ft1ls03p_08

ft1ls03p_08 - MIT OpenCourseWare http://ocw.mit.edu 8.323...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 8.323 Relativistic Quantum Field Theory I Spring 2008 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 ������ ����������� ������� ����� ������ � ��� ���� ����� ����� �� � ��� ���� — ���� ���� Eq. (55) was corrected on 5/6/08 ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� ��������������� �� ��� ���� ����� Let us assume that we are trying 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)| 1 electron, p = 0 . Under rotations the state(s) on the right transform under the spin- 1 represen­ 2 tation, so ψa (x) must contain this representation, or else the matrix element vanishes (i.e. the only spin that can be added to spin- 1 to get spin-0 is spin- 1 ). 2 2 But ψa (x) must transform under some representation of the Lorentz group, generated by 1 J = J+ +J− J + = (J + iK ) 2 (41) 1 J − = (J − iK ) K = i J− −J+ . 2 The simplest allowed representations are (j+ , j− ) = both. ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� 1 2, 0 or 0, 1 . We will use 2 –1– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 2. ������ �� 1 , 0 ��� 0, 1 2 2 Note that under a parity transformation, K → −K J →J , so J + = 1 (J + iK ), J − = 1 (J − iK ), implies that 2 2 J + ←→ J − . In addition, recall that the 4-vector rep is 1 , 1 . So if χ transforms under, say, the 22 1 , 0 rep, then ∂µ χ must belong to the 1 , 1 × 1 , 0 rep, which contains 2 22 2 the 0, 1 rep but not the 1 , 0 . Thus, ∂µ χ must interchange these two reps. 2 2 –2– ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� ��� ������� ����� ��� �����C� From the 1 , 0 or 0, 1 representations, one can see the relation between the 2 2 proper ortho chronous Lorentz group L↑ and SL(2,C). + 1 For the 2 , 0 rep, 1 1 i J − = 0 =⇒ J = σ , K =− σ , J+ = σ , 2 2 2 where the σi are the (traceless) Pauli spin matrices. Exponentiation of the σi with imaginary co efficients produces all unitary determinant one 2 × 2 matrices, or SU(2). Since e−2πiJz = −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 co efficients, so unitarity is lost. Exponentiation generates SL(2,C), the group of complex 2 × 2 matrices with determinant 1. One still has e−2πiJz = −1 and the resulting 2:1 relationship, so L↑ = SL(2,C)/Z2 . + ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� (42) –3– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 3. ��� ���� ����� Let ψL (x) be a 2-component field, represented as a 1 × 2 column vector, transforming according to the 1 , 0 rep. 2 Let ψR (x) be a 2-component field, represented as a 1 × 2 column vector, transforming according to the 0, 1 rep. 2 The Dirac 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) ψ (x) ψa (x) = 2 = ψ3 (x) ψ4 (x) ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� ψ L ( x) ψR (x) . (43) –4– ��� ���� ����� �� Dirac discovered that one can construct the 1 , 0 + 0, 1 representation of the 2 2 Lorentz group by starting with four 4 × 4 “Dirac” matrices γ µ , chosen to satisfy {γ µ , γ ν } = 2g µν , (44) where the curly brackets denote the anticommutator, and the right-hand side is multiplied by an implicit 4 × 4 identity matrix. Given Eq. (44), the matrices S µν = iµν [γ , γ ] 4 (45) automatically have the Lorentz group commutation relations, [S µν , S ρσ ] = 4i {g ρν S µσ } µν , ρσ where the pairs of subscripts denote antisymmetrizations as defined by Eq. (21), 4/13/06. ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –5– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 4. ���� � �� ���� ������ ����������� Following Peskin and Schro eder, we will use 0 1 γ0 = 1 0 0 −σ i γi = , σi 0 , (46) where the entries in γ 0 represent 2 × 2 blo cks of zeros or the 2 × 2 identity matrix, and the σ i represent the Pauli spin matrices, σ1 = 0 1 1 0 0 −i i0 σ2 = , , σ3 = 1 0 0 −1 . (47) This is called the Weyl or the chiral representation. Another popular choice is to diagonalize γ 0 . ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –6– ���������� �� ���� ���� � �� ���� ����� �� With some calculation, one finds that Jk = 1 2 km S m = 1 2 σk 0 0 σk , K k = S 0k = − i 2 σk 0 0 −σ k . (48) This gives k J+ = 1k 1 (J + iK k ) = 2 2 ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� σk 0 0 0 , k J− = 1k 1 (J − iK k ) = 2 2 0 0 0 . σk (49) –7– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 5. ��������� �� ������ ��� ��� ���� ��� ����� A one-particle state must remain a one-particle state under a Poincar´ transforma­ e tion, so the space of one-particle states forms a representation of the Poincar´ e group. e P 2 ≡ Pµ P µ is a Casimir operator of the Poincar´ group, so a single particle has a unique value of P 2 , P 2 = m2 , where m is called the mass of the particle. Particles seem to also have a definite spin. Is this a Casimir operator? Answer: Yes. The operator is 1 νλ σ P. (50) µνλσ J 2 Wµ is called the Pauli-Lubanski pseudovector. In the rest frame of P µ one has W 2 ≡ Wµ W µ , where Wµ = 2 W 2 = −m2 J . One knows that W 2 , J µν = 0, since Wµ transforms as a vector under Lorentz transformations. It is also translation-invariant, since J νλ , P ρ = i g ρλ P ν − g ρν P λ , and the sum with µνλσ will vanish when two indices are contracted with P . ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –8– ��������� �� ��������������� �� ��� ���� ��� ����� 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, the little group is the rotation group. But we know about the rotation group! The particle must have spin j , an integer or half-integer, with 2j + 1 spin states described by s ≡ J z , with s = −j, −j + 1, . . . , j . (The eigenvalue of J z is often called m, but m = mass.) ˆ n p = 0, s U R(ˆ, θ ) p = 0, s = e−iθn·J ss ≡ Ds s R(ˆ, θ ) . n (51) –9– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 6. ������ ���� ������� P 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 bo ost that transforms the rest vector to the specified momentum p . Bp is not uniquely defined by this criterion. For Eq. (52) to be unitary, we must be using the covariant normalization: p , s |p , s = 2Ep (2π )3 δs s δ (3) (p − p ) . ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –10– ������ �� ��� ���� ��� ����� There are two standard ways to cho ose Bp , and therefore to define a basis for the Hilbert space of 1-particle states: ������ ��� boost in direction of p : (c) ˆ Bp = B p, ξ (|p |) = e−iξp·K , ˆ (53) where ξ (|p |) is the rapidity asso ciated with p , tanh ξ = |p |/E . ���� ���� boost in positive z -direction, then rotate in the z -p plane (along ˆ the shorter of the two options): (h) Bp = R(ˆ) Bz ξ (|p |) . p (54) This pro cess preserves the helicity, the component of the spin in the direction of the momentum. Thus, s = helicity. –11– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 7. ������� ��������������� �� ����� ������ 