Representations of the Poincaré Group

Transcription

1 Representations of the Poincaré Group Ave Khamseh Supervisor: Professor Luigi Del Debbio, University of Edinburgh Last updated: July 28, 23

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3 Contents The Poincaré group. Characterisation of the Poincaré group The Poincaré algebra Connection with one-particle states Little group Massive case Massless case Fields with spin, /2, 3 2. Representations of the Lorentz group using SU(2) SU(2) Left and right handed spinor (Weyl) fields Manipulating spinor indices The Dirac equation Supersymmetry Extension to graded algebras Introducing supersymmetric algebras Irreducible representations of Supersymmetry algebra Massless case Massive case iii

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5 Chapter The Poincaré group. Characterisation of the Poincaré group In this section, we will study the characterisation and the properties of the Poincaré group. We start from µ,theminkowskimetric, µ = B A = µ (.) and work with natural units i.e. c = ~ = Any transformation that satisfies the following condition is a Lorentz transformation The Lorentz condition can also we written as µ µ = (.2) µ = µ (.3) Proof: Multiply equation.2 by µ µ ( )= = = = µ µ = µ µ (.4) which one may relabel in order to get equation.3. The aim here is to show that these transformations form a group, i.e. we need to prove the group properties; closure and the existence of the identity element and that of the inverse.

6 2 CHAPTER. THE POINCARÉ GROUP Closure: First perform a coordinate transformation x µ = µ x + a µ where satisfies the Lorentz condition.2 and then a second one as follows x µ = µ x +ā µ = µ ( x + a )+ā µ = ( µ ) {z } x + ( µ x +ā µ ) {z } Lorentz transf another translation (.5) When µ and µ both satisfy the Lorentz condition, so does their product and hence we have the composition rule for these transformations: Identity: The identity transformation is T (, ). Proof: Using the composition rule.6 we have as required. Inverse: The inverse trasformation is T (, a) Proof: Again using equation.6 as required. T (, ā)t (,a)=t (, a +ā) (.6) T (, )T (,a)=t (,a) (.7) T (, a)t (,a)=t (, a a)=t (, ) (.8) To every transformation T, there corresponds a unitary linear operator that acts on vectors in Hilbert space such that it takes! U(,a) and obeys the composition rule With identity and inverse elements are U(, ā)u(,a)=u(, a +ā) (.9) U(, ) (.) and U(, a) (.) respectively. The whole group of such transformation is known as the inhomogeneous Lorentz group or Poincaré group. Now we will consider two di erent subgroups of this group. The first one is the group of transformations with no translations present i.e. a µ =. This group is known as the homogeneous Lorentz group or simply the Lorentz group.

7 .. CHARACTERISATION OF THE POINCARÉ GROUP 3 Using the Lorentz condition.2, the property of the product of determinants and the fact that det( ) = we have (det ) 2 =) det = ± (.2) For µ and µ we have According to equations.2 and.3 ( ) = (.3) µ µ = =+=) ( ) 2 =+ i i =+ i i (.4) So ( )2 which in turn implies that either + or apple. Also, note that the length of the 3-vector (, 2, 3 )isp ( )2 according to equation.4. Similarly (, 2, 3 ) has length p ( )2 and so the scalar product of the two 3-vectors is bounded, q 3 3 apple ( q q )2 q ( )2 ) ( ) apple ( )2 ( )2 ) q q (.5) ( )2 ( )2 apple ( ) q q ) ( ) ( )2 ( )2 We know the identity is where = and so in order to preserve continuity we need to take ( )2 (see Figure.) and det =. This is to ensure that they have the same sign as the identity since we want to be able to obtain any Lorentz transformation from the identity by a continuous change of parameters. Hence it forms a subgroup of Lorentz transformations know as the proper orthochronous Lorentz group.

8 4 CHAPTER. THE POINCARÉ GROUP Figure.: Allowed region.2 The Poincaré algebra In this section we aim to find information about the symmetries of the Poincaré group using the properties of group elements near the identity. For this purpose, consider the infinitesimal Poincaré transformation µ = µ +! µ, a µ = µ (.6) Using the Lorentz condition.2 to first order in! we must have = µ ( µ +! µ )( +! )= +! +! + O(! 2 ) (.7) but since is symmetric with respect to the change of its two indices! +! =)! =! (.8) i.e.! µ is a rank-2 antisymmetric tensor in four dimensions and so it has 6 independent components with zeros on the diagonal. On the other hand, µ is a 4-vector so it has 4 components. Therefore in order to specify a Poincaré transformation we need independent parameters. For an infinitesimal Lorentz transformation one can make a Taylor expansion as follows U( +!, ) = 2 i! µ M µ + i µ P µ +... (.9) where M µ and P µ are Hermitian operators so that U is a unitary operator. Here is a good point to stop and explore these operators in more detail.

9 .2. THE POINCARÉ ALGEBRA 5 We can make an infinitesimal change in position and time coordinates of a quantum state and then Taylor expand as follows (x +,t+ ) = (x,t)+i ( ir ) {z } ˆP i (i d ) dt {z } Ĥ (.2) In other words, the momentum operator and the Hamiltonian correspond to space and time translations respectively. Putting them together, we form the momentum-energy four vector P µ which corresponds to translations in space-time. By the same analogy, the angular momentum operator is the generator of space rotations (See later (.77) for a more rigorous proof). Now consider the general Lorentz transformations from one frame to another i.e. a boost. We will now prove that boosts are rotations in Minkowski space, i.e. hyperbolic rotations. Consider a boost in the x direction: 8 t vx = (t ) c >< 2 x = (x vt) y (.2) = y >: z = z Since we are working in natural unit, c = and so we can define = v c = v. Therefore the transformation can be written as follows: t t Bx y A = B C Bx ya (.22) z z Since y and z do not change in this transformation, for simplicity, we may only consider t and x from now on. Define the rapidity variable such that tanh =,so = p = p tanh 2 = cosh (.23) tanh = sinh cosh Hence t cosh x = sinh ) sinh = (.24) sinh cosh t x (.25)

10 6 CHAPTER. THE POINCARÉ GROUP which is a hyperbolic rotation, as required. Anicetrick:Consider two contracted tensors! µ M µ. It is possible to prove that when! µ is symmetric (antisymmetric), M µ has to be symmetric (antisymmetric) as well. Let Now contract this with! µ, M µ = 2 (M µ + M µ ) + {z } 2 (M µ M µ ) {z } Symm part M µ AntiSymm part M µ (s) (a) (.26)! µ is symmetric: The antisymmetric part of M vanishes! µ M µ! µ M µ =! µ M µ! µ M µ =! µ M µ! µ M µ = (.27)! µ is antisymmetric: The symmetric part of M vanishes! µ M µ +! µ M µ =! µ M µ +! µ M µ =! µ M µ! µ M µ = (.28) Therefore, going back to equation.9 and knowing that! is antisymmetric, it is immediately implied that M is antisymmetric with respect to the exchange its two indices and so it has 6 independent components. M µ = M µ (.29) We now know that p µ is the 4-momentum operator since it corresponds to translations. Also, M µ is a rotation operator which contains the usual angular momentum 3-vector as well is the boost 3-vector i.e. 6 independent components, as expected. Consider the product U(,a)U( +!, )U (,a)=u( ( +!),! a) (.3) where we have used the composition rule.9 and.. Expanding to first order in! and, U(,a)[ 2! µ M µ + µ P µ ]U (,a)= 2 (! ) M +(! a) P (.3)

