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Let's consider a complex scalar field $\phi$ with potential $\frac{\lambda }{4!}(|\phi|^2-v^2)^2$ (Goldstone model).

Applying the Schwinger-Dyson equations, I obtain $\langle \phi \rangle=0$. This result is intuitive because if we have an interaction involving even powers of the field, we will never be able to draw a tadpole of the type

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But it is known that if we compute the expected value of $\phi$ at the true vacuum instead of the Fock vacuum, we would get $\langle \phi \rangle \propto v$. I don't see how we can represent this result with Feynman diagrams. Is it even possible?

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You have to rewrite the Lagrangian in terms of the field $\varphi = \phi - v$. Then quantize perturbatively around $\varphi = 0$. This will give you (by Schwinger-Dyson eq) $\left< \varphi \right> = 0$, which in turn means that $\left< \phi \right> = v$.

Or are you asking something completely different?

P.S. Exploring the physical vacuum state through the perturbative expansions around $\phi = 0$ is a bad idea. In some cases it might give you something physically meaningful, but generally lots of interesting physics gets lost. This is because perturbation theory is approximate, and each perturbative expansion has an (analogue of) a radius of convergence, which most of the times doesn't allow you to pass between different vacuums without getting something completely nonsensical.

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  • $\begingroup$ Oh, I already knew about the field redefinition method, I was asking how can we draw the $\langle \phi \rangle$ using Feynman diagrams. I now realize that if you redefine the field you can get tadpoles, although they are obtained from the cubic interaction of $\varphi$. I guess this means that one cannot use Feynman diagrams of $\phi$, but should work with $\varphi$. Also, somewhat related: if I consider a mass term of $\phi$, I get the contribution $v\rho$. What is the interpretation of this term? $\endgroup$ – jinawee Mar 24 '17 at 11:36
  • $\begingroup$ @jinawee I don't understand the question. $\left<\phi\right> = \left<\varphi + v\right> = v$, that's the point. This is by definition of the Fock vacuum. Field redefinition is just a convenient way of saying that we want $\left<\phi\right>=v$ to hold, which is equivalent to specifying the physical vacuum state. What is $\rho$ exactly? $\endgroup$ – Prof. Legolasov Mar 24 '17 at 11:59
  • $\begingroup$ Nevermind the last question, I figured it out. The initial question is more on the line that the one-point function is to order $\lambda$, $\langle \phi \rangle = v + \frac{\lambda v^2}{3} \Delta_{12}\Delta_{22}$. But this involves a cubic interaction, not a quartic one as we started with. Are there any comments you can add to make this more intuitive? Like deriving the Feynman rules from a Lagrangian whose minimum is not when the fields are zero is not a good idea or things like that. $\endgroup$ – jinawee Mar 24 '17 at 12:08

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