Symmetry groups in physics

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 group

• Poincare group

• Symplectic group

• Lie group

• Lie algebra

• Lie algebra representation

• sl(2,R) algebra

• UIRs of su(2) algebra

G.J. — Symmetry groups in physics Contents 2/24

Lorentz group

Now we consider spacetime symmetry groups in relativistic mechanics.The spacetime geometry here is given by Minkowski space, which is a4d space with coordinates xµ, µ = (0, 1, 2, 3), and the metric tensor

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 betweeninertial systems. Mathematically it is given by linear transformations

xµ 7→ xµ = Λµ n xν ,

which preserves the metric structure of Minkowski space

Λα µ gαβ Λβ ν = gµν .

The matrix 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

det[Λ] = ±1 and (Λ00)2 − Λi 0 Λi 0 = 1.

Therefore, Lorentz group splits in four non-connected parts

1. (det[Λ] = 1, Λ00 ≥ 1), 2. (det[Λ] = 1, Λ0

0 ≤ 1),

3. (det[Λ] = −1, Λ00 ≥ 1), 4. (det[Λ] = −1, Λ0

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 subgroups and they are obtainedby 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 rotation and ΛB is a boost corresponding to thetransformation to a moving inertial system.For a given Λµ ν , the boost matrix ΛB has the following block structure

ΛB =

(Λ0

0 Λ0l

Λ0k δkl + Λ0

k Λ0l

1+Λ00

).

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

A rotation matrix ΛR has the form

ΛR =

(1 0l0k Rkl

),

where 0k and 0l denote null 3-vectors written as a column and as a row,respectively, and Rkl is given by

Rkl = Λk l − Λk0 Λ0

l1+Λ0

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.

Exercise 29. Check that ΛB is indeed a Lorentz matrix, i.e. ΛTB gΛB = g.

Exercise 30. Check that the Rkl is an orthogonal matrix.

Hint: use the relations Λαµ Λαν = δµν = Λνα Λµα,which follow from the definition of Lorentz group.

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

ΛB = exp[ηlKl] ,

where Kl are the boost generator matrices

K1 =

0 1 0 01 0 0 00 0 0 00 0 0 0

, K2 =

0 0 1 00 0 0 01 0 0 00 0 0 0

, K3 =

0 0 0 10 0 0 00 0 0 01 0 0 0

,and ηk is related to Λ0

k. Find this relation and express the velocity ofmotion 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 ofLorentz 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→ x′ = Λx+ a .

Two consecutive transformations then define the multiplication rule

(Λ′, a′) · (Λ, a) = (Λ′ · Λ, Λ′a+ a′) .

Minkowski space has a natural generalization for d-dimensional (d > 1)spacetime with metric tensor gµν = diag(−1, 1, ..., 1).This space is denoted by R1,d−1.

Generalization of Poincare group for R1,d−1 is straightforward.

Exercise 32. Check that Poincare group in R1,d−1 is d(d+1)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

STΩS = Ω , with Ω =

(0n In−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 calledsymplectic.

Exercise 35. Prove that the number of rows of a symplectic matrix is even.

Thus, Ω is a symplectic matrix.

Note that the symplectic group Sp(n) describes linear canonicaltransformations in 2n-dimensional phase space.

G.J. — Symmetry groups in physics Symplectic group 8/24

Lie Groups

Lie group is a group that is also a manifold, 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 amanifold) 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 dimension.

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 conditionsf i(u, 0) = ui = f i(0, u) ,

and, therefore, their expansion in powers of u and v reads

f i(u, v) = ui + vi +Aijkujvk + ... (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

R(u) = I + uiRi + 12u

iujRij + ... (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

RiRj = AkijRk +Rij .

From Rij = Rji then follows

RiRj −Rj Ri = Ckij Rk ,

with Ckij = Akij −Akji.The operators Ri are called Lie group generators.

Their commutators 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 capitalletters. 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 isbilinear, antisymmetric and satisfies the Jacobi identity.The product of two vectors A ∈ G and B ∈ G is called the Lie bracketand it is denoted by [A , B ].Thus, [A , B ] ∈ G and it satisfies the conditions

[λA+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 andcomplex algebras one uses the letters R and C, respectively.

Examples:

1. R3 is a Lie algebra under the vector product ~a×~b.2. The set of smooth functions on a phase space is a Lie algebra withrespect to Poisson brackets.G.J. — Symmetry groups in physics Lie algebras 11/24

Lie algebra representation

A representation of G is a linear map of G to a space of linear operatorsA 7→ OA , such that

[A , B ] 7→ OA OB − OB OA .

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 theadjoint representation.

Another example of Lie algebra representation is given by Lie groupgenerators 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

[ em , en ] = Cmnl el ,

with some constants Cmnl , which are called the structure constants of G.

Check that the structure constants satisfy the relations

Cmnl = −Cnm l ,

Cmnk Ckl

j + Cnlk Ckm

j + Clmk Ckn

j = 0 .

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

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

Check that(adem)l n = Cmn

l ,

and relate the quadratic identity for the structure constants to thecommutator 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)

T0 =

(0 1−1 0

), T1 =

(0 11 0

), T2 =

(1 00 −1

).