11 .2. THE POINCARÉ ALGEBRA 7 Therefore for! on the RHS we have, 2 µ! µ ( ) M µ! µ ( ) a P = 2 µ! µ ( ) M 2 [ µ! µ ( ) a P µ! µ ( ) a P ]= 2 ( µ! µ M + µ! µ a P! µ µ a P )= 2 ( µ! µ M + µ! µ a P! µ µ a P ) (.32) Where, on the second line we have used the fact that w is antisymmetric, on the third line we have relabelled µ and and and on the fourth line. Equating coe cients of! and on both side of the equation we get, U(,a)M µ U (,a)= µ (M a P + a P ) (.33) U(,a)P µ U (,a)= µ P (.34) Note: The equation above is simply stating how the operator P transforms under U, writing it as a linear combination of operators on the right hand side. using. to first order in! and we have, U (,a)=u(, a)=u(( +!), ( +!) ) = U(!, (!) ) =U(!, ) (.35) the LHS of equation.34 becomes and the RHS Setting LHS = RHS, i ( 2! M + i P )P µ ( + i 2! M i P ) =P µ i 2! M P µ + i P P µ + i 2! P µ M i P µ P =P µ + i( 2! M P µ + P P µ + 2! P µ M P µ P ) =P µ + i[ 2! M + i P,P µ ] i[ µ P =( (.36) µ +! µ )P (.37) 2! M + i P,P µ ]=! µ P (.38)

12 =M µ + 2! [ µ M µ M M µ + M µ ] µ P + P µ (.4) 8 CHAPTER. THE POINCARÉ GROUP Similarly for M µ the of equation.33 LHS is, and the RHS, M µ + i[ 2! M + i P,M µ ] (.39) µ M =( µ +! µ )( +! )(M P + P ) =M µ +! µ M +! M µ µ P + P µ =M µ + µ! M +! M µ µ P + P µ =M µ + 2 [ µ! M + µ! M +! M µ +! M µ ] µ P + P µ =M µ + 2 [ µ! M µ! M! M µ +! M µ ] µ P + P µ Equating we get, i[ 2! M + P,M µ ]= 2! [ µ M µ M M µ + M µ ] µ P + P µ (.4) Finally, we extract the coe cients of! and and use the fact that M µ is antisymmetric, [M µ,m ]= i( µ M µ M M µ + M µ ) (.42) and [P,M µ ]=i( µ P P µ ) (.43) On the other hand, equation.38 implies the following, [P µ,p ]= (.44) So we have the momentum, angular momentum and boost 3-vectors, P = P,P 2,P 3 (.45) J = M 23,M 3,M 2 (.46) K = M,M 2,M 3 (.47)

13 .2. THE POINCARÉ ALGEBRA 9 The above commutation relations imply, [J i,j j ]=i ijk J k (.48) [J i,k j ]=i ijk K k (.49) [K i,k j ]= i ijk J k (.5) [J i,p j ]=i ijk P k (.5) [K i,p j ]= ih ij (.52) [J i,h]=[p i,h]=[h, H] = (.53) Now we define Pauli-Lubanski pseudo-vector by [K i,h]= ip i (.54) which has the following properties, W µ = 2 µ M P = 2 µ P M (.55) Proof: Use the fact the P s commute W µ P µ = (.56) W µ P µ = 2 µ M P P µ = 2 µ M P µ P = 2 µm P P µ = 2 µ M P P µ = W µ P µ ) W µ P µ = (.57) [W µ,p ]= (.58)

14 CHAPTER. THE POINCARÉ GROUP Proof: [W µ,p ]= 2 µ ([P,P ]M + P [M,P ]) = 2 µ ([P,P ]M + P [M,P ]) = i 2 µ P ( P P )= i 2 ( µ P P µ P P ) = i 4 ( µ P P + µ P P µ P P µ P P ) = i 4 ( µ P P + µ P P µ P P µ P P )= (.59) Where on the fourth line we have used the fact that P s commute and then relabelled the indices for the second and fourth terms. This commutation relation implies that the Pauli- Lubanski vector is invariant under translations. Now define M µ = 2 µ M (.6) So that W µ = P Mµ. Multiplying both sides of equation.6 by µ one gets µ Mµ = M = 2M (.6) where we have used the identity for the four dimensional Levi-Civita tensor. Therefore, M = 2 µ Mµ (.62) Let us determine how M µ transforms under a Lorentz transformation by considering [ M µ,m ]= 2 [M,M ]= 2 µ [M,M ] = i 2 µ M M M + M = i 2 µ M µ M + µ M µ M = i µ M µ M = i µ ( 2 apple M apple ) µ ( 2 apple M apple ) = i 2 µ apple µ apple M apple = i 2 [ µ ( apple apple ) µ ( apple apple ) µapple ( ) $ ] M apple = i 2 [ 2 µ M Mµ + Mµ M µ + M µ $ ] = i[ µ M µ M + Mµ Mµ ] (.63)

15 .2. THE POINCARÉ ALGEBRA which is the same result as equation.42. In other words, Mµ behaves like a second rank tensor under a Lorentz transformation. Note that on the fifth line we have used the identity for 4-dimensional Levi-Civita tensors: µ apple = µ ( apple apple ) µ ( apple apple ) µapple ( ) (.64) We will use this result to prove the equation below [W µ,m ]= i( µ W µ W ) (.65) Proof: [W µ,m ]=[P Mµ,M ]=P [ M µ,m ]+ [P,M ] M µ = ip ( µ M µ M + Mµ Mµ ) +i P M µ i P Mµ (.66) =i µ W i µ W ip M µ + ip Mµ + ip M µ ip Mµ =i µ W i µ W In other words, W µ transforms as a 4-vector under homogeneous Lorentz transformations. Using the equation above, [W µ,w ]= i µ W P (.67) Proof: [W µ,w ]= 2 [W µ,m P ]= 2 [W µ,m ]P + M [W µ,p ] = 2 apple [W µ,m apple ]P = i 2 apple ( µ W k appleµ W ) p = i µ W µ W P = i µ W + µ W P 2 2 = i µ W P (.68)