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 = U (BΛp ) U −1 (BΛp ) U (Λ) U (Bp ) |p = 0, s . Now define the Wigner rotation −1 RW (Λ, p ) ≡ BΛp Λ Bp . (55) Note that RW maps a rest vector to a rest vector, so it is a rotation. But we know about rotations! U ( B Λ p ) p = 0, s U (Λ) |p , s = p = 0, s U RW (Λ, p ) p = 0, s . s ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –12– U (Λ) |p , s = U ( B Λ p ) p = 0, s p = 0, s U RW (Λ, p ) p = 0, s . s Λp , s Ds s RW (Λ, p ) . U (Λ) |p , s = (56) s ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –13– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 8. ����� ������ ��� ������� �� ������� ��� � (Review, Massive Particles) Basis States in Rest Frame: ˆ p = 0, s U R(ˆ, θ ) p = 0, s = e−iθn·J n ss n ≡ Ds s R(ˆ, θ ) . (51) Basis States in Arbitrary Frame: |p , s ≡ U (Bp ) |p = 0, s , (52) where Bp is a bo ost (canonical or helicity) that boosts a rest vector to p . ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –14– Λp , s Ds s RW (Λ, p ) , U (Λ) |p , s = ������� �������������� �� ����� ������ (56) s where −1 RW (Λ, p ) ≡ BΛp Λ Bp . (55) is called the Wigner rotation. Recall: Bp is the standard boost, in either the canonical or helicity basis. ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –15– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 9. ���������� �� ������ Free particle Fock space basis vectors: |p 1 s1 , . . . , 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 . We will learn (so on!) that for spin- 1 particles, |ψ1 = ± |ψ2 , depending on 2 whether the permutation is even (+) or o dd (-). Lorentz Transformations: each particle transforms independently. U (Λ) |p 1 s1 , . . . , p N sN = |Λp 1 s1 , . . . , Λp N sN (57) {s j } × Ds1 s1 RW (Λ, p 1 ) . . . DsN sN RW (Λ, p N ) . ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –16– �������������� �� �������� ��� ������������ ��������� U (Λ)a†N (p N ) s {s j } 2Ep N |p 1 s1 , . . . , p N −1 sN −1 = a† (Λp N ) s 2EΛp N Λp 1 s1 , . . . , Λp N −1 sN −1 N × Ds1 s1 RW (Λ, p 1 ) . . . DsN sN RW (Λ, p N ) = sN so a† (Λp N ) s 2EΛp N U (Λ) |p 1 s1 , . . . , p N −1 sN −1 DsN sN RW (Λ, p N ) , N U (Λ)a†N (p N ) s sN 2Ep N |ψ = a† (Λp N ) s N ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� 2EΛp N DsN sN RW (Λ, p N ) U (Λ) |ψ . –17– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 10. U (Λ)a†N (p N ) s sN 2Ep N |ψ = a† (Λp N ) s N 2EΛp N DsN sN RW (Λ, p N ) U (Λ) |ψ . So, U (Λ) a† (p ) U −1 (Λ) = s EΛp Ep s a† (Λp ) Ds s RW (Λ, p ) . s (58) − Dss1 RW (Λ, p ) as (Λp ) , (59) Taking the adjoint: U (Λ) as (p ) U −1 (Λ) = EΛp Ep s where I used the unitarity of Ds s : − ∗ Ds s RW (Λ, p ) = Dss1 RW (Λ, p ) . ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –18– �������� ����� ��� 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 ≡ K 1 − J 2 = −J 10 + J 13 (60) M 2 ≡ K 2 + J 1 = −J 20 + J 23 . Algebra of little group: [M 1 , M 2 ] = 0 J 3 , M 1 = iM 2 (61) J 3 , M 2 = −iM 1 . The little group is E(2), the Euclidean group (rotations and translations in R2 ) ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –19– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 11. ��������������� �� ���� 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. Finite-dimensional (1D) representations with M 2 = 0. When M 2 = 0, J 3 becomes a Casimir operator, since J 3 , M i is proportional to M j . So in this case M1 = M2 = 0 J 3 = constant ∈ Z/2 . 