g)=S D(g)S,thentheyhave the same characters because

1 1 D0 (g)=TrD0(g)=Tr S D(g)S =Tr D(g)SS =TrD(g)=D(g). (10.32) If two group elements g1 and g2 are in the same conjugacy class, that is, 424 Group theory if g = gg g 1 for all g G, then they have the same character in a given 2 1 2 representation D(g) because 1 1 D(g2)=TrD(g2)=TrD(gg1g )=Tr D(g)D(g1)D(g ) 1 =Tr D(g1)D (g)D(g) =TrD(g1)=D(g1). (10.33)

10.9 Direct products Suppose D(a)(g)isak-dimensional representation of a group G, and D(b)(g) is an n-dimensional representation of the same group. Then their product (a,b) (a) (b) Dim,j`(g)=Dij (g) Dm`(g) (10.34) is a (kn)-dimensional direct-product representation of the group G.Direct products are also called tensor products. They occur in quantum systems that have two or more parts, each described by a di↵erent space of vectors. Suppose the vectors i for i =1...k are the basis vectors of the k- | i (a) dimensional space Vk on which the representation D (g) acts, and that the vectors m for m =1...n are the basis vectors of the n-dimensional space | i (b) Vn on which D (g) acts. The kn vectors i, m are basis vectors for the | i (a,b) kn-dimensional tensor-product space Vkn. The matrices D (g) defined as

i, m D(g)(a,b) j, ` = i D(a)(g) j m D(b)(g) ` (10.35) h | | i h | | ih | | i act in this kn-dimensional space Vkn and form a representation of the group G; this direct-product representation usually is reducible. Many tricks help one to decompose reducible tensor-product representations into direct sums of irreducible representations (Georgi, 1999; Zee, 2016). Example 10.13 (Adding Angular Momenta). The addition of angular mo- menta illustrates both the direct product and its reduction to a direct sum of irreducible representations. Let D(j1)(g) and D(j2)(g)respectivelybethe (2j + 1) (2j + 1) and the (2j + 1) (2j + 1) representations of the 1 ⇥ 1 2 ⇥ 2 rotation group SU(2). The direct-product representation D(j1,j2)

(j1,j2) (j1) (j2) m0 ,m0 D m ,m = m0 D (g) m m0 D (g) m (10.36) h 1 2| | 1 2i h 1| | 1ih 2| | 2i is reducible into a direct sum of all the irreducible representations of SU(2) (j1+j2) ( j1 j2 ) from D (g)downtoD | | (g) in integer steps:

(j1,j2) (j1+j2) (j1+j2 1) ( j1 j2 +1) ( j1 j2 ) D = D D D | | D | | (10.37) ··· each irreducible representation occurring once in the direct sum. 10.10 Finite Groups 425

Example 10.14 (Adding Two Spins). When one adds j1 =1/2toj2 =1/2, one finds that the tensor-product matrix D(1/2,1/2) is equivalent to the direct sum D(1) D(0) D(1)(✓)0 D(1/2,1/2)(✓)=S 1 S (10.38) 0 D(0)(✓) ✓ ◆ where the matrices S, D(1), and D(0) are 4 4, 3 3, and 1 1. ⇥ ⇥ ⇥

10.10 Finite Groups A finite group is one that has a finite number of elements. The number of elements in a group is the order of the group.

Example 10.15 (Z2). The group Z2 consists of two elements e and p with multiplication rules ee = e, ep = pe = p, and pp = e. (10.39) Clearly, Z is abelian, and its order is 2. The identification e 1 and 2 ! p 1 gives a 1-dimensional representation of the group Z in terms of ! 2 1 1 matrices, which are just numbers. ⇥ It is tedious to write the multiplication rules as individual equations. Nor- mally people compress them into a multiplication table like this:

ep ⇥ eep (10.40) ppe

A simple generalization of Z2 is the group Zn whose elements may be represented as exp(i2⇡m/n) for m =1,...,n. This group is also abelian, and its order is n.

Example 10.16 (Z3). The multiplication table for Z3 is

eab ⇥ eeab (10.41) aabe bbea 426 Group theory which says that a2 = b, b2 = a, and ab = ba = e.

10.11 The Regular Representation For any finite group G we can associate an orthonormal vector g with | ii each element g of the group. So g g = . These orthonormal vectors i h i| ji ij g form a basis for a vector space whose dimension is the order of the group. | ii The matrix D(gk) of the regular representation of G is defined to map any vector g into the vector g g associated with the product g g | ii | k ii k i D(g ) g = g g . (10.42) k | ii | k ii Since group multiplication is associative, we have D(g )D(g ) g = D(g ) g g = g (g g ) = (g g )g ) = D(g g ) g . j k | ii j | k ii | j k i i | j k i i j k | ii (10.43) Because the vector g was an arbitrary basis vector, it follows that | ii D(gj)D(gk)=D(gjgk) (10.44) which means that the matrices D(g) satisfy the criterion (10.5) for their being a representation of the group G. The matrix D(g) has entries [D(g)] = g D(g) g . (10.45) ij h i| | ji The sum of dyadics g g over all the elements g of a finite group G is | `ih `| ` the unit matrix g g = I (10.46) | `ih `| n g G X`2 in which n is the order of G, that is, the number of elements in G.Soby taking the m, n matrix element of the multiplication law (10.44), we find [D(g g )] = g D(g g ) g = g D(g )D(g ) g (10.47) j k m,n h m| j k | ni h m| j k | ni = g D(g ) g g D(g ) g = [D(g )] [D(g )] . h m| j | `ih `| k | ni j m,` k `,n g G g G X`2 X`2 Example 10.17 (Z3’s Regular Representation). The regular representation of Z3 is 100 001 010 D(e)= 010,D(a)= 100,D(b)= 001 (10.48) 0 1 0 1 0 1 001 010 100 @ A @ A @ A so D(a)2 = D(b), D(b)2 = D(a), and D(a)D(b)=D(b)D(a)=D(e). 10.12 Properties of Finite Groups 427 10.12 Properties of Finite Groups In his book (Georgi, 1999, ch. 1), Georgi proves the following theorems:

1. Every representation of a finite group is equivalent to a unitary represen- tation. 2. Every representation of a finite group is completely reducible. 3. The irreducible representations of a finite abelian group are one dimen- sional. 4. If D(a)(g) and D(b)(g) are two unitary irreducible representations of di- mensions na and nb of a group G of N elements g1,...,gN ,thenthe functions n a D(a)(g) (10.49) N jk r are orthonormal and complete in the sense that

N na (a) (b) D ⇤(g )D (g )= . (10.50) N ik j `m j ab i` km Xj=1 5. The order N of a finite group is the sum of the squares of the dimensions of its inequivalent irreducible representations

2 N = na. (10.51) a X

Example 10.18 (ZN ). The abelian cyclic group ZN with elements

2⇡ij/N gj = e (10.52) has N one-dimensional irreducible representations

(a) 2⇡iaj/N D (gj)=e (10.53) for a =1, 2,...,N. Their orthonormality relation (10.50) is the Fourier formula N 1 2⇡iaj/N 2⇡ibj/N e e = . (10.54) N ab Xj=1 2 The na are all unity, there are N of them, and the sum of the na is N as required by the sum rule (10.51). 428 Group theory 10.13 Permutations

The permutation group on n objects is called Sn. Permutations are made of cycles that change the order of some of the n objects. For instance, the permutation (1 2) = (2 1) is a 2-cycle that means x x x ; the unitary 1 ! 2 ! 1 operator U((1 2)) that represents it interchanges states like this:

U((1 2)) +, = U((1 2)) +, 1 , 2 = , 1 , +, 2 = , + . (10.55) | i | i| i | i | i | i The 2-cycle (3 4) means x x x , it changes (a, b, c, d)into(a, b, d, c). 3 ! 4 ! 3 The 3-cycle (1 2 3) = (2 3 1) = (3 1 2) means x x x x , it changes 1 ! 2 ! 3 ! 1 (a, b, c, d)into(b, c, a, d). The 4-cycle (1 3 2 4) means x x x x 1 ! 3 ! 2 ! 4 ! x and changes (a, b, c, d)into(c, d, b, a). The 1-cycle (2) means x x and 1 2 ! 2 leaves everything unchanged. The identity element of Sn is the product of 1-cycles e = (1)(2) ...(n). The inverse of the cycle (1324) must invert x x x x x , 1 ! 3 ! 2 ! 4 ! 1 so it must be (1 4 2 3) which means x x x x x so that it 1 ! 4 ! 2 ! 3 ! 1 changes (c, d, b, a) back into (a, b, c, d). Every element of Sn has each integer from 1 to n in one and only one cycle. So an arbitrary element of Sn with `k k-cycles must satisfy n k`k = n. (10.56) Xk=1

10.14 Compact and Noncompact Lie Groups Imagine rotating an object repeatedly. Notice that the biggest rotation is by an angle of ⇡ about some axis. The possible angles form a circle; the ± space of parameters is a circle. The parameter space of a compact group is compact—closed and bounded. The rotations form a compact group. Now consider the translations. Imagine moving a pebble to the Sun, then moving it to the next-nearest star, then moving it to the nearest galaxy. If space is flat, then there is no limit to how far one can move a pebble. The parameter space of a noncompact group is not compact. The translations form a noncompact group. We’ll see that compact Lie groups possess unitary representations, with n n unitary matrices D(↵), while noncompact ones don’t. Here ↵ stands ⇥ for the parameters ↵1,...,↵n that label the elements of the group, three for the rotation group. The ↵’s usually are real, but can be complex. 10.15 Generators 429 10.15 Generators To study continuous groups, we will use calculus and algebra, and we will focus on the simplest part of the group—the elements g(↵) for ↵ 0which ⇡ are near the identity e = g(0) for which all ↵a = 0. Each element g(↵) of the group is represented by a matrix D(↵) D(g(↵)) in the D representa- ⌘ tion of the group and by another matrix D (↵ ) D (g(↵)) in any other D 0 0 ⌘ 0 0 representation of the group. Every representation must respect the multi- plication law of the group. So if g(↵0)g(↵)=g(↵00), then the matrices of the D representation must satisfy D(↵0)D(↵)=D(↵00), and those of any other representation D0 must satisfy D0(↵0)D0(↵)=D0(↵00). A generator ta of a representation D is the partial derivative of the matrix D(↵) with respect to the component ↵a of ↵ evaluated at ↵ =0 @D(↵) t = i . (10.57) a @↵ a ↵=0 When all the parameters ↵ are infinitesimal, ↵ 1, the matrix D(↵)is a | a|⌧ very close to the identity matrix I

D(↵) I + i ↵ t . (10.58) ' a a a X As n , this relation becomes exact !1 ↵ ↵ D = I + i a t (10.59) n n a a ⇣ ⌘ X and by exponentiating it we get the matrix D(↵) that represents the group element g(↵)

n ↵a D(↵)= lim I + i ta =exp i ↵a ta (10.60) n n !1 a ! a ! X X in the exponential parametrization. The i’s appear in these equations so that when the generators ta are hermitian matrices, (ta)† = ta, and the ↵’s are real, the matrices D(↵) are unitary

1 D (↵)= exp i ↵ t = D†(↵)=exp i ↵ t . (10.61) a a a a a ! a ! X X Compact groups have finite-dimensional, unitary representations with her- mitian generators. 430 Group theory 10.16 Lie algebra

Suppose tb and tc are any two generators of a representation D. Let ↵✏,a be the infinitesimal exponential parameters ↵✏,a =(0,...,✏,...,0) which is all zeros except for an ✏ in position a and ↵✏,b which is all zeros with ✏ in position b. The multiplication law of the group determines the parameters ↵(✏, a, b) of the product

g(↵ )g(↵ )g( ↵ )g( ↵ )=g(↵(✏, a, b)). (10.62) ✏,b ✏,a ✏,b ✏,a

The matrices of any two representations D with generators ta and D0 with generators ta0 must obey the same multiplication law

