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Lecture 6: The Poincaré Group Sept. 23, 2013 for any two vectors V and W , from which we can conclude
c
Copyright 2005 by Joel A. Shapiro gµν Λµρ Λν σ = gρσ . (2)

Last time we saw that for a scalar field φ(x), for every Poincaré transfor- In matrix language this is ΛT gΛ = g, a pseudo-orthogonality condition, where
mation Λ : xµ 7→ Λµν xν + cµ , there is a unitary operator U(Λ) which trans- g replaces the identity in the usual orthogonality condition OT O = 1I, with
forms the field by U(Λ)φ(x)U −1 (Λ) = φ(Λx). As for the individual particle OT the transpose of the matrix O.
states, we expect there may be sets of fields which, rather than transforming Examining the (00) component of this equation, we have
as scalars, transform within themselves, with 3
(Λ00 )2 − (Λj 0 )2 = 1
X
−1
U(Λ)φa (x)U (Λ) = Mba (Λ) φb(Λx). j=1

we see that, because Λµν is real, (Λ00 )2 ≥ 1. This divides the Lorentz trans-
Two successive Poincaré transformations must transform as their composite,
formations into those with Λ00 ≥ 1, which are called orthochronous because
U(Λ2 Λ1 ) φa (x) U −1 (Λ2 Λ1 ) = Mca (Λ2 Λ1 ) φc (Λ2 Λ1 x) they preserve the direction of time, and those with negative Λ00 , which do
not. We may also take the determinant of (2) to conclude (det Λ)2 = 1,
= U(Λ2 ) U(Λ1 ) φa (x) U −1 (Λ1 ) U −1 (Λ2 )
so this divides the Lorentz transformations into those with determinant +1
= U(Λ2 ) Mba (Λ1 ) φb (Λ1 x) U −1 (Λ2 ) and those with determinant −1. Only the proper orthochronous Lorentz
= Mcb (Λ2 )Mba (Λ1 ) φc (Λ2 Λ1 x) transformations, those with positive Λ00 and positive determinant, can arise
from a continuous acceleration or rotation, and it is only these which are es-
which requires that Λ 7→ M(Λ) is a representation, sential for any high energy theory. The others involve parity or time-reversal,
and are still interesting, but we will delay discussion of them.
Mca (Λ2 Λ1 ) = Mcb (Λ2 ) Mba (Λ1 ),
Any proper orthochronous Lorentz transformation can be written as the
in the same way as we found for the action on states. repeated application of an infinitesimal one. Thus we can write Λ = eaℓ L̃ℓ ,
So to discuss fields like the electromagnetic field, which we expect will not where aℓ are some continuous real parameters describing the group element,
transform like a scalar, we need to understand the possible representations and each L̃ℓ·· is a real 4 × 4 real matrix. As gντ is a constant, differentiating
of the Poincaré group, and in particular the Lorentz subgroup. So it is time (2) with respect to aℓ at aℓ = 0 gives
to discuss this group in detail. gµν L̃ℓµρ δσν + gµν δρµ L̃ℓν σ = 0, or L̃ℓ σρ + L̃ℓ ρσ = 0,
For particle theorists a primary requirement of a quantum field theory is
that it be invariant under the Poincaré group, which consists of the proper that is, L̃ℓ ρσ is a real antisymmetric matrix. There are six linearly indepen-
orthochronous Lorentz transformations and the translations: dent 4 × 4 antisymmetric real matrices, corresponding to the three rotations
and three directions for Lorentz boosts, and thus there are 6 components to
Λ : xµ → x′ µ = Λµν xν + cµ , (1) aℓ .
For a continuous (Lie) group which is connected, as the proper ortho-
with cµ an arbitrary constant vector. The Lorentz condition on the real chronous Lorentz transformation group is, most1 of the group properties are
matrix Λ is that it preserve the Minkowski product: if V ′ µ = Λµν V ν and
1
W ′ µ = Λµν W ν , then We will see later that some global properties are not determined. An example you
already know comes from the rotation group SO(3) and the group of unitary 2×2 matrices
µ of determinant 1, SU(2). The generators of these have the same familiar algebra, [Li , Lj ] =
V µ Wµ = V ′ Wµ′ = gµν Λµρ Λν σ V ρ W σ iǫijk Lk , but a rotation through 2π gives the identity for SO(3) but it takes a rotation by
= gρσ V ρ W σ 4π to reach the identity in SU(2). The latter is called the covering group of SO(3).
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determined by the commutation relations of the generators, that is, by the have written things as mathematicians do, but physicists generally prefer to
derivatives of the group elements with respect to the group parameters, eval- write their group elements as
uated at the identity element. The generators form a vector space which is P
aℓ Lℓ
a Lie algebra. In our case, that means we will learn what we need from the Λ = e−i ℓ ,
commutators of the L̃ℓ matrices, together with their commutators with the
translations. with the extra −i in the exponent, because they expect the group transfor-
The index ℓ describing the six independant antisymmetric 4 × 4 real mation to act as a unitary operator and would like the generators to act as
matrices is most conveniently described by a pair of 4 dimensional indices, hermitean operators. We may therefore write the physicists’ generators as
with the understanding that L̃αβ = −L̃βα . Please note that L̃αβ , for each Lαβ = iL̃αβ , with
pair α, β, is a matrix, not a matrix element. We may define a basis for the (Lαβ )µν = iδαµ gβν − iδβµ gαν ,
vector space of generators [Lαβ , Lγζ ]µν = −igαγ (Lβζ )µν + igβγ (Lαζ )µν + igαζ (Lβγ )µν − igζβ (Lαγ )µν
µν
or as matrices,

