Why a massless spin-2 field transforms with a diffeomorphism

Introduction

In the previous post we saw that gauge invariance is not just a decorative feature of electromagnetism. Classically, the electromagnetic potential has the redundancy

$$A_\mu \to A_\mu+\partial_\mu\chi,$$

and in quantum field theory that redundancy is forced on us if a local Lorentz-covariant vector field is supposed to describe a massless helicity-1 particle.

The same logic has a spin-2 version.

If a massless spin-2 particle is described by a local Lorentz-covariant symmetric tensor field $h_{\mu\nu}(x)$, then under Lorentz transformations this field cannot transform as a strict rank-two tensor on the physical Hilbert space. It must transform as a tensor plus a redundancy of the form

$$h_{\mu\nu}(x)\to h_{\mu\nu}(x) +\partial_\mu \xi_\nu(x)+\partial_\nu \xi_\mu(x).$$

This is exactly the linearized form of diffeomorphism invariance. If the full metric is written as

$$g_{\mu\nu}=\eta_{\mu\nu}+\kappa h_{\mu\nu},$$

then an infinitesimal coordinate transformation produces, at leading order around flat spacetime,

$$\delta h_{\mu\nu} =\partial_\mu\xi_\nu+\partial_\nu\xi_\mu,$$

up to sign conventions for active versus passive transformations.

So the question is:

Why should quantum mechanics force a spin-2 field to have the same redundancy that the metric has under infinitesimal coordinate transformations?

The answer is again the little group of a massless particle.

One important scope point: the argument below is about a symmetric rank-two field used to describe a pure massless spin-2 particle. It is not a statement about every possible two-index Lorentz tensor.

From the point of view of Lorentz representations, a generic rank-two tensor is reducible. It decomposes into three natural pieces:

$$T_{\mu\nu} = \left(T_{(\mu\nu)}-\frac{1}{4}\eta_{\mu\nu}T^\rho{}_\rho\right) +T_{[\mu\nu]} +\frac{1}{4}\eta_{\mu\nu}T^\rho{}_\rho.$$

These are:

• the symmetric traceless part, which contains the spin-2 representation;
• the antisymmetric part, which is a two-form representation;
• the trace, which is a scalar representation.

Each sector should be analyzed according to the physical massless representation it is meant to describe. The antisymmetric two-form has a different gauge redundancy, for example

$$B_{\mu\nu}\to B_{\mu\nu} +\partial_\mu\Lambda_\nu-\partial_\nu\Lambda_\mu,$$

not the symmetric diffeomorphism-like transformation. The transformation

$$h_{\mu\nu}\to h_{\mu\nu} +\partial_\mu\xi_\nu+\partial_\nu\xi_\mu$$

belongs to the symmetric tensor field $h_{\mu\nu}$ that represents the metric perturbation and carries the physical helicities $\pm2$. In the rest of this post we restrict attention to that symmetric spin-2 sector.

The quantum-mechanical starting point

In quantum mechanics, symmetries must preserve transition probabilities. Wigner’s theorem says that such symmetries are represented on the Hilbert space by either unitary or antiunitary operators. Following Weinberg’s construction, for continuous spacetime symmetries connected to the identity we use unitary representations of the Poincare group.

One-particle states may be labelled as

$$|p,\sigma\rangle,$$

where $p^\mu$ is the four-momentum and $\sigma$ is the internal label. For a massive particle $\sigma$ is ordinary spin. For a massless particle $\sigma$ is helicity.

As before, choose a standard null momentum

$$k^\mu=(\kappa,0,0,\kappa),$$

and choose a Lorentz transformation $L(p)$ such that

$$p^\mu={L(p)^\mu}_\nu k^\nu.$$

Given an arbitrary Lorentz transformation $\Lambda$, the part that acts on the internal labels is

$$W(\Lambda,p)=L^{-1}(\Lambda p)\,\Lambda\,L(p).$$

This transformation leaves $k$ fixed:

$$W(\Lambda,p)k=k.$$

Therefore $W(\Lambda,p)$ belongs to the little group. For a massless particle the little group is

$$ISO(2).$$

Its rotation subgroup measures helicity. Its translation subgroup is the source of gauge redundancy.