D(↵✏,b) D(↵✏,a) D( ↵✏,b) D( ↵✏,a)=D(↵(✏, a, b)) (10.63) D0(↵ ) D0(↵ ) D0( ↵ ) D0( ↵ )=D0(↵(✏, a, b)) ✏,b ✏,a ✏,b ✏,a 2 with the same parameters ↵✏,a, ↵✏,a, and ↵(✏, a, b). To order ✏ ,theproduct of the four D’s is

i✏t i✏ta i✏t i✏ta D(↵ ) D(↵ ) D( ↵ ) D( ↵ )=e b e e b e ✏,b ✏,a ✏,b ✏,a 2 2 ✏ 2 ✏ 2 (1 + i✏tb tb )(1 + i✏ta ta) ⇡ 2 2 (10.64) ✏2 ✏2 (1 i✏t t2)(1 i✏t t2) ⇥ b 2 b a 2 a 1+✏2(t t t t )=1+✏2[t ,t ]. ⇡ a b b a a b The other representation must give the same result with primes

2 D0(↵ ) D0(↵ ) D0( ↵ ) D0( ↵ ) 1+✏ [t0 ,t0 ]. (10.65) ✏,b ✏,a ✏,b ✏,a ⇡ a b These products (10.64 & 10.65) represent the same group element g(↵(✏, a, b)) which is near the identity. So the matrix D(↵(✏, a, b)) must be a linear com- bination of generators of order ✏2

n D(↵(✏, a, b)) 1+i✏2 f c t (10.66) ⇡ ab c c=1 X and the matrix D0(↵(✏, a, b)) must be the same linear combination of the generators tc0 n 2 c D0(↵(✏, a, b)) 1+i✏ f t0 . (10.67) ⇡ ab c c=1 X 10.16 Lie algebra 431 The product formulas 10.64 and 10.66 for the representation D and 10.65 and 10.67 for the representation D0 now give n 2 2 c 1+✏ [ta,tb]=1+i✏ fab tc c=1 Xn (10.68) 2 2 c 1+✏ [ta0 ,tb0 ]=1+i✏ fab tc0 c=1 X which in turn imply the Lie algebra formulas n n c c [ta,tb]= fab tc and [ta0 ,tb0 ]= fab tc0 . (10.69) c=1 c=1 X X Thus the commutator of any two generators is a linear combina- c tion of all the generators. The coecients fab are the structure constants of the group. They are the same for all representations of the group. Unless the parameters ↵a are redundant, the generators are linearly in- dependent. They span a vector space, and any linear combination may be called a generator. By using the Gram-Schmidt procedure (section 1.10), we may make the generators ta orthogonal with respect to the inner product (1.91)

(ta,tb)=Tr ta† tb = kab (10.70) in which k is a nonnegative normalization⇣ ⌘ constant that depends upon the representation. We can’t normalize the generators, making k unity, because c the structure constants fab are the same in all representations. By multiplying both sides of the first of the two Lie algebra formulas

(10.69) by td† and using the orthogonality (10.70) of the generators, we find

c c d Tr [ta,tb] td† = ifab Tr tc td† = ifab kcd = ik fab (10.71) ⇣ ⌘ ⇣ ⌘ c which implies that the structure constant fab is the trace

c i f = Tr [t ,t ] t† . (10.72) ab k a b c ⇣ ⌘ Because of the antisymmetry of the commutator [ta,tb], structure constants are antisymmetric in their lower indices f c = f c . (10.73) ab ba

From any n n matrix A, one may make a hermitian matrix A + A and ⇥ † 432 Group theory an antihermitian one A A†. Thus, one may separate the nG generators (h) (ah) into a set that are hermitian ta and a set that are antihermitian ta . The exponential of any imaginary linear combination of n n hermitian (h) ⇥ generators D(↵)=exp i↵ ta is an n n unitary matrix since a ⇥ ⇣ ⌘ (h) (h) 1 D†(↵)=exp i↵ t† =exp i↵ t = D (↵). (10.74) a a a a ⇣ ⌘ ⇣ ⌘ A group with only hermitian generators is compact and has finite-dimensional unitary representations. On the other hand, the exponential of any imaginary linear combination (ah) of antihermitian generators D(↵)=exp i↵a ta is a real exponential of (ah) their hermitian counterparts ita whose⇣ squared⌘ norm

2 (ah) D(↵) =Tr D(↵)†D(↵) =Tr exp 2↵ it (10.75) k k a a h i h ⇣ ⌘i grows exponentially and without limit as the parameters ↵ .A a !±1 group with some antihermitian generators is noncompact and does not have finite-dimensional unitary representations. (The unitary representa- tions of the translations and of the Lorentz and Poincar´egroups are infinite dimensional.) Compact Lie groups have hermitian generators, and so the structure- constant formula (10.72) reduces in this case to

c f =( i/k)Tr [t ,t ] t† =( i/k)Tr ([t ,t ] t ) . (10.76) ab a b c a b c ⇣ ⌘ Now, since the trace is cyclic, we have

f b =( i/k)Tr ([t ,t ] t )=( i/k)Tr (t t t t t t ) ac a c b a c b c a b =( i/k)Tr (t t t t t t ) b a c a b c =( i/k)Tr ([t ,t ] t )=f c = f c . (10.77) b a c ba ab Interchanging a and b, we get

f a = f c = f c . (10.78) bc ab ba Finally, interchanging b and c in (10.77) gives

f c = f b = f b . (10.79) ab ca ac Combining (10.77, 10.78, & 10.79), we see that the structure constants of a compact Lie group are totally antisymmetric

f b = f b = f c = f c = f a = f a . (10.80) ac ca ba ab bc cb 10.16 Lie algebra 433 Because of this antisymmetry, it is usual to lower the upper index

f f c (10.81) abc ⌘ ab and write the antisymmetry of the structure constants of compact Lie groups more simply as

f = f = f = f = f = f . (10.82) acb cab bac abc bca cba For compact Lie groups, the generators are hermitian, and so the struc- ture constants fabc are real, as we may see by taking the complex con- jugate of the formula (10.76) for fabc

f ⇤ =(i/k)Tr (t [t ,t ]) = ( i/k)Tr ([t ,t ] t )=f . (10.83) abc c b a a b c abc

It follows from (10.69 & 10.81–10.83) that the commutator of any two generators of a Lie group is a linear combination

c [ta,tb]=ifab tc (10.84) of its generators t , and that the structure constants f f c are c abc ⌘ ab real and totally antisymmetric if the group is compact.

Example 10.19 (Gauge Transformation). The action density of a Yang- Mills theory is unchanged when a spacetime dependent unitary matrix U(x) changes a vector (x) of matter fields to 0(x)=U(x) (x). Terms like † are invariant because †(x)U †(x)U(x) (x)= †(x) (x), but how can i kinetic terms like @i † @ be made invariant? Yang and Mills introduced matrices Ai of gauge fields, replaced ordinary derivatives @i by Covariant derivatives D @ + A , and required that D = UD or that i ⌘ i i i0 0 i

@i + Ai0 U = @iU + U@i + Ai0 U = U (@i + Ai) . (10.85)

Their nonabelian gauge transformation is

A0 (x)=U(x)A (x)U †(x) (@ U(x)) U †(x). (10.86) i i i One can write the unitary matrix as U(x)=exp( ig ✓ (x) t )inwhichg a a is a coupling constant, the functions ✓a(x) parametrize the gauge transfor- mation, and the generators ta belong to the representation that acts on the vector (x) of matter fields. 434 Group theory 10.17 The Rotation Group The rotations and reflections in three-dimensional space form a compact group O(3) whose elements R are 3 3 real matrices that leave invariant ⇥ the dot product of any two three vectors (Rx) (Ry)=xTRTR y = xTIy = x y. (10.87) · · These matrices therefore are orthogonal (1.180) RTR = I. (10.88) Taking the determinant of both sides and using the transpose (1.208) and product (1.221) rules, we have (det R)2 = 1 (10.89) whence det R = 1. The group O(3) contains reflections as well as rotations ± and is disjoint. The subgroup with det R = 1 is the group SO(3). An SO(3) element near the identity R = I + ! must satisfy (I + !) T (I + !)=I. (10.90) Neglecting the tiny quadratic term, we find that the infinitesimal matrix ! is antisymmetric !T = !. (10.91) One complete set of real 3 3 antisymmetric matrices is ⇥ 00 0 001 0 10 ! = 00 1 ,!= 000,!= 100 (10.92) 1 0 1 2 0 1 3 0 1 01 0 100 000 @ A @ A @ A which we may write as

[!b]ac = ✏abc (10.93) in which ✏abc is the Levi-Civita symbol which is totally antisymmetric with ✏123 = 1 (Tullio Levi-Civita 1873–1941). The !b are antihermitian, but we make them hermitian by multiplying by i

tb = i!b so that [tb]ac = i✏abc (10.94) and R = I i✓ t . b b The three hermitian generators ta satisfy (exercise 10.15) the commutation relations

[ta,tb]=ifabc tc (10.95) 10.17 The Rotation Group 435 in which the structure constants are given by the Levi-Civita symbol ✏abc

fabc = ✏abc (10.96) so that

[ta,tb]=i✏abc tc. (10.97) They are the generators of the defining representation of SO(3) (and also of the adjoint representation of SU(2) (section 10.24)). Physicists usually scale the generators by ~ and define the angular-momentum generator La as

La = ~ ta (10.98) so that the eigenvalues of the angular-momentum operators are the physical values of the angular momenta. With ~, the commutation relations are

[La,Lb]=i ~ ✏abc Lc. (10.99) The matrix that represents a right-handed rotation (of an object) by an angle ✓ = ✓ about an axis ✓ is | | i✓ t i✓ L/~ D(✓)=e · = e · . (10.100) By using the fact (1.289) that a matrix obeys its characteristic equation, one may show (exercise 10.17) that the 3 3 matrix D(✓) that represents ⇥ a right-handed rotation of ✓ radians about the axis ✓ is the matrix (10.7) whose i, jth entry is D (✓) = cos ✓ sin ✓✏ ✓ /✓ +(1 cos ✓) ✓ ✓ /✓2 (10.101) ij ij ijk k i j in which a sum over k = 1, 2, 3 is understood. Example 10.20 (Demonstration of Commutation Relations). Take a big sphere with a distinguished point and orient the sphere so that the point lies in the y-direction from the center of the sphere. Now rotate the sphere by a small angle, say 15 degrees or ✏ = ⇡/12, right-handedly about the x-axis, then right-handedly about the y-axis by the same angle, then left- handedly about the x-axis and then left-handedly about the y-axis. Using the approximations (10.64 & 10.66) for the product of these four rotation matrices and the definitions (10.98) of the generators and of their structure constants (10.99), we have ~ta = L1 = Lx, ~tb = L2 = Ly, ~fabctc = ✏12cLc = L3 = Lz, and 2 2 ✏ ✏ 2 i✏Ly/~ i✏Lx/~ i✏Ly/~ i✏Lx/~ i✏ Lz/~ e e e e 1+ [Lx,Ly]=1+i Lz e ⇡ ~2 ~ ⇡ (10.102) 436 Group theory which is a left-handed rotation about the (vertical) z-axis. The magnitude of that rotation should be about ✏2 =(⇡/12)2 0.069 or about 3.9 degrees. ⇡ Photographs of an actual demonstration are displayed in Fig. 10.1. The demonstrated equation (10.102) shows (exercise 10.16) that the gen- erators Lx and Ly satisfy the commutation relation

[Lx,Ly]=i~Lz (10.103) of the rotation group.

10.18 Rotations and reflections in 2n dimensions The orthogonal group O(2n) of rotations and reflections in 2n dimensions is the group of all real 2n 2n matrices O whose transposes OT are their ⇥ inverses OT O = OOT = I (10.104) in which I is the 2n 2n identity matrix. Those with unit determinant, ⇥ det O = 1, constitute the subgroup SO(2n) of rotations in 2n dimensions. Asymmetricsum A, B = AB + BA is called an anticommutator. { } Complex fermionic variables i obey the anticommutation relations

i, † = ~ ik, i, k =0, and †, † =0. (10.105) { k} { } { i k} Their real xi and imaginary yi parts 1 1 xi = ( i + †) and yi = ( i †) (10.106) p2 i ip2 i obey the anticommutation relations

xi,xk = ~ ik, yi,yk = ~ ik, and xi,yk =0. (10.107) { } { } { } More simply, the anticommutation relations of these 2n hermitian variables v =(x1,...,xn,y1,...,yn) are

vi,vk = ~ ik. (10.108) { } If the real linear transformation v = L v +L v + +L v preserves i0 i1 1 i2 2 ··· i2n 2n these anticommutation relations, then the matrix L must satisfy 2n 2n 2n ~ ik = v0,v0 = Lij Lk` vj,v` = Lij Lk` ~ j` = ~ Lij Lkj { i k} { } j,X`=1 j,X`=1 Xj=1 (10.109) 10.18 Rotations and reflections in 2n dimensions 437 Physical Demonstration of the Commutation Relations

Figure 10.1 Demonstration of equation (10.102) and the commutation re- lation (10.103). Upper left: black ball with a white stick pointing in the y-direction; the x-axis is to the reader’s left, the z-axis is vertical. Upper right: ball after a small right-handed rotation about the x-axis. Center left: ball after that rotation is followed by a small right-handed rotation about the y-axis. Center right: ball after these rotations are followed by a small left-handed rotation about the x-axis. Bottom: ball after these rotations are followed by a small left-handed rotation about the y-axis. The net e↵ect is approximately a small left-handed rotation about the z-axis.

which is the statement that it is orthogonal, LLT = I. Thus the group O(2n) is the largest group of linear transformations that preserve the anticommu- 438 Group theory tation relations of the 2n hermitian real and imaginary parts vi of n complex fermionic variables i.