L̃αβ = δαµ δβν − δβµ δαν .

Matrix multiplication requires lowering one matrix element index, [Lαβ , Lγζ ] = −igαγ Lβζ + igβγ Lαζ + igαζ Lβγ − igζβ Lαγ (4)
µ
We will not be needing the L̃αβ any more, so from now on the ˜ will be

L̃αβ = δαµ gβν − δβµ gαν
ν reserved for other meanings.
and the commutator is therefore Because sometimes we think in terms of space and time, and not always
iµ µ  ρ µ  ρ
in four dimensions, it is also useful to divide the six generators of the general
Lorentz transformations into three spatial ones2 and the three space-time
h  
L̃αβ , L̃γζ = L̃αβ L̃γζ − L̃γζ L̃αβ
ν ρ ν ρ ν
ones:
1
  
= δαµ gβρ − δβµ gαρ δγρ gζν − δζρ gγν Jℓ = ǫℓjk Ljk , Kj = L0j .
   2
− δγµ gζρ − δζµ gγρ δαρ gβν − δβρ gαν
We can find their commutators from (4), keeping in mind that gij = −δij ,
= δαµ gβρ δγρ gζν − δαµ gβρ δζρ gγν − δβµ gαρ δγρ gζν + δβµ gαρ δζρ gγν and noting that Ljk = ǫjkℓ Jℓ :
−δγµ gζρδαρ gβν + δγµ gζρδβρ gαν + δζµ gγρ δαρ gβν − δζµ gγρ δβρ gαν
1
= δαµ gβγ gζν − δαµ gβζ gγν − δβµ gαγ gζν + δβµ gαζ gγν [Jj , Jk ] = ǫjℓm ǫkpq [Lℓm , Lpq ]
4
−δγµ gζα gβν + δγµ gζβ gαν + δζµ gγα gβν − δζµ gγβ gαν i
    = ǫjℓm ǫkpq (δℓp Lmq − δmp Lℓq − δℓq Lmp + δmq Lℓp )
= gαγ δζµ gβν − δβµ gζν − gβγ δζµ gαν − δαµ gζν 4
i
= (ǫjpm ǫkpq Lmq − ǫjℓp ǫkpq Lℓq − ǫjℓm ǫkpℓ Lmp + ǫjℓm ǫkpm Lℓp )
   
−gαζ δγµ gβν − δβµ gγν + gζβ δγµ gαν − δαµ gγν 4
 µ  µ  µ  µ
i
= −gαγ L̃βζ + gβγ L̃αζ + gαζ L̃βγ − gβζ L̃αγ = − [−δjk δmq + δmk δjq ]Lmq + [δjq δℓk − δℓq δjk ]Lℓq
ν ν ν ν 4
(3)

