On the LORENTZ GROUP

Fys3120: On the LORENTZ GROUP

DEFINITION OF A GROUP:

A group G = gi is a set (collection of elements) with the following properties: { } * multiplication such that gigj G ; * ∃ a unity element g = 1 such that∈ g g = g g = g ; ∃ 0 i 0 0 i i * (g1g2) g3 = g1 (g2g3) (Associative law) * an inverse· (g )·−1 for all g G ∃ i i ∈ LIE GROUPS In theoretical physics one most often need to study continuous groups, called Lie groups. A Lie group is a group where the elements are functions of continuous parameters:

g(α1, α2, ...., αn) , (1) where all αi might be chosen real. Important groups are for instance

U(1) = eiα ; α R , (2) ∈ n o i.e. the collection of all phase factors, and

SU(n)= U = n n matrix (n =2, 3, ..) ; U C , U †U = 1 , detU =1 × kl ∈ n (3)o Here R is the set of all real numbers, C is the set of all complex numbers, and k and l run from 1 to n. One may show that the group elements of SU(n) have (n2 1) inde- − pendent real parameters. later we will use the group SU(2) to study three- dimensional rotations.

THE LORENTZ GROUP Most often, we think of a Lorentz transformation as a boost (special Lorentz transformation) along an axis, for instance the x-axis, with the following relations between the following position and time coordinates:

x′ = γ (x v t) , t′ = γ (t v x/c2) ; γ = (1 v2/c2) , (4) − − − between the inertial frames S and S′. The easiest way to obtain the transformations in (4) might be the following: First we write down the relations

x′ = γ (x v t) ; x = γ (x′ + v t′) , (5) − where γ is an unknown constant. Here we have assumed that the relations between the position and time coordinates are linear (A linear movement in one inertial frame is also linear in another frame.) Further, x′ must be

1 proportional to (x vt) because the origo x′ = 0 of S′ is moving with constant velocity v in− the frame S,- i.e. x′ = 0 must imply (x v t) = 0. − Then this plus linearity implies the ﬁrst equation in (5). The second equation in (5) follows from the symmetry of the problem: The frame S moves with velocity ( v) with respect to S′. Next we consider a light signal travelling − to the right, starting at time t = t′ = 0 in both frames (the clocks in S and S′are assumed to be synchronized when they pass each other at x = x′ = 0 and t = t′ = 0). Then we use the postulate that the velocity of light is the same in both frames, such that

x = c t ; x′ = c t′ . (6)

From the equations in (5) and (6) one easily finds the value for γ and t′ in equation(4). But Lorentz transformations are defined much wider than in (5). Lorentz transformations are all coordinate transformations which conserve the quadrate µ µ 2 2 2 of a four vector x , i.e. conserve x x x xµ = (~x) c t , where 0 · ≡ − 2 x = x0 = c t. Among such are also rotations (which conserve (~x) sepa- rately)− a subgroup. We will first discuss the Rotation group,- and afterwards study the boosts. The rotation group A 3-dimensional rotation may be expressed as:

x x′ = R x . (7) i → i ij j Here and hereafter sum over repeated indices is understood. (Einstein’s sum convention). The matrix R has the property that the length of the position vector is conserved under the transformation:

(x~′)2 = (~x)2 . (8)

The rotation group may be considered to be the set of all 3 3 matrices R where the inverse R−1 is the same as the transpose RT : ×

O(3) = R ; R R ; RT R = 1 . (9) kl ∈ · n o

Simple spesial cases are rotations R~n(φ) an angle φ around (the same) axis ~n (where ~n = 1), for example around the x-, y-, or the z-xis: | | 1 0 0 c 0 s c s 0 φ − φ φ φ R = 0 c s , R = 01 0 , R = s c 0 (10) x  φ φ  y   z  − φ φ  0 s c s 0 c 0 01  φ φ   φ φ     −      where cφ cosφ og sφ sinφ. Rotations which are continous with the unit element (as≡ the transformations≡ in (10)) are in the group SO(3), diﬀerent from O(3) which also contains space invertion, ~x ~x. → −

2 Rotations around an axis ~n constitute a subgroup

R (φ ) R (φ ) = R (φ + φ ) , (11) ~n 1 · ~n 2 ~n 1 2 which follows from

cosφ cosφ sinφ sinφ = cos(φ + φ ) , 1 2 − 1 2 1 2 cosφ1 sinφ2 + sinφ1 cosφ2 = sin(φ1 + φ2) . (12)

Such rotations about the same axis commute with each other and we say that they constitute an Abelian sub-group. In general, rotations around diﬀerent axes will not commute.