10.19 The defining representation of SU(2) The smallest positive value of angular momentum is ~/2. The spin-one-half angular momentum operators are represented by three 2 2 matrices ⇥ ~ S = (10.110) a 2 a in which the a are the Pauli matrices 01 0 i 10 = ,= , and = (10.111) 1 10 2 i 0 3 0 1 ✓ ◆ ✓ ◆ ✓ ◆ which obey the multiplication law

i j = ij + i✏ijk k (10.112) summed over k from 1 to 3. Since the symbol ✏ijk is totally antisymmetric in i, j, and k, the Pauli matrices obey the commutation and anticommutation relations

[i,j] ij ji =2i✏ijk k ⌘ (10.113) , + =2 . { i j}⌘ i j j i ij The Pauli matrices divided by 2 satisfy the commutation relations (10.97) of the rotation group 1 1 1 2 a, 2 b = i✏abc 2 c (10.114) and generate the elements⇥ of the group⇤ SU(2) ✓ ✓ exp i ✓ = I cos + i ✓ˆ sin (10.115) · 2 2 · 2 ⇣ ⌘ in which I is the 2 2 identity matrix, ✓ = p✓2 and ✓ˆ = ✓/✓. ⇥ It follows from (10.114) that the spin operators (10.110) satisfy

[Sa,Sb]=i ~ ✏abc Sc. (10.116)

10.20 The Lie Algebra and Representations of SU(2) The three generators of SU(2) in its 2 2 defining representation are the ⇥ Pauli matrices divided by 2, ta = a/2. The structure constants of SU(2) 10.20 The Lie Algebra and Representations of SU(2) 439 are fabc = ✏abc which is totally antisymmetric with ✏123 =1 1 1 1 [ta,tb]=ifabctc = 2 a, 2 b = i✏abc 2 c. (10.117) For every half-integer ⇥ ⇤ n j = for n =0, 1, 2, 3,... (10.118) 2 there is an irreducible representation of SU(2)

(j) i✓ J (j) D (✓)=e · (10.119)

(j) (j) in which the three generators ta Ja are (2j + 1) (2j + 1) square ⌘ (j) ⇥ hermitian matrices. In a basis in which J3 is diagonal, the matrix elements (j) (j) (j) of the complex linear combinations J J1 iJ2 are ± ⌘ ± (j) (j) J1 iJ2 = s0,s 1 (j s)(j s + 1) (10.120) ± s0,s ± ⌥ ± h i p where s and s run from j to j in integer steps and those of J (j) are 0 3 (j) J3 = ss0,s. (10.121) s0,s h i Borrowing a trick from section 10.25, one may show that the commutator of the square J (j) J (j) of the angular momentum matrix commutes with (j·) (j)2 (j) every generator Ja .ThusJ commutes with D (✓) for every element of the group. Part 2 of Schur’s lemma (section 10.7) then implies that J (j)2 must be a multiple of the (2j + 1) (2j +1) identity matrix. The coecient ⇥ turns out to be j(j + 1) J (j) J (j) = j(j + 1) I. (10.122) · Combinations of generators that are a multiple of the identity are called Casimir operators. (2) (2) Example 10.21 (Spin 2). For j = 2, the spin-two matrices J+ and J3 are 02 0 0 0 200 0 0 00p60 0 010 0 0 (2) 0 1 (2) 0 1 J+ = 00 0p60 and J3 = 000 0 0 B 00 0 0 2C B 000 10C B C B C B 00 0 0 0C B 000 0 2C B C B C @ A @ (10.123)A (2) † and J = J+ . ⇣ ⌘ 440 Group theory The tensor product of any two irreducible representations D(j) and D(k) of SU(2) is equivalent to the direct sum of all the irreducible representations D(`) for j k ` j + k | |  j+k D(j) D(k) = D(`) (10.124) ⌦ `= j k M| | each D(`) occurring once.

Example 10.22 (Addition theorem). The spherical harmonics Y`m(✓,)= ✓, `, m of section 8.13 transform under the (2` + 1)-dimensional repre- h | i sentation D` of the rotation group. If a rotation R takes ✓, into the vector ✓ ,, so that ✓ , = U(R) ✓, ,thensummingoverm from ` to `,we 0 0 | 0 0i | i 0 get

Y ⇤ (✓0,0)= `, m ✓0,0 = `, m U(R) ✓, `,m h | i h | | i ` = `, m U(R) `, m0 `, m0 ✓, = D (R)m,m Y ⇤ (✓,). h | | ih | i 0 `,m0 Suppose now that a rotation R maps ✓ , and ✓ , into ✓ , = | 1 1i | 2 2i | 10 10 i U(R) ✓ , and ✓ , = U(R) ✓ , . Then summing over the repeated | 1 1i | 20 20 i | 2 2i indices m, m , and m from ` to `,wefind 0 00 ` ` Y (✓0 ,0 )Y ⇤ (✓0 ,0 )=D (R)⇤ Y (✓ , )D (R) Y ⇤ (✓ , ). `,m 1 1 `,m 2 2 m,m0 `,m0 1 1 m,m00 `,m00 2 2 ` In this equation, the matrix element D (R)⇤ is m,m0 ` ` 1 D (R)⇤ = `, m U(R) `, m0 ⇤ = `, m0 U †(R) `, m = D (R )m ,m. m,m0 h | | i h | | i 0 Thus since D` is a representation of the rotation group, the product of the two D`’s in (10.22) is

` ` ` 1 ` D (R)⇤ D (R) = D (R ) D (R) m,m0 m,m00 m0,m m,m00 ` 1 ` = D (R R)m0,m00 = D (I)m0,m00 = m0,m00 .

So as long as the same rotation R maps ✓1,1 into ✓10 ,10 and ✓2,2 into ✓20 ,20 ,thenwehave ` `

Y`,m(✓10 ,10 )Y`⇤,m(✓20 ,20 )= Y`,m(✓1,1)Y`⇤,m(✓2,2). m= ` m= ` X X We choose the rotation R as the product of a rotation that maps the unit vectorn ˆ(✓2,2)intoˆn(✓20 ,20 )=zˆ =(0, 0, 1) and a rotation about the z axis that mapsn ˆ(✓1,1)intoˆn(✓10 ,10 )=(sin✓, 0, cos ✓)inthex-z plane where it makes an angle ✓ withn ˆ(✓20 ,20 )=zˆ.WethenhaveY`⇤,m(✓20 ,20 )= 10.21 How a field transforms under a rotation 441

Y`⇤,m(0, 0) and Y`,m(✓10 ,10 )=Y`,m(✓, 0) in which ✓ is the angle between the unit vectorsn ˆ(✓10 ,10 ) andn ˆ(✓20 ,20 ), which is the same as the angle between the unit vectorsn ˆ(✓1,1) andn ˆ(✓2,2). The vanishing (8.107) at ✓ = 0 of the associated Legendre functions P for m = 0 and the definitions (8.4, `,m 6 8.100, & 8.111–8.113) say that Y`⇤,m(0, 0) = (2` + 1)/4⇡m,0, and that Y`,0(✓, 0) = (2` + 1)/4⇡P`(cos ✓). Thus, ourp identity (10.22) gives us the for the spherical harmonics addition theorem (8.122) p ` 2` +1 P (cos ✓)= Y (✓ , )Y ⇤ (✓ , ). ` 4⇡ `,m 1 1 `,m 2 2 m= ` X

10.21 How a field transforms under a rotation (j) Under a rotation R,afield s(x) that transforms under the D represen- tation of SU(2) responds as

1 (j) 1 U(R) s(x) U (R)=D (R ) s (Rx). (10.125) s,s0 0

Example 10.23 (Spin and Statistics). Suppose a, m and b, m are any | i | i eigenstates of the rotation operator J3 with eigenvalue m (in units with ~ = c = 1). If u and v are any two space-like points, then in some Lorentz frame they have spacetime coordinates u =(t, x, 0, 0) and v =(t, x, 0, 0). Let U be the unitary operator that represents a right-handed rotation by ⇡ about the 3-axis or z-axis of this Lorentz frame. Then

im⇡ 1 im⇡ U a, m = e a, m and b, m U = b, m e . (10.126) | i | i h | h | And by (10.125), U transforms a field of spin j with x (x, 0, 0) to ⌘ 1 (j) 1 i⇡s U(R) s(t, x) U (R)=D (R ) s (t, x)=e s(t, x). (10.127) ss0 0 1 Thus by inserting the identity operator in the form I = U U and using both (10.126) and (10.127), we find, since the phase factors exp( im⇡) and exp(im⇡) cancel,

1 1 b, m (t, x) (t, x) a, m = b, m U (t, x)U U (t, x)U a, m h | s s | i h | s s | i = e2i⇡s b, m (t, x) (t, x) a, m .(10.128) h | s s | i Now if j is an integer, then so is s, and the phase factor exp(2i⇡s)=1is 442 Group theory unity. In this case, we find that the mean value of the equal-time commutator vanishes b, m [ (t, x), (t, x)] a, m =0. (10.129) h | s s | i Thus fields of integral spin commute at space-like separations. They rep- resent bosons. On the other hand, if j is half an odd integer, that is, j =(2n + 1)/2, where n is an integer, then the phase factor exp(2i⇡s)= 1 is minus one. In this case, the mean value of the equal-time anticommutator vanishes b, m (t, x), (t, x) a, m =0. (10.130) h |{ s s }| i Thus fields of half-odd-integral spin anticommute at space-like separations. They represent fermions. While not a proof of the spin-statistics theorem, this argument shows that the behavior of fields under rotations does deter- mine their equal-time commutation or anticommutation relations

2j (t, x) (t, x0)+( 1) (t, x0) (t, x) = 0 (10.131) s s0 s0 s and their statistics.

