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Home , Dirac fermion, Matrix (mathematics), Spinor, Weyl equation

2 Lecture 1: , their properties and prodcuts

Consider a of a single massless Dirac . The Lagrangian is

= ¯ i@ˆ . (2.1) L ⇣ ⌘ The is i@ˆ =0, (2.2) which, in space becomes

pUˆ (p)=0, pVˆ (p)=0, (2.3) depending on whether we take positive-energy(particle) or negative-energy (anti-particle) solutions of the Dirac equation. Therefore, in the massless case no di↵erence appears in equations for paprticles and anti-particles. Finding one solution is therefore sucient. The is simplified if we take matrices in Weyl repreentation where

µ µ 0 = µ . (2.4) " ¯ 0 # and µ =(1,~) and ¯µ =(1, ~ ). The are 01 0 i 10 = ,= ,= . (2.5) 1 10 2 i 0 3 0 1 " # " # " #

The 5 is taken to be 10 5 = . (2.6) " 01#

We can use the matrix 5 to construct projection operators on to upper and lower parts of the four-component spinors U and V . The projection operators are 1 1+ Pˆ = 5 , Pˆ = 5 . (2.7) L 2 2 Let us write u (p) U(p)= L , (2.8) uR(p) ! where uL(p) and uR(p) are two-component spinors. Since

µ 0 pµ pˆ = µ , (2.9) " pµ¯ 0(p) # andpU ˆ (p) = 0, the two-component spinors satisfy the following (Weyl) equations

µ µ pµ uR(p)=0,pµ¯ uL(p)=0. (2.10)

–3– Suppose that we have a left handed spinor uL(p) that satisfies the . We can use it to construct a spinor that satisfies the Weyl equation for the right-handed spinor. Indeed, let us take u˜R(p)=i2uL(p)⇤. (2.11) Then,

µ µ 0=i2 [pµ¯ uL(p)]⇤ = i2pµ¯ ⇤u⇤ (p) L (2.12) µ µ2 µ µ = ip ¯ ( 1) u (p)⇤ = p i u (p)⇤ = p u˜ (p), µ 2 L µ 2 L µ R and we conclude thatu ˜R(p) is a right-handed spinor. To get some insight into what left- and right-handiness means, we write Weyl equations in component form

~p ~ ~p ~ uR(p)= uR(p),uL(p)= uL(p). (2.13) p0 p0 For a massless particle, p = p~ . Hence, for positive p , u (p) describes an incoming particle | 0| | | 0 R with along the direction of its momentum and uL(p) describes an incoming particle with spin in the direction that is opposite to its momentum. Incoming particles can also be viewed as outgoing anti-particles. We choose uL(p) to describe outgoing right-handed anti-particles and uR(p) to describe lef-handed outgoing anti-particles. Outgoing particles are described by Dirac-conjugate spinors, as usual. We now construct the four-component spinors from the two-component ones

uL(p) 0 UL(p)=Np ,UR(p)=Np , (2.14) 0 ! uR(p) ! where Np is the normalization constant that we will determine later. We now introduce the following notations for the four-component spinors

UL(p)=p ],UR(p)=p , i (2.15) U (p)= p, U (p)=[p. L h R Very often, for simplicity of notation, we will replace the momentum label in the spinor by its label , e.g. p ] i], etc. The conjugate spinors are obtained in the standard way. We find i !

U (p)=N 0,u (p)† , U (p)=N u (p)†, 0 . (2.16) L p L R p R ⇣ ⌘ ⇣ ⌘ It is now easy to derive first results for the spinor products.

U (p)U (q)= pq]=0, U (p)U (q)=[pq =0, (2.17) L L h R R i However, pq and [pq] spinor products do not need to vanish and we next compute them. h i For this, we will need explicit expressions for left- and right-handed spinors.

