## Why do people say CSB is non-perturbative?

For example, in Srednicki's QFT book, he said: "because $$\langle\bar{\psi}\psi\rangle$$ vanishes at tree level, perturbative corrections then also vanish, because of the chiral flavor symmetry of the lagrangian. Thus the value of vev is not accessible in perturbation theory."

I don't understand, because what he said seems to be true even for the simplest $$Z_2$$-breaking phi-4 theory. Here we can say $$\langle\phi\rangle$$ vanishes because if we use the original field to do perturbation calculations, no Feynman diagram can be drawn that contributes to this vev, exactly because of the $$Z_2$$ symmetry.

So is it true that any SSB is non-perturbative? Or am I missing something?
 Recognitions: Gold Member Homework Help Science Advisor In SSB with a scalar field, there is a tree-level potential for the scalar field (consisting of mass term and $$\phi^4$$ interaction). By breaking $$\phi$$ up into a classical and quantum part, the effective field theory of the components leads directly to the value of the VEV. This EFT can be explicitly related to Feynman diagrams. For chiral QCD with vanishing quark masses, there is no tree-level potential that could lead to the $$\langle \bar{\psi}_i\psi_i\rangle$$ VEV. Therefore fermionic condensates are produced by nonperturbative physics. Another way to understand this is that $$\langle \bar{\psi}_i\psi_i\rangle$$ is a dimensionful quantity. The only scale in the theory is the QCD scale, which is defined at a point where perturbative physics is strictly not valid. So observables that depend on this scale are always tied to nonperturbative phenomena.

 Quote by fzero In SSB with a scalar field, there is a tree-level potential for the scalar field (consisting of mass term and $$\phi^4$$ interaction). By breaking $$\phi$$ up into a classical and quantum part, the effective field theory of the components leads directly to the value of the VEV. This EFT can be explicitly related to Feynman diagrams. For chiral QCD with vanishing quark masses, there is no tree-level potential that could lead to the $$\langle \bar{\psi}_i\psi_i\rangle$$ VEV. Therefore fermionic condensates are produced by nonperturbative physics. Another way to understand this is that $$\langle \bar{\psi}_i\psi_i\rangle$$ is a dimensionful quantity. The only scale in the theory is the QCD scale, which is defined at a point where perturbative physics is strictly not valid. So observables that depend on this scale are always tied to nonperturbative phenomena.
I'm sorry but I don't get it. What do you mean by breaking phi up into a classical and quantum part? What kind of diagrams do you relate to the vev? Can you explain it in more detail? Thank you, fzero.

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## Why do people say CSB is non-perturbative?

If we have the potential

$$V(\phi) = \frac{m^2}{2} \phi^2 + \lambda \phi^4$$

the classical solution $$\upsilon$$ satisfies

$$\left. \frac{\delta V}{\delta \phi} \right|_{\phi = \upsilon} =0.$$

$$\upsilon$$ is by definition the VEV of $$\phi$$, namely $$\langle \phi\rangle =\upsilon$$. Small fluctuations around the VEV are parameterized by the quantum field $$\varphi$$ satisfying

$$\phi = \upsilon + \varphi. (*)$$

The effective potential for $$\varphi$$ can be obtained by direct substitution as $$V(\upsilon + \varphi)$$. We can either expand the polynomial or note the Taylor expansion

$$V_\text{eff} (\varphi) = V(\upsilon) + \frac{1}{2} \left. \frac{\delta^2 V(\phi)}{\delta\phi^2} \right|_{\phi=\upsilon} \varphi^2 + \cdots . (**)$$

However we can also compute the potential from Feynman diagrams. I don't want to try to draw diagrams, but you can probably find some if you search around for "background field method." The idea is that the Feynman diagrams are a diagrammatic representation of correlation functions:

$$\langle \phi(p_1) \phi(p_2) \phi(p_3) \phi(-p_1-p_2-p_3)\rangle \sim \lambda.$$

If we expand these correlation functions using (*), we'll find terms like

$$\upsilon^2 \langle \varphi(p) \varphi(-p) \rangle \sim \lambda \upsilon^2 .$$

This expression shows how the $$\phi^4$$ term generates a contribution to the effective mass for the $$\varphi$$ field, in correspondence with the 2nd term in (**). Now it turns out that we can also obtain these terms from the Feynman diagrams for the $$\phi^4$$ theory where we count the different ways that we can insert $$\upsilon$$ or $$\varphi$$ on the legs instead of $$\phi$$.

The upshot of all this, which involves no nonperturbative physics at all, is that, for certain values of $$m^2, \lambda$$, the potential for $$\varphi$$ will not be invariant under $$\varphi\rightarrow -\varphi$$. So the original symmetry is spontaneously broken by the VEV.