16 = 2 M M P 2 + M M P P (.69) 2 CHAPTER. THE POINCARÉ GROUP W 2 = W µ W µ = 4 µ M P µ M P = 4 ( µ µ )M P M P = 4 [ ] M P M P = 4 [ M P M P M P + M P + M P + M P ]M P = 2 M P M P + 2 M P M P + 2 M P M P = 2 M P M P + M P M P = 2 M P M P + M P M P apple h i = 2 M + M M P + i( P + P ) P = 2 M M P 2 i 2 (M P P + M P P ) +M M P P + i(4m P P + M P P ) on the sixth and tenth lines we have made use of the antisymmetry of M and equation.43 on the eighth. W 2 is Lorentz invariant (since it is the length of a 4-vector) and hence commutes with all the other generators. The other Lorentz invariant entity is the mass operator P 2 = P µ P µ which again, commutes with all generators. Therefore, W 2 and P 2 are the Casimir operators for the Poincaré algebra. For a pure translation we have, U(,a )U(,a)=U(,a + a) (.7) Therefore, a total translation can be made using many infinitesimal ones i.e. U(,a)=U(,N a) =U(, a)u(, a)...u(, a) =[U(, a)] N (.7) {z } Ntimes Now, we let N!and expand the right hand side of the equation to first order in a using equation.9 to get U(,a)= lim N! [ + i N P µ a µ ] N (.72)

17 .2. THE POINCARÉ ALGEBRA 3 Figure.2: Rotation Hence the operator for pure translations acting on the Hilbert space is We know that (see Figure.2) U(,a)=e ip µ a µ (.73) x = x +(n x) (.74) is the active rotation of vector x by an angle about axis n. Now consider the case where the coordinate system is rotated by angle rather than the vector itself. This is known as the passive transformation. This transformation is equivalent to an active rotation of the vector by. So for the case of the rotation operator acting on the state vector we have: (x) =U(R( )) (x) = (R ( )x) (.75) and so for an infinitesimal rotation we do a Taylor expansion using the formula below for small a, X (x + a) = n! (a r)n (x) =exp(a r) (x) (.76) n= (x +(n x) ) =exp( (n x) r) (x) =exp(i (n x) ( ir)) (x) =exp(i (n x) p) (x) =exp(i (x p) n) (x) =exp(i n J) (x) =exp(i J) (x) (.77)

18 4 CHAPTER. THE POINCARÉ GROUP where we have used a =(n x). Knowing that any rotation is a combination of many infinitesimal ones, we have Therefore, U( )U( )...U( ) = U(N ) =U( ) (.78) {z } Ntimes U(R( ), ) = e ij (.79).3 Connection with one-particle states The task here is to classify the one-particle states by their transformation under the Poincaré group. First, we need to choose a label for the physical state vector. Since P 2 and W 2 are the Casimir operators for the algebra and so commute with all the other generator, it is possible to label the states by p and where the latter, as we will see later on, represents the spin of the particle. So for the energy-momentum operator P µ p, = p µ p, (.8) According to equation.73, the states transform under translations as U(,a) p, = e ip a p, (.8) Using equation.34, we have P µ U( ) p, {z } =U( )[U ( )P µ U( )] p, = U( )(( ) µ P ) p, = µ p U( ) p, {z } (.82) Therefore, by comparing it to equation.8, one can write U( ), as a linear combination U( ) p, = X C (,p) p, (.83) In general, it may be possible for C to be made block diagonal so that each block furnishes a representation of the Poincaré group.therefore, our goal from now on is to find the irreducible representations of the Poincaré group. Consider the forward light cone, for any 4-vector q µ V + = {q µ : q,q 2 } (.84)

19 .3. CONNECTION WITH ONE-PARTICLE STATES 5 Then Proof: q 2 V + ) q µ = µ q 2 V +, 8 µ 2 SO(3, ) (.85) q µ q µ = µ q µ q =( ) µ µ q q = q q = q q = q 2 (.86) and so the second condition is satisfied. As for the first one, q = q µ = q + q + 2q 2 + 3q 3 (.87) We have the 3-vectors, (, 2, 3 ) which has length p ( )2 by equation.4, and (q,q 2,q 3 ) which has length q. Taking the scalar product q q + 2q 2 + 3q 3 apple q ( )2 ) q q q apple q ( )2 ) q (.88) q ( )2 apple q q ) q q q ( )2 apple q but since we are dealing with proper orthochronous subgroup we must have so > p ( )2. Also, for light-like 4-vector invariant length q 2 =(q ) 2 q 2 and given q by equation.84, it is implied that q q. Therefore, apple q and so the first condition is also satisfied. Q.E.D Note that p 2 and the sign of p (for p 2 ) are left invariant by all proper orthochronous Lorentz transformations. So one can choose a standard 4-momentum k µ and apply a Lorentz boost such that p µ = L µ (p)k µ (.89) and define states with momentum p p, N(p)U(L(p)) k, (.9) where N(p) is a normalisation factor, i.e. a number. When we act on p, with a homogeneous Lorentz transformation U( ) and use the composition rule for the RHS, we get U( ) p, =N(p)U( L(p)) k, =N(p)U(L( p))u(l ( p) L(p)) k, (.9) According to equation.89, the transformation L ( p) L(p), acting on k, takes k! p! p! k (.92)

20 6 CHAPTER. THE POINCARÉ GROUP i.e. this transformation is a member of the subgroup of the homogeneous Lorentz group consisting of Lorentz transformations W µ that leave the standard 4-momentum invariant: This subgroup is known as the little group..3. Little group For any W satisfying equation.93, we can write W µ k = k µ (.93) U(W ) k, = X D (W ) k, (.94) Since W leaves k invariant, the only degree of freedom that can change is. Also, for W,W 2 little group, we have X D ( WW) k, = U( WW) k, = U( W )U(W ) k, =U( W ) X D (W ) k, = X D (W )[U( W ) k, ] = X X D (W )D ( W ) k, (.95) Comparing the LHS with the RHS D ( WW)= X D ( W )D (W ) (.96) This is precisely the definition of the representation of a group. So it can be said that D(W ) furnish a representation of the little group. So for a particular element of the little group we have, using.9 W (,p) L ( p) L(p) (.97) U( ) p, =N(p)U(L( p))u(w (,p)) k, =N(p) X D (W (,p))u(l( p)) k, N(p) X = D (W (,p)) p, N( p) (.98)

21 .3. CONNECTION WITH ONE-PARTICLE STATES 7 where on the third line, we have used equation.9. Remember that the aim is to find the coe cients C. So we need to find the representation of the little group i.e. D(W ) and the normalisation factors. In other words, the representations of the Poincaré group are being derived using the representations of the little group. This method is known as the method of induced representations. Table. summarises di erent classes of 4-momenta and the corresponding little group element. Table.: Momenta and Little Group Particle Type Condition Standard k µ Little Group (a)massive p 2 = M 2 >,p > (M,,, ) SO(3) (b)massive p 2 = M 2 >,p < ( M,,, ) SO(3) (c)massless p 2 =,p > (apple,,, apple) ISO(2) (d)massless p 2 =,p < ( apple,,, apple) ISO(2) (e)tachyonic p 2 = N 2 < (,,,N) SO(, 2) (f)vacuum p µ = (,,, ) SO(, 3) We will now briefly discuss each case. The connection between the little group and the Pauli- Lubanski vector defined in section.55 and a more detailed explanation of cases (a) and (c) will be presented in the next two sections. Case (a): Since we are in the rest frame of a massive particle, all ordinary three dimensional rotations leave k µ invariant. Therefore the little group is all rotations in the three spacial dimensions i.e. SO(3). Case (b): This is not a physical state because the energy is negative. Since state of lowest energy is always more favourable, we keep decreasing the energy down to which is not physical. So we define the vacuum state to be at zero. The rotation group is again, SO(3) for the same reason as above. Case (c): Covers particles of mass zero, since the mass operator p 2 =. The energy p is more than zero. Therefore it represents a physical state. As well as that, the little group is the group of Euclidean Geometry consisting of rotations and translations in two dimensions, i.e. ISO(2).

