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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
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Eq. (55) was corrected on 5/6/08
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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- 12 represen­
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- 12 to get spin-0 is spin- 12 ).
But ψa (x) must transform under some representation of the Lorentz group,
generated by
1 )
+ + J −
+ iK
J = J
J + = (J
2
(41)
1
− iK
)
= i J − − J
+ .
J − = (J
K
2
The simplest allowed representations are (j+ , j− ) = 12 , 0 or 0, 12 . We will use
both.
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–1–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 2.
1 1
������ �� 2 , 0 ��� 0, 2
Note that under a parity transformation,
→ −K
K
→ J ,
J
+ = 1
(J
+ iK
), J − = 1 (J
− iK
), implies that
so J
2
2
+ ←→ J
− .
J
1
1
under, say, the
In addition,
recall
that
the
4-vector
rep
is
, . Soif χtransforms
2
2 1
1
1
1
2
, 0
rep,
then ∂µ χ must belong
to the 2
, 2 × 2
, 0 rep, which contains
1
1
the 0, 2 rep but not the 2
, 0 . Thus, ∂µ χ must interchange these two reps.
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–2–
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From the 12
, 0 or 0, 12 representations, one can see the relation between the
proper orthochronous Lorentz group L↑+ and SL(2,C).
, 0 rep,
For the 1
2
1
= − i
σ ,
= 1 σ ,
J − = 0 =⇒ J
K
J + = 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,
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 coefficients, 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 .
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(42)
–3–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 3.
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Let ψL (x) be a 2-component field,
represented
as a 1 × 2 column vector,
transforming according to the 12 , 0 rep.
represented
as a 1 × 2 column vector,
Let ψR (x) be a 2-component field,
transforming according to the 0, 12 rep.
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)
ψL (x)
 ψ2 (x) 
ψa (x) = 
.
=
ψ3 (x)
ψR (x)
ψ4 (x)
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(43)
–4–
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Dirac discovered that one can construct the 12 , 0 + 0, 12 representation of the
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 =
i µ ν
[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.
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–5–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 4.
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Following Peskin and Schroeder, we will use
0
0
1
γ =
1
0
i
,
γ =
0
−σ i
σi
0
,
(46)
where the entries in γ 0 represent 2 × 2 blocks of zeros or the 2 × 2 identity matrix,
and the σ i represent the Pauli spin matrices,
1
σ =
0
1
1
0
2
,
σ =
0 −i
i 0
,
3
σ =
1
0
0
−1
.
(47)
This is called the Weyl or the chiral representation. Another popular choice is to
diagonalize γ 0 .
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–6–
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With some calculation, one finds that
1 km m
1 σk 0
k
,
J = S =
2
2 0 σk
k
K =S
0k
i
=−
2
σk
0
0
−σ k
.
(48)
This gives
k
J+
1
1
= (J k + iK k ) =
2
2
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σk
0
0
0
,
k
J−
1
1
= (J k − iK k ) =
2
2
0
0
0
.
σk
(49)
–7–
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 Poincaré transforma­
tion, so the space of one-particle states forms a representation of the Poincaré
group.
P 2 ≡ Pµ P µ is a Casimir operator of the Poincaré 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
µνλσ J νλ P σ .
(50)
2
Wµ is called the Pauli-Lubanski pseudovector. In the rest frame of P µ one has
2
2
2 W = −m J . One knows that W 2 , J µν = 0, since Wµ transforms as a
vector under Lorentz transformations. It is also translation-invariant, since
νλ
J , 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 .
W 2 ≡ Wµ W µ , where Wµ =
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–8–
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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, 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.)
p = 0, s U R(n̂, θ) p = 0, s = e−iθn̂·J
s s
≡ Ds s R(n̂, θ) .
(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 boost 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 ) .
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–10–
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There are two 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)
Bp = B p,
ˆ ξ(|p|)
= e−iξˆp·K ,
(53)
where ξ(|p |) is the rapidity associated with p , tanh ξ = |p |/E.
