Superficial degree of divergence for scalar theories

[CAF123]

I have a few questions regarding the derivation of the degree of divergence for feynman diagrams. The result is $$D = [g_E] - \sum_{n=3}^{\infty} V_n [g_n]$$ (following notation in Srednicki, $P118$)

I am trying to understand what $[g_E]$ is here? Since in this set up we are summing over all possible scalar theories $n \in [3, \infty)$, we will have a tree level local interaction diagram corresponding to the case where $E=i$ where $i$ is some element of the set $[3,\infty)$. So is it right to say that $[g_E]$ denotes a particular element of the set $[3,\infty)$?

If that is correct, then in the theory $$\mathcal L = \frac{1}{2} Z_{\phi} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2}Z_m m^2 \phi^2 - \frac{1}{k!} Z_g g \phi^k$$ (i.e a theory where we now include only one of the elements in the above set) the formula for $D$ now becomes $D = [g_E] - v_k [g]$ and in this case $[g_E] = [g]$, with $v_k$ the number of times a vertex shows up in some diagram? Is that correct understanding?

 

[nrqed]

I have a few questions regarding the derivation of the degree of divergence for feynman diagrams. The result is $$D = [g_E] - \sum_{n=3}^{\infty} V_n [g_n]$$ (following notation in Srednicki, $P118$)

I am trying to understand what $[g_E]$ is here? Since in this set up we are summing over all possible scalar theories $n \in [3, \infty)$, we will have a tree level local interaction diagram corresponding to the case where $E=i$ where $i$ is some element of the set $[3,\infty)$. So is it right to say that $[g_E]$ denotes a particular element of the set $[3,\infty)$?

If that is correct, then in the theory $$\mathcal L = \frac{1}{2} Z_{\phi} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2}Z_m m^2 \phi^2 - \frac{1}{k!} Z_g g \phi^k$$ (i.e a theory where we now include only one of the elements in the above set) the formula for $D$ now becomes $D = [g_E] - v_k [g]$ and in this case $[g_E] = [g]$, with $v_k$ the number of times a vertex shows up in some diagram? Is that correct understanding?

Hi,

Well, I would not say that we are "summing over all possible scalar theories", that's misleading. We are consider one diagram in a single theory. Now, this diagram has a certain number of external lines, let's call it "k", and $[g_E]$ is the dimension of the coefficient of the term with k powers of the scalar field in the potential. For example, if there are 7 external lines, $[g_E]$ is the dimension of the coefficient of $\phi^7$. Then the sum is over all the vertices in the diagram.

 

[CAF123]

Hi nqred,

Srednicki begins by writing down $$\mathcal L = - \frac{1}{2} Z_{\phi} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2}Z_m m^2 \phi^2 - \sum_{n=3}^{\infty} \frac{1}{n!} Z_n g_n \phi^n$$ which seems to me to be a summation over all possible scalar theories, i.e in such a theory we have a $\phi^3, \phi^4...$ etc interaction. Is it not? Suppose we have a diagram with E external legs. Then in $D = [g_E] - \sum_{n=3}^{\infty} V_n [g_n]$, the first term $[g_E]$ corresponds to the tree level local interaction vertex that occurs when $E=i$ for some i in the interval [3,infinity]. So we can write $$D = [g_E] - V_i[g_i] - \sum_{n \neq i}^{\infty} V_n [g_n] = (1-V_i)[g_E] - \sum_{n \neq i}^{\infty} V_n [g_n]$$ Each of the $[g_n]$ have a different mass dimensionality in a fixed number of space time dimensions.

Am I thinking about this wrongly? Edit: The last equality I wrote must be wrong because it doesn't reproduce the correct D for some diagrams. I was thinking of the one loop box correction to the $\phi^3$ local vertex interaction. In d=6, the coupling has null mass dimension and we only have one theory (phi^3) so Srednicki's formula reduces to $D = [g_E] - V_3 [g_3]$. In this case, $V_3 = 4$ and $[g_E]$ = 0. But this doesn't recover the D=-2 you get by power counting.

Thanks!

 

[nrqed]

Hi nqred,

Srednicki begins by writing down $$\mathcal L = - \frac{1}{2} Z_{\phi} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2}Z_m m^2 \phi^2 - \sum_{n=3}^{\infty} \frac{1}{n!} Z_n g_n \phi^n$$ which seems to me to be a summation over all possible scalar theories, i.e in such a theory we have a $\phi^3, \phi^4...$ etc interaction. Is it not?

Hi again!

Well, it is just a question of terminology. The way I see it is that we are considering one theory (because we are dealing with a single lagrangian) but this theory may contain an infinite number of terms. Even if there is an infinite number of terms it is still one theory.

Suppose we have a diagram with E external legs. Then in $D = [g_E] - \sum_{n=3}^{\infty} V_n [g_n]$, the first term $[g_E]$ corresponds to the tree level local interaction vertex that occurs when $E=i$ for some i in the interval [3,infinity]. So we can write $$D = [g_E] - V_i[g_i] - \sum_{n \neq i}^{\infty} V_n [g_n] = (1-V_i)[g_E] - \sum_{n \neq i}^{\infty} V_n [g_n]$$ Each of the $[g_n]$ have a different mass dimensionality in a fixed number of space time dimensions.

Am I thinking about this wrongly?

Yes, we can write this.

Edit: The last equality I wrote must be wrong because it doesn't reproduce the correct D for some diagrams. I was thinking of the one loop box correction to the $\phi^3$ local vertex interaction. In d=6, the coupling has null mass dimension and we only have one theory (phi^3)

Not necessarily! One can also have $\phi^4,\phi^5 \ldots$ terms! The theory will be nonrenormalizable but it is still perfectly well defined as an effective field theory

so Srednicki's formula reduces to $D = [g_E] - V_3 [g_3]$. In this case, $V_3 = 4$ and $[g_E]$ = 0. But this doesn't recover the D=-2 you get by power counting.

Thanks!

Watch out. This box diagram has four external lines. Therefore $[g_E]$ is the dimension of the coefficient of an $\phi^4$ interaction would have (note that it does not matter if there is such an interaction in the Lagrangian or not, $[g_E]$ is the dimension of the coefficient of such an interaction if it was present) and for a $\phi^4$ interaction in 6 dimensions, the coefficient would have dimension -2. So for your diagram, $[g_E]=-2$ and therefore D=-2.