–4– Let us choose the left-handed spinor to be

uL(p) a p~ ~ a a UL(p)=Np ,uL(p)= , = . (2.18) 0 ! b ! p0 b ! b ! where a, b are two complex . We assume that they are normalized in the standard way u+(p)u (p)=1 a 2 + b 2 =1. (2.19) L L !|| | | To construct the right-handed spinor, we write

0 UR(p)=Np , (2.20) i2uL(p)⇤ ! and since 01 b⇤ i = we find i u (p)⇤ = . (2.21) 2 10 2 L a ! ⇤ ! We now have all the spinors fixed and can compute the spinor products. We will need Dirac- conjugate spinors as well. They are

U (p)=N (0, 0,a⇤,b⇤) , U (p)=N (b, a, 0, 0) , (2.22) L p R p We now compute the normalization condition using the completeness relation

2 uL(p) + 0 + U(p) U (p)=Np 0,uL (p) + uR(p), 0 ⌦ 0 ⌦ uR(p) ⌦ (L,R) " ! ! # 2X (2.23) 0 u u+(p) = N 2 L ⌦ L . p u u+(p)0 " R ⌦ R #

To proceed further, we need the of the two-component spinors uL and uR. Since those spinors describe normalized quantum mechanical states with spin projections ± on the axis ~n = p/p~ , it follows that u (p) u+(p) and u (p) u+(p) are projection operators 0 L ⌦ L R ⌦ R whose explicit expression is known from (spin one-half) quantum mechanis

µ + 1 ~n~ p0 p~ ~ pµ uL(p) uL (p)= = = ⌦ 2 2p0 2p0 µ (2.24) + 1+~n~ p0 + p~ ~ pµ¯ uR(p) uR(p)= . = = . ⌦ 2 2p0 2p0 Hence, we find

2 µ 2 Np 0 pµ Np µ U(p) U (p)= µ = pµ . (2.25) ⌦ 2p0 pµ¯ 0 2p0 (L,R) " # 2X

–5– . Since the density matrix for a massless with momentum p should be equal to pˆ, we conclude that the normalization constant should be choosen as Np = p2p0. To construct spinors explicitly, we need to solve the equation

a a p~ ~n~ = ,~n= . (2.26) b b p~ ! ! | | which is equivalent to finding the of the spin 1/2 state polarized along ~n axis. The solutions of this problem are given in any book on and we just borrow them from there. So, writing the vector ~n as

~n =(sin✓ cos , sin ✓ sin , cos ✓) (2.27) we find ✓ i/2 a sin 2 e uL(p)= = ✓ i/2 . (2.28) b ! cos 2 e !

This solution is, of course, not unique since the phase of uL(p) is arbitrary. But, once we choose uL(p) and with the rules for constructing uR(p) from the complex-conjugate uL, all phases for spinor products are determined. We are now in to compute spinor products and discuss some relations between them. We will start with

+ 0 + pq = U L(p)UR(q)=NpNq 0,uL (p) = NpNquL (p)uR(q) h i uR(q) ! (2.29) T T = NpNq(uL(p))⇤uR(q)=NpNq(uR(q)uL⇤ (p)) + ⇤ = NpNq(uR(q)uL(p))⇤ = U R(q)UL(p) =[qp]⇤.

Next, let us consider pq [qp]. It reads⇥ ⇤ h i pq [qp]=U (p)U (q)U (q)U (p)=Tr U (p) U (p)U (q) U (p) . (2.30) h i L R R L L ⌦ L R ⌦ R The two matrices that appear in that formula are⇥ ⇤

µ 0 pµ 00 UL(p) U L(p)= ,UR(q) U R(Q)= µ . (2.31) ⌦ " 00# ⌦ " qµ¯ 0 #

As the result Tr U (p) U (p)U (q) U (p) =2p qµ, (2.32) L ⌦ L R ⌦ R µ where we have used ⇥ ⇤ Tr [µ¯⌫]=2gµ⌫. (2.33) Therefore, we find pq [qp]=2pq. (2.34) h i

–6– Since [qp]= pq ,wefind h i⇤

2 2 ipq ipq pq = [qp] =2pq or pq = 2pq e , [qp]= 2pq e . (2.35) |h i| | | h i | | | | p p This formula is usually refered to as the statement that spinor products are square roots of products. Next property of spinor products that we want to discuss is the anti-. To see it, consider