22 8 CHAPTER. THE POINCARÉ GROUP Case (d): Again, it does not represent a physical state since the energy p <. Case (e): This is the space-like case and so it does not represent a physical state. The little group is the group of rotations that that keep the third direction fixed i.e. 2+ dimensional rotations keeping the third axis fixed SO(, 2) Case (f): This is a physical state and it corresponds to the vacuum. The standard 4-vector is at the origin and so the all rotations in space-time leave k µ invariant. Hence the little group is SO(, 3). Before we proceed to building the representations of the Poincaré group for the massive (a) and massless (c) cases, let us consider the normalisation of the states. We choose the states with standard momentum k µ to be orthonormal. Form quantum mechanics, we know that the scalar product of the eigenstates of a Hermitian operator with eigenvalues k and k is h k, k, i = 3 (k k) (.99) Therefore, using.94 and.98, it can be said that the representations of the little group are unitary: D (W )=D (W ) (.) However, for arbitrary momenta the scalar product is h p, p, i.9 = h p, N(p)U(L(p)) k, i = N(p)hU (L(p)) p, k, i =N(p)hU (L(p)) p, k, i = N(p)hU(L (p)) p, k, i N (p ) = N(p) N (L (p)p ) hx D (W (L (p),p )) L (p)p, k, i =N(p) N (p ) X N (k D (W (L (p),p ))h k ), k, i = N(p)N (p ) X D (W (L (p),p )) (k k) (.) =N(p)N (p )D (W (L (p),p )) (k k) where on the fifth line we take N (k) = and on the sixth line the fact that the delta function only picks up the term = from the sum. For p = p the transformation W (L (p),p) = and the scalar product of the two states is h p, p, i = N(p) 2, (k k) (.2) The remaining task now is to find the proportionality factor relating (k k) and (p p).

23 .3. CONNECTION WITH ONE-PARTICLE STATES 9 Consider the integral of an arbitrary scalar function f(p) over 4-momenta with p 2 = M 2 and p > i.e. (a) and (c), Z Z d 4 p (p 2 M 2 ) (p )f(p) = d 3 p dp ((p ) 2 p 2 M 2 ) (p )f(p, p) Z = d 3 p dp ((p ) 2 ( p p 2 + M 2 ) 2 ) (p )f(p, p) Z = d 3 p dp 2 p p 2 + M [ p 2 (p p 2 + M 2 )+ (p + p p 2 + M 2 )] (p )f(p, p) (.3) Z = d 3 p dp 2 p p (p p 2 + M 2 )f(p, p) p 2 + M 2 Z = d 3 p 2 p p 2 + M 2 f(p p 2 + M 2, p) where on the second line we have used the identity (x 2 x 2 )= 2 x ( (x x )+ (x + x )) (.4) the appearance of the delta function (p 2 M 2 ) is due to the condition p 2 = M 2. As well as that, the step function (p ) is required to impose the condition p > since (x) = for x, (x) = for x apple. This integral is invariant under Lorentz transformations i.e. it is a Lorentz scalar. The reason for this is that f(p) is a scalar and so Lorentz invariant. Also, d 4 d 4 p (.5) {z } is a Lorentz invariant. Finally, the step function (p ) also remains invariant under Lorentz transformations since the sign of p > does not change. p 2 M 2 is known as the mass shell and the invariant volume element is d 3 p p p 2 + M 2 (.6) By the definition of the delta function Z F (p) = F (p ) 3 (p p )d 3 p Z = F (p )[ p p 2 + M 2 3 (p d 3 p (.7) p)] p p 2 + M 2

24 2 CHAPTER. THE POINCARÉ GROUP We know that the LHS is invariant and so must be the RHS. We also know that the volume element is invariant, and so the term in the brackets i.e. p p 2 + M 2 3 (p p)=p 3 (p p) (.8) does not change from going from one frame to another, then p 3 (p p)=k 3 (k k) (.9) Hence, substituting into equation.99 h p, p, i = N(p) 2 p k 3 (p p) (.) So in order to normalise, we take N(p) tobe N(p) = s k p (.) In the two sections that follow, we will derive the representations of Poincaré group for massive and massless cases (a,c) and find the connection between the corresponding little group and the Pauli-Lubanski vector..3.2 Massive case Going back to the definition of the Pauli-Lubanksi vector.55, in the rest frame of the particle we have the standard momentum k µ =(M,,, ) and so W =; W i = MJ i (.2) i.e. the spacial components of W are proportional to J i which are in turn the generators of the little group SO(3). In general, J = orbital angular momentum + intrinsic angular momentum (spin). However, since we are now in the rest frame of the particle i.e. p =, J is in fact the total spin of the particle. W 2 = W µ W µ = MJ 2 = Ms(s + ) (.3) when acting on a state with total spin s and mass M. The representation is therefore labelled by (M,s). We will write J S and s =, 2,,... and = s,..., s. There are 2s + possible values for. The set of 2s + states { s, i} is called a multiplet.