The physical spin-2 representation

The Lie algebra of the massless little group is generated by

$$J_3,\qquad N_1,\qquad N_2,$$

with

$$[J_3,N_1]=iN_2,\qquad [J_3,N_2]=-iN_1,\qquad [N_1,N_2]=0.$$

For ordinary massless particles of fixed helicity, such as photons or gravitons, we take the discrete-helicity representation. In this representation the translation generators act trivially on physical one-particle states:

$$N_1|k,\sigma\rangle=N_2|k,\sigma\rangle=0.$$

The rotation part acts by a phase:

$$U(R(\theta))|k,\sigma\rangle =e^{i\sigma\theta}|k,\sigma\rangle.$$

This equation is the definition of helicity in this frame. The number $\sigma$ is the eigenvalue of the generator of rotations around the direction of motion. Equivalently, if a state or polarization picks up the phase $e^{i\sigma\theta}$ under a rotation by $\theta$ around the momentum axis, then its helicity is $\sigma$.

For the spin-2 field we will see below that the physical transverse tensor polarizations pick up the phases

$$e^{+2i\theta},\qquad e^{-2i\theta}.$$

That is where the helicities $+2$ and $-2$ come from. If parity is imposed, both helicities appear. The little-group argument itself does not require parity; it only tells us how each helicity representation must fit inside a Lorentz-covariant field.

Vector polarizations first

For the standard null momentum $k^\mu=(\kappa,0,0,\kappa)$, choose the usual transverse spin-1 polarization vectors

$$\epsilon_+^\mu(k)=\frac{1}{\sqrt{2}}(0,1,i,0), \qquad \epsilon_-^\mu(k)=\frac{1}{\sqrt{2}}(0,1,-i,0).$$

They satisfy

$$k_\mu\epsilon_\pm^\mu(k)=0.$$

Under the rotation part of the little group,

$$R(\theta)\epsilon_\pm^\mu(k) =e^{\pm i\theta}\epsilon_\pm^\mu(k).$$

Under the translation part of the little group, the polarization vector does not remain exactly the same. To first order in the little-group translation parameters, it shifts by a multiple of the null momentum:

$$S(\alpha,\beta)\epsilon_\pm^\mu(k) =\epsilon_\pm^\mu(k)+c_\pm(\alpha,\beta)k^\mu +O(\alpha^2,\beta^2,\alpha\beta).$$

This is the spin-1 seed of gauge invariance: the physical Hilbert-space state does not change under little-group translations, but the Lorentz vector polarization does change by a longitudinal term.

Build spin-2 polarizations

A helicity-$\pm2$ polarization tensor may be built from the symmetric product of two helicity-$\pm1$ polarization vectors:

$$\epsilon_{\pm}^{\mu\nu}(k) =\epsilon_\pm^\mu(k)\epsilon_\pm^\nu(k).$$

This tensor is symmetric,

$$\epsilon_{\pm}^{\mu\nu}=\epsilon_{\pm}^{\nu\mu},$$

transverse,

$$k_\mu\epsilon_{\pm}^{\mu\nu}=0,$$

and traceless,

$${\eta_{\mu\nu}}\epsilon_\pm^{\mu\nu}=0.$$

Under a rotation around the direction of motion,

$$R(\theta)\epsilon_{\pm}^{\mu\nu}(k) =e^{\pm 2i\theta}\epsilon_{\pm}^{\mu\nu}(k).$$

This is the promised derivation of the helicity label. Each vector polarization carries helicity $\pm1$, so each one picks up a phase $e^{\pm i\theta}$. The tensor polarization is a product of two equal-helicity vector polarizations, so the phases multiply:

$$e^{\pm i\theta}e^{\pm i\theta}=e^{\pm 2i\theta}.$$

Comparing with

$$U(R(\theta))|k,\sigma\rangle=e^{i\sigma\theta}|k,\sigma\rangle,$$

we identify these tensor polarizations as helicity $+2$ and helicity $-2$. This is what the word “spin-2” means for a massless particle: the physical one-particle representation has helicity magnitude $2$. The symmetric tensor field has more components than these two physical polarizations, and the remaining components are removed by the gauge redundancy derived below.

Now apply a little-group translation. Since each vector polarization shifts by a term proportional to $k^\mu$, the tensor polarization transforms as

$$\epsilon_\pm^\mu\epsilon_\pm^\nu \to \left(\epsilon_\pm^\mu+c_\pm k^\mu\right) \left(\epsilon_\pm^\nu+c_\pm k^\nu\right).$$

Keeping only first order in the little-group translation parameters gives

$$\epsilon_{\pm}^{\mu\nu}(k) \to \epsilon_{\pm}^{\mu\nu}(k) +k^\mu \zeta_\pm^\nu(k) +k^\nu \zeta_\pm^\mu(k),$$

where

$$\zeta_\pm^\mu(k)=c_\pm(\alpha,\beta)\epsilon_\pm^\mu(k).$$

This is the crucial spin-2 analogue of the spin-1 result.