+[δmk δjp − δjk δmp ]Lmp − [δjk δℓp − δℓk δjp ]Lℓp
As must be the case, the commutator of two generators is a linear com- 2
If you need a refresher (or remediation) on the Levi-Civita tensor ǫijk , please see the
bination of generators, for any Lie algebra is closed under commutation. We the supplementary note on “ǫijk and Cross Products in 3-D Euclidean space”.
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i 1 Including the translations

= − (Lkj + Lkj + Lkj + Lkj )
4
= i Ljk = iǫjkℓ Jℓ . (5) The full Poincaré group contains, in addition to the Lorentz transformations,
1 i translations T in four dimensions, generated by the 4-momentum operator
[Jj , Kk ] = ǫjpq [Lpq , L0k ] = ǫjpq (δpk L0q − δqk L0p ) = +iǫjkq L0q
2 2 Pµ ,
µ
[Jj , Kk ] = iǫjkq Kq . (6) T = e−ic Pµ : xµ → x′ µ = xµ + cµ .
[Kj , Kk ] = [L0j , L0k ] = −iLjk Clearly translations by amounts cµ1 and cµ2 commute with each other, so
= −iǫjkℓ Jℓ . (7)
[Pµ , Pν ] = 0.
Eqs. (5) and (6) are the usual commutation relations for the rotation operator
with any vector, [Jj , Vk ] = iǫjkℓ Vℓ . So J~ and K
~ rotate as vectors ought to,
But the translations
µ do not commute with the Lorentzµtransformations:
~ Eq. (7) is something

under the action of the angular momentum generators J. (Λ Tx)µ = e−iaℓ Lℓ (xν + cν ), while (T Λx)µ = e−iaℓ Lℓ xν + cµ , so
ν ν
else however, a somewhat surprising statement that Lorentz boosts do not
commute but rather their commutator is a generator of a rotation. (In 504,  µ
([Λ, T]x)µ = e−iaℓ Lℓ − 1I cν .
E&M II, we see that this gives rise to the Thomas precession and g = 2 for ν
the electron.) µP

The algebra of these six generators is simplified if we consider the complex As e−ic µ
is the operator which implements T : xµ → x′ µ = xµ + cµ , we
linear combinations Lj ± := 21 (Jj ± iKj ), which satisfy the commutators have  µ