25 .3. CONNECTION WITH ONE-PARTICLE STATES 2 Acting with S i (or more generally J i ) on this orthonormal set of vectors, results in a linear combination of { s, i} vectors, i.e what we have here is an irreducible representation which is 2s + dimensional. In other words, the unitary representation of the little group SO(3) can be broken up into a direct sum of the irreducible unitary representations D (j) (R). These representations are built up using matrices for infinitesimal rotations R ik = ik + ik where ik is antisymmetric with respect to the exchange of its two indices. From Quantum Mechanics, and We also know that D (s) ( + ) = and the factors N ± can be found to be Hence, S (s) ± = S (s) 2 S (s) 23 ± S(s) 3 i 2 ik(s (s) ik ) (.4) S 2 s, i = s(s + ) (.5) S 3 s, i = s, i (.6) S ± s, i = N ± s, i (.7) N ± = p (s )(s ± ± ) (.8) = S (s) 3 = hs, S 3 s, i = (.9) p = hs, S ± S 2 s, i =, ± (s )(s ± + ) (.2) Hence the representation of the group elements obtained by taking the exponential of the generators is D (s) ( ) = hs, e i S s, i (.2) For a particle of mass M> and spin s, equation.98 now reads U( ) p, = s ( p) p X D (j) (W (,p)) p, (.22) where W (,p)=l ( p) L(p) is known as the Wigner rotation. In order to calculate this rotation, one can apply a standard boost L(p) p p p ˆp 2 ˆp2 2 ˆp3 2 p ˆp 2 L(P )= +( )ˆp 2 B p ( )ˆp ˆp 2 ( )ˆp ˆp ˆp 2 2 ( )ˆp2 ˆp +( )ˆp 2 C (.23) p 2 ( )ˆp 2 ˆp 3 A ˆp 2 3 ( )ˆp3 ˆp ( )ˆp 3 ˆp 2 +( )ˆp 2 3

26 22 CHAPTER. THE POINCARÉ GROUP Figure.3: direction q such that ˆp i = p i p with = p 2 +M 2 M. (For a more detailed discussion of how L(p) is actually derived, see the proof below.) Note that when µ is an arbitrary 3-dimensional rotation R, the Wigner rotation W (,p)= R, 8p. Proof: We would like to go from a frame where the particle has momentum k, i.e. the rest frame, to a frame where the particle is viewed to have momentum p. We first make a boost, say in the third direction: p 2 B( p ) = B p A (.24) 2 where we have used the fact that = p ) p 2 =. 2 However, in order to get to the more general form of L we need to apply a 3-dimensional spacial rotation that takes us from the third direction into the direction of ˆp according to Figure.3,

27 .3. CONNECTION WITH ONE-PARTICLE STATES 23 L(p) =R(ˆp )B( p )R (ˆp ) (.25) with R(ˆp ) of the form R(ˆp )= R 3 3 (ˆp i ) C A (.26) i.e. a rotation by some angle about axis n, normal to the plane of z and ˆp with n = ẑ p p and cos = ẑ p p. Then for an arbitrary rotation = R, one can use equation.97 W (R,p)=L (Rp)RL(p) = R(Rˆp )B( p )R (Rˆp ) RR(ˆp )B( p )R (ˆp ) =R(Rˆp )B ( p ) R (Rˆp )RR(ˆp ) B( p )R (ˆp ) {z } R( ) (.27) Note that R (Rˆp )RR(ˆp ) takes the 3-axis into the direction of ˆp then to Rˆp and finally back to the 3-axis. Since the 3-axis remains fixed under this transformation, it must be equivalent to a rotation by some angle about the 3-axis i.e. R (Rˆp )RR(ˆp )=R( ) = B cos sin sin cos A (.28) It can easily be verified that R( ) commutes with B( p ) which implies that W (R,p)=R(Rˆp )B ( p )R( )B( p )R (ˆp ) =R(Rˆp )R( )R (ˆp ).28 = R (.29) Hence W (R,p)=R (.3) Q.E.D

28 24 CHAPTER. THE POINCARÉ GROUP Therefore, it can be concluded that under rotations, the states of a moving massive particle have the same transformations as that of the non-relativistic case. This can also be extended to multi-particle states..3.3 Massless case Consider the standard 4-momentum k µ =(,,, ) satisfying equation.93. When such a Lorentz transformation acts on a time-like 4-vector t µ =(,,, ), it results in a 4-vector Wt such that its Length: (Wt) µ (Wt) µ = t µ t µ = Scalar product with k: (Wt) µ k µ = t µ k µ = The second condition is satisfied by any 4-vector of the form The first condition then implies that (Wt) µ =(+,,, ) (.3) = (.32) Therefore, it can be said that W µ acting on t has the same e ect as the Lorentz transformation S µ (, ), where S is S µ (, )= + C A (.33) But the question is how this matrix is actually derived. From equation.3, it can easily be observed that the first column of such a 4 4 matrix must be as above. Also when S acts on k it must leave it invariant and hence the last column. Now it remains to find the other two columns. We will use the fact that S is a Lorentz transformation µ S µ S = (.34) and as a result we may impose two conditions. The first one is that the length of each column must be invariant and equal to. The second one is that the columns must be orthonormal to each other. We may also choose the block in the middle to be the identity and calculate the rest of the entires based on that. So are task is to find x, y, u, v

29 .3. CONNECTION WITH ONE-PARTICLE STATES 25 For example, one can try = S µ (, )= + x? u? y? v? = 2 Then, using the first condition C A (.35) =S 2S 2 S 2S 2 S 2 2S 2 2 S 3 2S 3 2 =u v 2 ) u = ±v (.36) Similarly by taking = = one gets x = ±y. Imposing the second condition and setting =2, = 3 and then =, = 3 we get x = y = and u = v =. Hence the result. We mentioned that acting on t, this matrix has the same e ect as W. This does not mean that they are equal but it does imply that S (, )W is a Lorentz transformation that leaves the vector (,,, ) invariant. Therefore, it can only be a pure rotation. As shown above, S µ also leaves (,,, ) invariant, i.e. the 3-axis remains fixed and so S (, )W must be a rotation by an arbitrary angle about the 3-axis where S (, )W = R( ) (.37) R µ ( ) = B cos sin sin cos A (.38) From equation.37, it can be concluded that the most general element of the little group has the form W (,, )=S(, )R( ) (.39) The aim now is to identify this group. It can be checked that for those transformations with =wehaver = and and for those with = =) S =, S(, )S(, )=S( +, + ) (.4) R( )R( ) =R( + ) (.4) which is also additive. These transformations forming the two subgroups above are Abelian i.e. their elements commute with each other. As well as that, this first subgroup mentioned

30 26 CHAPTER. THE POINCARÉ GROUP is invariant such that its elements are transformed into other elements of the same subgroup by any element of the group R( )S(, )R ( ) =S( cos + sin, sin + cos ) (.42) which can easily be verified by simple matrix multiplication. Therefore the product of any group elements can be calculated using the above equations. These multiplication rules can be identified as those of the little group ISO(2) i.e. the group of Euclidean Geometry consisting of rotations (by angle ) and translations (by vector (, )) in two dimensions. Here is good point to stop and turn our attention to the Pauli-Lubanksi vector. Starting from the standard 4-momentum k µ =(apple,,, apple) and the Pauli-Lubanski vector we have, W = 2 µ 3M µ k 3 = applem 2 = applej 3 = W 3 (.43) W = apple(m 23 + M 2 )= apple(j + K 2 ) (.44) W 2 = apple(m 3 + M )= apple(j 2 K ) (.45) from which the following can be verified: [W,W 2 ] = (.46) [J 3,W ]=apple[j 3, J K 2 ]=iapple( J 2 + K )=+iw 2 (.47) [J 3,W 2 ]=apple[j 3, J 2 + K ]=iapple(j + K 2 )= iw (.48) which is in fact the algebra of the little group ISO(2) for the massless case. Groups with no invariant Abelian subgroups are called semi-simple groups. They have certain simple properties, hence the name. ISO(2), like the Poincaré group, is not semi-simple. We will reproduce the Lie Algebra.46 to.48 using the general little group element W and infinitesimal transformations. W (,, ) µ = µ +! µ, (.49) On the other hand, W (,, ) = S(, )R( ) sousing.33 and.38 and making the approximation cos( ), sin( ) and = for infinitesimal,,, we get! µ = C A (.5)

