Homework 2 – Solutions

Massless spin-3 particles

Consider a massless particle with helicity ±3.

1. Construct the field operator and discuss its behavior under Lorentz transformations.

2. What is the gauge invariance we need in order to construct a Lorentz invariant theory?

3. What is the “field-strength” operator in this case—the analogue of F µν , which is gauge-invariant and therefore Lorentz-covariant?

4. By S-matrix arguments, we have seen that in the low-energy limit, massless spin-1 particles(“photons”) can only couple to conserved charges, and massless spin-2 particles (“gravitons”)can only couple to the four-momentum. If we go through the same argument for our masslessspin-3 particles, what do we get?

Solution

1. As usual, the polarization tensors with helicity h = ±3 are

eµνρ±3 (�p) ≡ eµ

±1(�p)eν±1(�p)eρ

±1(�p) , (1)

where eµ±1(�p) is the polarization tensor for a photon of momentum �p and helicity ±1. The

field operator for our spin-3 particle is

Cµνρ(x) ≡ 1

(2π)3/2

∑h=±3

∫d3p√2p0

[eµνρ

h (�p)ah(�p) eip·x + eµνρ ∗h (�p)a†

h(�p) e−ip·x]

, (2)

Under Lorentz transformations everything transforms nicely but the polarization tensors:

Λµµ′Λν

ν′Λρρ′e

µ′ν′ρ′±3 (�p) = · · · = e±3iθ

(eµνρ±3 (Λp) + (Λp)µvνρ + permutations of µ, ν, ρ

), (3)

where the phase θ and the symmetric tensor vµν depend on the transformation Λ as well ason �p. As a result, the field operator transforms as

U(Λ)Cµνρ(x)U−1(Λ) = (Λ−1)µµ′(Λ−1)ν

ν′(Λ−1)ρρ′ C

µνρ(Λx) (4)

+ ∂µξνρΛ (x) + ∂νξρµ

Λ (x) + ∂ρξµνΛ (x) , (5)

where ξµνΛ (x) is a symmetric linear combination of creation and annihilation operators that

depends on the applied Lorentz transformation Λ.

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2. The above transformation law tells us that in order to have Lorentz invariant dynamics, ourLagrangian should have a Lorentz invariant looking structure and also be invariant underthe gauge transformation

Cµνρ(x) → Cµνρ(x) + ∂µξνρ(x) + ∂νξρµ(x) + ∂ρξµν(x) , (6)

for a generic symmetric tensor function ξµν(x).

3. For the photon, we know that we can get a true tensor field if we use the polarization tensor

Eµν±1(�p) ≡ (

ipµ eν±1(�p) − ipν eµ

±1(�p))

, (7)

which still has helicity ±1, in place of eµ±1(�p). So here we just have to construct a field

operator out of the product of three such polarization tensors,

Eµνρστω±3 (�p) ≡ Eµν

±1(�p)Eρσ±1(�p)Eτω

±1(�p) (8)

Dµνρστω ≡ 1

(2π)3/2

∑h=±3

∫d3p√2p0

[Eµνρστω

h (�p)ah(�p) eip·x + h.c.]

(9)

To get the explicit expression in terms of derivatives of the original Cµνρ(x), we can use themnemonic rule we discussed for the Riemann tensor:

Dµνρστω ∼ F µνF ρσF τω (10)

=(∂µAν − ∂νAµ

)(∂ρAσ − ∂σAρ

)(∂τAω − ∂ωAτ

)(11)

→ (∂∂∂C

)([µν][ρσ][τω]), (12)

where by the last expression we mean a tensor built out of triple derivatives acting on C,antisymmetric under exchanging the indices within each of the bracketed pairs, and totallysymmetric under exchanging the pairs. The overall normalization of course is a matter ofconvention. Given the symmetries of D, one can check that it is invariant under the gaugetransformation (6), and it is therefore a true Lorentz tensor.

4. Let’s call c the coupling constant of a given particle to a very soft massless spin-3 particle.That is, following the same notation of the lecture notes, let’s assume that in the very softlimit the emission amplitude is

M = 2i (2π)4 c pµpνpρ · e∗µνρ(q) · δσσ′ , for qµ → 0 . (13)

(This is our elementary vertex amplitude.) Then, the amplitude for emitting a very softspin-3 particle in a generic α → β process is

M → Mαβ

∑n

ηn cn pµnpν

npρn

pn · q · e∗µνρ(q) (14)

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Lorentz invariance tells us that this expression should vanish if we replace the emittedparticle’s polarization tensor with qµ. We get

Mαβ

∑n

ηn cn pµnpν

n = 0 (15)

That is, processes with non-vanishing Mαβ must conserve the tensor

Oµν =∑

cnpµnpν

n . (16)

But, as we discussed, the only kinematical constraint obeyed by all non-trivial scatteringprocesses is the total four-momentum conservation law. For instance, by applying the con-servation of Oµν to a 2 → 2 elastic scattering one gets that either the two particles do notinteract at all, or the coupling constants cn must vanish. We are thus led to the conclusionthat

cn = 0 (17)

for all particles. That is, our massless spin-3 particle cannot have any interactions thatsurvive in the very soft limit, q → 0. The same conclusion applies to higher spins. Noticethat this does not mean that high spin massless particles do not exist. It means that, ifthey do exist, they cannot have interactions that survive at low energies. In particular, theycannot mediate long range 1/r2 forces like Coulomb’s or Newton’s.

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