A common procedure to determine the spin of the excitations of a quantum field is to first determine the conserved currents arising from quasi-symmetries via Noether's theorem. **For example**, in the case of the Dirac field, described by the Lagrangian,

$$\mathcal{L}=\bar{\psi}(i\gamma^\mu \partial_\mu -m)\psi $$

the associated conserved currents under a translation are,

$$T^{\mu \nu} = i \bar{\psi}\gamma^\mu \partial^\nu \psi - \eta^{\mu \nu} \mathcal{L}$$

and the currents corresponding to Lorentz symmetries are given by,

$$(\mathcal{J}^\mu)^{\rho \sigma} = x^\rho T^{\mu \sigma} - x^\sigma T^{\mu \rho}-i\bar{\psi}\gamma^\mu S^{\rho \sigma} \psi$$

where the matrices $S^{\mu \nu}$ form the appropriate representation of the Lorentz algebra. After canonical quantization, the currents $\mathcal{J}$ become operators, and acting on the states will confirm that, in this case, the excitations carry spin $1/2$. In gravity, we proceed similarly. The metric can be expanded as,

$$g_{\mu \nu} = \eta_{\mu \nu} + f_{\mu \nu}$$

and we expand the field $f_{\mu \nu}$ as a plane wave with operator-valued Fourier coefficients, i.e.

$$f_{\mu \nu} \sim \int \frac{\mathrm{d}^3 p}{(2\pi)^3} \frac{1}{\sqrt{\dots}} \left\{ \epsilon_{\mu \nu} a_p e^{ipx} + \dots\right\}$$

**We only keep terms of linear order** $\mathcal{O}(f_{\mu \nu})$, compute the conserved currents analogously to other quantum field theories, and once promoted to operators as well act on the states to determine the excitations indeed have spin $2$. An argument as to why a massless spin $2$ particle must be a graviton is given by Feynman et al., and I present verbatim a paraphrasing by Professor D. Tong:

To summarize, theories of massless spin 2 fields only make sense if there is a gauge symmetry to remove the negative norm states. In general relativity, this gauge symmetry descends from diffeomorphism invariance. The argument of Feynman and Weinberg now runs this logic in reverse. It goes as follows: suppose that we have a massless, spin 2 particle. Then, at the linearized level, it must be invariant under the gauge symmetry $f_{\mu \nu} \to f_{\mu \nu} + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu$ in order to eliminate the negative norm states. Moreover, this symmetry must survive when interaction terms are introduced. But the only way to do this is to ensure that the resulting theory obeys diffeomorpism invariance. That means the theory of any interacting, massless spin 2 particle is Einstein gravity, perhaps supplemented by higher derivative terms.

**Supplementary Information**

The negative norm states that Tong refers to are ghost states which the Polyakov action initially suffers from. They arise because of the commutation relations,

$$[a_{m}^\mu,a^\nu_n] = m\delta_{m+n,0} \eta^{\mu \nu}$$

Regardless of the metric convention, either timelike or spacelike oscillators give rise to the negative norms. They are unphysical, violate unitarity, and hence a consistent theory requires their removal.

**Counting physical degrees of freedom**

The graviton has spin $2$, and as it is massless only two degrees of freedom. We can verify this in gravitational perturbation theory. We know $h^{ab}$ is a symmetric matrix, and only $d(d+1)/2$ distinct components. In de Donder gauge, $$\nabla^{a}\bar{h}^{ab} = \nabla^a\left(h^{ab}-\frac{1}{2}h g^{ab}\right) = 0$$

which provides us $d$ gauge constraints. There is also a residual gauge freedom, providing that infinitesimally, we shift by a vector field, i.e.

$$X^\mu \to X^\mu + \xi^\mu$$

providing $\square \xi^\mu + R^\mu_\nu \xi^\nu = 0$, which restricts us by $d$ as well. Therefore the total physical degrees of freedom are,

$$\frac{d(d+1)}{2}-2d = \frac{d(d-3)}{2}$$

If $d=4$, the graviton indeed has only two degrees of freedom.

**Important Caveat**

Although we often find a field with a single vector index has spin one, with two indices spin two, and so forth, it is not always the case, and determining the spin should be done systematically. Consider, for example, the Dirac matrices, which satisfy the Clifford algebra,

$$\{ \Gamma^a, \Gamma^b\} = 2g^{ab}$$

On an $N$-dimensional Kahler manifold $K$, if we work in local coordinates $z^a$, with $a = 1,\dots,N$, and the metric satisfies $g^{ab} = g^{\bar{a} \bar{b}} = 0$, the expression simplifies:

$$\{ \Gamma^a, \Gamma^b\} = \{ \Gamma^{\bar{a}}, \Gamma^{\bar{b}}\} = 0$$
$$\{ \Gamma^a, \Gamma^{\bar{b}}\} = 2g^{ab}$$

Modulo constants, we see that we can think of $\Gamma^a$ as an annihilation operator, and $\Gamma^{\bar{b}}$ as a creation operator for fermions. Given that we define $\lvert \Omega \rangle$ as the Fock vacuum, we can define a general spinor field $\psi$ on the Kahler manifold $K$ as,

$$\psi(z^a,\bar{z}^{\bar{a}}) = \phi(z^a,\bar{z}^{\bar{a}}) \lvert \Omega \rangle + \phi_{\bar{b}}(z^a,\bar{z}^{\bar{a}}) \Gamma^{\bar{b}} \lvert \Omega \rangle + \dots$$

Given that $\phi$ has no indices, we would expect it to be a spinless field, but it can interact with the $U(1)$ part of the spin connection. Interestingly, we can only guarantee that $\phi$ is neutral if the manifold $K$ is Ricci-flat, in which case it is Calabi-Yau manifold.

This post imported from StackExchange Physics at 2015-03-04 16:10 (UTC), posted by SE-user JamalS