31 .3. CONNECTION WITH ONE-PARTICLE STATES 27 which is antisymmetric and has 6 independent components. and so the corresponding operator U(W (,, )) = i A i B i J 3 (.5) where A = J 3 + J = J 2 + K (.52) are Hermitian operators. Then using equations.48 to.5 B = J 23 + J 2 = J + K 2 (.53) [J 3,A]=+iB (.54) [J 3,B]= ia (.55) [A, B] = (.56) which is precisely the same algebra expected using the Pauli-Lubanksi vector mentioned above, i.e. the algebra of ISO(2). Because operators A and B commute, they can be simultaneously diagonalised. A k,a,b = a k,a,b B k,a,b = b k,a,b (.57) Consider the transformation of A and B, by observing equation.42 and remembering that the coe cients of A and B in equation.5 are and respectively, we can find the transformation of A and B to be as follows U[R( )]AU [R( )] = A cos B sin U[R( )]BU [R( )] = A sin + B cos (.58) This can also be derived directly by using the matrix form of J 3, A and B, i.e. J 3 = B i i A (.59) i A = B i A + B i A i {z } {z } J 2 K (.6)

32 28 CHAPTER. THE POINCARÉ GROUP Similarly i B = B ia + B i A i {z } {z } J K 2 and simply going through the multiplication U[R( )]AU [R( )] i B cos sin C B i i C B cos sin sin cos sin cos A = A cos i as expected. The calculation is similar for B. (.6) B sin (.62) Define k,a,b U (R( )) k,a,b. Then for A, we have Similarly U[R( )]AU [R( )] k,a,b =(A cos B sin ) k,a,b ) U[R( )]A k,a,b =(acos b sin ) k,a,b ) (.63) A k,a,b =(acos b sin ) k,a,b B k,a,b =(a sin + b cos ) k,a,b (.64) The problem is that if one finds one set of non-zero eigenvalues of A and B, one finds a whole continuum. However, this is in contrast with the observations that suggest massless particles have no continuous degree of freedom. As a result other restrictions need to be imposed. We must require the physical states to be eigenvectors of A and B with eigenvalues a=b=. A k, = B k, = (.65) Note that here we have labeled the states by k, since a and b are now fixed. Now, in order to distinguish these states from each other, we need to consider the eigenvalues of the remaining generator J 3. J 3 k, = k, (.66) Since the standard 4-momentum k =(,,, ), the 3-momentum vector k =(,, ) is in the the 3-direction which means that is the component of angular momentum in the direction of motion and it is called helicity.

33 .3. CONNECTION WITH ONE-PARTICLE STATES 29 Therefore, it can be said that the states are identified by the momentum in the 3-direction and the angular momentum in the direction of motion. Since U(W (,, )) = i A i B i J 3 for infinitesimal,,, for finite and we have and for finite U(S(, )) = e i( A+ B) (.67) U(R( )) = e ij 3 The little group element W can then be written as follows using equation.39 (.68) U(W ) k, = e i( A+ B) e i J 3 k, = e i k, (.69) where we have used the fact that the A and B have zero eigenvalues when acting on. Hence, according to equation.94 D (W )=e i (.7) Then, using equations.98 and. the Lorentz transformation rule for a massless particle with helicity is given by s ( p) U( ) p, = p e i (,p) p, (.7) where (,p)isdefinedby W (,p) L p L(p) S( (,p), (,p))r( (,p)) (.72) The allowed values of are restricted to integers and half integers due to topological considerations. In order to calculate the little group element W we need to set a standard Lorentz transformation that takes k µ =(apple,,, apple)! p µ. We choose it such that it has the following form L(p) =R(ˆp )B( p /apple)r (ˆp ) (.73) with B(u) being a pure boost in the 3-direction: (u 2 + )/2u (u 2 )/2u B(u) (u 2 )/2u (u 2 + )/2u C A (.74)

34 3 CHAPTER. THE POINCARÉ GROUP One can check that this is matrix a a Lorentz transformation (boost) by comparing it with the general boost along the 3-axis C A (.75) where we must have =. In our case, we may set u2 + 2u and u2 2u,then it is easy to verify that the required relation ( u2 + 2u )2 ( u2 2u )2 = holds as required by a Lorentz transformation. On the other hand R(ˆp ) is a pure rotation that takes the 3-axis into the direction of motion ˆp. For example in polar coordinates ˆp =(sin cos, sin sin, cos ) (.76) then taking R(ˆp ) to be a rotation by angle about the 2-axis, taking (,, )! (sin,, cos ), followed by a rotation about the 3-axis by angle : U(R(ˆp )) = e i J 3 e i J 2 (.77) with apple apple and apple < 2. Note that the reason we give U(R(ˆp )) rather than R(ˆp )is that shifting or by 2, leads to the same result for R(ˆp ) but picks up a minus sign for U(R(ˆp )) when it acts on states with half-integer spins. Since.77 is a rotation that takes the 3-axis into direction ˆp, any other choice of R(ˆp )willbedi erent from this one by at most an initial rotation about the 3-axis (because it keeps this axis fixed) which means that one-particle states merely di er by a phase factor. It is very important to note that helicity is Lorentz invariant i.e. a massless particle of helicity, say, looks the same (aside from its momentum) in all inertial reference frames. The reason lies in equation.7. It can be observed that remains invariant. This is in contrast with the massive case.22 were s mix under the Lorentz transformation and hence it is not invariant and it transforms among the 2s + possible values.

