Physical meaning of quartic observable in abelian Higgs model

Consider the $$\mathrm{U}(1)$$ gauge-Higgs model defined by the lagrangian $$$$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+D^\mu\phi^\dagger D_\mu\phi-V(\phi^\dagger\phi),$$$$ where $$V$$ is the usual (ssb inducing) Higgs potential and $$\phi$$ is just a complex scalar field.

If we are working on a space-time lattice, we can get the Higgs mass $$m_H$$ from the two-point correlation function of the observable $$$$\langle\phi^\dagger(x)\phi(x)\rangle.$$$$ My question now is how to interpret the observable $$$$\langle\phi^\dagger(x)\phi(x)\phi^\dagger(y)\phi(y)\rangle$$$$ and what physical quantities can be derived from it. Here $$x,y$$ are two arbitrary lattice points.

• Are you sure the spacetime variables are supposed to be the same in the n-point functions? I would expect the full 2-point function (giving you the effective mass) to be $\langle\phi^\dagger(x)\phi(y)\rangle$. Commented Jun 1, 2021 at 11:58
• Sorry for the late answer! Yes, I am sure. Commented Jun 1, 2021 at 20:03
• if it were non-relativistic I would say they are density and density density correlation respectively. Commented Jun 2, 2021 at 0:24

In a general sense, all correlation functions that involve an arbitrary field $$\mathcal O(x)$$ contain singularities in momentum space, at the position of the masses of all states created by $$\mathcal O(x)$$. "Polology" as referred to by Weinberg, cf. ref1§10.2.

For the specific example $$\mathcal O=\phi$$, the statement is that all correlation functions of $$\phi$$ allow you to measure the mass $$m_H$$. The two-point function is the minimal such object hence the most convenient, but the four-point function works too.

In this sense, as far as your interpretation of $$\langle\phi^2\rangle$$ goes, the interpretation of $$\langle\phi^4\rangle$$ is identical.

That being said, it is more common to utilize $$\langle\phi^4\rangle$$ to define a general notion of strength of interaction: we typically define the four-body interaction coupling constant as (cf.ref2§16) $$\lambda_4:=\lim_{p^2\to\mu^2}\langle\phi^4(p)\rangle_c$$ where $$\bullet(p)$$ denotes the Fourier transform of $$\bullet(x)$$ and $$\mu$$ is some convenient mass scale, such as $$\mu=m_H$$ or sometimes also $$\mu=0$$; the choice is yours. Also, the subscript $$c$$ denotes connected part, the idea being that in the free case $$\langle\phi^4\rangle_c=0$$ and so $$\lambda_4=0$$, consistent with lack of interactions.

The object $$\lambda_4$$ is a definition so not precisely useful in developing intuition for what $$\langle\phi^4\rangle$$ does for you. But anyway, this is a general issue (or feature) of QFTs. Correlation functions are renormalization-dependent so they are not directly observable. Because of how divergences are organized, the ambiguities are often fixed once you make a finite number of choices. And the most convenient choices (both from the computational point of view, and from the experimental point of view) involve the lowest order correlation functions. For example, we typically use the $$i$$-th point function to define the $$i$$-th point vertex (with $$i=2$$ denoting the mass), and after fixing a few of these, you can use any other correlation function to predict stuff. Healthy theories only require finitely many choices. (Also, keep in mind that this general description is only one of the many possible approaches. All correlation functions are sensitive to all parameters of your theory, so you can use whatever you want to fix/measure anything. In the example above, you could use $$\langle\phi^4\rangle$$ to fix the mass $$m_H$$, and $$\langle\phi^2\rangle$$ to fix the strength $$\lambda_4$$. Both correlation functions are sensitive to both parameters.)

References.

1. Weinberg - Quantum theory of fields, Vol.1. Foundations.

2. Srednicki - Quantum Field Theory.

• Thank you very much for your answer! I will take a closer at those references. Commented Jun 2, 2021 at 6:03