0 + 0 0 0 1 pq = U L(p)UR(q)=NpNq 0,uL(p) = NP Nq 0, 0,ap⇤,bp⇤ h i uR(q) ! bq⇤ (2.36) B C B a⇤ C B q C @ A = N N a⇤b⇤ b⇤a⇤ =( 1)N N a⇤b⇤ b⇤a⇤ = qp . p q p q p q p q q p q p h i Therefore, we conclude that spinor products satisfy the following equations

pq = qp , [pq]= [qp], (2.37) h i h i where the latter relation can be proved in a similar manner. In practical computations, we often need to compute matrix elements of (products) of Dirac matrices between di↵erent spinors. For those cases there are a few identities that can be used. For example, there is a relation

µ µ U R(p) UR(q)=U L(q) UL(p) (2.38) that we will now prove. Using explicit representation for left- and right-handed spinors, we obtain

µ + µ U L(q) UL(p)=NpNquL(q) ¯ uL(p), µ + µ (2.39) U R(p) UR(q)=NpNquR(p) uR(q).

We will now use a relation between two-component left- and right-handed spinors uR(q)= µ i2uL⇤ (q), to rewrite U R(p) UR(q) as

µ + µ T µ U R(p) UR(q)=NpNquR(p) uR(q)=NP NquL(p) 2 2 uL⇤ (q). (2.40)

Since µ µ =0 µ µ µ T µ2 µ 2 2 = µ =1, 3 , ( ) =( 1) , (2.41) 8 <> µ µ =2 we can write :> µ µ T 2 2 =(¯ ) . (2.42)

–7– Then,

T µ T µ T + µ NP NquL(p) 2 2 uL⇤ (q)=NpNquL(p)(¯ ) uL⇤ (q)=NpNquL(q) ¯ u(p). (2.43)

We conclude [pµq = qµp]. (2.44) i h Further relations between di↵erent spinor products are obtained using Fiertz identities for µ matrices. We start by writing

µ ( ) = 0 0 ~ ~ = ~ ~ . (2.45) ab µ cd ab cd ab cd ab cd ab cd To simplify the second term, we write

~ ~ = A B . (2.46) ab cd ad cb ab cd i Using the fact that abab = 0 and that ~ ab~ bc =3ac, we find two equations for A and B

0=A 2B, 3=2A B, B =1,A=2. (2.47) ) For the -matrices, the result reads

(µ) ( ) =(¯µ) (¯ ) =2( )=2✏ace✏ = 2(i ) (i ) . (2.48) ab µ cd ab µ cd ab cd ad bc bde 2 ac 2 bd The significance of this equation is that the order of spinor indices, as they appear on the left- and the right-hand sides is di↵erent and this can be used for simplifications of spinor products. Indeed, consider

pµq] k l]=U (p)µU (q)U (k) U (l) h h µ L L L µ L + µ + = NpNqu (p)¯ uL(q) NkNlu (k)¯µuL(l) L L (2.49) = NpNqNkNl[uL(p)]a⇤ (¯µ)ab [uL(q)]b[uL(k)]c⇤ (¯µ)cd [uL(l)]d 2 2 =2NpNqNkNl [uL(p)]a⇤iac[uL(k)]c⇤ [uL(q)]bibd[uL(l)]d

To simplify this expression, we use the relation between left- and right-handed spinors

i2uL⇤ = uR, (2.50) to write

2 + NpNk [uL(p)]⇤i [uL(k)]⇤ = NpNk[uL(p)]⇤[uR(k)]a = NpNkuL(p) uR(k)= pk , a ac c a h i (2.51) N N [u (q)] i2 [u (l)] = u+(q)u (l)=[lq]. q l L b bd L d R L Hence, we find an identity pµq] k l]=2pk [lq]. (2.52) h h µ h i

–8– Similarly [kµl p q] = 2[kq] pl . (2.53) ih µ h i Finally, spinor products obey Schouten identities of the following form

ij kl + ik lj + il jk =0, h ih i h ih i h ih i (2.54) [ij][kl]+[ik][lj]+[il][jk]=0.