35 Chapter 2 Fields with spin, /2, The aim of the chapter is to construct Lorentz-invariant field theories using fields of spin, /2 and in connection with the representation of the Lorentz group. We will then discuss spinor fields in more detail and learn how to manipulate spinor indices. 2. Representations of the Lorentz group using SU(2) SU(2) Let U( ), just as before, a be unitary operator corresponding to a Lorentz transformation. A scalar field operator transforms under a Lorentz transformation as follows The derivative of this field transforms as U( )'(x)u ( ) ='( x) (2.) U( )@ µ '(x)u ( ) = '( x) (2.2) where we put a bar at the top to indicated that it is the derivative with respect the argument x = x This implies that a vector field A µ and a tensor field B µ transform as U( )A (x)u ( ) = µ A ( x) (2.3) U( )B µ (x)u ( ) = µ B ( x) (2.4) respectively. Note that the symmetry and antisymmetry properties of B µ are preserved under Lorentz 3

36 32 CHAPTER 2. FIELDS WITH SPIN, /2, transformations. Proof: Let B µ = B µ then B µ! B µ = µ B = µ B = µ B = B µ (2.5) and so the symmetry is preserved. The antisymmetric case is similar. Q.E.D We define the trace of B to be T (x) µ B µ. This entity transforms like a scalar field under Lorentz: U( )T (x)u ( ) =T ( x) (2.6) Proof: Using the Lorentz condition, µ = µ, T (x) = µ B µ = µ B µ = B (2.7) Therefore, one can write an arbitrary tensor B µ as B µ (x) =A µ (x)+s µ (x)+ 4 µ T (x) (2.8) where A µ is an antisymmetric tensor and hence traceless, S µ is symmetric and traceless i.e. taking µ S µ =. It is important to notice that the fields A µ,s µ and T do not mix with one another under Lorentz transformations. This is because we just proved that the symmetry properties are preserved under Lorentz transformations and T is a scalar while the other two are tensors. The main question now is that whether this can be further decomposed into smaller irreducible representations. We will come back to this question later on in this chapter. Consider a field ' A (x) wherea is a Lorentz index. It transforms under Lorentz according to U( )' A (x)u ( ) =LA B ( )' B ( x) (2.9) where LA B ( ) is a finite dimensional matrix that depends on so that it results in a linear combination on the RHS of the equation. The group composition rule is obeyed L B A ( )LB C ( ) =LA C ( ) (2.) Therefore, LA B ( ) furnish the representation of the homogeneous Lorentz group, as seen before. Again, for and infinitesimal transformation µ = µ +! µ U( +!) = i 2! µ M µ (2.)

37 2.. REPRESENTATIONS OF THE LORENTZ GROUP USING SU(2) SU(2) 33 with M µ being the generators of the Lorentz group and! µ being antisymmetric. As shown in the first chapter, the Lie algebra of the Lorentz group is specified by [M µ,m ]= i( µ M µ M M µ + M µ ) (2.2) From which the relations (.48 to.5) between the angular momenta J i 2 ijkm jk and boosts K i M i are derived. For an infinitesimal trasformation, Then, equation 2.9 implies LA B ( +!) = A B i 2! µ (S µ ) B A (2.3) [M µ, ' A (x)] = L µ ' A (x)+(s µ ) B A ' B (x) (2.4) where L µ = i(x µ ). Note that when µ and are the spacial coordinates, L µ corresponds to orbital angular momentum. Proof: The LHS of 2.9 to first order in! reads ( i 2! µ M µ )' A (x)( + i 2! µ M µ )=' A (x)+ i 2! µ [' A (x),m µ ] (2.5) On the RHS of 2.9, we need make a taylor expansion: ' B (x!x) =' B (x)! µ µ ' B (x), then [ B A =' A (x) We now make use of the antisymmetry of! µ i 2! µ (S µ ) A B ](' B (x)! µ µ ' B (x))! µ µ i ' A (x) 2! µ (S µ ) A B ' B (x) (2.6)! µ µ = 2 [! µ µ! µ x ]= 2 [ µ + x ]! µ (2.7) Therefore, comparing the coe we have cients of! on the left and the right hand sides of the equation, [' A (x),m µ ]= i µ )' A (x) (S µ ) B A ' B (x) (2.8) Substituting for L µ gives the result. Q.E.D It is not hard to show that both the di erential operator and matrices S µ satisfy the same

38 34 CHAPTER 2. FIELDS WITH SPIN, /2, commutation relations as generators M µ, i.e. equation 2.2. Proof: We first find the commutator then ]=(x ) )(x ) [L µ, L ]= [x ] = x + x µ x = x µ (2.9) = ( x µ µ + µ x + µ + µ µ ) = { (x µ ) µ ) µ )+ µ x )} = i{ L µ µ L µ L + L µ } = i( µ L µ L + L µ L µ ) (2.2) We know that the algebra.48 to.5 form the Lie algebra of SO(3). Consider equation.48, from quantum mechanics we know that we can find three (2j + ) (2j + ) hermitian matrices corresponding the the angular momenta in each direction. For example the matrix of J 3 is a diagonal matrix corresponding to an infinitesimal rotation by some angle about the third axis. The eigenvalues of J 3 are m = j, j +,...,j. The allowed values of j are, /2,,.... For a finite rotation, and some j (as seen in the previous chapter) we have, d (j) ( ) m m = e im m m =) d (j) (2 ) =e i(2 m) = cos(2 m) =( ) 2m =( ) 2j (2.2) for m = j. Then,ifj is an integer, rotation by 2 is equivalent to no rotation. However, when j is a half-integer, rotation by 2 picks up a minus sign. Therefore, the representations of the Lie algebra of SO(3) is di erent with that of the group SO(3). So we require a group that satisfies the Lie algebra given, but also solves the problem with the minus sign mentioned above. This group is known as the Special Unitary group or SU(2). The properties of the two groups are summarised below: We now define the operators N and N as the generators of SU(2) group: N i 2 (J i ik i ) (2.22) N i 2 (J i + ik i ) (2.23) and so In other words, and J i = N i + N i (2.24) K i = i(n i N i ) (2.25)

39 2.. REPRESENTATIONS OF THE LORENTZ GROUP USING SU(2) SU(2) 35 Table 2.: SO(3) and SU(2) properties SO(3) SU(2) group of 3 3 orthogonal matrices group of 2 2 unitary matrices det = det = Real elements Complex elements OO T = UU = Equations.48 to.5 then imply that the two operators satisfy the following [N i,n j ]=i ijk N k (2.26) [N i,n j ]=i ijkn k (2.27) [N i,n j ] = (2.28) Note that we have two di erent SU(2) algebras, one being the hermitian conjugate of the other. Hence the representation of the Lorentz group the 4 space-time dimensions is specified by two integers or half-integers, n and n. These representations are labelled by (n, n ). The dimension of the representation is then determined by its number of components i.e. (2n + )(2n + ). Since J i = N i + N i, given an n, one can determine the allowed values of j using the rule for addition of angular momenta: j = n n, n n +,...,n n. The most important of these representations are given below: (, ) scalar or singlet (2.29) (, ) left-handed spinor (2.3) 2 (, ) right-handed spinor (2.3) 2 ( 2, ) vector (2.32) 2 It is interesting to know why ( 2, 2 ) is in fact the vector representation. Firstly, we know under a Lorentz transformation, the components of a 4-vector mix with one another and so the representation is irreducible. Secondly, a 4-vector is, of course, four dimensional so (2n + )(2n + ) = 4. The only possibilities are ( 3 2, ), (, 3 2 ) and ( 2, 2 ). However, we also know that rotating a 3-vector by 2 is equivalent to no rotation and hence it correspond to

