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Home , Lorentz group, Symplectic group

groups in Part 2

George Jorjadze Free University of Tbilisi

Zielona Gora - 25.01.2017

G.J.— Symmetry groups in physics Lecture 3 1/24 Contents

• Lorentz • Poincare group • • • • Lie algebra representation • sl(2, R) algebra • UIRs of su(2) algebra

G.J.— Symmetry groups in physics Contents 2/24 Now we consider symmetry groups in relativistic mechanics. The spacetime geometry here is given by , which is a 4d space with coordinates xµ, µ = (0, 1, 2, 3), and the

gµν = diag(−1, 1, 1, 1). We set c = 1 (see Lecture 1) and x0 is associated with time. Lorentz group describes transformations of spacetime coordinates between inertial systems. Mathematically it is given by linear transformations µ µ µ ν x 7→ x˜ = Λ n x , which preserves the metric structure of Minkowski space α β Λ µ gαβ Λ ν = gµν . The form of these equations reads ΛT g L = g . This relation provides 10 independent equations, since g is symmetric. Hence, Lorentz group is 6 parametric.

G.J.— Symmetry groups in physics Lorentz group 3/24 From the definition of Lorentz group follows that 0 2 i i det[Λ] = ±1 and (Λ 0) − Λ 0 Λ 0 = 1. Therefore, Lorentz group splits in four non-connected parts 0 0 1. (det[Λ] = 1, Λ 0 ≥ 1), 2. (det[Λ] = 1, Λ 0 ≤ 1), 0 0 3. (det[Λ] = −1, Λ 0 ≥ 1), 4. (det[Λ] = −1, Λ 0 ≤ 1). The matrices of the first part are called proper Lorentz transformations. They form a group which is denoted by SO↑(1, 3). Geometrically it is a connected domain like SO(2). However, it is not compact and needs 6 parameters. Other three parts of Lorenz group are not and they are obtained by the products

Λ2 · SO↑(1, 3), Λ3 · SO↑(1, 3), Λ4 · SO↑(1, 3), where the matrices Λ2, Λ3 and Λ4 can be choose as follows

Λ2 = diag(−1, −1, 1, 1) , Λ3 = diag(1, −1, 1, 1) , Λ4 = diag(−1, 1, 1, 1) . They correspond to reflections of the space-time axes.

Thus, it suffices to investigate the group SO↑(1, 3).

G.J.— Symmetry groups in physics Lorentz group 4/24 Below we show that Λ ∈ SO↑(1, 3) can be represented as

Λ = ΛR · ΛB, where ΛR is a and ΛB is a boost corresponding to the transformation to a moving inertial system. µ For a given Λ ν, the boost matrix ΛB has the following block structure

0 0 ! Λ 0 Λ l 0 0 ΛB = 0 Λ k Λ l . Λ k δkl + 0 1+Λ 0

µ So, it is defined by the first row of Λ ν only.

A rotation matrix ΛR has the form   1 0l ΛR = , 0k Rkl where 0k and 0l denote null 3-vectors written as a column and as a row, respectively, and Rkl is given by k 0 k Λ 0 Λ l Rkl = Λ l − 0 . 1+Λ 0

G.J.— Symmetry groups in physics Lorentz group 5/24 Exercise 28. Check that the given ΛR and ΛB indeed provide ΛR · ΛB = Λ. T Exercise 29. Check that ΛB is indeed a Lorentz matrix, i.e. ΛB g ΛB = g.

Exercise 30. Check that the Rkl is an orthogonal matrix. αµ µ µα Hint: use the relations Λ Λαν = δν = Λνα Λ , which follow from the definition of Lorentz group.

Exercise 31. Check that the boost ΛB can be represented as the exponent

ΛB = exp[ηl Kl] , where Kl are the boost generator matrices

 0 1 0 0   0 0 1 0   0 0 0 1  1 0 0 0 0 0 0 0 0 0 0 0 K =  ,K =  ,K =  , 1  0 0 0 0  2  1 0 0 0  3  0 0 0 0  0 0 0 0 0 0 0 0 1 0 0 0

0 and ηk is related to Λ k. Find this relation and express the velocity of motion between the inertial systems through ηk.

G.J.— Symmetry groups in physics Lorentz group 6/24 Poincare group Poincare group is given by the set of pairs (Λ, a), where Λ is a matrix of Lorentz group and a is a 4-vector, a = (a0, a1, a2, a3). Thus, Poincare group is 10 parametric. A group element (Λ, a) creates the transformation of Minkowski space x 7→ x0 = Λx + a . Two consecutive transformations then define the multiplication rule (Λ0, a0) · (Λ, a) = (Λ0 · Λ, Λ0a + a0) . Minkowski space has a natural generalization for d-dimensional (d > 1) spacetime with metric tensor gµν = diag(−1, 1, ..., 1). 1,d−1 This space is denoted by R . 1,d−1 Generalization of Poincare group for R is straightforward. 1,d−1 d(d+1) Exercise 32. Check that Poincare group in R is 2 parametric.