To prove the Schouten identities note that, due to the antisymmetry of a spinor ij = ji , the left hand sides of the above equations are antisymmetric w.r.t. j, k and l. h i h i However, a fully anti-symmetric combination of three two-components object ( well, e↵ectively two-component) is zero. Indeed, suppose we write a right-handed spinor (so e↵ectively two component object) as = j kl + k lj + l jk . (2.55) i ih i ih i ih i This spinor can be decomposed as

= a ⇠ + a ⌘ , (2.56) i ⇠ i ⌘ i where ⇠ and ⌘ are two spinors. It is easy to see that j = k = l =0(this h i h i h i follows from jj = 0 and the antisymmetry of spinor products). These equations imply h i a j⇠ + a j⌘ =0, (2.57) ⇠h i ⌘h i and similar equations for other spinors. The solution a = a = 0 which implies that = 0. ⇠ ⌘ i The Schouten identities then follows. Another useful equation is a relative of the Gordon identity, but for massless spinors

µ µ µ 0 ¯ 0 p p]=U L(p) µ UL(p)=Tr UL(p) UL(p) µ h ¯ 0 ! " ⌦ ¯ 0 !# (2.58) ⌫ µ 0 p⌫ 0 ⌫ µ µ =Tr µ = p⌫Tr ( ¯ )=2p . " 00! ¯ 0 !# There is a relation between matrix elements of products of -matrices that allows us to reverse the order of elements of the matrix prodcuts. Indeed,

p µ....µ2N+1 q]=U¯ (p)µ....µ2Nn1 U (q) h | 1 | L 1 L µ1 µ2n+1 + 0 .... uL(q) = NP Nq 0,uL (p) µ µ ¯ 1 ....¯ 2n+1 0 ! 0 ! (2.59)

+ µ1 µ2n+1 T µ2n+1,T µ1,T = NpNquL (p)¯ ....¯ uL(q)=NpNquL(q)¯ ...¯ uL⇤ (p) T µ2n+1 µ1 + µ2n+1 µ1 = NpNq uL(q)2 ... 2uL⇤ (p)=NpNquR(q) ... uR(p),

µ µ T where we used 2¯ 2 =( ) . Hence, we obtain

p µ....µ2n+1 q]=[q µ2n+1 ....µ1 p . (2.60) h | 1 | | | i

–9– Similarly [p µ....µ2n+1 q = q µ2n+1 ....µ1 p]. (2.61) | 1 i h | A relation for even of matrices reads

[p µ1 ....µ2n q]= [q µ2n ...µ1 p], | | | (2.62) p µ1 ....µ2n q = q µ2n ...µ1 p . h | | i h | i Finally, consider pµq] . This is a matrix, such that it depends on the spinors con- h µ structed out of p and q momenta and that has the property that it must contain i [j and i] j, i h where i = p, q and j = p, q since µ is helicity conserving. Hence, we can write

pµq] = A q] p + A p [q + A p] q + A q [p . (2.63) h µ 1| h | 2| i | 3| h | 4| i |

The coecients A1..4 can be constrained by considering matrix elements of the right-hand side and the left hand side with various spinors. Taking the matrix element with respect to p h and q], we find A = 0. Taking the matrix element with respect to [q and p ,wefindA = 0. 4 i 3 Hence, pµq] = A q] p + A p [q . (2.64) h µ 1| h | 2| i | To find A and A , we consider matrix elements with repspect k and l], where k and l are 1 2 h | independent momenta. We find

A kp [ql]= pµq] k l]=2pk [lq], (2.65) 2h i h h µ h i which means that A2 = 1. We find A1 in a similar way. We obtain

pµq] =2(q] p + p [q ) . (2.66) h µ | h | | i | This concludes our discussion of the spinor algebra.

– 10 – 3 Lecture 2: massless spin-one particles

A vector ✏ of a massless particle with momentum k is a four-vector that satisfies the following conditions

µ µ µ, 0=✏ k ,✏r =0,✏✏ ⇤ = 1. (3.1) µ µ µ The first condition is “transversality”, the second condition is the gauge choice ( r2 = µ 0,rµA = 0 and the last condition is normalization. The sum over two polarizations reads µ ⌫ ⌫ µ µ ⌫ µ⌫ k r + k r ✏ ✏⇤ = g + . (3.2) k r (1,2) 2X · Given the two massless vectors k and r, it is easy to construct a four-vector that satisfies the transversality and the gauge choice conditions. In fact, we can write down two independent vectors ⌘µ =[rµk ,⌘µ = rµk]. (3.3) 1 i 2 h It is obvious that, thanks to Dirac equation for massless spinors, the transversality and the gauge conditions are fullfilled. To claim that the two ⌘ vectors can be choosen as polarization vectors for massless gauge bosons, we will have to normalize them, check their and make sure that the sum over polarizations works out correctly. To do this, we need to find vectors. To this end, consider

