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2 Lecture 1: Spinors, Their Properties and Spinor Prodcuts 2 Lecture 1: spinors, their properties and spinor prodcuts Consider a theory of a single massless Dirac fermion . The Lagrangian is = ¯ i@ˆ . (2.1) L ⇣ ⌘ The Dirac equation is i@ˆ =0, (2.2) which, in momentum 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 sufficient. The algebra is simplified if we take γ matrices in Weyl repreentation where µ µ 0 σ γ = µ . (2.4) " σ¯ 0 # and σµ =(1,~σ) andσ ¯µ =(1, ~ ). The Pauli matrices are − 01 0 i 10 σ = ,σ= − ,σ= . (2.5) 1 10 2 i 0 3 0 1 " # " # " − # The matrix γ5 is taken to be 10 γ5 = − . (2.6) " 01# We can use the matrix γ5 to construct projection operators on to upper and lower parts of the four-component spinors U and V . The projection operators are 1 γ 1+γ Pˆ = − 5 , Pˆ = 5 . (2.7) L 2 R 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 pµσ pˆ = µ , (2.9) " pµσ¯ 0(p) # andpU ˆ (p) = 0, the two-component spinors satisfy the following (Weyl) equations µ µ pµσ uR(p)=0,pµσ¯ uL(p)=0. (2.10) –3– Suppose that we have a left handed spinor uL(p) that satisfies the Weyl equation. 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)=iσ2uL(p)⇤. (2.11) Then, µ µ 0=iσ2 [pµσ¯ uL(p)]⇤ = iσ2pµσ¯ ⇤u⇤ (p) L (2.12) µ δµ2 µ µ = ip σ σ¯ ( 1) u (p)⇤ = p σ iσ u (p)⇤ = p σ u˜ (p), µ 2 − L µ 2 L µ R and we conclude thatu ˜R(p) is a right-handed spinor. To get some physics 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 spin 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~ ~σ a a UL(p)=Np ,uL(p)= , = . (2.18) 0 ! b ! p0 b ! − b ! where a, b are two complex numbers. 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) iσ2uL(p)⇤ ! and since 01 b⇤ iσ = we find iσ 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 + Uλ(p) U λ(p)=Np 0,uL (p) + uR(p), 0 ⌦ 0 ⌦ uR(p) ⌦ λ (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 density matrix 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~σ p0 p~ ~σ pµσ uL(p) uL (p)= − = − = ⌦ 2 2p0 2p0 µ (2.24) + 1+~n~σ p0 + p~ ~σ pµσ¯ uR(p) uR(p)= . = = . ⌦ 2 2p0 2p0 Hence, we find 2 µ 2 Np 0 pµσ Np µ Uλ(p) U λ(p)= µ = pµγ . (2.25) ⌦ 2p0 pµσ¯ 0 2p0 λ (L,R) " # 2X –5– . Since the density matrix for a massless Dirac fermion 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~σ = ,~n= . (2.26) b − b p~ ! ! | | which is equivalent to finding the wave function of the spin 1/2 state polarized along ~n − axis. The solutions of this problem are given in any book on quantum mechanics and we just borrow them from there. So, writing the vector ~n as ~n =(sin✓ cos φ, sin ✓ sin φ, cos ✓) (2.27) we find ✓ iφ/2 a sin 2 e− uL(p)= = − ✓ iφ/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 position 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 pµσ 00 UL(p) U L(p)= ,UR(q) U R(Q)= µ . (2.31) ⌦ " 00# ⌦ " qµσ¯ 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 [σµσ¯⌫]=2gµ⌫. (2.33) Therefore, we find pq [qp]=2pq. (2.34) h i –6– Since [qp]= pq ,wefind h i⇤ 2 2 iφpq iφpq 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 scalar products. Next property of spinor products that we want to discuss is the anti-symmetry. 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)γ UR(q)=U L(q)γ 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)γ UL(p)=NpNquL(q) σ¯ uL(p), µ + µ (2.39) U R(p)γ UR(q)=NpNquR(p) σ uR(q). We will now use a relation between two-component left- and right-handed spinors uR(q)= µ iσ2uL⇤ (q), to rewrite U R(p)γ UR(q) as µ + µ T µ U R(p)γ UR(q)=NpNquR(p) σ uR(q)=NP NquL(p) σ2σ σ2 uL⇤ (q).
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