New answers tagged renormalization
2
The theory is (even classically) not scale invariant. Just by dimensional analysis, you can note that the scalar field has scaling dimension 1, and the mass (as the name suggests) must also have a scaling dimension of 1. So $m^2$ has a ascaling dimension of 2, which suggests the RG equation which you've written in the question. That essentially says that the ...
0
If you solve the equation to second order you obtain:
$g=\frac{g_0}{1-a g_0 log\lambda}=g_0 +\sum_n g_0^{n+1}a^nlog^n \lambda$
the second part is the contribution of powers of the first loop diagram, for instance try to write g at 1-loop first and then sum all the powers of the first order diagram.
2
I am not sure that one will appreciate this answer, but I will try.
I think that the punchline may not always be the case. It seems that you want some symmetry which forbids some counterterm. This means that this counterterm would have some unappreciated behaviour under the symmetry. Then you want the same symmetry to forbid the divergence. Why? Naively, ...
1
The pole corresponds to an on-shell particle going from one point to another. Then, the residue effectively tells you how many of those particles are being transmitted. Since in your physical/renormalized theory, the propagator should correspond to $1$ quantum of the renormalized field being transmitted, you set the residue at the pole to $1$.
4
Running the RGEs in reverse should be valid so long as you don't integrate over a scale where degrees of freedom enter/leave the theory. If you integrated out the electrons in QED, you'd have irrevocably lost that information in your low energy description of interacting photons. You'd see some non-renormalizable theory with interacting corrections to pure ...
2
Yes, suppose $[g] = \delta$. By dimensional analysis only we can write that a loop diagram contributes
$$
\sim g^{n} \int \frac{d^4 k}{k^{4-n\delta}}
$$
If $\delta=0$, this diverges logarithmically, but can be re-normalized. If $\delta$ is less than zero, it diverges by simple power counting.
This is VERY informal. Technically, you should study the ...
1
The OS condition that
$$
\frac{\partial\Sigma}{\partial p^2}|_{p^2=-m^2} = 0
$$
implies that the residue in the propagator remains equal to one.
Suppose that we used a different renormalization scheme in which our counter-terms contain no finite parts (e.g. MS scheme). In the OS scheme, we removed finite parts which were logarithmic in our artificial ...
2
First of all is the definition of coupling constant:
In physics, a coupling constant, usually denoted g, is a number that determines the strength of the force exerted in an interaction. Usually, the Lagrangian or the Hamiltonian of a system describing an interaction can be separated into a kinetic part and an interaction part. The coupling constant ...
0
This is different from the possible time-variation of the low-energy fine-structure constant, but the same considerations apply when you try to attribute any such variation in $\alpha$ to a variation in $e$, $\hbar$, or $c$. You can't. See Duff, "Comment on time-variation of fundamental constants," http://arxiv.org/abs/hep-th/0208093 . I also don't see why ...
0
People spent a lot of time trying to do this kind of thing ca. 1910, i.e., after SR but before quantum mechanics. To make the electrostatic self-energy no greater than the observed mass of the electron, you have to create some kind of model of an electron as an extended object, with a size that's at least on the order of the classical electron radius.
You ...
1
Firstly, if I'm correct in understanding your question, you are asking why one needs to regulate field theories since we renormalize them anyway?
Assuming this, here's my understanding:
Quantum field theory in general has several divergences, which we must deal with. We get around such divergences by dealing with renormalized quantities. But how does one ...
2
Ok I'll answer my own question. I asked my QFT professor, he said different methods of regularization will give different answers. but at the end of the day it doesn't matter because you're going to cancel whatever divergences from that integral from whatever method you use, with the mass counterterm anyway, to impose desired renormalization conditions. so ...
1
You hit upon the answer yourself: they are not talking about IR divergences. The renormalisation program is about absorbing unknown (not necessarily divergent) short distance physics into effective local operators. The beauty of this is we do not need a full understanding or rigorous formulation of the short distance physics if we are only interested on much ...
1
The physical masses should be independent of the renormalization scale. We, however, only calculate a finite number of loop corrections, resulting in a scale dependence in the physical mass. This scale dependence can be used to estimate the error in the mass calculation from the missing higher orders.
In principle, one could calculate the sparticle mass ...
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