G.J.— Symmetry groups in physics Poincare group 7/24 Symplectic group Sp(n) This group is given by 2n × 2n real matrices S, which satisfy the condition

 0 I  ST ΩS = Ω , with Ω = n n . −In 0n

Here, 0n and In are n × n zero and unit matrices, respectively. Exercise 33. Check that Sp(n, R) is n(2n + 1) parametric group. Exercise 34. Check that Sp(1, R) = SL(2, R). Definition. A non-degenerated and antisymmetric matrix is called symplectic. Exercise 35. Prove that the number of rows of a is even. Thus, Ω is a symplectic matrix. Note that the symplectic group Sp(n) describes linear canonical transformations in 2n-dimensional .

G.J.— Symmetry groups in physics Symplectic group 8/24 Lie Groups Lie group is a group that is also a , where the group operations (product and inversion) are given by smooth functions. In other words, a Lie group has two structures (of a group and of a manifold) and these structures are compatible. All considered matrix groups and Poincare group are Lie groups. A Lie group is parameterized by some coordinates. The number of coordinates is called group . A Lie group element g ∈ G is given by g = g(u), with u = (u1, . . . , un). Without loss of generality one can set g(0) = e. The group product defines functions f i(u, v), i = (1, . . . , n) obtained from g(u) · g(v) = g(f(u, v)) . These functions obviously satisfy the conditions f i(u, 0) = ui = f i(0, u) , and, therefore, their expansion in powers of u and v reads i i i i j k f (u, v) = u + v + Ajku v + ... (1)

G.J.— Symmetry groups in physics Lie Group 9/24 Let us consider a representation of G with operators Rg = R(u). These operators have a standard Taylor expansion i 1 i j R(u) = I + u Ri + 2 u u Rij + ... (2) and they obey the identity R(u) · R(v) = R(f(u, v)) . (3) Exercise 36. Using the expansions (1)-(2) and comparing the coefficients of uivj in (3), check that k Ri Rj = AijRk + Rij . From Rij = Rji then follows k Ri Rj − Rj Ri = Cij Rk , k k k with Cij = Aij − Aji. The operators Ri are called Lie group generators. Their create an algebra, which is called Lie algebra. Thus, a Lie group is related to its Lie algebra. The Lie algebra of a Lie group is denoted similarly, only without capital letters. For example, su(n) is the Lie algebra for SU(n) group. G.J.— Symmetry groups in physics Lie group 10/24 Lie algebras A Lie algebra G is a linear space with a multiplication rule, which is bilinear, antisymmetric and satisfies the Jacobi identity. The product of two vectors A ∈ G and B ∈ G is called the Lie bracket and it is denoted by [ A,B ]. Thus, [ A,B ] ∈ G and it satisfies the conditions

[λ A + B,C ] = λ [ A,C ] + [ B,C ] ,

[ A,B ] = − [ B,A ] , [[ A,B ] ,C ] + [ [ B,C ] ,A ] + [ [ C,A ] ,B ] = 0 . The numbers λ are real for real algebras and to distinguish the real and complex algebras one uses the letters R and C, respectively. Examples: 3 1. R is a Lie algebra under the vector product ~a ×~b. 2. The set of smooth functions on a phase space is a Lie algebra with respect to Poisson brackets.

G.J.— Symmetry groups in physics Lie algebras 11/24 Lie algebra representation A representation of G is a of G to a space of linear operators A 7→ OˆA , such that

[ A,B ] 7→ OˆA OˆB − OˆB OˆA .

Let us consider the operator adA acting on G by

adA(B) = [ A,B ] .

The Jacobi identity provides that

ad[ A,B ] = adA adB − adB adA .

Thus, the map A 7→ adA defines a representation of G, which is called the . Another example of Lie algebra representation is given by Lie group generators discussed above.

G.J.— Symmetry groups in physics Lie algebra representation 12/24 Exercise 37. Let en be a basis of a Lie algebra G. Since [ em , en ] ∈ G, one has

l [ em , en ] = Cmn el ,

l with some constants Cmn , which are called the structure constants of G. Check that the structure constants satisfy the relations

l l Cmn = −Cnm , k j k j k j Cmn Ckl + Cnl Ckm + Clm Ckn = 0 .

m Exercise 38. Let (adA) n be the matrix associate with adA in a basis en

l adA(en) = el (adA) n .