10.22 The addition of two spin-one-half systems The spin operators (10.110) ~ S = (10.132) a 2 a obey the commutation relation (10.116)

[Sa,Sb]=i ~ ✏abc Sc. (10.133) The raising and lowering operators

S = S1 iS2 (10.134) ± ± have simple commutators with S3

[S3,S ]= ~ S . (10.135) ± ± ± This relation implies that if the state 1 ,m is an eigenstate of S with | 2 i 3 eigenvalue ~m, then the states S 1 ,m either vanish or are eigenstates of ±| 2 i S3 with eigenvalues ~(m 1) ± 1 1 1 1 S3S ,m = S S3 ,m ~S ,m = ~(m 1)S ,m . (10.136) ±| 2 i ± | 2 i± ±| 2 i ± ±| 2 i 10.22 The addition of two spin-one-half systems 443 Thus the raising and lowering operators raise and lower the eigenvalues of S3. The eigenvalues of S3 = ~3/2 are ~/2. So with the usual sign and ± normalization conventions

S+ = ~ + and S + = ~ (10.137) |i | i | i |i while

S+ + = 0 and S =0. (10.138) | i |i The square of the total spin operator is simply related to the raising and lowering operators and to S3

2 2 2 2 1 1 2 S = S1 + S2 + S3 = S+S + S S+ + S3 . (10.139) 2 2 2 2 But the squares of the Pauli matrices are unity, and so Sa =(~/2) for all three values of a.Thus 2 3 2 S = ~ (10.140) 4 is a Casimir operator (10.122) for a spin one-half system. Consider two spin operators S(1) and S(2) as defined by (10.110) acting on two spin-one-half systems. Let the tensor-product states , = = (10.141) |± ±i |±i1|±i2 |±i1 ⌦|±i2 (1) (2) be eigenstates of S3 and S3 so that

(1) ~ (2) ~ S3 +, = +, and S3 , + = , + | ±i 2 | ±i |± i 2 |± i (10.142) (1) ~ (2) ~ S , = , and S , = , 3 | ±i 2 | ±i 3 |± i 2 |± i The total spin of the system is the sum of the two spins S = S(1) + S(2),so 2 2 (1) (2) (1) (2) S = S + S and S3 = S3 + S3 . (10.143) ⇣ ⌘ The state +, + is an eigenstate of S3 with eigenvalue ~ | i (1) (2) ~ ~ S3 +, + = S +, + + S +, + = +, + + +, + = ~ +, + . | i 3 | i 3 | i 2| i 2| i | i (10.144) So the state of angular momentum ~ in the 3-direction is 1, 1 = +, + . | i | i Similarly, the state , is an eigenstate of S3 with eigenvalue ~ | i (1) (2) ~ ~ S3 , = S , + S , = , , = ~ , | i 3 | i 3 | i 2| i 2| i | i (10.145) 444 Group theory and so the state of angular momentum ~ in the negative 3-direction is 1, 1 = , . The states +, and , + are eigenstates of S with | i | i | i | i 3 eigenvalue 0

(1) (2) ~ ~ S3 +, = S3 +, + S3 +, = +, +, =0 | i | i | i 2| i 2| i (10.146) (1) (2) ~ ~ S , + = S , + + S , + = , + + , + =0. 3| i 3 | i 3 | i 2| i 2| i To see which states are eigenstates of S2, we use the lowering operator for the combined system S = S(1) + S(2) and the rules (10.120, 10.137, & 10.138) to lower the state 1, 1 | i (1) (2) S +, + = S + S +, + = ~ ( , + + +, )=~ p2 1, 0 . | i | i | i | i | i ⇣ ⌘ Thus the state 1, 0 is | i 1 1, 0 = ( +, + , + ) . (10.147) | i p2 | i | i The orthogonal and normalized combination of +, and , + must be | i | i the state of spin zero 1 0, 0 = ( +, , + ) (10.148) | i p2 | i | i with the usual sign convention. To check that the states 1, 0 and 0, 0 really are eigenstates of S2,we | i | i use (10.139 & 10.140) to write S2 as 2 2 (1) (2) 3 2 (1) (2) S = S + S = ~ +2S S 2 · ⇣3 2 (1) ⌘(2) (1) (2) (1) (2) = ~ + S+ S + S S+ +2S3 S3 . (10.149) 2 (1) (2) (1) (2) Now the sum S+ S + S S+ merely interchanges the states +, and | i , + and multiplies them by ~2,so | i 2 3 2 2 2 2 S 1, 0 = ~ 1, 0 + ~ 1, 0 ~ 1, 0 | i 2 | i | i4 | i 2 2 =2~ 1, 0 = s(s + 1)~ 1, 0 (10.150) | i | i which confirms that s = 1. Because of the relative minus sign in formula (10.148) for the state 0, 0 ,wehave | i 2 3 2 2 1 2 S 0, 0 = ~ 0, 0 ~ 1, 0 ~ 1, 0 | i 2 | i | i2 | i 2 2 =0~ 1, 0 = s(s + 1)~ 1, 0 (10.151) | i | i 10.23 The Jacobi Identity 445 which confirms that s = 0.

Example 10.24 (Two equivalent representations of SU(2)). The identity exp i✓ ⇤ = exp i✓ (10.152) · 2 2 · 2 2 h ⇣ ⌘i ⇣ ⌘ shows that the defining representation of SU(2) (section 10.19) and its com- plex conjugate are equivalent (10.8) representations. To prove this identity, we expand the exponential on the right-hand side in powers of its argument

1 1 n exp i✓ = i✓ (10.153) 2 · 2 2 2 n! · 2 2 "n=0 # ⇣ ⌘ X ⇣ ⌘ and use the fact that 2 is its own inverse to get

1 1 n exp i✓ = i✓ . (10.154) 2 · 2 2 n! 2 · 2 2 n=0 ⇣ ⌘ X h ⇣ ⌘ i Since the Pauli matrices obey the anticommutation relation (10.113), and since both and are real, while is imaginary, we can write the 2 2 1 3 2 ⇥ matrix within the square brackets as i✓ = i✓ 1 + i✓ 2 i✓ 3 = i✓ ⇤ (10.155) 2 · 2 2 1 2 2 2 3 2 · 2 ⇣ ⌘ ⇣ ⌘ which implies the identity (10.152)

1 1 n exp i✓ = i✓ ⇤ = exp i✓ ⇤ . (10.156) 2 · 2 2 n! · 2 · 2 n=0 ⇣ ⌘ X h⇣ ⌘ i h ⇣ ⌘i

10.23 The Jacobi Identity Any three square matrices A, B, and C satisfy the commutator-product rule

[A, BC]=ABC BCA = ABC BAC + BAC BCA =[A, B]C + B[A, C]. (10.157)

Interchanging B and C gives

[A, CB]=[A, C]B + C[A, B]. (10.158) 446 Group theory Subtracting the second equation from the first, we get the Jacobi identity

[A, [B,C]] = [[A, B],C]+[B,[A, C]] (10.159) and its equivalent cyclic form

[A, [B,C]] + [B,[C, A]] + [C, [A, B]] = 0. (10.160)

Another Jacobi identity uses the anticommutator A, B AB + BA { }⌘ [A, B],C + [A, C],B +[ B,C ,A]=0. (10.161) { } { } { }

10.24 The Adjoint Representation

Any three generators ta, tb, and tc satisfy the Jacobi identity (10.160)

[ta, [tb,tc]] + [tb, [tc,ta]] + [tc, [ta,tb]] = 0. (10.162) By using the structure-constant formula (10.84), we may express each of these double commutators as a linear combination of the generators [t , [t ,t ]] = [t ,ifd t ]= f d f e t a b c a bc d bc ad e [t , [t ,t ]]] = [t ,ifd t ]= f d f e t (10.163) b c a b ca d ca bd e [t , [t ,t ]] = [t ,ifd t ]= f d f e t . c a b c ab d ab cd e So the Jacobi identity (10.162) implies that

d e d e d e fbcfad + fcafbd + fabfcd te = 0 (10.164) ⇣ ⌘ or since the generators are linearly independent

d e d e d e fbcfad + fcafbd + fabfcd =0. (10.165)

If we define a set of matrices Ta by

c (Tb)ac = ifab (10.166) then, since the structure constants are antisymmetric in their lower indices, we may write the three terms in the preceding equation (10.165) as

f d f e = f d f e =( T T ) (10.167) bc ad cb da b a ce

f d f e = f d f e =(T T ) (10.168) ca bd ca db a b ce 10.25 Casimir operators 447 and f d f e = if d (T ) (10.169) ab cd ab d ce or in matrix notation c [Ta,Tb]=ifab Tc. (10.170)

So the matrices Ta, which we made out of the structure constants by the rule c (Tb)ac = ifab (10.166), obey the same algebra (10.69) as do the generators ta. They are the generators in the adjoint representation of the Lie algebra. If the Lie algebra has N generators ta,thentheN generators Ta in the adjoint representation are N N matrices. ⇥

10.25 Casimir operators For any compact Lie algebra, the sum of the squares of all the generators

N C = t t t t (10.171) a a ⌘ a a a=1 X commutes with every generator tb

[C, tb]=[tata,tb]=[ta,tb]ta + ta[ta,tb]

= ifabctcta + taifabctc = i (fabc + fcba) tcta =0 (10.172) because of the total antisymmetry (10.82) of the structure constants. This sum, called a Casimir operator, commutes with every matrix

[C, D(↵)] = [C, exp(i↵ata)] = 0 (10.173) of the representation generated by the ta’s. Thus by part 2 of Schur’s lemma (section 10.7), it must be a multiple of the identity matrix

C = tata = cI. (10.174) The constant c depends upon the representation D(↵) and is called the quadratic Casimir 2 C2(D)=Tr ta /TrI. (10.175) Example 10.25 (Quadratic Casimirs of SU(2)). The quadratic Casimir C(2) of the defining representation of SU(2) with generators ta = a/2 (10.111) is

3 1 2 2 3 2 3 C (2)=Tr 1 /Tr (I)= · · 2 = . (10.176) 2 2 a 2 4 a=1 ! X 448 Group theory That of the adjoint representation (10.94) is

3 3 2 i✏abci✏cba C2(3)=Tr tb /Tr (I)= =2. (10.177) ! 3 Xb=1 a,b,cX=1

The generators of some noncompact groups come in pairs ta and ita, and so the sum of the squares of these generators vanishes, C = t t t t = 0. a a a a

10.26 Tensor operators for the rotation group (j) Suppose Am is a set of 2j + 1 operators whose commutation relations with the generators Ji of rotations are

(j) (j) (j) [Ji,Am ]=As (Ji )sm (10.178) in which the sum over s runs from j to j.ThenA(j) is said to be a spin-j tensor operator for the group SU(2).

Example 10.26 (A Spin-One Tensor Operator). For instance, if j = 1, (1) then (Ji )sm = i~✏sim, and so a spin-1 tensor operator of SU(2) is a vector (1) Am that transforms as

(1) (1) (1) [Ji,Am ]=As i ~ ✏sim = i ~ ✏ims As (10.179) under rotations.

Let’s rewrite the definition (10.178) as

(j) (j) (j) (j) JiAm = As (Ji )sm + Am Ji (10.180)

(j) (j) and specialize to the case i = 3 so that (J3 )sm is diagonal, (J3 )sm = ~msm

(j) (j) (j) (j) (j) (j) (j) J3Am = As (J3 )sm + Am J3 = As ~msm + Am J3 = Am (~m + J3) (10.181) Thus if the state j, s, E is an eigenstate of J3 with eigenvalue ~s,thenthe (j) | i state Am j, s, E is an eigenstate of J3 with eigenvalue ~(m + s) | i (j) (j) (j) J3A j, s, E = A (~m + J3) j, s, E = ~ (m + s) A j, s, E . (10.182) m | i m | i m | i (j) The J eigenvalues of the tensor operator Am and the state j, s, E add. 3 | i 10.27 Simple and semisimple Lie algebras 449 10.27 Simple and semisimple Lie algebras (i) An invariant subalgebra is a set of generators ta whose commutator with (i) every generator tb of the group is a linear combination of the generators tc of the invariant subalgebra

(i) (i) [ta ,tb]=ifabctc . (10.183)

The whole algebra and the null algebra are trivial invariant subalgebras. An algebra with no nontrivial invariant subalgebras is a simple algebra. A simple algebra generates a simple group. An algebra that has no nontrivial abelian invariant subalgebras is a semisimple algebra. A semisimple algebra generates a semisimple group.

Example 10.27 (Some Simple Lie Groups). The groups of unitary ma- trices of unit determinant SU(2), SU(3), . . . are simple. So are the groups of orthogonal matrices of unit determinant SO(n)(exceptSO(4), which is semisimple) and the groups of symplectic matrices Sp(2n) (section 10.31).

