Tag Info

New answers tagged

4

Your way of thinking is, essentially, correct. When it comes to this $\tilde\lambda\Lambda^2$, there is this famous quote (citing from memory, don't remember which book it's from, but it's famous), "Even though it is infinitely large, we will assume that it is finite, and that is furthermore infinitely small." In most QFTs the perturbative series diverges ...


2

In the second approach, you do have the option to treat the short distance physics in a variety of ways. Choosing point interactions will give you some particular values for $ \lambda _{ bare}$ and some particular rules about how to calculate $\delta \lambda $, and if you choose a different "UV completion" you will get a different value and a different rule. ...


0

What you describe is usually called regularization, as distinct from renormalization, although the terms are related. It could help to cite the paper that you are reading, but in any case, it often happens that long wavelength physics do not depend exactly on the details of short distance physics. For example, if you are scattering a particle off of a ...


2

Quantum dynamics is commonly known to be generated by self-adjoint operators. Therefore in order to properly define the dynamics of a system it is necessary to introduce a suitable self-adjoint Hamiltonian operator. In quantum field theories, this task is extremely difficult, because the formal operators that emerge quantizing a classical field theory ...


1

Consider the partition function $$Z = \int D\phi ~ e^{-S_0 - S_I},$$ where $S_0$ is the Gaussian/free part and $S_I$ is the interaction part of the action. Within a perturbative framework we may aim to systematically include the contributions of fast modes to the (effective) action for slow modes. For this we expand in the interaction strength as $$Z = \int ...


0

The unrenormalized vertex diagram is the dimensionally regularized Feynman integral corresponding to the following single Feynman diagram: It is called ‘unrenormalized’ because it is not accompanied by a diagram in which the momentum loop is replaced by a counterterm vertex generated by the QED Lagrangian. Adding a second Feynman diagram with the ...


1

I believe that this may be what you’re looking for. I don’t claim to be an expert in renormalization, so all members of the Physics Stack Exchange Community are welcome to correct my answer. To make the relationship between the two more precise, let’s call the on-shell renormalization scheme the ‘on-shell subtraction scheme’ and the BPHZ renormalization ...



Top 50 recent answers are included