Lorentz transformations A Lorentz transformation of a four vector xµ =(x0, ~x) is in analogy with (7) given by xµ x′µ = Lµ xν , (13) → ν where

(x′)2 x′µx′ = xµx x2 L µLα = gµ (L−1) ν = Lν . (14) ≡ µ µ ≡ ⇒ α ν ν ⇒ µ µ The Lorentz-group may be intepreted as the set of all 4 4 matrices where × the inverse is equal to its transpose:

O(3, 1) = L ; Lµ R ; LT L = 1 , (15) L ≡ ν ∈ · n o Special cases are the boosts L~e(χ) an hyperbolic angle χ along an axis, for instance the x-, y-, or z -axis:

c s 0 0 c 0 s 0 c 0 0 s χ − χ χ − χ χ − χ  sχ cχ 0 0   0 1 0 0   0 10 0  Lx = − , Ly = , Lz = (16) 0 0 10 sχ 0 cχ 0 0 01 0    −     0 0 01   0 0 0 1   s 0 0 c       χ χ       −  where c coshχ og s sinhχ, and further χ ≡ χ ≡ coshχ = γ = (1 β2)−1/2 ; sinhχ = βγ, β v/c , (17) − ≡ where v is the speed for the boost and c is the speed of light. Transformations which exclude space and time inversion are within the more restictive group SO(3, 1). For boosts along the same axis along ~e is

L (χ ) L (χ ) = L (χ + χ ) , (18) ~e 1 · ~e 2 ~e 1 2 Which follows from addition of hyperbloic angles.

coshφ1 coshχ2 + sinhχ1 sinhχ2 = cosh(χ1 + χ2) ,

coshχ1 sinhχ2 + sinhχ1 coshχ2 = sinh(χ1 + χ2) . (19)

3 The speed v for the total boost corresponding to the hyperbolic angle χ = χ1 + χ2 will be β + β v + v v = c β = c tanh(χ + χ ) = c 1 2 = 1 2 (20) · · 1 2 1+ β β 1+ v v /c2 1 · 2 1 2 Boosts along the same axis commute (They constitute an Abelian sub group). In general, boosts along diﬀerent axes will not commute. The set of all boosts is not a group in the mathematical sense, that is, the product of two boosts is not a pure boost. Rotations constitute (in the matematical sense) a sub-set of the Lorenzt- group. As 4 4 matrices, they are given by × 0 0 k k k L 0 =1 , L k = L 0 = 0 , L l = R l = Rkl (21)

Lie Algebra The group elements in a Lie group may be written

X = exp i αk Mk , (22) ( ) Xk where αj ; j =1, 2, ..., N is a set real parameters. and the generators

∂X M = i , (23) k − ∂α k !{αj =0} satisﬁy a Lie algebra (sum over m is understood):

[Mk , Ml] = ifklmMm , (24) where fklm is a set coeﬃcients (structure constants of the group) completely antisymmetric in the indices klm. If Mk is a set of generators for a group, we may obtain a new set of generators{ } M ′ (-which also satisﬁes the Lie algebra in (24)) by means of a { k} similarity transformation:

M M ′ = S M S−1 , (25) k → k k where S is an n n matrix (with existing inverse). Because × X = eY detX = eTrY , (26) ⇒ we obtain from (when) detX = 1 that

T rMk = 0 , (27) for all k =1, 2, ..., N.