Example 10.28 (Unification and Grand Unification). The symmetry group of the standard model of particle physics is a direct product of an SU(3) group that acts on colored fields, an SU(2) group that acts on left-handed quark and lepton fields, and a U(1) group that acts on fields that carry hypercharge. Each of these three groups, is an invariant sub- group of the full symmetry group SU(3) SU(2) U(1) , and the last c ⌦ ` ⌦ Y one is abelian. Thus the symmetry group of the standard model is neither simple nor semisimple. In theories of grand unification, the strong and electroweak interactions unify at very high energies and are described by a simple group which makes all its charges simple multiples of each other. Georgi and Glashow suggested the group SU(5) in 1976 (Howard Georgi, 1947– ; Sheldon Glashow, 1932– ). Others have proposed SO(10) and even bigger groups. 450 Group theory 10.28 SU(3) The Gell-Mann matrices are

010 0 i 0 100 = 100,= i 00,= 0 10, 1 0 1 2 0 1 3 0 1 000 000 000 @001A @00 iA @000A = 000,= 00 0 ,= 001, 4 0 1 5 0 1 6 0 1 100 i 00 010 @00 0A @ A10 0@ A 1 = 00 i , and = 01 0 . (10.184) 7 0 1 8 p 0 1 0 i 0 3 00 2 @ A @ A The generators t of the 3 3 defining representation of SU(3) are these a ⇥ Gell-Mann matrices divided by 2

ta = a/2 (10.185)

(Murray Gell-Mann, 1929–). The eight generators ta are orthogonal with k =1/2

1 Tr (t t )= (10.186) a b 2 ab and satisfy the commutation relation

[ta,tb]=ifabc tc. (10.187)

The trace formula (10.72) gives us the SU(3) structure constants as

f = 2i Tr ([t ,t ]t ) . (10.188) abc a b c

They are real and totally antisymmetric with f123 = 1, f458 = f678 = p3/2, and f = f = f = f = f = f =1/2. 147 156 246 257 345 367 While no two generators of SU(2) commute, two generators of SU(3) do. In the representation (10.184,10.185), t3 and t8 are diagonal and so commute

[t3,t8]=0. (10.189)

They generate the Cartan subalgebra (section 10.30) of SU(3). 10.29 SU(3) and quarks 451 10.29 SU(3) and quarks The generators defined by Eqs.(10.185 & 10.184) give us the 3 3represen- ⇥ tation

D(↵)=exp(i↵ata) (10.190) in which the sum a =1, 2,...8 is over the eight generators ta.Thisrepre- sentation acts on complex 3-vectors and is called the 3. Note that if

D(↵1)D(↵2)=D(↵3) (10.191) then the complex conjugates of these matrices obey the same multiplication rule

D⇤(↵1)D⇤(↵2)=D⇤(↵3) (10.192) and so form another representation of SU(3). It turns out that (unlike in SU(2)) this representation is inequivalent to the 3;itisthe3. There are three quarks with masses less than about 100 MeV/c2—the u, d, and s quarks. The other three quarks c, b, and t are more massive; mc =1.28 GeV, mb =4.18 GeV, and mt = 173.1 GeV. Nobody knows why. Gell-Mann and Zweig suggested that the low-energy strong interac- tions were approximately invariant under unitary transformations of the three light quarks, which they represented by a 3, and of the three light antiquarks, which they represented by a 3. They imagined that the eight 0 + light pseudoscalar mesons, that is, the three pions ⇡,⇡ ,⇡ , the neutral 0 + 0 ⌘, and the four kaons K ,K ,K K , were composed of a quark and an antiquark. So they should transform as the tensor product 3 3 = 8 1. (10.193) ⌦ They put the eight pseudoscalar mesons into an 8. They imagined that the eight light baryons — the two nucleons N and P , 0 + the three sigmas ⌃, ⌃ , ⌃ , the neutral lambda ⇤, and the two cascades 0 ⌅ and ⌅ were each made of three quarks. They should transform as the tensor product 3 3 3 = 10 8 8 1. (10.194) ⌦ ⌦ They put the eight light baryons into one of these 8’s. When they were writ- ing these papers, there were nine spin-3/2 resonances with masses somewhat 2 heavier than 1200 MeV/c — four ’s, three ⌃⇤’s, and two ⌅⇤’s. They put these into the 10 and predicted the tenth and its mass. In 1964, a tenth spin-3/2 resonance, the ⌦, was found with a mass close to their prediction 452 Group theory of 1680 MeV/c2, and by 1973 an MIT-SLAC team had discovered quarks inside protons and neutrons. (George Zweig, 1937–)

10.30 Cartan Subalgebra

In any Lie group, the maximum set of mutually commuting generators Ha generate the Cartan subalgebra

[Ha,Hb] = 0 (10.195) which is an abelian subalgebra. The number of generators in the Cartan subalgebra is the rank of the Lie algebra. The Cartan generators Ha can be simultaneously diagonalized, and their eigenvalues or diagonal elements are the weights H µ, x, D = µ µ, x, D (10.196) a| i a| i in which D labels the representation and x whatever other variables are needed to specify the state. The vector µ is the weight vector.Theroots are the weights of the adjoint representation.

10.31 The Symplectic Group Sp(2n) The real symplectic group Sp(2n, R) is the group of linear transformations that preserve the canonical commutation relations of quantum mechanics

[qi,pk]=i~ik, [qi,qk]=0, and [pi,pk] = 0 (10.197) for i, k =1,...,n. In terms of the 2n vector v =(q1,...,qn,p1,...,pn) of quantum variables, these commutation relations are [vi,vk]=i~Jik where J is the 2n 2n real matrix ⇥ 0 I J = (10.198) I 0 ✓ ◆ in which I is the n n identity matrix. The real linear transformation ⇥ 2n

vi0 = Ri` v` (10.199) X`=1 will preserve the quantum-mechanical commutation relations (10.197) if

2n 2n 2n

[vi0,vk0 ]= Ri` v`, Rkm vm = i~ Ri` J`m Rkm = i~Jik (10.200) " m=1 # X`=1 X `X,k=1 10.31 The Symplectic Group Sp(2n) 453 which in matrix notation is just the condition RJRT = J (10.201) that the matrix R be in the real symplectic group Sp(2n, R). The transpose and product rules (1.208 & 1.218) for determinants imply that det(R)= 1, ± but the condition (10.201) itself implies that det(R) = 1 (Zee, 2016, p. 281). In terms of the matrix J and the hamiltonian H(v)=H(q, p), Hamilton’s equations have the symplectic form

2n @H(q, p) @H(q, p) dvi @H(v) q˙i = andp ˙i = are = Ji` . (10.202) @pi @qi dt @v` X`=1 A matrix R = et obeys the defining condition (10.201) if tJ = JtT (exercise 10.22) or equivalently if JtJ = tT. It follows (exercise 10.23) that the generator t must be bs t = 1 (10.203) s bT ✓ 2 ◆ in which the matrices b, s1,s2 are real, and both s1 and s2 are symmetric. The group Sp(2n, R) is noncompact.

Example 10.29 (Squeezed states). A coherent state ↵ is an eigenstate of | i the annihilation operator a =(q + ip/)/p2~ with a complex eigenvalue ↵ (2.146), a ↵ = ↵ ↵ . In a coherent state with = pm!, the variances are | i | i (q)2 = ↵ (q q¯)2 ↵ = ~/(2m!) and (p)2 = ↵ (p p¯)2 ↵ = ~m!/2. h | | i h | | i Thus coherent states have minimum uncertainty, q p = ~/2. A squeezed state ↵ 0 is an eigenstate of a0 =(q0 + ip0/)/p2~ in which | i q0 and p0 are related to q and p by an Sp(2) transformation q ab q q d b q 0 = with inverse = 0 (10.204) p cd p p ca p ✓ 0◆ ✓ ◆✓ ◆ ✓ ◆ ✓ ◆✓ 0◆ in which ad bc = 1. The standard deviations of the variables q and p in the squeezed state ↵ are | 0i ~ d2 ~ c2 q = + m!b2 and p = + m!a2. (10.205) r2 rm! r2 rm! Thus by making b and d tiny, one can reduce the uncertainty q by any factor, but then p will increase by the same factor since the determinant of the Sp(2) transformation must remain equal to unity, ad bc = 1. 454 Group theory

Example 10.30 (Sp(2, R)). The matrices (exercise 10.27) cosh ✓ sinh ✓ T = (10.206) ± sinh ✓ cosh ✓ ✓ ◆ are elements of the noncompact symplectic group Sp(2, R). A dynamical map that takes a 2n vector v =(q ,...,q ,p ,...,p ) M 1 n 1 n from v(t1)tov(t2) has a jacobian (section 1.21)

@za(t2) Mab = (10.207) @zb(t1) in Sp(2n, R) if and only if its dynamics are hamiltonian (10.202, section 16.1) (Carl Jacobi 1804–1851, William Hamilton 1805–1865). The complex symplectic group Sp(2n, C) consists of all 2n 2n complex ⇥ matrices C that satisfy the condition

CJCT = J. (10.208)

The group Sp(2n, C) also is noncompact. The unitary symplectic group USp(2n) consists of all 2n 2n complex ⇥ unitary matrices U that satisfy the condition

UJUT = J. (10.209)

It is compact.

10.32 Quaternions If z and w are any two complex numbers, then the 2 2 matrix ⇥ zw q = (10.210) w z ✓ ⇤ ⇤◆ is a quaternion. The quaternions are closed under addition and multiplica- tion and under multiplication by a real number (exercise 10.21), but not under multiplication by an arbitrary complex number. The squared norm of q is its determinant

q 2 = z 2 + w 2 =detq. (10.211) k k | | | | The matrix products q q and qq are the squared norm q 2 multiplied by † † k k the 2 2 identity matrix ⇥ 2 q†q = qq† = q I (10.212) k k 10.32 Quaternions 455 The 2 2 matrix ⇥ 01 i = (10.213) 2 10 ✓ ◆ provides another expression for q 2 in terms of q and its transpose qT k k qTi q = q 2 i . (10.214) 2 k k 2 Clearly q =0impliesq = 0. The norm of a product of quaternions is the k k product of their norms q q = det(q q )= det q det q = q q . (10.215) k 1 2k 1 2 1 2 k 1kk 2k The quaternions thereforep formp an associative division algebra (over the real numbers); the only others are the real numbers and the complex numbers; the octonions are a nonassociative division algebra. One may use the Pauli matrices to define for any real 4-vector x a quater- nion q(x) as q(x)=x + i x = x + i x 0 k k 0 · x + ix x + ix = 0 3 2 1 (10.216) x + ix x ix ✓ 2 1 0 3◆ with squared norm q(x) 2 = x2 + x2 + x2 + x2. (10.217) k k 0 1 2 3 The product rule (10.112) for the Pauli matrices tells us that the product of two quaternions is q(x) q(y)= (x + i x)(y + i y) (10.218) 0 · 0 · = x y + i (y x + x y) i (x y) x y 0 0 · 0 0 ⇥ · · so their commutator is [q(x),q(y)] = 2 i (x y) . (10.219) ⇥ ·

Example 10.31 (Lack of Analyticity). One may define a function f(q) of a quaternionic variable and then ask what functions are analytic in the sense that the (one-sided) derivative 1 f 0(q)= lim f(q + q0) f(q) q0 (10.220) q 0 0! exists and is independent of the direction⇥ through⇤ which q 0. This space 0 ! of functions is extremely limited and does not even include the function f(q)=q2 (exercise 10.24). 456 Group theory 10.33 Quaternions and symplectic groups This section is optional on a first reading. One may regard the unitary symplectic group USp(2n) as made of 2n 2n ⇥ matrices W that map n-tuples q of quaternions into n-tuples q0 = Wq of quaternions with the same value of the quadratic quaternionic form

2 2 2 2 2 2 2 2 q0 = q0 + q0 + ...+ q0 = q + q + ...+ q = q . k k k 1k k 2k k nk k 1k k 2k k nk k k (10.221) By (10.212), the quadratic form q 2 times the 2 2 identity matrix I is k 0k ⇥ equal to the hermitian form q0†q0

2 q0 I = q0†q0 = q0†q0 + ...+ q0†q0 = q†W †Wq (10.222) k k 1 1 n n and so any matrix W that is both a 2n 2n unitary matrix and an n n ⇥ ⇥ matrix of quaternions keeps q 2 = q 2 k 0k k k 2 2 q0 I = q†W †Wq = q†q = q I. (10.223) k k k k The group USp(2n) thus consists of all 2n 2n unitary matrices that also ⇥ are n n matrices of quaternions. (This last requirement is needed so that ⇥ q0 = Wq is an n-tuple of quaternions.) The generators t of the symplectic group USp(2n) are 2n 2n direct- a ⇥ product matrices of the form

I A, S , S , and S (10.224) ⌦ 1 ⌦ 1 2 ⌦ 2 3 ⌦ 3 in which I is the 2 2 identity matrix, the three ’s are the Pauli matrices, ⇥ i A is an imaginary n n anti-symmetric matrix, and the S are n n real ⇥ i ⇥ symmetric matrices. These generators ta close under commutation

[ta,tb]=ifabctc. (10.225)

Any imaginary linear combination i↵ata of these generators is not only a 2n 2n antihermitian matrix but also an n n matrix of quaternions. Thus ⇥ ⇥ the matrices D(↵)=ei↵ata (10.226) are both unitary 2n 2n matrices and n n quaternionic matrices and so ⇥ ⇥ are elements of the group Sp(2n).

