# Why connected correlation function depends on (x-y) only when <Phi(x)> equal constant

Why the connected correlation functions depend on the x-y(the difference of two space-time coordinates of two points) only in case the expectation of field(<phi(x)>) is constant(the translation invariance of vacuum state).
Thank you very much in advanced.

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vanhees71
Gold Member
2019 Award

You've already answered the question yourself. The correlation or Green's function in vacuum quantum field theory are expectation values of field-operator product with respect to the vacuum state. If the theory is invariant with respect to space-time translations, i.e., if the Lagrangian doesn't depend on space-time coordinates explicitly, the correlation functions are translation invariant, i.e., if you add an arbitrary four-vector to each of its arguments it doesn't change its value, and thus in case of the two-point function (single-particle Green's function or self-energies) it must be a function of the only translation-invariant combination of its two space-time argument you can build, namely the difference of these space-time arguments.

If you have a scalar condensate (e.g., in the linear sigma model or in electroweak symmetry) due to spontaneous symmetry breaking, translation invariance is only preserved if it is space-time dependent. Otherwise the physics at one space-time point is different from that at another (e.g., the pion-decay constant is different at different space-time points if you have a space-time dependent vacuum expectation value of the sigma field), i.e., space-time translation invariance is broken, and your Green's functions are changing under space-time translations.

Thank you very much for your useful answer.By the way,please teach me what is the scalar condensation?

vanhees71
Gold Member
2019 Award

The most simple example is the linear sigma model. It consists of four real scalar fields,

$$\Phi=(\sigma,\vec{\phi}).$$

$$L=\frac{1}{2} (\partial_{\mu} \Phi) \cdot (\partial}_{\mu} \Phi) + \frac{\mu^2}{2} \Phi \cdot \Phi - \frac{\lambda}{4} (\Phi \cdot \Phi)^2.$$

I'm using the west-coast metric (+---), and thus the plus sign in front of the quadratic term is just of the opposite sign to be a proper mass term. What's going on is most easily seen already in the classical limit: The potential has a maximum rather than a minimum at $$\Phi=0$$, and thus you can not have $$\Phi=0$$ as a stable equilibrium condition. Quantizing the model, you cannot do perturbation theory around $$\langle \Phi \rangle=0.$$

You have to do perturbation theory around a minimum. This is continuous-fold degenerated since the model is symmetric under O(4) rotations in $$\Phi$$ space, but you can chose any of these minima and do perturbation theory around that one. The usual convention is to take $$\langle \sigma \rangle=\sigma_0$$ and $$\langle \vec{\phi} \rangle=0.$$

The solution for the minimum for $$\sigma_0=\text{const}$$ is given by

$$\mu^2 \sigma_0-\lambda \sigma_0^3=0, \Rightarrow \sigma_0=\frac{\mu}{\lambda}.$$

Now, you plug the ansatz

$$\Phi=(\sigma_0+\tilde{\sigma},\vec{\phi})$$

into the Lagrangian. You'll find that you get a theory, which describes one particle with positive squared mass $$m^2=2 \mu^2$$ and three particles with 0 mass.

This is one of the most simple examples of the spontaneous breaking of a global gauge symmetry. The model described above is the most simple effective model to describe pions, which appear here as the massless states. The vacuum is not symmetric under the full group O(4) anymore, but only under the O(3) rotations of the three last components of $$\Phi$$. The masslessness of the pions in this socalled chiral limit is exact to all orders in perturbation theory and is an example for the famous Nambu-Goldstone theorem: There are as many massless "Nambu-Goldstone bosons" in a theory with spontaneous symmetry breaking as the dimension of the symmetry group of the vacuum.

For more details on the linear sigma model, see my notes on QFT

http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf (p. 187ff)

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DrDu

Ok, but your order parameter is a vector, not a scalar, as asked for by ndung.

vanhees71
Gold Member
2019 Award

Here, "scalar" refers to the space-time properties. The four fields in the linear sigma model are scalar fields. Together they build a real vector in a (real) representation of the chiral group, $$\mathrm{SU}(x)_L \times \mathrm{SU}(2)_R,$$ in terms of the SO(4), which is locally isomorphic to the chiral group.

BTW: The symmetry breaking I described is in fact the breaking from the chiral group to the isovector symmetry $$\mathrm{SU}(2)_V,$$ here represented as SO(3), acting on the pion fields.

DrDu