µ µ µ + µ ⌘⇤ =([r k )⇤ = U (p) U (k) ⇤ = u (r) u (k)N N ⇤ 1 i R R R R r k T µ T T µ (3.4) = u (r) ⇤u⇤ (k)N N = u (r) ⇤i ⇤ ( i)⇤u (k)N N R R r k L 2 2 L r k Using = , T = and µ =¯µ,wefind 2⇤ 2 2 2 2 ⇤ 2 T T µ + µ µ u (r) ⇤i ⇤ ( i)⇤u (k)N N = N N u (r) ¯ u (k)= r k], (3.5) L 2 2 L r k r k L L h which means µ µ µ µ ⌘1⇤ = ⌘2 and ⌘2 ⇤ = ⌘1 . (3.6) Therefore, µ µ ⌘⇤ ⌘ = ⌘ ⌘ = r k] r k] rr [kk]=0, (3.7) 1 · 2 2 · 2 h h ⇠h i which implies that the two ⌘ vectors are indeed orthogonal. To normalize the ⌘-vectors, we need to compute

µ µ ⌘1 ⌘1⇤ = ⌘1 ⌘2 =[r k rµk]= r ([r k µ ) k] · · ih h i | (3.8) =2r ( r] k + k [r ) k]=2rk [rk]= p2[rk] p2[rk] ⇤ . h | h | | i | | h i ⇣ ⌘⇣ ⌘ Therefore, for normalized vectors we choose ( the signs are choosen for convenience) [rµk rµk] ✏µ = i ,✏µ = h , (3.9) 1 p2[rk] 2 p2 rk h i

– 11 – To understand what these vectors correspond to, we will evaluate them in a familiar kinematic case. Consider a photon propagating in the +z direction, so that its momentum is kµ = E (1, 0, 0, 1). The vector rµ is taken to be rµ = E (1, 0, 0, 1). Then 0 k]=UL(k)=Nk ,= , | 0 ! 1 ! (3.10) 0 0 1 k = UR(k)=Nk = Nk ,+ = , i i2⇤ ! + ! 0 ! Next, we need similar formulas for r-spinors. We find

r = U (r)=N 0,T , h | L r + T (3.11) [r = U R(r)=Nr ,0 . | Therefore,

T [rk]= NrNk = NrNk, rk = NrNk. (3.12) h i Then,

µ T µ [r k = NrNk + = NrNk (0, 1,i,0) , | i µ µ (3.13) r k]=([r k )⇤ = N N (0, 1, i, 0) . h | i r k Hence, we find 1 ✏µ = (0, 1, i, 0) . (3.14) 1,2 p2 ± We will consider calculations of scattering amplitudes assuming that all particles are outgoing. For this, we need complex-conjugates of actual polarization vectors. Hence, we write µ µ r k] 1 1 µ ✏ = h = [0, 1, i, 0] = [0, 1,i,0]⇤ = ✏⇤ , etc. (3.15) 2 p2 rk p2 p2 R h i So, to summarize, we will use polarization vectors for outgoing massless vector bosons µ µ µ r k] µ [r k ✏⇤ = h ,✏⇤ = i . (3.16) R p2 rk L p2[rk] h i In what follows, I will skip the complex-conjugate notation for the sake of simlpicity. The polarization vectors have peculiar transformation properties with respect to changing the reference vector. Indeed, consider a di↵erence of two polarization vectors with di↵erent reference vectors 1 rµk] sµk] ( rµk] sk sµk] rk ) ✏µ (k, r) ✏µ (k, s)= h h = h h ih h i . (3.17) R R p2 rk sk p2 rk sk ✓ h i h i ◆ h ih i To simplify this further, we use the following equation

pˆ = p] p + p [p . (3.18) | h | | i |

– 12 –