4 Lie-algebra for the rotation group We ﬁnd the generators for the rotation group from the expression for an inﬁnitesimal rotation around an axis along the unit vektor ~n: R (δφ) = 1 + δφ i ~n J~ + ((δφ)2) (28) ~n · O From this we obtain the generators for rotations around the x, y- and the z-axis: 00 0 0 0 i 0 i 0 − J1 =  0 0 i  ,J2 =  0 00  ,J3 =  i 0 0  (29) 0 i −0 i 0 0 0 0 0          −    which can also be written: . (J ) = iε , (30) k ln − kln which gives the well known algebra which is the same as for angular momenta h¯J~: [Jk ,Jl]= iεkln Jn . (31) A rotation around the unit vector ~n may be written: R = exp(iφ ~n J~) = 1 + i sinφ ~n J~ +(cosφ 1)(~n J~)2 (32) · · − · where we have expanded the exponential function , used (~n J~)3 = ~n J~ , (33) · · and then summed the series. The above expressions for Ji are not those usually used in quantum me- chanics, where one most often uses the spherical notation with a diagonal s J3 = J3 , which can be obtained by means of a transformations as in (25):

0 1 0 0 i 0 10 0 1 1 − J s = 1 0 1 ,J s = i 0 i ,J s = 00 0 ,(34) 1 √   2 √   3   2 0 1 0 2 0 i −0 0 0 1            −  which will also satisfy the Lie-algebra.

Lie-algebra for the Lorentz group We find the generators for boosts from the expressions for an infinitesimal boost along the unit vector ~e: R (δχ) = 1 + δχ i~e K~ + ((δχ)2) . (35) ~e · O which gives the generators for boosts along the x, y- and z-xis: 0 i 0 0 0 0 i 0 0 0 0 i i 0 0 0 0000 0000 K =   ,K =   ,K =   (36) x 0000 y i 0 0 0 z 0000        0000   0000   i 0 0 0              5 Multiplying these we find the Lie-algebra for the complete Lorentz group:

[J , J ]= iǫ J , [J , K ]= iǫ K , [K , K ]= iǫ J , (37) k l klm m k l klm m k l − klm m where Jk are the rotation generators and Kl are the boost generators (k,l,m = 1, 2, 3). In the Minkowski representation, the rotasjon generator are in prin- ciple 4x4 matrices, but with only zeros for all time components. A pure boost along the unit vector ~e may be written

L = B = exp(iχ~e K~ ) = 1 + i sinhχ~e K~ (coshχ 1)(~e K~ )2 (38) ~e · · − − · where we have used

(~e K~ )3 = ~e K~ ; coshχ = (1 v2/c2)1/2 (39) · − · − From and (38) we recover the transformation matrices in (16).

Connection between SU(2) and spatial rotations Given an arbitrary three vector ~x. We deﬁne the 2x2 matrix

S ~x ~σ detS = ~x2 . (40) ≡ · ⇒ − (The determinant is easily found by means of the standard expressions for the Pauli matrices). Now we let S transform after Uǫ SU(2) :

S S′ = USU † . (41) → Then S′ = x~′ ~σ detS′ = (x~′)2. (42) · ⇒ − (One might think that S′ also might contain a term proportional to the unit matrix; but because the Pauli matrices are traceless and Tr(S′) = Tr(U † US)= Tr(S) this is excluded). On the other hand,

detS′ = detU detS detU † = detU 2 detS = detS , (43) · · | | · such that the transformation (41) generates a rotation of ~x, conserving its lenght: 2 x~′ = ~x2 x′ = R x . (44) → i ij j From (7), (40) and (42) and Tr(σiσj)=2δij we ﬁnd the coeﬃcients 1 R = Tr σ Uσ U † . (45) ij 2 i j Note that +U and U give the same rotation, showing that SU(2) and SO(3) are not completely− equivalent. The connection between SU(2) and the rotation group means that gruppa betyr at

U U = U R(U ) R(U ) = R(U ) , (46) 1 · 2 3 ⇒ 1 · 2 3 6 which shows that SO(3) is a representation of SU(2). If we write i U = exp( θ ~n ~σ)= cos(θ/2) + i~n ~σsin(θ/2) , (47) 2 · · this represents a rotation an angle θ around the direction ~n (we have ~n = 1). | | In this case we ﬁnd

R = cosθ δ + (1 cosθ) n n + sinθ ǫ n , (48) ij ij − i j ijk k which is the rotation matrix in three spatial dimensions. Further, if χ is a two component spinor, ~v χ†~σχ will transform as a vector, i.e. ≡ v v′ = R v when χ χ′ = Uχ , (49) i → i ij j →