Example 10.32 (USp(2) = SU(2)). There is no 1 1 anti-symmetric ⇠ ⇥ matrix, and there is only one 1 1 symmetric matrix. So the generators t ⇥ a of the group Sp(2) are the Pauli matrices ta = a, and Sp(2) = SU(2). The 10.33 Quaternions and symplectic groups 457 elements g(↵) of SU(2) are quaternions of unit norm (exercise 10.20), and so the product g(↵)q is a quaternion

g(↵)q 2 =det(g(↵)q)=det(g(↵)) det q =detq = q 2 (10.227) k k k k with the same squared norm.

Example 10.33 (SO(4) = SU(2) SU(2)). If g and h are any two elements ⇠ ⌦ of the group SU(2), then the squared norm (10.217) of the quaternion q(x)= x + i x is invariant under the transformation q(x )=gq(x) h 1, that is, 0 · 0 x 2 + x 2 + x 2 + x 2 = x2 + x2 + x2 + x2.Sox x is an SO(4) rotation 0 10 20 30 0 1 2 3 ! 0 of the 4-vector x. The Lie algebra of SO(4) thus contains two commuting invariant SU(2) subalgebras and so is semisimple.

Example 10.34 (USp(4) ⇠= SO(5)). Apart from scale factors, there are three real symmetric 2 2 matrices S = , S = I, and S = and one ⇥ 1 1 2 3 3 imaginary anti-symmetric 2 2 matrix A = . So there are 10 generators ⇥ 2 of USp(4) = SO(5)

0 iI 0 k t1 = I 2 = ,tk1 = k 1 = ⌦ iI 0 ⌦ k 0 ✓ ◆ ✓ ◆ (10.228) 0 0 t = I = k ,t = = k k2 k ⌦ 0 k3 k ⌦ 3 0 ✓ k◆ ✓ k◆ where k runs from 1 to 3.

Another way of looking at USp(2n) is to use (10.214) to write the quadratic form q 2 as k k q 2 = qT q (10.229) k k J J in which the 2n 2n matrix has n copies of i on its 2 2 diagonal ⇥ J 2 ⇥ i2 000... 0 0 i 00... 0 0 2 1 00i2 0 ... 0 = B C (10.230) J B 000i2 ... 0 C B C B ...... C B . . . . . 0 C B C B 00000i2C B C @ A and is the matrix J (10.198) in a di↵erent basis. Thus any n n matrix of ⇥ quaternions W that satisfies

W T W = (10.231) J J 458 Group theory also satisfies Wq 2 = qTW T Wq = qT q = q 2 (10.232) k k J J J k k J and so leaves invariant the quadratic form (10.221). The group USp(2n) therefore consists of all 2n 2n matrices W that satisfy (10.231) and that ⇥ also are n n matrices of quaternions. ⇥

10.34 Compact Simple Lie Groups Elie´ Cartan (1869–1951) showed that all compact, simple Lie groups fall into four infinite classes and five discrete cases. For n =1, 2,..., his four classes are A = SU(n + 1) which are (n + 1) (n + 1) unitary matrices with unit • n ⇥ determinant, B = SO(2n + 1) which are (2n + 1) (2n + 1) orthogonal matrices with • n ⇥ unit determinant, C = Sp(2n) which are 2n 2n symplectic matrices, and • n ⇥ D = SO(2n) which are 2n 2n orthogonal matrices with unit determi- • n ⇥ nant.

The five discrete cases are the exceptional groups G2, F4, E6, E7, and E8. The exceptional groups are associated with the octonians

a + b↵i↵ (10.233) where the ↵-sum runs from 1 to 7; the eight numbers a and b↵ are real; and the seven i↵’s obey the multiplication law i i = + g i (10.234) ↵ ↵ ↵ in which g↵ is totally antisymmetric with

g123 = g247 = g451 = g562 = g634 = g375 = g716 =1. (10.235) Like the quaternions and the complex numbers, the octonians form a divi- sion algebra with an absolute value 1/2 a + b i = a2 + b2 (10.236) | ↵ ↵| ↵ that satisfies AB = A B (10.237) | | | || | 10.35 Group Integration 459 but they lack associativity. The group G2 is the subgroup of SO(7) that leaves the g↵’s of (10.234) invariant.

10.35 Group Integration Suppose we need to integrate some function f(g) over a group. Naturally, we want to do so in a way that gives equal weight to every element of the group. In particular, if g0 is any group element, we want the integral of the shifted function f(g0g) to be the same as the integral of f(g)

f(g) dg = f(g0g) dg. (10.238) Z Z Such a measure dg is said to be left invariant (Creutz, 1983, chap. 8). Let’s use the letters a = a1,...,an, b = b1,...,bn, and so forth to label the elements g(a), g(b), so that an integral over the group is

f(g) dg = f(g(a)) m(a) dna (10.239) Z Z in which m(a) is the left-invariant measure and the integration is over the n-space of a’s that label all the elements of the group. To find the left-invariant measure m(a), we use the multiplication law of the group g(a(c, b)) g(c) g(b) (10.240) ⌘ and impose the requirement (10.238) of left invariance with g g(c) 0 ⌘ f(g(b)) m(b) dnb = f(g(c)g(b)) m(b) dnb = f(g(a(c, b))) m(b) dnb. Z Z Z (10.241) We change variables from b to a = a(c, b) by using the jacobian det(@b/@a) which gives us dnb =det(@b/@a) dna

f(g(b)) m(b) dnb = f(g(a)) det(@b/@a) m(b) dna. (10.242) Z Z Replacing b by a = a(c, b) on the left-hand side of this equation, we find

m(a)=det(@b/@a) m(b) (10.243) or since det(@b/@a)=1/ det(@a(c, b)/@b)

m(a(c, b)) = m(b)/ det(@a(c, b)/@b). (10.244) 460 Group theory So if we let g(b) g(0) = e, the identity element of the group, and set ! m(e) = 1, then we find for the measure

m(a)=m(c)=m(a(c, b)) =1/ det(@a(c, b)/@b) . (10.245) |b=0 |b=0 Example 10.35 (The Invariant Measure for SU(2)). A general element of the group SU(2) is given by (10.115) as ✓ ✓ exp i ✓ = I cos + i ✓ˆ sin . (10.246) · 2 2 · 2 ⇣ ⌘ Setting a0 = cos(✓/2) and a = ✓ˆ sin(✓/2), we have g(a)=a + i a (10.247) 0 · in which a2 a2 + a a = 1. Thus, the parameter space for SU(2) is the ⌘ 0 · unit sphere S3 in four dimensions. Its invariant measure is

2 4 2 2 4 2 1/2 3 (1 a ) d a = (1 a a ) d a = (1 a ) d a (10.248) 0 Z Z Z or 2 1/2 1 m(a)=(1 a ) = . (10.249) cos(✓/2) | | We also can write the arbitrary element (10.247) of SU(2) as

g(a)= 1 a2 + i a (10.250) ± · and the group-multiplication lawp (10.240) as

1 a2 + i a = 1 c2 + i c 1 b2 + i b . (10.251) · · · p ⇣p ⌘⇣p ⌘ Thus, by multiplying both sides of this equation by i and taking the trace, we find (exercise 10.28) that the parameters a(c, b) that describe the product g(c) g(b) are

a(c, b)= 1 c2 b + 1 b2 c c b. (10.252) ⇥ To compute the jacobian ofp our formulap (10.245) for the invariant measure, we di↵erentiate this expression (10.252) at b = 0 and so find (exercise 10.29)

2 1/2 m(a)=1/ det(@a(c, b)/@b) =(1 a ) (10.253) |b=0 as the left-invariant measure in agreement with (10.249). 10.36 The Lorentz group 461 10.36 The Lorentz group The Lorentz group O(3, 1) is the set of all linear transformations L that leave invariant the Minkowski inner product xy x y x0y0 = xT⌘y (10.254) ⌘ · in which ⌘ is the diagonal matrix 1000 0100 ⌘ = 0 1 . (10.255) 0010 B 0001C B C @ A So L is in O(3, 1) if for all 4-vectors x and y (Lx) T ⌘L y = xTLT ⌘Ly= xT⌘y. (10.256) Since x and y are arbitrary, this condition amounts to LT⌘L= ⌘. (10.257) Taking the determinant of both sides and using the transpose (1.208) and product (1.221) rules, we have (det L)2 =1. (10.258) So det L = 1, and every Lorentz transformation L has an inverse. Multi- ± plying (10.257) by ⌘,wefind ⌘LT⌘L = ⌘2 = I (10.259)

1 which identifies L as 1 T L = ⌘L ⌘. (10.260) The subgroup of O(3, 1) with det L = 1 is the proper Lorentz group SO(3, 1). The subgroup of SO(3, 1) that leaves invariant the sign of the time compo- nent of timelike vectors is the proper orthochronous Lorentz group SO+(3, 1). To the Lie algebra of SO+(3, 1), we consider a Lorentz matrix L = I + ! that di↵ers from the identity matrix I by a tiny matrix ! and require it to satisfy the condition (10.257) for membership in the Lorentz group I + !T ⌘ (I + !)=⌘ + !T⌘ + ⌘!+ !T! = ⌘. (10.261) Neglecting!T!,wehave !T⌘ = ⌘! or since ⌘2 = I !T = ⌘!⌘. (10.262) 462 Group theory This equation says (exercise 10.31) that under transposition the time-time and space-space elements of ! change sign, while the time-space and space- time elements do not. That is, the tiny matrix ! is for infinitesimal ✓ and a linear combination ! = ✓ R + B (10.263) · · of three antisymmetric space-space matrices 000 0 0000 00 0 0 000 0 0001 00 10 R1 = 0 1 R2 = 0 1 R3 = 0 1 000 1 0000 01 0 0 B001 0C B0 100C B00 0 0C B C B C B C @ A @ A @ (10.264)A and of three symmetric time-space matrices 0100 0010 0001 1000 0000 0000 B1 = 0 1 B2 = 0 1 B3 = 0 1 0000 1000 0000 B0000C B0000C B1000C B C B C B C @ A @ A @ (10.265)A all of which satisfy condition (10.262). The three R are 4 4 versions of ` ⇥ the rotation generators (10.92); the three B` generate Lorentz boosts. If we write L = I + ! as

L = I i✓ iR i iB I i✓ J i K (10.266) ` ` ` ` ⌘ ` ` ` ` then the three matrices J` = iR` are imaginary and antisymmetric, and therefore hermitian. But the three matrices K` = iB` are imaginary and symmetric, and so are antihermitian. The 4 4 matrix L =exp(i✓ J i K ) ⇥ ` ` ` ` is not unitary because the Lorentz group is not compact. One may verify (exercise 10.32) that the six generators J` and K` satisfy three sets of commutation relations:

[Ji,Jj]=i✏ijkJk (10.267)

[Ji,Kj]=i✏ijkKk (10.268) [K ,K ]= i✏ J . (10.269) i j ijk k

The first (10.267) says that the three J` generate the rotation group SO(3); the second (10.268) says that the three boost generators transform as a 3-vector under SO(3); and the third (10.269) implies that four canceling in- finitesimal boosts can amount to a rotation. These three sets of commutation relations form the Lie algebra of the Lorentz group SO(3, 1). Incidentally, 10.36 The Lorentz group 463 one may show (exercise 10.33) that if J and K satisfy these commutation relations (10.267–10.269), then so do

J and K. (10.270) The infinitesimal Lorentz transformation (10.266) is the 4 4 matrix ⇥

1 1 2 3 1 1 ✓3 ✓2 L = I + ! = I + ✓`R` + `B` = 0 1 . (10.271) ✓ 1 ✓ 2 3 1 B ✓ ✓ 1 C B 3 2 1 C @ A a a b It moves any 4-vector x to x0 = Lx or in components x0 = L b x

