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  Why the sum of one-particle-irreducible graphs $\Pi^*$ can be written in this form?

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I am studying the basics of renormalisation. In 10.3 of Weinberg QFT book, motivated by the renomalisation of field, he express the bare Lagrangian as two part

\(\mathscr{L=L}_0+\mathscr{L}_1\),

\(\mathscr{L}_0=-\frac{1}{2}\partial_{\mu}\Phi\partial^{\mu}\Phi-\frac{1}{2}m^2\Phi^2\),

\(\mathscr{L}_1=-\frac{1}{2}(Z-1)(\partial_{\mu}\Phi\partial^{\mu}\Phi+m^2\Phi^2)+\frac{1}{2}Z\delta m^2\Phi^2-V(\Phi)\),

where

\(V(\Phi)\equiv V(\sqrt{Z}\Phi_B)\)

After expressing the corrected propagator as a geometric series of uncorrected propagator times the sum of one-particle-irreducible graphs $\Pi^*$, the only thing need to calculate is the $\Pi^*$, but he states:

In calculating $\Pi^*$, we encounter a tree graph arising from a single insertion of vertices corresponding to the terms in $\mathscr{L}_1$ proportional to $ \partial_{\mu}\Phi\partial^{\mu}\Phi$ and $\Phi^2$, plus a term $\Pi^*_{LOOP}$ arising from loop graphs like that in Figure 10.4(a):

\(\Pi^*(q^2)=-(Z-1)[q^2+m^2]+Z\delta m^2+\Pi^*_{LOOP}(q^2).\)

What's the tree graph arising from a single insertion of vertices that corresponds to the terms in $\mathscr{L}_1$ ? And why they are proportional?

asked May 15, 2014 in Theoretical Physics by coolcty (125 points) [ revision history ]
edited May 23, 2014 by coolcty

Hi coolcty, this looks like a promising new question :-)

To write centered LaTex equations and also get a preview of the maths, you can use the TEX-button of the editor. Applying this button, you do not need the $s.

You need to consider \(\mathscr{L}_0\) as the free Lagrangian and \(\mathscr{L}_1\) as the interaction, then can you derive the Feynman rules for the interaction term \(\mathscr{L}_1\)?

@JiaYiyang

Thanks for your hint. The tree is just a single line, but I thought it has branches.

Hi @coolcty, as you have figured out the answer, it would be nice if you could write a short answer to your own question (which is completely allowed) if you have time. (just a piece of advice, its completely fine if you don't). 

There is no $m_B$ in this formula, only physical $m$. Also, the third line is $\mathscr{L}_1$.

Let me take advantage of this question and explain you my position. Here $\mathscr{L}_0$ is supposed to describe a physical field, i.e., a dressed one. It can be free, as we can see. Now, let us suppose that all subtractions within $\mathscr{L}_1$ can be done exactly rather than in each perturbative order so that no kinetic terms, no $Z$ and $\delta m$ are present any more in $\mathscr{L}_1$. Then this interaction may only change the occupation numbers of physical quanta $\Phi$. That's it. This can be called a reformulated, physical theory. No renormalization is necessary. This is what I tried to explain to Dilaton and to others in a previous post. Our inability to guess the right $\mathscr{L}_1$ makes us discard wrong perturbative contributions of the wrong $V(\Phi)$. This discarding is renormalization.

Vladimir Kalitvianski: the exact $\mathcal{L}_{1}$ necessarely contains $Z$,... : there can be no simplication between the terms quadratic in $\Phi$ and the higher order terms in $V(\Phi)$. The cancellations appear in the computation of the physical amplitudes. But it is true that the set of physical amplitudes defined a "reformulated, physical theory" which does not contain $Z$,... If "no renormalization" is necessary, you have to explain how to compute these amplitudes directly.

@40227: I explained it here, for example (and in many other places).

By the way, the exact $\mathcal{L}_{1}$ does not contain $Z$, $\delta m$, ect. because the same terms, but of the opposite sign, are contained in $V(\Phi)$. It is not clearly visible because $\mathcal{L}_{1}$ is an operator, but it is so. As you correctly noticed, solutions of the reformulated theory do not contain them at all. It can only be in case of their factual absence in $\mathcal{L}_{1}$.

Vladimir Kalitvianski: I do not understand the meaning of "it is not clearly visible because $\mathcal{L}_1$ is an operator". As already mentionned, as $V(\phi)$ does not contain any term quadratic in $\phi$, there can be no cancellation of Z,...This is not a contradiction with the fact that the physical amplitudes does not contain Z...: the explicit Z,... dependence in $\mathcal{L}_{1}$ gives contributions to the physical amplitudes which cancel some loop contributions of the interaction $V(\phi)$.

@40227: If you make a difference between a function and its Taylor series, it is exactly zero although it is not clearly visible. Also, if your function is a solution to a differential equation, then some non trivial combination of this function and its derivatives can make zero. Zero does not depend on $x$. $Z$-dependence of $\mathcal{L}_1$ is of that sort. You see an expression involving $Z$ explicitly and implicitly, implicitly via $\Phi$, but this combination does not depend on $Z$.

Loops do contribute to the solutions $\Phi$ (they contribute to Green's functions because Green's functions are expressed via solutions).

Vladimir Kalitvianski: I was simply considering $\mathcal{L}_1$ as a function of the variable $\Phi$. When you say that $\Phi$ depends of $Z$, I guess you are talking about some specific $\Phi$, which one? A solution of the equation of motion? 

I agree that loops contribute, I have written cancellation of "some" loop contributions.

@40227:  Yes, a solution of the equation of motion. Instead of $\mathcal{L}_1$ we can better consider the equation for $\Phi$ where contribution of $\mathcal{L}_1$ is also $Z$-independent.

In fact, $\mathcal{L}_1$ is somewhat ambiguous since it is determined up to a full derivative of something. We should keep it in mind.

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