40 36 CHAPTER 2. FIELDS WITH SPIN, /2, integer values of j. This implies that the first two candidates are not allowed for a 4-vector whose space components are a 3-vector. The ( 2, 2 ) representation contains j =, and so it is the only correct 4-vector representation. 2.2 Left and right handed spinor (Weyl) fields The ( 2, ) representation of the Lie algebra of the Lorentz group contains the left-handed spinor field a (x). This is also called the left-handed Weyl field. Theindexa runs from to 2 and is known as the left-handed spinor index. a transforms under a Lorentz transformation according to U( ) a (x)u ( ) =La b ( ) b (x)( x) (2.33) Note that La b ( ) is a matrix in the ( 2, ) representation and satisfies the group composition rule La b ( )L c c ( ) =La ( ) (2.34) Again, for an infinitesimal rotation we have µ = b L b a ( +!) = b a µ +! µ and so i 2! µ (S µ L ) b a (2.35) Since! µ is antisymmetric, (S µ L ) a b = (S µ L ) a b also forms a set of 2 2 antisymmetric matrices. By the same methods seen before, it can be shown that they satisfy the same commutation relations as the generators M µ : Using and equation 2.33 we have [S µ L,S L ]= i( µ S L µ S L S µ L + S µ L ) (2.36) U( +!) = i 2! µ M µ (2.37) [M µ, a (x)] = L µ a(x)+(s µ L ) b a b (x) (2.38) where L µ = i(x µ ), as before. We will evaluate the fields at the origin of the space-time, x µ =, so that L µ =. Recall that for the spacial components of M µ we have, M ij = ijk J k with J k being the angular momentum operator. So ijk [J k, a ()] = (S ij L ) b a b () (2.39) In the ( 2, ) representation, the only allowed value of angular momentum is j = 2. For a spin- 2 operator, we have (S ij L ) a b = 2 ijk k (2.4)

41 2.2. LEFT AND RIGHT HANDED SPINOR (WEYL) FIELDS 37 where k, k =, 2, 3 are the familiar Pauli matrices, which are in fact the infinitesimal generators of SU(2): i =, =, 2 =, 3 = (2.4) i Note that the indices a, b run form to 2 and identify each element of the 2 2 matrix, whereas i, j indices in S ij are merely there to label the type of rotation we are considering. For example,(sl 2 ) means rotation in the -2 plane, i.e. about the 3-axis. Then (SL 2) a b = 2 2k k = 2 3.Then,e.g. (SL 2) =+ 2 and so on. Now we need to identify the matrices for the boost operator K k = M k in the ( 2, ) representation. We also know that in this representation, N k is zero since n =. Equations 2.24 and 2.25 then imply that K k = ij k and so (S k L ) b a = 2 i k (2.42) As we have already mentioned, hermitian conjugation swaps the two SU(2) Lie algebras. Hence if one takes the hermitian conjugate of a field a (x) inthe( 2, ) representation, it will result in a field in the (, 2 ) representation. This field is called the right-handed spinor (Weyl) field. We will adopt the notation for such a field. It transforms under Lorentz according to [ a (x)] = ȧ (x) (2.43) U( ) ȧ (x)u ( ) =R ḃ ȧ ( ) ḃ (x)( x) (2.44) with Rȧ ḃ being a matrix in the (, 2 ) representation. The group composition rule satisfied by these matrices is Rȧ ḃ ( )R ċ ċ ( ) =Rȧ ( ) (2.45) For an infinites rents transformation µ = where (S µ R ) ȧ ḃ = (S µ R ) ȧ ḃ the generators M µ.then ḃ R ḃ ȧ ( +!) = µ +! µ we have, just as before, ḃ ȧ i 2! µ (S µ R ) ḃ ȧ (2.46) are 2 2 matrices that obey the same commutation relation as [M µ, ȧ ()] = (Sµ R ) ȧ ḃ () (2.47) ḃ taking the hermitian conjugate, knowing that M µ is a hermitian operator, we get [ a (),M µ ]=[(S µ R ) ḃ ȧ ] b() (2.48)

42 38 CHAPTER 2. FIELDS WITH SPIN, /2, Comparing with [M µ, a ()] = (S µ L ) b a b () derived before, implies that (S µ R ) ḃ ȧ = [(S µ L ) b a ] (2.49) Let us now examine the properties of spinor indices in more detail. For this purpose, consider a field C ab (x) which carries two spinor indices. Under a Lorentz transformation U( )C ab U ( ) =La c Lb d C cd( x) (2.5) The question is whether the four components of C ab can be broken into smaller part that do no mix under the Lorentz transformations. To address this question we need to refer back to quantum mechanics. We know that two particles, each with spin 2, can be in a state of total spin zero or one. Then Total spin j=: The antisymmetric combination, i = p 2 ( "#i #"i) (2.5) Total spin j=: The symmetric combination 8, i = ##i ><, i = p 2 ( "#i + #"i) (2.52) >:, i = ""i In other words, 2 2 = A S. Note that the dimension of the left hand side is equal to 2 2 = 4 which is the same as that on the right hand side i.e. + 3 = 4. Therefore, for the Lorentz group we have Hence C ab (x) can be written as ( 2, ) ( 2, ) {z } dim = 2 2=4 =(, ) A (, ) S {z } dim = + 3 = 4 (2.53) C ab (x) = ab D(x)+G ab (x) (2.54) where G ab = G ba, D(x) is a scalar field (corresponding to spin=) and ab = antisymmetric constant defined such that 2 = 2 = +. Under a Lorentz transformation, the RHS of the above equation becomes ba is an U( )[ ab D(x)+G ab (x)] U ( ) = ab D( x)+l c a L d b G cd( x) (2.55)

43 2.2. LEFT AND RIGHT HANDED SPINOR (WEYL) FIELDS 39 where we have used fact that D(x) transforms as a scalar and is a constant. On the other hand, using equation 2.5 and then 2.54 itself, we have Comparing the two, we get L c a L d b C cd( x)=l c a L d b cdd( x)+l c a L d b G cd( x) (2.56) L c a L d b cd = ab (2.57) which implies that ab is an invariant under Lorentz transformations. In a sense, it is similar to the metric µ which is also a Lorentz invariant: µ = µ (2.58) In precisely the same way that we use µ and µ to raise and lower vector indices, one can use ab and ab to raise and lower left-handed spinor indices. We define ab = = ab (2.59) Then ab bc = c a (2.6) ab bc = a c (2.6) One can easily check these by simple matrix multiplication. So we may define As well as that, Equation 2.62 may also be written as a (x) ab b(x) (2.62) a = ab b = ab bc c = c a c (2.63) a = ab b = ba b = b ba = b ab (2.64) It is important to be careful with the signs when contracting indices, e.g. a a = ab b a = ba b a = b b Using the same method, one can find an invariant symbol ȧḃ = factor, i.e. for a right-handed spinor field ḃȧ (2.65) for the second SU(2) (, 2 ) (, 2 )=(, ) A (, ) S (2.66)