8.323 Relativistic Quantum Field Theory I Spring 2008

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For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 1.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Physics Department

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Eq. (55) was corrected on 5/6/08

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Let us assumethat wea retrying to construct a free ﬁeld ψa(x) for electrons which, like the free ﬁeld φ(x) for scalar particles, is linear in creation and annihilation operators. Then we expect a nonzero value for 0 |ψa(x)| 1electron, p =0 . 1 Under rotations the state(s) on the right transform under the spin- 2 represen tation, so ψa(x) must contain this representation, or else the matrix element 1 1 vanishes (i.e. the only spin that can be added to spin- 2 to get spin-0 is spin- 2 ). But ψa(x) must transform under some representation of the Lorentz group, generated by 1 J + = (J + iK ) J = J + + J − 2 (41) 1 J − = (J − iK ) K = i J − − J + . 2 ( )= 1 0 0 1 The simplest allowed representations are j+,j− 2 , or , 2 . Wewill use

both.

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� �� –1– Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 2.

1 1

0 �� 0 �� 2, , 2

Notetha t under a parity transformation,

K →−K

J → J,

= 1 ( + ) = 1 ( − ) so J + 2 J iK , J − 2 J iK , implies that

J + ←→ J − .

In addition, recall that the 4-vector rep is 1 , 1 .Soifχ transforms under, say, the 2 2 1 0 1 1 × 1 0 2 , rep, then ∂µχ mustbelong to the 2 , 2 2 , rep, which contains 1 1 0 0 µ

the , 2 rep but not the 2 , . Thus, ∂ χ must interchange these two reps.

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1 0 0 1 From the 2 , or , 2 representations, one can see the relation between the ↑ proper orthochronous Lorentz group L+ and SL(2,C). 1 0 For the 2 , rep, 1 1 i J + = σ, J − =0 =⇒ J = σ, K = − σ, 2 2 2 where the σi are the (traceless) Pauli spin matrices. Exponentiation of the σi with imaginary coeﬃcients produces all unitary determinant one 2× 2 matrices, −2πiJ or SU(2). Since e z = −1 , there are 2 matrices in SU(2) for every rotation group element, so the rotation group is SU(2)/Z2. The Lorentz group includes the exponentiation of the σi matrices with arbitrary complex coeﬃcients, so unitarity is lost. Exponentiation generates SL(2,C), thegrou p of complex 2× 2 −2πiJ matrices with determinant 1. One still has e z = −1 and there sulting 2:1 relationship, so

↑ = C Z (42)

L+ SL(2, )/ 2 .

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Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 3.

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( ) 1 × 2 Let ψL x be a 2-component fiel d, rep resented as a column vector, 1 0 transforming according to the 2 , rep. ( ) 1 × 2 Let ψR x be a 2-component fiel d, rep resented as a column vector, 0 1 transforming according to the , 2 rep. TheDirac field ψa(x) is 4-component field, represented as a 1 × 4 column vector, constructed by placing ψL(x) on top of ψR(x):

  ( ) ψ1 x  ψ2(x)  ψL(x) ψa(x)=   = . (43) ψ3(x) ψR(x)

ψ4(x)

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1 0 + 0 1 Dirac discovered that one can construct the 2 , , 2 representation of the Lorentz group by starting with four 4×4 “Dirac” matrices γµ,chosent os atisfy

{γµ,γν } =2gµν , (44)

wherethecurly brackets denotetheanticommutato r, and theright-hand side is multiplied by an implicit 4 × 4 identity matrix. Given Eq. (44), the matrices

i Sµν = [γµ ,γν ] (45) 4 automatically havetheLo rentz group commutation relations,

µν ρσ ρν µσ [S ,S ]=4i {g S } µν , ρσ

where the pairs of subscripts denote antisymmetrizations as deﬁned by

Eq. (21), 4/13/06.

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Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 4.

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Following Peskin and Schroeder, we will use i 0 01 i 0 σ γ = , γ = , (46) 10 −σi 0

where the entries in γ0 represent 2 × 2 blocks of zeros or the 2 × 2 identity matrix, and the σi represent the Pauli spin matrices, 1 01 2 0 −i 3 1 0 σ = , σ = , σ = . (47) 10 i 0 0 −1

This is called the Weyl or the chiral representation. Another popular choice is to

diagonalize γ0 .