3 (since e4πiJ = 1 in SL(2, C)) So, for m = 0 the little group do es not mix different s values. The helicity h of a massless particle is Lorentz-invariant, so the photon 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 were massless, they could have only h = − 1 (“left-handed” only). 2 ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –20– ���� ������ ��� ����� �������� � We assume that 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. The field must satisfy: µ µ eiPµ x ψa (y ) e−iPµ x = ψa (x + y ) , U −1 (Λ)ψa (x)U (Λ) = Λ 1 ,ab ψb (Λ−1 x) , 2 where i Λ 1 ,ab ≡ Mab = e− 2 ωµν S 2 ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� µν , with S µν = iµν [γ , γ ] . 4 (62) (63) –21– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 12. µ µ ψa (x) = eiPµ x ψa (0) e−iPµ x . But ψa (0) is linear in electron annihilation operators as (p ) and positron creation operators b† (p ), so write s d3 p ( 2π )3 ψa (0) = But 1 2E p s s as (p ) us (p ) + b† (p ) va (p ) a s µ µ µ µ . µ eiPµ x as (p ) e−iPµ x = e−ipµ x as (p ) , µ eiPµ x b† (p ) e−iPµ x = eipµ x b† (p ) , s s so ψa (x) = d3 p ( 2π )3 1 2E p µ s µ s as (p ) us (p ) e−ipµ x + b† (p ) va (p ) eipµ x a s , (64) where p0 = + |p |2 + m2 . The choice p0 > 0 will be important. ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� –22– ������� �������������� �� ψa(x) U −1 (Λ)ψa (x)U (Λ) = Λ 1 ,ab ψb (Λ−1 x) , 2 so U (Λ)ψa (x)U −1 (Λ) = Λ−,1ab ψb (Λ x) 1 2 = d3 p ( 2π )3 1 2E p EΛp −1 Dss RW (Λ, p ) as (Λp )us (p )e−ip·x + . . . a Ep s,s . Now let q ≡ Λp , and use d3 p 1 = ( 2π )3 2E p d3 q 1 , so ( 2π )3 2E q U (Λ)ψa (x)U −1 (Λ) = d3 q ( 2π )3 1 2E q s,s ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� − Dss1 RW (Λ, p ) as (q )us (p )e−i(Λ a −1 q )·x +... . –23– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 13. U (Λ)ψa (x)U −1 (Λ) = d3 q ( 2π )3 1 2E q − Dss1 RW (Λ, p ) as (q )us (p )e−i(Λ a s,s −1 q )·x +... . But this must equal 1 Λ−,ab ψb (Λ x) = 1 2 d3 q ( 2π )3 1 2E q 1 as (q )Λ−,ab us (q )e−iq ·(Λx) + . . . 1 b 2 s . These two expressions match if 1 Λ−,ab us (Λp ) = 1 b 2 s − us (p )Dss1 RW (Λ, p ) , a or equivalently, − us (Λp ) = Λ 1 ,ab us (p )Ds 1 RW (Λ, p ) . a b s 2 ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� (65) –24– ��������� �� ���� ����� �p = 0): − us (Λp ) = Λ 1 ,ab us (p )Ds 1 RW (Λ, p ) . a b s 2 (65) In the rest frame p = 0, Λ = R (rotation matrix), so Λp = 0, and Bp = BΛp = I , −1 so RW = BΛp ΛBp = R. So −1 us (p = 0) = Λ 1 ,ab (R)us (p = 0) Ds s (R) . a b 2 Lo ok separately at upper and lower spinor of ψ : ψ1 (x) ψ (x) ψa (x) = 2 = 3 (x) ψ4 (x) ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� ψ L ( x) ψR (x) . (43) –25– Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 14. We know that for rotations, Λ 1 ,ab are just the matrices generated by 2 Ji = 1 2 σi 0 0 σi , and the Ds s (R) matrices are the same! So the above equation can be rewritten as − uL,a s = Dab (R) uL,b s Ds 1 (R) , or uL D(R) = D(R) uL s for every R. Since the spin- 1 representation is irreducible, the only matrices that 2 commute with all D(R) are multiples of the identity. Cho ose us ,a (p = 0) = L √ s mδa . (66) Bo osting from the rest frame: For p = 0, let Λ = Bq , so Λp = q . Then −1 − RW (Λ, p ) = BΛp ΛBp = Bq 1 Bq I = I , So us (q ) = Λ 1 ,ab (Bq ) us (p = 0) . a b 2 ���� ���� ����� ������� ��������� �� �� ������� ������ ����� �� � ��� ���� (67) –26– ...
View Full Document

Ask a homework question - tutors are online