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boost in positive z-direction, then rotate in the z-p̂ plane (along
the shorter of the two options):
(h)
Bp
= R(p̂) Bz ξ(|p |) .
(54)
This process 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.
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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
= U (BΛp ) U −1 (BΛp ) U (Λ) U (Bp ) |p = 0, s .
Now define the Wigner rotation
RW (Λ,p ) ≡ BΛ−p1 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 (BΛp ) p = 0, s p = 0, s U RW (Λ,p ) p = 0, s .
U (Λ) |p , s =
s
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–12–
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 Ds s RW (Λ,p ) .
(56)
s
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–13–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 8.
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(Review, Massive Particles)
Basis States in Rest Frame:
p = 0, s U R(n̂, θ) p = 0, s = e−iθn̂·J
s s
≡ Ds s R(n̂, θ) .
(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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–14–
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U (Λ) |p , s =
Λp , s Ds s RW (Λ,p ) ,
(56)
s
where
RW (Λ,p ) ≡ BΛ−p1 Λ Bp .
(55)
is called the Wigner rotation.
Recall: Bp is the standard boost, in either the canonical or helicity basis.
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–15–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 9.
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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 (soon!) that for spin- 12 particles, |ψ1 = ± |ψ2 , depending on
whether the permutation is even (+) or odd (-).
Lorentz Transformations: each particle transforms independently.
U (Λ) |p 1 s1 , . . . ,p N sN =
|Λp 1 s1 , . . . , Λp N sN {sj }
(57)
× Ds1 s1 RW (Λ,p 1 ) . . . DsN sN RW (Λ,p N ) .
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–16–
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U (Λ)a†sN (p N ) 2Ep N |p 1 s1 , . . . ,p N−1 sN−1 =
†
as (Λp N ) 2EΛp N Λp 1 s1 , . . . , Λp N−1 sN −1
N
{sj }
× Ds
1 s1 RW (Λ,p 1 ) . . . DsN sN RW (Λ,p N )
†
=
as (Λp N ) 2EΛp N U (Λ) |p 1 s1 , . . . ,p N−1 sN−1 DsN sN RW (Λ,p N ) ,
sN
so
N
U (Λ)a†sN (p N ) 2Ep N |ψ =
†
as (Λp N ) 2EΛp N DsN sN RW (Λ,p N ) U (Λ) |ψ .
sN
N
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–17–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 10.
U (Λ)a†sN (p N ) 2Ep N |ψ =
†
as (Λp N ) 2EΛp N DsN sN RW (Λ,p N ) U (Λ) |ψ .
N
sN
So,
U (Λ) a†s (p ) U −1 (Λ) =
EΛp †
as (Λp ) Ds s RW (Λ,p ) .
Ep
(58)
EΛp −1 Dss RW (Λ,p ) as (Λp ) ,
Ep
(59)
s
Taking the adjoint:
U (Λ) as (p ) U −1 (Λ) =
s
where I used the unitarity of Ds s :
−1
p) .
Ds∗ s RW (Λ,p ) = Dss
RW (Λ,
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–18–
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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 ≡ 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
3
J , M 1 = iM 2
3
J , M 2 = −iM 1 .
(61)
The little group is E(2), the Euclidean group (rotations and translations in R2 )
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–19–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 11.
��������������� �� ����
2 = (M 1 )2 + (M 2 )2
Casimir Operator: M
2=
could then be rotated by any amount, so the rep is
Reps with M
0.M
infinite dimensional. This would involve an infinite number of states with the
same momentum, and does not correspond to anything known physically.
3
2
2
Finite-dimensional (1D) representations
3 with
M = 0. When Mj = 0, J
i
becomes a Casimir operator, since J , M is proportional to M . 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 does 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 = − 12 (“left-handed” only).
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–20–
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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 12 ,ab ψb (Λ−1 x) ,
where
i
Λ 12 ,ab ≡ Mab = e− 2 ωµν S
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������ ����� �� � ��� ����
µν
, with S µν =
i µ ν
[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†s (p ), so write
d3 p
1 s
†
s
a
(
p
)
u
(
p
)
+
b
(
p
)
v
(
p
)
.