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With someca lculation, one ﬁnds that 1 1 k k k k m m σ 0 k 0k i σ 0 J = S = k ,K= S = − k . 2 2 0 σ 2 0 −σ (48) This gives 1 1 k 1 1 k k k σ 0 k k k 0 0 J+ = (J + iK )= ,J− = (J − iK )= k . 2 2 0 0 2 2 0 σ

(49)

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Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 5.

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A one-particle state must remain a one-particle state under a Poincarétransfo rma tion, so the space of one-particle states forms a representation of the Poincaré group. 2 µ P ≡ PµP is a Casimir operator of the Poincarég roup, so a singlepa rticleha s a uniquev alue of P 2 , P 2 = m2,where m is called the mass of thep article. Particles seem to also have a definite spin. Is this a Casimir operator? Answer: Yes. The operator is 1 2 ≡ µ = νλ σ (50) W WµW , where Wµ 2µνλσ J P . µ Wµ is called the Pauli-Lubanski pseudovector. In the rest frame of P oneh as 2 2 2 2 µν W = −m J . Onek nows that W ,J =0,since Wµ transforms as a vector under Lorentz transformations. It is also translation-invariant, since J νλ ,Pρ = i g ρλ P ν − g ρν P λ ,

and thesu m with µνλσ will vanish when two indices are contracted with P .

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Consider ﬁrst P 2 > 0, i.e., massive particles. Diagonalize P µ and consider a particle in its rest frame, P µ =(m, 0, 0, 0). The subgroup of the Lorentz group that leaves P µ invariant is called the little group. In this case, thelittle group is therotation group. But wekn ow about therotation group! Thep article must havesp in j, an integer or half-integer, with 2j +1 spin states described by s ≡ J z,withs = −j, −j +1,... ,j. (The eigenvalue of J z is often called m, but m = mass.)

−iθnˆ·J p =0,s U R(ˆn, θ) p =0,s = e ≡ Dss R(ˆn, θ) . ss (51) –9– Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 6.

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By Wigner’s theorem, we know that we should expect to construct a unitary representation of the Poincare group in the Hilbert space. (Anti-unitary is excluded here, since the group is continuous.) We can therefore describe states with p =0 by boosting the p =0 states. So we can deﬁne

|p,s≡U(Bp ) |p =0,s , (52)

where Bp is a boost that transforms the rest vector to the speciﬁed momentum p. Bp is not uniquely deﬁned by this criterion. For Eq. (52) to beuni tary, we must beusi ng the covariant normalization:

3 (3)

p ,s |p,s =2Ep (2π) δss δ (p − p) .

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Therea ret wo standard ways to choose Bp , and therefore to deﬁne a basis for the

Hilbert space of 1-particle states:

�� boost in direction of p: (c) = ˆ (| |) = −iξpˆ· K (53) Bp B p, ξ p e ,

This process preserves the helicity, theco mponent of thespi n in the direction of the momentum. Thus, s = helicity. –11–

Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 7.

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Translations are no problem, since these are eigenstates of momentum. To apply a Lorentz transformation, use

U(Λ) |p,s = U(Λ) U(Bp ) |p =0,s −1 = U(BΛp ) U (BΛp ) U(Λ) U(Bp ) |p =0,s . Now deﬁne the Wigner rotation

(Λ ) ≡ −1 Λ (55) RW ,p BΛp Bp .

Notethat RW maps a rest vector to a rest vector, so it is a rotation. But we know about rotations! U(Λ) |p,s = U(BΛp ) p =0,s p =0,s U RW (Λ,p) p =0,s .

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U(Λ) |p,s = U(BΛp ) p =0,s p =0,s U RW (Λ,p) p =0,s . s

U(Λ) |p,s = Λp,s Dss RW (Λ,p) . (56)

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Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 8.

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(Review, Massive Particles)

Basis States in Rest Frame:

−iθnˆ·J p =0,s U R(ˆn, θ) p =0,s = e ≡ Dss R(ˆn, θ) . ss (51) Basis States in Arbitrary Frame:

|p,s≡U(Bp ) |p =0,s , (52)

where Bp is a boost (canonical or helicity) that boosts a rest vector to p.