Check that l l (adem ) n = Cmn , and relate the quadratic identity for the structure constants to the of the basis vectors in the adjoint representation.

G.J.— Symmetry groups in physics Lie algebra representation 13/24 sl(2, R) algebra This algebra is given by 2 × 2 real traceless matrices. The Lie bracket is the commutator [ A,B ] = AB − BA. Since the commutator of two matrices is traceless, [ A,B ] ∈ sl(2, R). One can use the following basis in sl(2, R)  0 1   0 1   1 0  T = ,T = ,T = . 0 −1 0 1 1 0 2 0 −1

The basis elements Tµ (µ = 0, 1, 2) satisfy the matrix relations λ Tµ Tν = ηµν I + µν Tλ ,

where I is the unit matrix, ηµν = diag(−1, 1, 1) form the metric tensor of 3d Minkowski space and µνλ is the Levi-Chivita tensor, with 012 = 1. The rising-lowering indices are provided by ηµν and one has λ [ Tµ,Tν ] = 2µν Tλ .

λ Thus, the numbers 2µν are the structure constants of sl(2, R).

G.J.— Symmetry groups in physics sl (2, R) algebra 14/24 Let us introduce the following scalar product in sl(2, R) 1 h A,B i = Tr(AB) . 2

It provides h Tµ ,Tν i = ηµν. µ µ µ The expansion in the basis, A = A Tµ, yields A = h A,T i and µ ν h A,B i = ηµν A B . Thus, sl(2, R) is isometric to 3d Minkowski space. The action of SL(2, R) group on sl(2, R) algebra is defined by A 7→ gAg−1, where A ∈ sl(2, R) and g ∈ SL(2, R). Exercise 39. Check that this action is a representation of SL(2, R) and it preserves the scalar product in sl(2, R). This representation is called adjoint representation. A Lie algebra scalar product which is invariant under the adjoint action of the group is called .

G.J.— Symmetry groups in physics sl (2, R) algebra 15/24 Exercise 40. Check that the adjoint transformation of A ∈ sl(2, R) µ µ µ µ ν transforms its coordinates A = h T A i by A 7→ Λ ν A , with

µ µ −1 Λ ν = h T g Tν g i .

Since sl(2, R) algebra is isometric to 3d Minkowski space, µ Λ ν is a matrix of 3d . Thus, one obtains a map from SL(2, R) to 3d Lorentz group. µ SL(2, R) is a connected manifold. Hence, Λ ν ∈ SO↑(1, 2). This map is a . µ Note that g and −g are mapped to the same Λ ν. Euclidean analog of sl(2, R) is su(2) algebra, which is the algebra of antihermitian and traceless matrices. Its basis is given by iσk, k = (1, 2, 3), where σk are . Below we construct finite dimensional UIR’s of this algebra which is known as quantization.

G.J.— Symmetry groups in physics sl (2, R) algebra 16/24 Spin quantization

Let us consider the algebra of angular momentum operators for ~ = 1 [Jm,Jn] = imnl Jl . These commutation relations are equivalent to

[J1,J2] = iJ3 , [J2,J3] = iJ1 , [J3,J1] = iJ2 . It is also called the spin algebra. Examples: 1. The standard realization is given by 1 Jm = 2 mkl Mkl , where Mkl = qkpl − qlpk and one has the canonical commutators

[qk, pl] = iδkl , [qk, ql] = [pk, pl] = 0 . Using the commutation relations

[AB, CD] = A[B,C]D + AC[B,D] + [A, C]BD + C[A, D]B,

we can check that the operators Jm provide the spin algebra.

G.J.— Symmetry groups in physics Spin quantizaton 17/24 2. The next example is given by Pauli matrices 1 Jm = 2 σm, which form the spin algebra (see Exercise 19 in Lecture 2, page 15). 3. In this example we choose the matrices

 0 0 0   0 0 i   0 −i 0  J1 =  0 0 −i  ,J2 =  0 0 0  ,J3 =  i 0 0  . 0 i 0 −i 0 0 0 0 0

They are obtained by dR (α) J = i m | , m dα α=0 where Rm(α) are the rotation matrices around the coordinate axes considered in the previous lecture (see Lecture 2, p.19). It is easy to check that these 3 × 3 matrices also form the spin algebra.

