Srednicki treated $$ \begin{aligned} L_{0}&=-\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^2\\ L_{1}&=\frac{1}{6}Z_{g}g\phi^3+L_\text{counterterm}\\ L_\text{counterterm}&=-\frac{1}{2}(Z_{\phi}-1)\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}(Z_{m}-1)\phi^2+Y\phi \end{aligned} $$ Firstly, he neglected the counterterm Lagrangian $L_\text{counterterm}$ and only considered $L_{1}=\frac{1}{6}Z_{g}g\phi^3$, and thus define $$Z_{1}(J)\propto e^{(i\int d^4x L_{1})}Z_{0}(J)$$ Then after computing using Feynman diagrams, we could easily normalize $Z_{1}(J)$: since $Z_{1}(0)=1$, we omit the vacuum diagrams (those with no sources). I have two questions:

  1. He later add the term "$Y\phi$" into $L_{1}$ to make $\langle0|\phi(x)|0\rangle=0$. So now is still $Z_{1}(0)=1$?
  2. By adjusting "$Y\phi$", we could make $\langle0|\phi(x)|0\rangle=0$, but when we continue to add the remaining counterterm "$-\frac{1}{2}(Z_{\phi}-1)\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}(Z_{m}-1)\phi^2$" to $L_{1}$, we get the final expression $$Z(J)=\exp\left[i\int d^4x -\frac{1}{2}(Z_{\phi}-1)\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}(Z_{m}-1)\phi^2\right]Z_{1}(J),$$ how can we know that now we still have $\langle0|\phi(x)|0\rangle=0$?

1 Answer 1


1) We fix $Z[0]=1$ by omitting vacuum diagrams, no matter what the interactions are. This means: we don't really care what terms there are in the lagrangian: as long as you only consider connected diagrams, you will have $Z[0]=1$. You can add whatever interaction you want to $\mathcal L$, and the normalisation remains unchanged, simply by following the rule "$W(J)$ is given by the sum of all connected diagrams".

2) The logical order is the other way around: we first consider the full lagrangian (what you call "the final expression"), and then we adjust $Y$ to make $\langle 0|\phi(x)|0\rangle=0$. Srednicki follows the inverse order for pedagogical reasons, but in practice, the counterterms are adjusted a posteriori, when we consider the full theory.

I hope this answers your questions.

  • $\begingroup$ Thank you very much for your answer. But I am still a little confused. In Srednicki's book, he adjusted $Y$ to make $<0|\phi(x)|0>=0$, and we can get the expression of $Y$, and $L_{1}=\frac{1}{6}Z_{g}g\phi^3+Y\phi$. Now we consider all counterterms into $L_{1}$, so $L_{1}=\frac{1}{6}Z_{g}g\phi^3-\frac{1}{2}(Z_{\phi}-1)\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}(Z_{m}-1)\phi^2+Y\phi$, and we still want to make $<0|\phi(x)|0>=0$ by adjusting $Y$, so now the expression of $Y$ will be different from the former one? $\endgroup$
    – youyou
    Commented Feb 20, 2016 at 19:25
  • $\begingroup$ Also, at the end of that section, Srednicki wrote "$W(J)$ is given by the sum of all connected diagrams with no tadpoles and at least two sources, and including the counterterm vertices just discussed." The "diagrams" here correspond to $L_{1}=\frac{1}{6}Z_{g}g\phi^3-\frac{1}{2}(Z_{\phi}-1)\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}(Z_{m}-1)\phi^2+Y\phi$, right? But all of the diagrams he drew are about $L_{1}=\frac{1}{6}Z_{g}g\phi^3$ or $L_{1}=\frac{1}{6}Z_{g}g\phi^3+Y\phi$. $\endgroup$
    – youyou
    Commented Feb 20, 2016 at 19:32
  • $\begingroup$ After all, the number of sources of a diagram will change if we further consider $-\frac{1}{2}(Z_{\phi}-1)\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}(Z_{m}-1)\phi^2$ $\endgroup$
    – youyou
    Commented Feb 20, 2016 at 19:34
  • $\begingroup$ The explicit expression for $Y$ is irrelevant: its only purpose is to ensure $\langle\phi\rangle=0$. The actual value of $Y$ changes from theory to theory, and changes if you include more interaction terms. But in the end, we dont use $Y$ for anything: its just there so that $\phi$ has zero vev. So yes, the expression for $Y$ will be different, but that is irrelevant. We cant measure $Y$, we only measure physical quantities, and these are independent of $Y$. As for your second comment: yes, the diagrams correspond to the full lagrangian, with every term. This means in general there are (1/2) $\endgroup$ Commented Feb 20, 2016 at 19:46
  • $\begingroup$ (2/2) more diagrams than just those discussed in that section. If you consider the full lagrangian, you'll get many different possible vertices (with one/two and three lines), so there are many diagrams in general. Anyway, my advice is: keep on reading the book and pay attention to all the examples Sred. discusses. All these matters can only be clarified through examples. In fact, I didn't really understand the basics of renormalisation (i.e., the meaning of the $Z_i$ factors) until the third section of the book, where there are detailed calculations on QED. You have to be a little patient :) $\endgroup$ Commented Feb 20, 2016 at 19:52

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