−ia −ic µ

−i e−iaℓ Lℓ cν P µ µ
i e ℓ Lℓ ,e P µ =e ν − e−ic Pµ .
[Lj + , Lk + ] = ǫjkℓ (Jℓ + iKℓ + iKℓ + Jℓ ) = iǫjkℓ Lℓ + ,
4
i Expanding to first order in aℓ and cν gives
[Lj + , Lk − ] = ǫjkℓ (Jℓ − iKℓ + iKℓ − Jℓ ) = 0,
4 [Lαβ , Pν ] = (Lαβ )µν Pµ = i(δαµ gβν − δβµ gαν )Pµ = −igαν Pβ + igβν Pα .
i
[Lj − , Lk − ] = ǫjkℓ (Jℓ − iKℓ − iKℓ + Jℓ ) = iǫjkℓ Lℓ − . (8)
4 Thus we have found the commutation relations which define the Lie algebra
Thus we have two sets of mutually commuting generators, so we can find the of the Poincaré group.
possible representations of fields by asking how they transform under each of
the two independant algebras, each of which has the commutation relations
of ordinary rotations, SO(3) or SU(2). We know the finite dimensional rep- 2 Casimir Operators
resentations from our quantum mechanics course — they are labelled by a
total spin which is a half integer. An operator C constructed from the Lie algebra generators which commutes
We have discussed the properties of the Lorentz transformations as if they with all the generators is called a Casimir operator. One easy example for
were simply matrices acting on coordinates, but of course we also have oper- the Poincaré group is C1 = P2 := Pµ Pµ , for it obviously commutes with all
ators which act on the states of our system, provided for us by the Noether Pν , but also
theorem. These operators need to have the same group properties as matri-
ces do, but to distinguish the more general operators, we will write them in [Lαβ , C1 ] = [Lαβ , Pµ ]Pµ + Pµ [Lαβ , Pµ ]
boldface, Λ, Lαβ , Jℓ , Kℓ , and Lℓ± being the operator versions of Λ, Lαβ , Jℓ , Kℓ , = iδαµ Pβ Pµ − iδβµ Pα Pµ + igαµ Pµ Pβ − igβµ Pµ Pα
and Lℓ± respectively. We need to make this distinction at this point to deal = i[Pβ , Pα ] + i[Pα , Pβ ] = 0
with translations, which do not act as matrices on the coordinate space.
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as the P’s commute. Actually, as you will find useful to prove for Homework The only null vectors whose products with a given non-zero null vector
#3, question 1 , the Lorentz product of any two vectors W µ Vµ commutes vanish are those proportional to that vector. To check that, go to the frame
with Lαβ , even if W and V do not commute with each other. where the given vector is (E, 0, 0, ±E). Then the other null vector V has
A less obvious Casimir operator is the square of the four-vector V0 = ±V3 and has no room for V1 or V2 . So Wµ |pi = hPµ |pi for some
number h, known as the helicity. As W and P both transform as vectors
1
Wµ = ǫµνρσ Pν Lρσ , under proper Lorentz transformations, we expect h to be invariant, and by
2 examining its value in the reference frame with P µ = (E, 0, 0, E) we see that
which is known as the Pauli-Lubanski vector. Here ǫµνρσ is the four-dimensional the helicity is the angular momentum in the direction of motion,
Minkowski space tensor, defined by being totally antisymmetric under inter-
change of any two indices, with ǫ0123 = 1. In Homework #3, question 1 we J~ · P~
h=− .
show that W2 also commutes with all the generators of the Poincaré group. |P~ |
Because any Casimir operator commutes with all the generators of the
group, any irreducible representation has the Casimir operator acting as a But J~ is a pseudo-vector, while P~ is a vector, that is, under parity ~x → −~x,
c-number on it. As we have seen, we expect single particle states to lie t → +t, P~ changes sign, while J~ does not. So the helicity changes sign under
in irreducible representations of the Poincaré group, so to have specific parity, and if we want a parity-invariant theory, our non-zero helicity states
numerical values for P2 and W2. must occur in pairs, while if we don’t care about parity that is not the case.
Of course we recognize P 2 = E 2 − P~ 2 = m2 as the square of the mass of a We shall see that this is an important issue in neutrino physics.
system. What is W 2 ? This is most easily understood classically by going to As we saw earlier, a field that is not a scalar will be part of a collection
rest frame of the system, where P~ = 0, P 0 = m. As W2 is Lorentz invariant φa (x) satisfying U(Λ)φa (x)U −1 (Λ) = Mba (Λ)φb (Λx) for some representation
(it commutes with all Lµν ) this is sufficient. Then Wµ = 12 m ǫµ0ρσ Lρσ = M of the Lorentz group. The books all write this differently,
1
2
m ǫµ0jk Ljk , which vanishes for µ = 0 and Wℓ = − 21 m ǫℓjk Ljk = −mJℓ , so
W 2 = −m2 J 2 = −m2 s(s + 1), where s is the quantum number for the total U(Λ)φa (x)U −1 (Λ) = Dab (Λ−1 )φb (Λx), (9)
angular momentum of the system in its rest frame. For a single particle that
is called the spin, and from quantum mechanics we know that s must take where D is a representation, but that is equivalent4 .
on only half-integer values. Now, as we have seen, the Lie algebra of the Lorentz transformations can
The above argument assumed our state had a positive P 2 . We might be broken up into two commuting SU(2) algebras. We know that the finite
wish to exclude from consideration states with P 2 < 0, which are tachyons 4
Here are some simple facts about representations:
moving faster than the speed of light, for which there are at least some If M : G → N × N complex matrices is a representation, (so M (g1 )M (g2 ) = M (g1 ◦ g2 )),
tricky problems in being consistent with causality and relativity, and for so are
which there is no experimental evidence, despite recent excitement. But we MC : g 7→ (M (g))∗
certainly cannot exclude massless particle states with P 2 = 0. Note that MI : g 7→ (M (g −1 ))T
in general Pµ W µ = 0 from the definition of W and the commutation of the MH : g 7→ (M (g −1 ))†
momentum. For massless states which might be the limit of those with mass For example,
without having s → ∞, we will have3 W 2 = 0. Then
MI (g1 )MI (g2 ) = (M (g1−1 ))T (M (g2−1 ))T = (M (g2−1 )M (g1−1 ))T = (M (g2−1 ◦ g1−1 ))T
Wµ Wµ |pi = Wµ Pµ |pi = Pµ Pµ |pi = 0. = (M ((g1 ◦ g2 )−1 ))T = MI (g1 ◦ g2 ).
3
States with m = 0, W 2 6= 0 do not seem to occur, though I am not sure what would In particular, by replacing the representation M (g) by D(g) = MI (g) = M (g −1 )T , we get
follow for them. the revised expression (9).
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dimensional representations of SU(2) are labelled by a half integer spin s, or