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U(Λ) |p,s = Λp,s Dss RW (Λ,p) , (56) s

where (Λ ) ≡ −1 Λ (55) RW ,p BΛp Bp .

is called the Wigner rotation.

Recall: Bp is the standard boost, in either the canonical or helicity basis.

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Alan Guth, 8.323 Lecture, April 24 & 29, 2008,p. 9.

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Freepa rticleF ock spacebasis vectors:

|p 1s1,... ,p N sN . If two kets |ψ1 and |ψ2 are identical except for the ordering of the particles, they represent the same physical state. For scalar particles, |ψ1 = |ψ2. will 1 | = ±| We learn (soon!) that for spin- 2 particles, ψ1 ψ2 , depending on whether the permutation is even (+) or odd (-). Lorentz Transformations: each particle transforms independently.

U(Λ) |p 1s1,... ,pN sN = |Λp 1s ,... ,ΛpN s 1 N (57) {s } j × s s W (Λ 1) s s W (Λ N )

D 1 1 R ,p ...D N N R ,p .

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† U(Λ)a (p N ) 2Ep |p 1s1,... ,pN−1sN−1 = sN N † (Λ ) 2 Λ Λ a s p N EΛp N p 1s1,... , p N−1sN−1 N {s } j × s s W (Λ 1) s s W (Λ N ) D 1 1 R ,p ...D N N R ,p † = (Λ N ) 2 Λp (Λ) | 1 1 N−1 N−1 s s W (Λ N ) a s p E N U p s ,... ,p s D N N R ,p , N s N so † (Λ) ( N ) 2 p | = U a sN p E N ψ † (Λ N ) 2 Λp s s W (Λ N ) (Λ) | a s p E N D N N R ,p U ψ . N s

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� �� –17– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 10.

† (Λ) ( N ) 2 p | = U a sN p E N ψ † (Λ N ) 2 Λp s s W (Λ N ) (Λ) | a s p E N D N N R ,p U ψ . N s N So, † −1 EΛp † (Λ) ( ) (Λ) = (Λ ) (Λ ) (58) U a s p U a s p Ds s RW ,p . Ep s

Taking theadjoint: −1 EΛp −1 (Λ) ( ) (Λ) = (Λ ) (Λ ) (59) U as p U Dss RW ,p as p , Ep s

where I used the unitarity of Dss: ∗ (Λ ) = −1 (Λ )

Ds s RW ,p Dss RW ,p .

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What is the“little group” when P 2 =0, so there is no rest frame? Let pµ =(ω, 0, 0,ω) Little group: generated by J 3 , M 1,and M 2,where J 3 = J 12 M 1 ≡ K1 − J 2 = −J 10 + J 13 (60) M 2 ≡ K2 + J 1 = −J 20 + J 23 .

Algebra of little group: [M 1,M2]=0 J 3,M1 = iM 2 (61) J 3,M2 = −iM 1 .

Thelittle group is E(2), theE uclidean group (rotations and translations in R2)

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Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 11.

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Casimir Operator: M 2 =(M 1)2 +(M 2)2 Reps with M 2 =0 . M could then be rotated by any amount, so the rep is inﬁnite dimensional. This would involve an inﬁnite number of states with the same momentum, and does not correspond to anything known physically. 2 =0 2 =0 3 Finite-dimensional (1D) representat ions with M .WhenM , J becomes a Casimir operator, since J 3 ,Mi is proportional to M j .Soi nt his case M 1 = M 2 =0 J 3 = constant ∈ Z/2 .

3 (since e4πiJ =1i n SL(2, C)) So, for m =0 thelittle group does not mix diﬀerent s values. The helicity h of a massless particle is Lorentz-invariant, so thepho ton can have h = ±1, but does not need an h =0 state. The h = ±1 states are related by parity, but not by Lorentz transformations. If neutrinos = − 1

were massless, they could have only h 2 (“left-handed” only).

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Weassumethat ψa(x) is linear in creation and annihilation operators. Electrons are charged. The field should have a definite charge, to construct charge-conserving interactions. So let ψa(x) contain annihilation operators for electrons. Then it must contain creation operators for antiparticles, i.e. † positrons. (This is like the charged scalar field.) Then ψa(x) will create electrons and annihilate positrons. Thefie ld must satisfy:

µ µ iPµx −iPµx e ψa(y) e = ψa(x + y) , (62) −1 −1 U (Λ)ψa(x)U(Λ) = Λ 1 ,abψb(Λ x) , 2

where i µν − ωµν S µν i µ ν Λ 1 ≡ = 2 = [ ] (63) ,ab Mab e , with S γ ,γ .

