# What does Lorentz index structure say about a full-fledged correlator?

I have a probably dumb question. Consider the following position space correlation function in a YM-theory (with or without matter fields):

$$f_{\mu_1\cdots \mu_n}^{a_1\cdots a_n}(x_1,\ldots,x_n)=\langle\Omega|T\left\{A_{\mu_1}^{a_1}(x_1)\cdots A_{\mu_n}^{a_n}(x_n)\right\}|\Omega \rangle \tag{1}$$

where $$A$$ is the usual gauge-connection (or gluon field, whatever you want to call it). What's crucial here is that $$|\Omega\rangle$$ is the full-fledged interacting vacuum state, which I believe by assumption/construction is invariant under Lorentz transformations (of course, acting on the states $$|\psi\rangle$$ under a suitable unitary representation, which I'll also take to not be projective for simplicity):

$$U(\Lambda)|\Omega\rangle=|\Omega\rangle \tag{2}$$

where $$\Lambda\in SO(3,1)$$ is a Lorentz transformation, and $$U(\Lambda)$$ is a unitary representation of it.

Now, I believe the only invariant tensors of $$SO(3,1)$$ in the defining representation (i.e. with the $$\mu$$ indices) are the Minkowski metric and combinations of them:

$$\eta^{\mu\nu}=\Lambda^\mu_\rho \Lambda ^\nu_\sigma\eta^{\rho\sigma} \tag{3}$$

Here's my question: With this in mind, can we say that all correlation functions with an odd number of Lorentz indices (living in the defining rep. of course) is equal to zero? For example, take the one-point function, and ignore the color indices:

\begin{align} a^\mu &= \langle \Omega | A^\mu (x) |\Omega \rangle \,\,\,\,\leftarrow(x \text{ independent by translation inv.})\\ &\\ &= \langle \Omega | \underset{\text{insert}}{\underbrace{U^\dagger (\Lambda) U(\Lambda)}}A^\mu (x) \underset{\text{insert}}{\underbrace{U^\dagger (\Lambda) U(\Lambda)}} |\Omega \rangle \\ &\\ &=\langle \Omega | \left[ U(\Lambda)A^\mu (x)U^\dagger (\Lambda) \right] |\Omega \rangle \\ &\\ &=\langle \Omega | \Lambda^\mu_\nu A^\nu (x) |\Omega \rangle \\ &\\ &=\Lambda^\mu_\nu a^\nu \tag{4} \end{align}

Thus $$a^\mu$$ must be proportional to an invariant tensor with only one free index. Since there is none, it's equal to zero. Similar arguments applied to higher-point functions would imply:

\begin{align} \langle A_\mu A_\nu \rangle &\sim \eta_{\mu\nu} \tag{5}\\ &\\ \langle A_\mu A_\nu A_\rho \rangle &\sim 0 \tag{6}\\ &\\ \langle A_\mu A_\nu A_\rho A_\sigma \rangle &\sim \eta_{\mu\nu}\eta_{\rho\sigma} + \text{ permutations of }(\mu\nu\rho\sigma) \tag{7} \end{align}

So the above reasoning would imply that any n-point correlator with $$n=$$odd is zero, but how can that be correct? The amplitude for $$1\rightarrow 2$$ gauge scattering is calculated from this exact same correlator:

\begin{align} \langle q_1,\lambda_1; \ldots \text{(out)}| p_1,\alpha_1; \ldots \text{(in)}\rangle &= \lim_{p,q\rightarrow \text{on-shell}} \prod_{p_i} \frac{(-i)\epsilon_{\mu_i}(p_i,\alpha_i)\Delta^{-1}(p_i)^{\mu_i\nu_i}}{(2\pi)^4} \prod_{q}\, \text{(similar thing...)}\\ &\\ &\times \int d^4x_i d^4y_i \,\,e^{ip_ix_i}e^{-iq_iy_i} \langle \Omega |T\left\{ A_{\nu_i}(x_i)\cdots A_{\rho_i}(y_i) \right\}|\Omega\rangle \tag{8} \end{align}

where $$\alpha,\lambda$$ are physical polarizations. If my previous arguments still hold, then that would mean $$1\rightarrow 2$$ scattering is impossible! But that just seems silly.

I'd also appreciate directions to resources that discuss this.

• The two-point function, for example, depends on $x-y$. So you have more Lorentz objects beyond the metric: $\alpha(x-y)\ \eta^{\mu\nu}+\beta(x-y)\ (x-y)^\mu (x-y)^\nu$ instead of just $\propto\eta^{\mu\nu}$ (with $\alpha,\beta\colon \mathbb R^d\to \mathbb C$ a pair of arbitrary functions). The same holds true for higher order correlators. – AccidentalFourierTransform Feb 25 at 19:44
• @AccidentalFourierTransform Yes but $(x-y)^\mu$ is not invariant under Lorentz transformations, whereas those correlators are (assuming my calculation in (4) is correct), so you can't have any $(x-y)^\mu$ terms. The Lorentz index structure must purely come from invariant tensors, and thus the dependence on the $x_i$'s must only come from the Lorentz invariant combinations of the $x_i$'s. – Arturo don Juan Feb 25 at 19:51
• Nope. $U(\Lambda)A^\mu (x)U^\dagger (\Lambda)=\Lambda^\mu{}_\nu A^\nu(\color{red}{\Lambda} x)$. The two-point function is covariant, not invariant. – AccidentalFourierTransform Feb 25 at 21:06
• @AccidentalFourierTransform Oh shoot thanks, forgot about that. Do you know how that will that affect my (false) results at and after (4)? – Arturo don Juan Feb 25 at 21:38