In the Ch.18, book of QFT by Mark Srednicki (p.118), it says the diagram have the same mass dimension with tree diagram with the same external lines, because both of them contribute to the same scattering amplitude. But I am confused of the reason of this argument.
$\begingroup$
$\endgroup$
0
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
One argument goes as follows:
Recall that the path-integral/partition function $Z[J]$ is the generator of all Feynman diagrams in the source picture. Similarly $W_c[J]$ is the generator of all connected Feynman diagrams in the source picture. All the Feynman diagrams in $Z[J]$ and $W_c[J]$ have mass-dimension zero.
Now we want to find the mass-dimension of an amputated Feynman diagram $D$ with $E_f$ external lines with a field $\phi_f$ of type $f$.
Hence we just have to strip the corresponding dimensionless Feynman diagram $D[J]$ in the source picture of
- $E_f$ sources $\tilde{J}_f$ in Fourier momentum space;
- $E_f$ free propagators $\widetilde{G}_{0f}$ in Fourier momentum space for the external legs of type $f$;
- $E_f$ $d$-dimensional momentum integrations;
- an overall $d$-dimensional Dirac delta function imposing total momentum conservation.
This leads to Srednicki's claim that the mass-dimension $[D]$ of the amputated Feynman diagram $D$ only depends on the number of external legs.
In fact, it is not difficult to compute that the mass-dimension of the amputated Feynman diagram $D$ is$^1$ $$\begin{align}[D]~=~&d - \sum_f E_f\left(\underbrace{[\tilde{J}_f]}_{=-[\phi_f]}+\underbrace{[\widetilde{G}_{0f}]+d}_{=[G_{0f}]=2[\phi_f]}\right)\cr ~=~&d-\sum_f E_f[\phi_f].\end{align}$$
If the amputated Feynman diagram $D$ is 1-particle-irreducible (1PI), there is an even simpler argument: Just functionally differentiate the dimensionless effective action $\Gamma[\phi_{\rm cl}]$ (which is the generator of amputated 1PI diagrams) with the appropriate number of Fourier-transformed fields $\tilde{\phi}_f$ and remove a Dirac delta function imposing total momentum conservation.
--
$^1$ It is implicitly assumed that the coefficients in front of the kinetic terms in the action are dimensionless.
answered Mar 25, 2020 at 21:48
-
$\begingroup$ Notes for later: Replace mass dimension with inverse length dimension. $\quad[\overline{D}]=d-\sum_f E_f[\overline{\phi}_f]$. A delta-function-stripped $E$-point scattering amplitude ${\cal M}=\hbar^{\frac{2-d}{2}E+d}\overline{\cal M}$ has $[{\cal M}]=M^{\frac{2-d}{2}E+d}$ and $[\overline{\cal M}]=L^{\frac{d-2}{2}E-d}$, cf. Schwartz eq. (5.18). $\endgroup$– Qmechanic ♦Commented Mar 6, 2022 at 15:47
$\begingroup$
$\endgroup$
This statement is correct.
What is a Feynman diagram? The most naive way to look into it is to say that it's a diagram that dictates all possible ways how a set of initial particles can transform into a set of final particles. The intermediate steps(the loops and internal lines) basically show one of the plausible ways allowed by the theory(these steps may not be possible on-shell, nevertheless are allowed off-shell).
S matrix is the probability amplitude for the whole process taking into consideration all quantum corrections. Therefore classically disallowed intermediate steps are also taken into account. But ultimately all the S-Matrix cares about at the end of the day is the initial and final particles. All the Feynman diagrams(i.e. all possible paths for the process) therefore must be added to get the answer. Since you can only add things that have same dimension, all these diagrams must have same mass dimension. And the dimension for each of these is just sum of dimension of the external states(initial+final).