The basis elements Tµ (µ = 0, 1, 2) satisfy the matrix relations

Tµ Tν = ηµν I + εµνλ Tλ ,

where I is the unit matrix, ηµν = diag(−1, 1, 1) form the metric tensor of3d Minkowski space and εµνλ is the Levi-Chivita tensor, with ε012 = 1.The rising-lowering indices are provided by ηµν and one has

[Tµ, Tν ] = 2εµνλ Tλ .

Thus, the numbers 2εµνλ 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)

〈A , B 〉 =1

2Tr(AB) .

It provides 〈Tµ , Tν 〉 = ηµν .The expansion in the basis, A = Aµ Tµ, yields Aµ = 〈A , Tµ 〉 and

〈A , B 〉 = ηµν AµBν .

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 itpreserves 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 ofthe group is called Killing form.G.J. — Symmetry groups in physics sl(2,R) algebra 15/24

Exercise 40. Check that the adjoint transformation of A ∈ sl(2,R)transforms its coordinates Aµ = 〈TµA 〉 by Aµ 7→ Λµ ν A

ν , with

Λµ ν = 〈Tµ g Tν g−1 〉 .

Since sl(2,R) algebra is isometric to 3d Minkowski space,Λµ ν is a matrix of 3d Lorentz transformation.

Thus, one obtains a map from SL(2,R) to 3d Lorentz group.

SL(2,R) is a connected manifold. Hence, Λµ ν ∈ SO↑(1, 2).

This map is a homomorphism.Note that g and −g are mapped to the same Λµ ν .

Euclidean analog of sl(2,R) is su(2) algebra, which is the algebra ofantihermitian and traceless matrices.Its basis is given by iσk, k = (1, 2, 3), where σk are Pauli matrices.

Below we construct finite dimensional UIR’s of this algebra which is knownas spin 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] = iεmnl 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

Jm = 12 εmklMkl ,

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

Jm = 12 σm,

which form the spin algebra (see Exercise 19 in Lecture 2, page 15).

3. In this example we choose the matrices

J1 =

0 0 00 0 −i0 i 0

, J2 =

0 0 i0 0 0−i 0 0

, J3 =

0 −i 0i 0 00 0 0

.

They are obtained by

Jm = idRm(α)

dα|α=0 ,

where Rm(α) are the rotation matrices around the coordinate axesconsidered 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

J2 = J21 + J2

2 + J23 .

Using the spin algebra and the commutation relations

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

one obtains

[J2, Jm] = 0.

An operator which commutes with all elements of Lie algebra is called theCasimir operator. Thus, J2 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.

The operators e−iθkJk 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

J2 = J23 + J3 + J−J+ = J2

3 − J3 + J+J− .

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

Exercise 41. Check that J+|m 〉 and J−|m 〉 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 tobe a state with some maximal value of m, say m = j, which is annihilatedby the operator J+. This state is called highest-weight vector and wedenote it by | j, j 〉, where the left j indicates what is the maximaleigenvalue of J3 and the right j stands for the value of m for this state.Thus, we have

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

where 0 is the null vector of the Hilbert space and Cj is a constant.

Exercise 42. Using the form of J2 in terms of J3 and J±, check that

J2| j, j 〉 = j(j + 1)| j, j 〉 .

Exercise 43. Check that J−| j, j 〉 is an eigenstate of J2 with the sameeigenvalue j(j + 1).

In general, a consecutive action of J− reduce the eigenvalue of J3, withoutchanging the eigenvalue of J2.G.J. — Symmetry groups in physics Highest-weight vector 21/24

By the same argument as before, there has to be a state | j, j′ 〉,which is annihilated by the operator J−.This state satisfies the conditions

J3 | j, j′ 〉 = j′ | j, j′ 〉 , J− | j, j′ 〉 = 0 , J2 | j, j′ 〉 = j(j + 1)| j, j′ 〉 .

Using again the form of J2 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 integer or half integer.

Assuming that the states | j, m 〉 have unit norms, one can fix thecoefficients Cm and Cm in the equations

J−| j, m 〉 = Cm| j, m− 1 〉, J+| j, m 〉 = Cm| j, m+ 1 〉.

Starting with m = j and using J†− = J+, together with the spin algebra,we obtain

C2j = 〈 j, j |J+ J−| j, j 〉 = 2j , ⇒ Cj =

√2j.

Then, Cj−1 = 〈 j, j |J+ | j, j − 1 〉 =√

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 thecoefficients one finally finds

J3 | j, m 〉 = m | j, m 〉 ,

J− | j, m 〉 =√

(j +m)(j −m+ 1) | j, m− 1 〉 ,

J+ | j, m 〉 =√

(j −m)(j +m+ 1) | j, m+ 1 〉 .

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 = 1are 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-compact group, 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 〉 with integer n ≥ 0 form an orthonormal basisand the operators

L0 |µ, n 〉 = (µ+ n) |µ, n 〉 ,

L−|µ, n 〉 =√

(2µ+ n− 1)n |µ, n− 1 〉 ,

L+|µ, n 〉 =√

(2µ+ n)(n+ 1) |µ, n+ 1 〉 ,

form sl(2,R) algebra

[L0, L±] = L± , [L−, L+] = 2L0 .

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