G.J.— Symmetry groups in physics Spin quantization 18/24 Casimir operator

The angular momentum square operator is defined by 2 2 2 2 J = J1 + J2 + J3 . Using the spin algebra and the commutation relations

[AB,C] = A[B,C] + [A, C]B,

one obtains 2 [J ,Jm] = 0. An operator which commutes with all elements of Lie algebra is called the Casimir operator. Thus, J 2 is the Casimir operator of the spin algebra. Our aim is to describe finite dimensional UIRs of the spin algebra. Unitarity here means hermiticity of the operators Jk. −iθ J The operators e k k are then unitary for real parameters θk (see Exersice 14 in Lecture 2, page 14).

G.J.— Symmetry groups in physics Casimir operator 19/24 Raising-lowering operators It is convenient to introduce the operators

J± = J1 ± iJ2 .

† Since J1 and J2 are assumed Hermitian, J+ = J−. The spin algebra in terms of these operators takes the form

[J3,J+] = J+ , [J3,J−] = J− , [J+,J−] = 2J3 ,

and the Casimir operator becomes

2 2 2 J = J3 + J3 + J−J+ = J3 − J3 + J+J− .

Let us denote by | m i an eigenstate of the operator J3, which corresponds to an eigenvalue m. The number m is real, since J3 is Hermitian.

Exercise 41. Check that J+| m i and J−| m i are also eigenstates of J3 with eigenvalues m + 1 and m − 1, respectively.

G.J.— Symmetry groups in physics Raising-lowering operators 20/24 Highest-weight vector Since we are looking for a finite dimensional representation, there has to be a state with some maximal value of m, say m = j, which is annihilated by the operator J+. This state is called highest-weight vector and we denote it by | j, j i, where the left j indicates what is the maximal eigenvalue of J3 and the right j stands for the value of m for this state. Thus, we have

J+| j, j i = 0 ,J−| j, j i = Cj | j, j − 1 i ,

where 0 is the of the and Cj is a constant. 2 Exercise 42. Using the form of J in terms of J3 and J±, check that J 2| j, j i = j(j + 1)| j, j i .

2 Exercise 43. Check that J−| j, j i is an eigenstate of J with the same eigenvalue j(j + 1).

In general, a consecutive action of J− reduce the eigenvalue of J3, without changing the eigenvalue of J 2.

G.J.— Symmetry groups in physics Highest-weight vector 21/24 By the same argument as before, there has to be a state | j, j0 i, which is annihilated by the operator J−. This state satisfies the conditions

0 0 0 0 2 0 0 J3 | j, j i = j | j, j i ,J− | j, j i = 0 ,J | j, j i = j(j + 1)| j, j i .

2 0 Using again the form of J in terms of J3 and J±, we find j = −j. Then, m takes 2j + 1 different values from j to −j, and we conclude that j has to be or half integer. Assuming that the states | j, m i have unit norms, one can fix the coefficients Cm and C˜m in the equations

J−| j, m i = Cm| j, m − 1 i,J+| j, m i = C˜m| j, m + 1 i.

† Starting with m = j and using J− = J+, together with the spin algebra, we obtain 2 √ Cj = h j, j |J+ J−| j, j i = 2j , ⇒ Cj = 2j. √ Then, C˜j−1 = h j, j |J+ | j, j − 1 i = 2j.

G.J.— Symmetry groups in physics Highest-weight vector 22/24 UIRs of spin algebra Using again the spin algebra and the recursive relations between the coefficients one finally finds

J3 | j, m i = m | j, m i , p J− | j, m i = (j + m)(j − m + 1) | j, m − 1 i , p J+ | j, m i = (j − m)(j + m + 1) | j, m + 1 i . These equations fix the finite dimensional UIRs of the spin algebra. They are characterized by j, which is integer or half-integer. For a given j, the dimension of the representation is 2j + 1. Exercise 44. Check that the constructed representations are unitary. Exercise 45. Check that these representations for j = 1/2 and j = 1 are equivalent to Example 2 and Example 3, respectively (see page 18).

G.J.— Symmetry groups in physics UIRs of spin algebra 23/24 UIRs of sl(2, R). sl(2, R) algebra has no finite dimensional UIR. Its UIRs are infinite dimensional. Note that SL(2, R) is a non-, while SU(2) is compact. In general, non-compact groups have no finite dimensional UIRs. Here we present a class of UIRs of sl(2, R). They are characterized by a positive number µ. The states | µ, n i with integer n ≥ 0 form an orthonormal basis and the operators

L0 | µ, n i = (µ + n) | µ, n i , p L−| µ, n i = (2µ + n − 1) n | µ, n − 1 i , p L+| µ, n i = (2µ + n)(n + 1) | µ, n + 1 i , form sl(2, R) algebra [L0,L±] = L± , [L−,L+] = 2L0 .

G.J.— Symmetry groups in physics UIRs of sl (2, R) 24/24