~J, φm (x) = − 1 ~σmm′ φm′ + i~x × ∇φ
~ m.
h i
and the two subscript indices are the mz indices which run from −s to s
by unit steps. Each index for the state under Lorentz transformations then 2
turns into a pair of indices, one for L+ and one for L− . We will call the spin Notice that this is perhaps opposite of what you would expect from non-
A for L+ and B for L− , and replace φb by φm1 m2 . Rewriting relativistic quantum mechanics, as we might have expected
~ ~ ~ ~ ~ ~ ~ ~
Λ = e−iθ · J − i~κ · K = e−iθ · (L+ + L− ) − ~κ · (L+ − L− ) ~ = ~r × p~ + 1 ~σ ,
L ~
(with p~ = −ih̄∇)
~ ~ ~ ~ 2
= e−iθ+ · L+ − iθ− · L−
for a spin 1/2 particle. But our expressions for H and P had the same
with reversed sign,
θ+ = θ − iκ, θ− = θ + iκ.
Then ∂
[H, φ] = −i φ,
~ ~ ~ ~
! ∂t
D(m1 m2 )(m′1 m′2 ) e+iθ+ · L+ + iθ− · L− ∂
[Pj , φ] = +i j φ,
! ! ∂x
~ ~ ~ ~
= A
Dm ′
1 m1
e+iθ+ · L+ Dm
B
′
2 m2
e+iθ− · L− . which are similarly opposite to H = ih̄∂/∂t and p~ = −ih̄∇ which we are
used to from Quantum Mechanics. This is connected to the reversed φ(Λx)
Evaluating U(Λ)φa (x)U −1 (Λ) − φa (x) from (9) to first order, we have5 at the end of Lecture 5.
−i ~θ · ~J + ~κ · K,
~ φm m (x) = D A ′ (i~θ+ · L
~ + )φm′ m
h i
1 2 m1 m1 1 2
(10)
~ ~ ~ ~µ ~ µ xν ∂µ φm m .
 
B
+Dm ′ (iθ− · L− )φm1 m′ + −iθ · J ν − i~
2 m2 2
κ·K ν 1 2

Let us first consider the ( 12 , 0) representation, that is, A = 12 , B = 0.

Then there are no m2 indices (or rather, there is only one value for it) and
D 0 = 1, while
1 1
D 2 (Li ) = σi ,
2
the familiar Pauli spin matrix. Thus
1
[Jℓ , φm (x)] = − σℓ mm′ φm′ + Jℓ µν xν ∂µ φm
2
1
= − σℓ mm′ φm′ − iǫℓjk xk ∂j φm
2
5
We are using the same notation D for the representation of the group elements,
D : g → D(g) for g ∈ G, and for the representation of the generators of infinitesimal
transformations, L ∈ G, where the Lie algebra G is the set of linear combinations of the
basis of generators. Hopefully you have already mastered this possibly confusing notation
in treating the rotation group in quantum mechanics. As the group elements g = eiθℓ Lℓ
can all be written as exponentials of the Lie algebra elements Lℓ , it is natural to define
D(g) = eiθℓ D(Lℓ ) .