0 0 1 2 3 x0 = x + 1x + 2x + 3x 1 0 1 2 3 x0 = 1x + x ✓3x + ✓2x 2 0 1 2 3 x0 = 2x + ✓3x + x ✓1x 3 0 1 2 3 x0 = x ✓ x + ✓ x + x . (10.272) 3 2 1 More succinctly with t = x0,thisis

t0 = t + x · x0 = x + t + ✓ x (10.273) ^ in which means cross-product. ^⌘⇥ For arbitrary real ✓ and , the matrices

i✓ J i K L = e · · (10.274) form the subgroup of O(3, 1) that is connected to the identity matrix I.The matrices of this subgroup have unit determinant and preserve the sign of the time of time-like vectors, that is, if x2 < 0, and y = Lx,theny0x0 > 0. This is the proper orthochronous Lorentz group SO+(3, 1). The rest of the (homogeneous) Lorentz group can be obtained from it by space ,time , P T and spacetime reflections. PT The task of finding all the finite-dimensional irreducible representations of the proper orthochronous homogeneous Lorentz group becomes vastly simpler when we write the commutation relations (10.267–10.269) in terms of the hermitian matrices 1 J ± = (J iK ) (10.275) ` 2 ` ± ` 464 Group theory which generate two independent rotation groups + + + [Ji ,Jj ]=i✏ijkJk

[Ji,Jj]=i✏ijkJk (10.276) + [Ji ,Jj]=0. Thus the Lie algebra of the Lorentz group is equivalent to two copies of the Lie algebra (10.117) of SU(2). The hermitian generators of the rotation subgroup SU(2) are by (10.275)

+ J = J + J . (10.277) The antihermitian generators of the boosts are (also by 10.275)

+ K = i J J . (10.278) + Since J and J commute, the finite-dimensional irreducible representa- tions of the Lorentz group are the direct products

i✓ J i K ( i✓ ) J + +( i✓+) J D(j, j0)(✓, )=e · · = e · · (10.279) ( i✓ ) J + ( i✓+) J = e · e · of the nonunitary representations

(j,0) ( i✓ ) J + (0,j ) ( i✓+) J D (✓, )=e · and D 0 (✓, )=e · (10.280) generated by the three (2j + 1) (2j + 1) matrices J + and by the three ⇥ ` (2j + 1) (2j + 1) matrices J . 0 ⇥ 0 ` Under a Lorentz transformation L,afield (j,j0) (x) that transforms under m,m0 the D(j,j0) representation of the Lorentz group responds as

(j,j0) 1 (j,0) 1 (0,j0) 1 (j,j0) U(L) (x) U (L)= D (L ) D (L ) (Lx). (10.281) m,m0 mm00 m0m000 m00,m000

The representation D(j,j0) describes objects of the spins s that can arise from the direct product of spin-j with spin-j0 (Weinberg, 1995, p. 231)

s = j + j0,j+ j0 1,..., j j0 . (10.282) | | For instance, D(0,0) describes a spinless field or particle, while D(1/2,0) and D(0,1/2) respectively describe left-handed and right-handed spin-1/2 fields or particles. The representation D(1/2,1/2) describes objects of spin 1 and spin 0—the spatial and time components of a 4-vector. The interchange of J + and J replaces the generators J and K with J and K, a substitution that we know (10.270) is legitimate. 10.37 Two-dimensional left-handed representation of the Lorentz group 465 10.37 Two-dimensional left-handed representation of the Lorentz group (1/2,0) The generators of the representation D with j =1/2 and j0 = 0 are + given by (10.277 & 10.278) with J = /2 and J = 0. They are 1 1 J = and K = i . (10.283) 2 2 The 2 2 matrix D(1/2,0) that represents the Lorentz transformation (10.274) ⇥

i✓ J i K L = e · · (10.284) is D(1/2,0)(✓, )=exp( i✓ /2 /2) . (10.285) · · And so the generic D(1/2,0) matrix is

(1/2,0) z /2 D (✓, )=e · (10.286) with =Rez and ✓ =Imz. It is nonunitary and of unit determinant; it is a member of the group SL(2,C) of complex unimodular 2 2 matrices. The ⇥ (covering) group SL(2,C) relates to the Lorentz group SO(3, 1) as SU(2) relates to the rotation group SO(3).

Example 10.36 (The standard left-handed boost). For a particle of mass m>0, the standard boost that takes the 4-vector k =(m, 0)top =(p0, p), where p0 = m2 + p2 is a boost in the pˆ direction. It is the 4 4 matrix ⇥ 0 1 pB(p)=R(pˆ) B (p ) R (pˆ)=exp(↵ pˆ B) (10.287) 3 · in which cosh ↵ = p0/m and sinh ↵ = p /m, as one may show by expand- | | ing the exponential (exercise 10.35). This standard boost is represented by D(1/2,0)(0, ), the 2 2 matrix (10.284), with = ↵ pˆ.Thepower-series ⇥ expansion of this matrix is (exercise 10.36)

(1/2,0) ↵pˆ /2 D (0,↵pˆ)=e · = I cosh(↵/2) pˆ sinh(↵/2) · = I (p0 + m)/(2m) pˆ (p0 m)/(2m) · (p0 + m)I p = p · p (10.288) 2m(p0 + m) in which I is the 2 2 identityp matrix. ⇥ Under D(1/2,0), the vector ( I,) transforms like a 4-vector. For tiny ✓ 466 Group theory and , one may show (exercise 10.38) that the vector ( I,) transforms as

(1/2,0) (1/2,0) D† (✓, )( I)D (✓, )= I + · (10.289) (1/2,0) (1/2,0) D† (✓, ) D (✓, )= +( I) + ✓ ^ which is how the 4-vector (t, x) transforms (10.273). Under a finite Lorentz transformation L, the 4-vector Sa ( I,) goes to ⌘ (1/2,0) a (1/2,0) a b D† (L) S D (L)=L bS . (10.290)

A massless field u(x) that responds to a unitary Lorentz transformation U(L)like 1 (1/2,0) 1 U(L) u(x) U (L)=D (L ) u(Lx) (10.291) is called a left-handed Weyl spinor. Its action density

(x)=iu†(x)(@ I ) u(x) (10.292) L` 0 r· is Lorentz covariant, that is

1 U(L) (x) U (L)= (Lx). (10.293) L` L` Example 10.37 (Why is Lorentz covariant). We first note that the L` derivatives @b0 in `(Lx) are with respect to x0 = Lx.Sincetheinverse 1 L 1 a 1a b matrix L takes x0 back to x = L x0 or in tensor notation x = L b x0 , the derivative @b0 is a @ @x @ 1a @ 1a @b0 = b = b a = L b a = @a L b. (10.294) @x0 @x0 @x @x Now using the abbreviation @ I @ Sa and the transformation 0 r· ⌘ a laws (10.290 & 10.291), we have

1 (1/2,0) 1 a (1/2,0) 1 U(L) `(x) U (L)=iu†(Lx)D †(L )( @aS )D (L ) u(Lx) L 1a b = iu†(Lx)( @aL bS ) u(Lx) b = iu†(Lx)( @0 S ) u(Lx)= (Lx) (10.295) b L` which shows that is Lorentz covariant. L` Incidentally, the rule (10.294) ensures, among other things, that the di- a vergence @aV is invariant

a a 1b a c b c b (@aV )0 = @a0 V 0 = @b L a L cV = @b c V = @b V . (10.296) 10.38 Two-dimensional right-handed representation of the Lorentz group 467 Example 10.38 (Why u is left handed). The spacetime integral S of the action density is stationary when u(x) satisfies the wave equation L` (@ I ) u(x) = 0 (10.297) 0 r· or in momentum space (E + p ) u(p)=0. (10.298) · Multiplying from the left by (E p ), we see that the energy of a particle · created or annihilated by the field u is the same as its momentum, E = p . | | The particles of the field u are massless because the action density has no L` mass term. The spin of the particle is represented by the matrix J = /2, so the momentum-space relation (10.298) says that u(p) is an eigenvector of pˆ J with eigenvalue 1/2 · 1 pˆ J u(p)= u(p). (10.299) · 2 A particle whose spin is opposite to its momentum is said to have negative helicity or to be left handed. Nearly massless neutrinos are nearly left handed. One may add to this action density the Majorana mass term

1 1 T (x)= mu†(x) u⇤(x) m⇤ u (x) u(x) (10.300) LM 2 2 2 2 which is Lorentz covariant because the matrices 1 and 3 anti-commute with 2 which is antisymmetric (exercise 10.41). This term would vanish if u1u2 were equal to u2u1. Since charge is conserved, only neutral fields like neutrinos can have Majorana mass terms. The action density of a left- handed field of mass m is the sum = + of the kinetic one (10.292) L L` LM and the Majorana mass term (10.300). The resulting equations of motion

0=i (@0 ) u m2u⇤ r· (10.301) 0= @2 2 + m 2 u 0 r | | show that the field u represents particles of mass m . | |

10.38 Two-dimensional right-handed representation of the Lorentz group (0,1/2) The generators of the representation D with j = 0 and j0 =1/2 are + given by (10.277 & 10.278) with J = 0 and J = /2; they are 1 1 J = and K = i . (10.302) 2 2 468 Group theory Thus 2 2 matrix D(0,1/2)(✓, ) that represents the Lorentz transformation ⇥ (10.274) i✓ J i K L = e · · (10.303) is D(0,1/2)(✓, )=exp( i✓ /2+ /2) = D(1/2,0)(✓, ) (10.304) · · which di↵ers from D(1/2,0)(✓, ) only by the sign of . The generic D(0,1/2) matrix is the complex unimodular 2 2 matrix ⇥ (0,1/2) z /2 D (✓, )=e ⇤· (10.305) with =Rez and ✓ =Imz. Example 10.39 (The standard right-handed boost ). For a particle of mass m>0, the “standard” boost (10.287) that transforms k =(m, 0)to p =(p0, p)isthe4 4 matrix B(p)=exp(↵ pˆ B) in which cosh ↵ = p0/m ⇥ · and sinh ↵ = p /m. This Lorentz transformation with ✓ = 0 and = ↵ pˆ | | is represented by the matrix (exercise 10.37)

(0,1/2) ↵pˆ /2 D (0,↵pˆ)=e · = I cosh(↵/2) + pˆ sinh(↵/2) · = I (p0 + m)/(2m)+pˆ (p0 m)/(2m) · (10.306) p0 + m + p = p · p 2m(p0 + m) in the third line of whichp the 2 2 identity matrix I is suppressed. ⇥ Under D(0,1/2), the vector (I,) transforms as a 4-vector; for tiny z

(0,1/2) (0,1/2) D† (✓, ) ID (✓, )=I + · (10.307) (0,1/2) (0,1/2) D† (✓, ) D (✓, )= + I + ✓ ^ as in (10.273). A massless field v(x) that responds to a unitary Lorentz transformation U(L) as 1 (0,1/2) 1 U(L) v(x) U (L)=D (L ) v(Lx) (10.308) is called a right-handed Weyl spinor. One may show (exercise 10.40) that the action density

(x)=iv†(x)(@ I + ) v(x) (10.309) Lr 0 r· is Lorentz covariant 1 U(L) (x) U (L)= (Lx). (10.310) Lr Lr 10.39 The Dirac Representation of the Lorentz Group 469 Example 10.40 (Why v is right handed). An argument like that of example (10.38) shows that the field v(x) satisfies the wave equation (@ I + ) v(x) = 0 (10.311) 0 r· or in momentum space (E p ) v(p)=0. (10.312) · Thus, E = p , and v(p) is an eigenvector of pˆ J | | · 1 pˆ J v(p)= v(p) (10.313) · 2 with eigenvalue 1/2. A particle whose spin is parallel to its momentum is said to have positive helicity or to be right handed. Nearly massless antineutrinos are nearly right handed. The Majorana mass term 1 1 T (x)= mv†(x) v⇤(x) m⇤ v (x) v(x) (10.314) LM 2 2 2 2 like (10.300) is Lorentz covariant. The action density of a right-handed field of mass m is the sum = + of the kinetic one (10.309) and this L Lr LM Majorana mass term (10.314). The resulting equations of motion

0=i (@0 + ) v m2v⇤ r· (10.315) 0= @2 2 + m 2 v 0 r | | show that the field v represents particles of mass m . | |