2 4

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� �� –21– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 12.

µ µ iPµx −iPµx ψa(x)= e ψa(0) e .

But ψa(0) is linear in electron annihilation operators as(p) andpositroncreation operators b† (p),sowrite s d3 1 p s † s a(0) = s( ) ( )+ ( ) ( ) ψ 3 a p u a p b s p va p . (2π) 2 p E s

But µ µ µ iPµx −iPµx −ipµx e as(p) e = e as(p) ,

µ µ µ iPµx † −iPµ x ipµ x † e bs(p) e = e bs(p) , so

3 d 1 µ µ p s −ipµx † s ipµx a( )= s( ) ( ) + ( ) ( ) ψ x 3 a p u a p e bs p va p e , (2π) 2 p E s

(64)

where p0 =+ |p|2 + m2 . Thechoice p0 > 0 will beimp ortant.

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�� ( ) �� ψa x

−1 −1 U (Λ)ψa(x)U(Λ) = Λ 1 ,abψb(Λ x) , 2 so (Λ) ( ) −1(Λ) = Λ−1 (Λ ) U ψa x U 1 ,abψb x 2 3 d p 1 EΛp −1 s −ip·x = (Λ ) (Λ ) ( ) + 3 Dss RW ,p as p u a p e ... . (2π) 2 Ep Ep s,s Now let q ≡ Λp, and use d3 1 d3 1 p = q 3 3 , so (2π) 2Ep (2π) 2Eq −1 U(Λ)ψa(x)U (Λ) = 3 d q 1 −1 s −i(Λ−1 q)·x D RW (Λ,p) as (q )u (p)e + ... . (2 )3 2 ss a

π Eq s,s

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� �� –23– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 13.

−1 U(Λ)ψa(x)U (Λ) = 3 d 1 −1 q −1 s −i(Λ q)·x D RW (Λ,p) as (q )u (p)e + ... . (2π)3 2 ss a Eq s,s But this must equal −1 Λ 1 ψb(Λ x)= 2 ,ab 3 d q 1 −1 s −iq·(Λx) ( )Λ ( ) + 3 as q 1 ,abu b q e ... . (2π) 2 2 Eq s These two expressions match if Λ−1 s (Λ )= s ( ) −1 (Λ ) 1 u b p u a p Dss RW ,p , 2 ,ab s or equivalently,

s s −1 u (Λp)=Λ1 u (p)D RW (Λ,p) . (65)

a 2 ,ab b s s

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�� p =0): ��

s s −1 u (Λp)=Λ1 u (p)D RW (Λ,p) . (65) a 2 ,ab b s s

In there st frame p =0, Λ=R (rotation matrix), so Λp =0,andB p = BΛp = I, = −1Λ = so RW BΛp Bp R.So

s s −1 u (p =0)= Λ1 (R)u (p =0)D (R) . a 2 ,ab b s s Look separately at upper and lower spinor of ψ:

  ( ) ψ1 x  ψ2(x)  ψL(x) ψa(x)=   = . (43) ψ3(x) ψR(x)

ψ4(x)

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� �� –25– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 14.

Wekn ow that for rotations, Λ 1 are just the matrices generated by 2 ,ab 1 i i σ 0 J = i , 2 0 σ

and the Dss(R) matrices are the same! So the above equation can be rewritten

as s s −1 uL,a = Dab(R) uL,b Dss(R) , or uL D(R)=D(R) uL 1 for every R. Sinceth e spin- 2 representation is irreducible, the only matrices that commutewith all D(R) aremu ltiples of theid entity. Choose

√ s ( =0)= s (66) u L,a p mδa .

Boosting from the rest frame: For p =0,letΛ= Bq ,soΛ p = q .Then (Λ )= −1 Λ = −1 = RW ,p BΛp Bp Bq Bq I I, So

s s u (q )=Λ1 (Bq ) u (p =0). (67)

a 2 ,ab b

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