The physical graviton state does not transform under the little-group translations, because we are in the discrete-helicity representation. But the Lorentz-covariant tensor polarization does transform. The mismatch must be a redundancy of the field description.

From polarization shift to field transformation

Write a free massless spin-2 field schematically as

$$h^{\mu\nu}(x) = \sum_{\sigma=\pm2} \int d\Pi_p \left[ \epsilon_\sigma^{\mu\nu}(p)a_\sigma(p)e^{ip\cdot x} +\epsilon_\sigma^{\mu\nu}(p)^*a_\sigma^\dagger(p)e^{-ip\cdot x} \right],$$

where $d\Pi_p$ is the Lorentz-invariant massless phase-space measure.

The little-group translation shifts the polarization tensor by

$$\epsilon^{\mu\nu}(p) \to \epsilon^{\mu\nu}(p)+p^\mu \zeta^\nu(p)+p^\nu \zeta^\mu(p).$$

In position space, multiplication by $p^\mu$ becomes a derivative. Therefore the field must be identified under

$$h^{\mu\nu}(x) \to h^{\mu\nu}(x) +\partial^\mu \xi^\nu(x) +\partial^\nu \xi^\mu(x).$$

Lowering indices,

$$h_{\mu\nu}(x) \to h_{\mu\nu}(x) +\partial_\mu \xi_\nu(x) +\partial_\nu \xi_\mu(x).$$

This is not an optional convention. It is the redundancy required if a local Lorentz-covariant symmetric tensor is to describe only the physical helicity $\pm2$ states of a massless particle.

Recovering linearized diffeomorphisms

Now compare this with the transformation of a metric under an infinitesimal diffeomorphism.

Let

$$x^\mu\to x^\mu-\xi^\mu(x).$$

The metric transforms by the Lie derivative:

$$\delta g_{\mu\nu} =\mathcal{L}_\xi g_{\mu\nu} =\xi^\rho\partial_\rho g_{\mu\nu} +g_{\rho\nu}\partial_\mu\xi^\rho +g_{\mu\rho}\partial_\nu\xi^\rho.$$

Now expand around flat spacetime:

$$g_{\mu\nu}=\eta_{\mu\nu}+\kappa h_{\mu\nu}.$$

At leading order in the fluctuation, the background metric is constant, so

$$\partial_\rho \eta_{\mu\nu}=0.$$

Keeping only the first-order transformation of the perturbation gives

$$\delta h_{\mu\nu} =\partial_\mu\xi_\nu+\partial_\nu\xi_\mu,$$

up to an overall normalization of $\xi$ and the sign convention for the diffeomorphism.

This is exactly the redundancy found from the quantum-mechanical little-group argument.

So at first order we have recovered linearized diffeomorphism invariance from the requirement that a massless helicity-2 particle be described by a Lorentz-covariant local tensor field without introducing unphysical states.

What this does and does not prove

The argument proves the linearized gauge redundancy:

$$\text{massless helicity-2} +\text{ Lorentz covariance} +\text{ unitarity} \Longrightarrow h_{\mu\nu}\sim h_{\mu\nu} +\partial_\mu\xi_\nu+\partial_\nu\xi_\mu.$$

This is the spin-2 analogue of the photon statement

$$A_\mu\sim A_\mu+\partial_\mu\chi.$$

It does not, by itself, derive the full nonlinear Einstein equations. To get there one must also ask how the massless spin-2 field can interact consistently. Requiring consistent coupling to energy and momentum, together with the preservation of the gauge redundancy, leads to the nonlinear diffeomorphism-invariant structure of general relativity.

But the first step is already remarkable: before invoking geometry as a classical starting point, quantum mechanics plus Lorentz symmetry already force the field that describes a massless spin-2 particle to carry the infinitesimal redundancy of a metric perturbation.

References

• Steven Weinberg, The Quantum Theory of Fields, Volume I, Chapter 5.
• Steven Weinberg, The Quantum Theory of Fields, Volume I, sections on massless particles and the little group.
• Steven Weinberg, The Quantum Theory of Fields, Volume II, chapters on general relativity as the theory of an interacting massless spin-2 field.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, sections on gravitons and linearized gravity.