10.39 The Dirac Representation of the Lorentz Group Dirac’s representation of SO(3, 1) is the direct sum D(1/2,0) D(0,1/2) of D(1/2,0) and D(0,1/2). Its generators are the 4 4 matrices ⇥ 1 0 i 0 J = and K = . (10.316) 2 0 2 0 ✓ ◆ ✓ ◆ Dirac’s representation uses the Cli↵ord algebra of the gamma matrices a which satisfy the anticommutation relation a,b a b + b a =2⌘abI (10.317) { }⌘ in which ⌘ is the 4 4 diagonal matrix (10.255) with ⌘00 = 1 and ⌘jj =1 ⇥ for j = 1, 2, and 3, and I is the 4 4 identity matrix. ⇥ Remarkably, the generators of the Lorentz group ij 0j J = ✏ijkJk and J = Kj (10.318) 470 Group theory may be represented as commutators of gamma matrices i J ab = [a,b]. (10.319) 4 They transform the gamma matrices as a 4-vector [J ab,c]= ia ⌘bc + ib ⌘ac (10.320) (exercise 10.42) and satisfy the commutation relations i[J ab,Jcd]=⌘bc J ad ⌘ac J bd ⌘da J cb + ⌘db J ca (10.321) of the Lorentz group (Weinberg, 1995, p. 213–217) (exercise 10.43). The gamma matrices a are not unique; if S is any 4 4 matrix with an ⇥ inverse, then the matrices a SaS 1 also satisfy the definition (10.317). 0 ⌘ The choice 01 0 0 = i and = i (10.322) 10 0 ✓ ◆ ✓ ◆ makes J and K block diagonal (10.316) and lets us assemble a left-handed spinor u and a right-handed spinor v neatly into a 4-component spinor u = . (10.323) v ✓ ◆ Dirac’s action density for a 4-spinor is = (a@ + m) (@ + m) (10.324) L a ⌘ 6 in which 0 01 i † = † = v u . (10.325) ⌘ 10 † † ✓ ◆ The kinetic part is the sum of the left-handed and right-handed action L` Lr densities (10.292 & 10.309) a @ = iu† (@ I ) u + iv† (@ I + ) v. (10.326) a 0 r· 0 r· If u is a left-handed spinor transforming as (10.291), then the spinor

v1 0 i u1† v = 2 u⇤ = (10.327) ⌘ v2 i 0 u† ✓ ◆ ✓ ◆ 2! transforms as a right-handed spinor (10.308), that is (exercise 10.44)

z⇤ /2 z /2 ⇤ e · 2 u⇤ = 2 e · u . (10.328) Similarly, if v is right handed, then u =⇣ v is⌘ left handed. 2 ⇤ 10.40 The Poincar´eGroup 471 The simplest 4-spinor is the Majorana spinor

u 2v⇤ 2 = = = i ⇤ (10.329) M u v M ✓ 2 ⇤◆ ✓ ◆ whose particles are the same as its antiparticles. (1) (2) If two Majorana spinors M and M have the same mass, then one may combine them into a Dirac spinor

(1) (2) 1 (1) (2) 1 u + iu uD D = M + i M = (1) (2) = . (10.330) p2 p2 v + iv vD ⇣ ⌘ ✓ ◆ ✓ ◆ The Dirac mass term

m = m v† u + u† v (10.331) D D D D D D ⇣ ⌘ conserves charge, and since exp(z /2) exp( z /2) = I it also is Lorentz ⇤ · † · invariant. For a Majorana field, it reduces to

1 1 1 T 2 m M M = 2 m v†u + u†v = 2 m u†2u⇤ + u 2u (10.332) 1 ⇣ ⌘T ⇣ ⌘ = m v† v⇤ + v v 2 2 2 ⇣ ⌘ a Majorana mass term (10.300 or 10.314).

10.40 The Poincar´eGroup The elements of the Poincar´egroup are products of Lorentz transformations and translations in space and time. The Lie algebra of the Poincar´egroup therefore includes the generators J and K of the Lorentz group as well as the hamiltonian H and the momentum operator P which respectively generate translations in time and space. Suppose T (y) is a translation that takes a 4-vector x to x + y and T (z)is a translation that takes a 4-vector x to x + z.ThenT (z)T (y) and T (y)T (z) both take x to x + y + z. So if a translation T (y)=T (t, y)isrepresented by a unitary operator U(t, y)=exp(iHt iP y), then the hamiltonian H · and the momentum operator P commute with each other [H, P j] = 0 and [P i,Pj]=0. (10.333)

We can figure out the commutation relations of H and P with the angular- momentum J and boost K operators by realizing that P a =(H, P )isa 4-vector. Let i✓ J i K U(✓, )=e · · (10.334) 472 Group theory be the (infinite-dimensional) unitary operator that represents (in Hilbert space) the infinitesimal Lorentz transformation L = I + ✓ R + B (10.335) · · where R and B are the six 4 4 matrices (10.264 & 10.265). Then because ⇥ P is a 4-vector under Lorentz transformations, we have 1 +i✓ J+i K i✓ J i K U (✓, )PU(✓, )=e · · Pe · · =(I + ✓ R + B) P · · (10.336) or using (10.307) (I + i✓ J + i K) H (I i✓ J i K)=H + P (10.337) · · · · · (I + i✓ J + i K) P (I i✓ J i K)=P + H + ✓ P . · · · · ^ Thus, one finds (exercise 10.44) that H is invariant under rotations, while P transforms as a 3-vector

[Ji,H] = 0 and [Ji,Pj]=i✏ijkPk (10.338) and that [K ,H]= iP and [K ,P ]= i H. (10.339) i i i j ij By combining these equations with (10.321), one may write (exercise 10.46) the Lie algebra of the Poincar´egroup as i[J ab,Jcd]=⌘bc J ad ⌘ac J bd ⌘da J cb + ⌘db J ca i[P a,Jbc]=⌘abP c ⌘acP b [P a,Pb]=0. (10.340)

Further reading The books Group theory in a Nutshell for Physicists (Zee, 2016), Lie Al- gebras in Particle Physics (Georgi, 1999), and Group theory and Its Ap- plication to the Quantum Mechanics of Atomic Spectra (Wigner, 1964) are excellent. For applications to molecular physics, see Chemical Applications of Group theory (Cotton, 1990); for applications to condensed-matter physics, see Group theory and Quantum Mechanics (Tinkham, 2003).

Exercises 10.1 Show that all n n (real) orthogonal matrices O leave invariant the 2⇥ 2 2 2 2 quadratic form x1 +x2 +...+xn, that is, that if x0 = Ox,thenx0 = x . Exercises 473 10.2 Show that the set of all n n orthogonal matrices forms a group. ⇥ 10.3 Show that all n n unitary matrices U leave invariant the quadratic ⇥ form x 2 + x 2 +...+ x 2, that is, that if x = Ux,then x 2 = x 2. | 1| | 2| | n| 0 | 0| | | 10.4 Show that the set of all n n unitary matrices forms a group. ⇥ 10.5 Show that the set of all n n unitary matrices with unit determinant ⇥ forms a group. (j) 10.6 Show that the matrix D (g)= j, m0 U(g) j, m is unitary because m0m h | | i the rotation operator U(g) is unitary j, m U (g)U(g) j, m = . h 0| † | i m0m 10.7 Invent a group of order 3 and compute its multiplication table. For extra credit, prove that the group is unique. 10.8 Show that the relation (10.23) between two equivalent representations is an isomorphism. 10.9 Suppose that D1 and D2 are equivalent, finite-dimensional, irreducible representations of a group G so that D (g)=SD (g)S 1 for all g G. 2 1 2 What can you say about a matrix A that satisfies D2(g) A = AD1(g) for all g G? 2 10.10 Find all components of the matrix exp(i↵A)inwhich 00 i A = 00 0 . (10.341) 0 1 i 00 @ A i↵A i↵A 10.11 If [A, B]=B,finde Be . Hint: what are the ↵-derivatives of this expression? 10.12 Show that the direct-product matrix (10.34) of two representations D and D0 is a representation. 10.13 Find a 4 4 matrix S that relates the direct-product representation ⇥ D(1/2,1/2) to the direct sum D(1) D(0). 10.14 Find the generators in the adjoint representation of the group with structure constants fabc = ✏abc where a, b, c run from 1 to 3. Hint: The answer is three 3 3 matrices t , often written as L . ⇥ a a 10.15 Show that the generators (10.94) satisfy the commutation relations (10.97). 10.16 Show that the demonstrated equation (10.102) implies the commuta- tion relation (10.103). 10.17 Use the Cayley-Hamilton theorem (1.289) to show that the 3 3 ⇥ matrix (10.100) that represents a right-handed rotation of ✓ radians about the axis ✓ is given by (10.101). 10.18 Verify the mixed Jacobi identity (10.161). 10.19 For the group SU(3), find the structure constants f123 and f231. 474 Group theory 10.20 Show that every 2 2 unitary matrix of unit determinant is a quater- ⇥ nion of unit norm. 10.21 Show that the quaternions as defined by (10.210) are closed under addition and multiplication and that the product xq is a quaternion if x is real and q is a quaternion. 10.22 Show that the square of the matrix (10.198) is J 2 = I,whereI is the 2n 2n identity matrix. Then by setting R =exp(✏t)with0<✏ 1, ⇥ ⌧ show that t|J = Jt and JtJ = t|. 10.23 Show that JtJ = t| implies that t is given by (10.203). 10.24 Show that the one-sided derivative f 0(q) (10.220) of the quaternionic function f(q)=q2 depends upon the direction along which q 0. 0 ! 10.25 Show that the generators (10.224) of Sp(2n) obey commutation rela- tions of the form (10.225) for some real structure constants fabc and a suitably extended set of matrices A, A0, . . . and Sk, Sk0 ,.... 10.26 Show that for 0 <✏ 1, the real 2n 2n matrix T =exp(✏JS)in ⌧ ⇥ which S is symmetric satisfies T TJT = J (at least up to terms of order ✏2) and so is in Sp(2n, R). 10.27 Show that the matrix T of (10.206) is in Sp(2,R). 10.28 Use the parametrization (10.250) of the group SU(2), show that the parameters a(c, b) that describe the product g(a(c, b)) = g(c) g(b) are those of (10.252). 10.29 Use formulas (10.252) and (10.245) to show that the left-invariant measure for SU(2) is given by (10.253). 10.30 In tensor notation, which is explained in chapter 11, the condition (10.262) that I + ! be an infinitesimal Lorentz transformation reads a !T = !a = ⌘ !c ⌘da in which sums over c and d from 0 b b bc d to 3 are understood. In this notation, the matrix ⌘ lowers indices ef and ⌘gh raises them, so that ! a = ! ⌘da. (Both ⌘ and ⌘gh b bd ef are numerically equal to the matrix ⌘ displayed in equation (10.255).) Multiply both sides of the condition (10.262) by ⌘ae = ⌘ea and use the relation ⌘da ⌘ = ⌘d d to show that the matrix ! with both ae e ⌘ e ab indices lowered (or raised) is antisymmetric, that is, ! = ! and !ba = !ab. (10.342) ba ab 10.31 Show that the six matrices (10.264) and (10.265) satisfy the SO(3, 1) condition (10.262). 10.32 Show that the six generators J and K obey the commutations rela- tions (10.267–10.269). 10.33 Show that if J and K satisfy the commutation relations (10.267– 10.269) of the Lie algebra of the Lorentz group, then so do J and K. Exercises 475 10.34 Show that if the six generators J and K obey the commutation re- + lations (10.267–10.269), then the six generators J and J obey the commutation relations (10.276). 10.35 Relate the parameter ↵ in the definition (10.287) of the standard boost B(p) to the 4-vector p and the mass m. 10.36 Derive the formulas for D(1/2,0)(0,↵pˆ) given in equation (10.288). 10.37 Derive the formulas for D(0,1/2)(0,↵pˆ) given in equation (10.306). 10.38 For infinitesimal complex z, derive the 4-vector properties (10.289 & 10.307) of ( I,)underD(1/2,0) and of (I,)underD(0,1/2). 10.39 Show that under the unitary Lorentz transformation (10.291), the action density (10.292) is Lorentz covariant (10.293). 10.40 Show that under the unitary Lorentz transformation (10.308), the action density (10.309) is Lorentz covariant (10.310). 10.41 Show that under the unitary Lorentz transformations (10.291 & 10.308), the Majorana mass terms (10.300 & 10.314) are Lorentz covariant. 10.42 Show that the definitions of the gamma matrices (10.317) and of the generators (10.319) imply that the gamma matrices transform as a 4- vector under Lorentz transformations (10.320). 10.43 Show that (10.319) and (10.320) imply that the generators J ab satisfy the commutation relations (10.321) of the Lorentz group. 10.44 Show that the spinor v = 2u⇤ defined by (10.327) is right handed (10.308) if u is left handed (10.291). 10.45 Use (10.337) to get (10.338 & 10.339). 10.46 Derive (10.340) from (10.321, 10.333, 10.338, & 10.339).