ψa (0) =
s
a
s
a
(2π)3 2Ep s
But
µ
µ
µ
µ
µ
eiPµ x as (p ) e−iPµ x = e−ipµ x as (p ) ,
µ
eiPµ x b†s (p ) e−iPµ x = eipµ x b†s (p ) ,
so
ψa (x) =
d3 p
1 s
−ipµ xµ
†
s
ipµ xµ
,
(
p
)
u
(
p
)
e
+
b
(
p
)
v
(
p
)
e
a
s
a
s
a
(2π)3 2Ep s
where p0 = + |p |2 + m2 . The choice p0 > 0 will be important.
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(64)
–22–
������� �������������� �� ψa(x)
U −1 (Λ)ψa (x)U (Λ) = Λ 12 ,ab ψb (Λ−1 x) ,
so
U (Λ)ψa (x)U −1 (Λ) = Λ−1
ψb (Λ x)
1
2 ,ab
3
EΛp −1 d p
1
s
−ip·x
=
Dss RW (Λ,p ) as (Λp )ua (p )e
+... .
(2π )3 2Ep
E
p
s,s
Now let q ≡ Λp , and use
d3 p 1
=
(2π)3 2Ep
d3 q 1
, so
(2π)3 2Eq
U (Λ)ψa (x)U −1 (Λ) =
1 −1 d3 q
s
−i(Λ−1 q)·x
D
R
(Λ,
p
)
a
(
q
)u
(
p
)e
+
.
.
.
.
W
s
a
ss
(2π)3 2Eq
s,s
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–23–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 13.
U (Λ)ψa (x)U −1 (Λ) =
1 −1 d3 q
s
−i(Λ−1 q)·x
R
(Λ,
p
)
a
(
q
)u
(
p
)e
+
.
.
.
.
D
W
s
a
ss
(2π)3 2Eq
s,s
But this must equal
ψb (Λ x) =
Λ−1
1
2 ,ab
d3 q
1 −1
s
−iq·(Λx)
a
(
q
)Λ
u
(
q
)e
+
.
.
.
.
1
s
b
2 ,ab
(2π)3 2Eq s
These two expressions match if
Λ−1
us (Λp ) =
1
,ab b
2
s
−1
usa (p )Dss
p) ,
RW (Λ,
or equivalently,
usa (Λp ) = Λ 12 ,ab usb (p )Ds−1
p) .
s RW (Λ,
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(65)
–24–
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p
= 0):
usa (Λp ) = Λ 12 ,ab usb (p )Ds− 1s RW (Λ,p ) .
(65)
In the rest frame p = 0, Λ = R (rotation matrix), so Λp = 0, and Bp = BΛp = I,
so RW = BΛ−p1 ΛBp = R. So
usa (p = 0) = Λ 12 ,ab (R)usb (p = 0) Ds− s1 (R) .
Look separately at upper and lower spinor of ψ:


ψ1 (x)
ψL (x)
 ψ2 (x) 
ψa (x) = 
.
=
ψ3 (x)
ψR (x)
ψ4 (x)
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(43)
–25–
Alan Guth, 8.323 Lecture, April 24 & 29, 2008, p. 14.
We know that for rotations, Λ 12 ,ab are just the matrices generated by
1 σi 0
i
,
J =
2 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
s (R) , or uL D(R) = D(R) uL
for every R. Since the spin- 12 representation is irreducible, the only matrices that
commute with all D(R) are multiples of the identity. Choose
usL,a (p = 0) =
√
mδas .
(66)
Boosting from the rest frame: For p = 0, let Λ = Bq , so Λp = q . Then
−1
−1
RW (Λ,p ) = BΛ
p = Bq Bq I = I ,
p ΛB
So
usa (q ) = Λ 12 ,ab (Bq ) usb (p = 0) .
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(67)
–26–
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