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In Peskin & Schroeder's book, they introduce the linear sigma model with a Lagrangian (chapter 11.1,p349)

$$ L = \frac{1}{2} (\partial_\mu \phi^i)^2 + \frac{1}{2} \mu^2 (\phi^i)^2 - \frac{\lambda}{4} \left[ (\phi^i)^2 \right]^2.\tag{11.5} $$

Taking the minimum of the corresponding Hamiltonian to derive the lowest-energy classical configuration and define a set of shifted fields from the classical confuguration, they derive the lagrangian for linear sigma model in terms of $\sigma$ field and $\pi$ field (chapter 11.1,p350):

\begin{align} L &= \frac{1}{2} (\partial_\mu \pi^k)^2 + \frac{1}{2} (\partial_\mu \sigma)^2 - \frac{1}{2} (2\mu^2) \sigma^2 \notag \\ &\quad - \sqrt{\lambda} \mu \sigma^3 - \sqrt{\lambda} \mu (\pi^k)^2 \sigma - \frac{\lambda}{4} \sigma^4 - \frac{\lambda}{2} (\pi^k)^2 \sigma^2 - \frac{\lambda}{4} \left[ (\pi^k)^2 \right]^2 \tag{11.9} \end{align}

and this give rise of a set of feynman rules shown in the figure 11.3:

the Feynman rules

Then they proceed to state(p355) that the tadpole diagram (the figure below) produce a vacuum expectation of $\sigma$, the exact quote goes like:

This amplitude produces a vacuum expectation value of $\sigma$ and so, since $\phi^N=v+\sigma$, shifts the vacuum expectation value of $\sigma$. Such a shift is quite acceptable, as long as it is finite after counterterms are properly added into the computation of the amplitude.

The figure for tadpole diagram:

the tadpole diagram

My question is: Why this tadpole diagram represent the vacuum expectation value of $\sigma$ field? I fail to understand it intuitively.

I have read this post but I still couldn't understand it because I don't understand how the vacuum expectation value is related to the tadpole diagram shown in my figure. I have also read this post but this does not answer how the vacuum expectation value is related to the tadpole Feynman diagram.

Qmechanic
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Interstellar
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1 Answers1

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$$\begin{align}\text{VEV}~=~&\underbrace{\langle\Omega|\sigma(x)|\Omega\rangle_{J=0}}\cr ~=~& \overbrace{\langle\sigma(x)\rangle_{J=0}}^{\text{1-pt function}} \cr\cr ~=~&\underbrace{\left.\frac{\delta W_c[J]}{\delta J(x)}\right|_{J=0}}\cr ~=~&\overbrace{\underbrace{\left(\fbox{conn.}==\stackrel{\sigma}{=}==\right)}_{\text{OP's figure}}}^{\text{conn. tadpole diagrams}}. \end{align} $$ The 3rd equality is eq. (4) in my Phys.SE answer here.

Qmechanic
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