In the fundamental representation of SU(2) the generators are (k=1,2,3): 1 L = σ S(1/2) , (50) k 2 k ≡ k which corresponds to spin 1/2. Higher representations (s = 1, 3/2, ...) have the same Lie algebra (s) (s) (s) Sk , Sl = i ǫklmSm , (51) h i where ǫklm is the Levi-Civita tensor in three dimensions. But other matrix 2 relations are different for different spin. For example, σj = 1 is only valid (1) 2 (1) (1) for spin 1/2. For spin 1, (Sj ) = 1, but (Sj ) = Sj . The spin 1 representation of SU(2) corresponds6 to the fundamental representation of the rotation group, often given in the cartesian representation. (S(1)) = iǫ . k ln − kln Connection between SL(2C) and Lorentz transformations The group SL(2C) is defined by

SL(2C)= A =2 2 matrix ; A C , detA =1 . (52) { × kl ∈ } The group SL(2C) is related to the Lorentz group in the same way as SU(2) is related to the rotation group. Let

T x σµ detT = xµx x2 , (53) ≡ µ ⇒ µ ≡ where we have deﬁned

σµ (1, ~σ) , and similarly one deﬁnes σµ ( 1, ~σ) . (54) ≡ ≡ − Now, let T transform as A SL(2C) : e ∈ T T ′ = AT A† . (55) → 7 Then we have T ′ = x′ σµ detT ′ =(x′)2. (56) µ ⇒ On the other hand,

detT ′ = detA detT detA† = detA 2 detT = detT , (57) · · | | · such that the norm (“length”) of the four vector is conserved:

(x′)2 = x2 . (58)

Then we ﬁnd in analogy with (45) : 1 Lµ = Tr σµAσ A† . (59) ν 2 ν The connection between SL(2C) ande the Lorentz group means that

A A = A L(A ) L(A ) = L(A ) , (60) 1 · 2 3 ⇒ 1 · 2 3 i.e the Lorentz group is a representation of SL(2C). Further, the quantity vµ χ†σµχ will transform as a four vector, i.e.. ≡ v v′ = L ν v when χ χ′ = Aχ , (61) µ → µ µ ν → and χ is a two component spinor. In the SL(2C) representation, the generators are given by 1 i J~ = ~σ and K~ = ~σ (62) 2 2 One can show that an arbitrary element of SL(2C) can be written as

A = H U where H† = H and U † = U −1 . (63) · which means that any Lorentz transformation might be written as a boost times a rotation. Furthermore, H can be written in analogy with U in (47): 1 H = exp( iψ ~e K~ ) = exp( ψ ~e ~σ)= cosh(ψ/2) + ~e ~σsinh(ψ/2) . (64) − · 2 · ·

One may show that all elements in SL(2C) may be factorized in a boost and a rotation. From the connection between SL(2C) and the Lorentz group, we obtain the Lorentz transformations L:

L(A) = L(H) L(U) , (65) · where L(H)= B(H) is a pure boost and L(U)= R(U) is a pure rotation.

8 Proof: From an arbitrary element A SL(2C) we deﬁne a new matrix F : ∈ F A A† F † =(A†)† A† = A A† = F . (66) ≡ ⇒ Because F is hermittean and positive deﬁnite it must have real and positive eigen values f1 og f2,and there must exist a matrix V which transform F to diagonal form:

f 0 F F ′ = VFV † = 1 . (67) → 0 f2 !

Thus the square root of this matrix must also be hermittean:

√f 0 H′ √F ′ = 1 , (68) ≡ 0 √f2 !

Transforming back we obtain:

H = √F = V † H′ V . (69)

We deﬁne

U H−1 A U U † = H−1 A A† H−1 = H−1 (HH) H−1 = 1. (70) ≡ ⇒ From the practical side, the point is that 2 2 matrices in SL(2C) are × simpler to handle than the 4 4 matrices L in the Minkowski representation, such that Lorentz transformations× may be factorized by meansofv SL(2C). To do this factorization explicite, we must of course ﬁnd H−1 from F .