# Weinberg-Witten theorem The Weinberg-Witten theorem(s) states: > **Theorem 1**: A theory that allows the construction of a Lorentz-covariant, conserved current $J^\mu$ cannot contain massless particles of spin gt; 1/2$ with non-vanishing value of $\int J^0 d^3x$ (i.e. charge). > > **Theorem 2**: A theory that allows a Lorentz-covariant, conserved $T^{\mu\nu}$ cannot contain massless particles of spin $j>1$. This forbids a $d$-dimensional gravitational theory to emerge from a $d$-dimensional theory without gravitons, but the theory does not say anything about a $(d+1)$-dimensional gravitational theory emerging from a $d$-dimensional field theory, i.e., [[0001 AdS-CFT|AdS/CFT]] is allowed! ## **Remarks** - the theorems apply to both elementary and composite particles - the theorem is compatible with QED - photons: spin 1 but no charge - electrons: spin 1/2 - [[0071 Yang-Mills|YM]] SU(2): - with $A^3_{\mu}, \,A^{\pm}_\mu=\frac{1}{\sqrt{2}}(A_\mu^1\pm iA_\mu^2)$, $A^{\pm}_\mu$ massless spin-1 charged under $U(1)$ subalgebra generated by $\sigma^3/2$ - BUT: there does not exist a conserved Lorentz-covariant current for this $U(1)$ - gauge-invariant and conserved current is not Lorentz-covariant - Lorentz-covariant and gauge invariant current is not conserved - these theorems do not forbid graviton from GR - In GR, there is no *conserved* Lorentz-covariant $T^{\mu\nu}$ - $\nabla_\mu T^{\mu\nu}=0$ <=> *covariantly conserved* (but not conserved) - *conserved* means ordinary derivatives - theorem 2 => none of renormalisable QFTs in Minkowski can have emergent gravity - hidden assumption: the particles live in the same spacetime in the original theory! ## Proof - Suppose there exist massless particles of spin-$j$. - one particle state: $|k^\mu,\sigma\rangle$, $\sigma=\pm j$ = helicity - consider rotation $\hat{R}(\theta,\hat k)|k,\sigma\rangle=e^{i\sigma \theta}|k,\sigma\rangle$ around its spatial direction - now conserved current $J^\mu$ => $Q=\int d^3 x J^0$ - $T^{\mu\nu}$=> $\hat P^\mu=\int T^{0\mu}$, $\hat P^\mu|k,\sigma\rangle=k^\mu|k,\sigma\rangle$ - if charged under $\hat Q$: $\hat Q|k,\sigma\rangle=q|k,\sigma\rangle$ - want to show: 1. if $q\ne0$, $j\le1/2$; 2. $j\le1$. - i) Lorentz symmetry: $\left\langle k, \sigma\left|J^{\mu}\right| k^\prime, \sigma\right\rangle$ -> $\frac{qk^\mu}{k^0}\frac{1}{(2\pi)^3}$ and $\left\langle k, \sigma\left|T^{\mu\nu}\right| k^\prime, \sigma\right\rangle$ -> $\frac{k^\mu k^\nu}{k^0}\frac{1}{(2\pi)^3}$ as $k\rightarrow k^\prime$ - intuition: when $\mu=0$, $|k,\sigma\rangle$ is eigenstate of $J^0$ with eigenvalue $q$; when $\mu\ne0$ is must carry a $\mu$ index and must reduce to $q$ when $\mu=0$ - ii) for massless particles, $k^2=0={k^\prime}^2$ => $k.k^\prime<0$, $k+k^\prime$ timelike. -> choose a frame s.t. $\vec k+\vec k^\prime=0$. so $k^\mu=(E,0,0,E),{k^\prime}^\mu=(E,0,0,-E)$ - iii) under rotation of $\theta$ around 3-direction, $\hat R(\theta)|k,j\rangle=e^{\mathrm{i}j\theta}|k,j\rangle$, $\hat R(\theta)|k^\prime j\rangle=e^{-\mathrm{i}j\theta}|k^\prime,j\rangle$. Consider $\langle k^\prime,j|\hat R^{-1}(\theta)J^\mu \hat R(\theta)|k,j\rangle$ => $e^{2 i j{\theta}}\langle k^\prime, j| J^{k}|k, j\rangle ={\Lambda^\mu}_\nu\langle k^\prime, j| J^{k}|k, j\rangle$ ==(4)==. - can do it for $T^{\mu\nu}$ similarly => ==(5)== - now $\Lambda$ can only have eigenvalue $e^{\pm i\theta}, 1$ - => $j\le1/2$ from (4) and $j\le1$ from (5). QED ## Refs - almost all the notes above are taken from [[Rsc0010 Hong Liu Lectures on holography]] ## Alternative derivation - [[BenedettiCasiniMagan2022]][](https://arxiv.org/pdf/2205.03412.pdf) - using generalised symmetry and Noether theorem techniques