Question about RG and scaling in qft

  • Thread starter Thread starter beacon
  • Start date Start date
  • Tags Tags
    Qft Scaling
beacon
Messages
8
Reaction score
0
In critical phenomena, we can enlarge the block size(momenta fluctuation) by Kadanoff transformation, say
k \rightarrow bk (b<=1), and scale the new Hamiltonian by k' = k/b, x'=bx to recover to the original block size.
In QFT, similarly integrating out the high momenta produces the effective Langrangian,

\int_{k<=b\Lambda} [D\phi] exp(iS_{eff}) = \int_{b\Lambda <k < \Lambda} [D\phi] exp(iS).

The parameters y in the effective langrangian S_{eff} should depend on b. We can also do a scaling k' = k/b, x'=bx in S_{eff} to get S'_{eff} whose path integral is now \int_{k' <= \Lambda}. The parameters y' also depend on b. My puzzle is that which are the so-called beta fuctions, dy \over db or dy' \over db
 
Last edited:
Physics news on Phys.org


In QFT, the concept of renormalization group (RG) plays a crucial role in understanding the behavior of a theory at different length scales. The idea is to integrate out high momenta modes and study how the theory changes as we vary the energy scale or the size of the block. This is analogous to the Kadanoff transformation in critical phenomena, where we enlarge the block size by a factor of b and then rescale the new Hamiltonian and coordinates to recover the original block size.

In the context of QFT, integrating out high momenta modes leads to an effective Lagrangian, which contains all the relevant information about the theory at a given energy scale. This effective Lagrangian depends on a parameter b, which is related to the scale at which we are studying the theory. As you correctly pointed out, we can also perform a scaling transformation on the effective Lagrangian to get a new effective Lagrangian, which corresponds to studying the theory at a different energy scale. This new effective Lagrangian will have different parameters, denoted by y', which also depend on b.

Now, the beta function is defined as the rate of change of a parameter with respect to the scale b. In other words, it tells us how a parameter changes as we study the theory at different energy scales. In the context of your question, the beta function would be given by dy/db or dy'/db, depending on which effective Lagrangian we are considering. Both of these beta functions are important in understanding the behavior of the theory at different scales.

To summarize, the beta function is a crucial tool in studying the behavior of a theory under scale transformations, and both dy/db and dy'/db are important in understanding the behavior of the theory at different energy scales.
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...

Similar threads

Replies
2
Views
3K
Replies
9
Views
3K
Replies
9
Views
6K
4
Replies
175
Views
25K
Replies
3
Views
2K
Replies
4
Views
4K
Back
Top