Suppose $g$ is the parameter set and $\Lambda\equiv\Lambda_0e^{-t}$ the momentum cutoff, then usually one finds the renormalization group equations to take the form $$\frac{dg(t)}{dt}=\beta(g).$$ My question is that why is there no explicit $t$-dependence (of course there is implicit dependence as $g$ is also a function of $t$) on the right hand side, or equivalently, in $\beta$?

I can understand that without $t$-dependence, the flow equations have the invariance under $$g(t)\rightarrow{g}(t+t_0),$$ meaning that the system does not know what is $t=0$. But my confusion is that should the system be defined on a lattice then $t=0$ has a clear meaning: it means we are looking at the `original' Hamiltonian and $t<0$ is ill-defined. In that case, shouldn't one have $\beta(g,t)$ instead of $\beta(g)$?

  • $\begingroup$ "the system does not know what is $t=0$" and neither do I... would you please define the symbols? $\endgroup$ – AccidentalFourierTransform Dec 5 '16 at 13:32
  • $\begingroup$ @AccidentalFourierTransform $t=-\log[\Lambda/\Lambda_0]$, where $\Lambda$ is the momentum cutoff and $\Lambda_0$ some constant. $\endgroup$ – Chen Fang Dec 5 '16 at 13:40

I don't fully understand what you mean. Of course, there is a time dependence on the RHS due to $t$-dependence of $g$.

To understand why there is no explicit $t$-dependence on the RHS, look for a derivation of this equation in every QFT book. Basically, it is because the matrix elements of bare operators, from which this RG equation is derived, don't know of your choice of the scale $\mu$ you prefer to work on, so they can't depend on it.

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  • $\begingroup$ I understand that this is true for any continuum theory. Is this also true if the original Hamiltonian is defined on a lattice? It is not obvious to me. $\endgroup$ – Chen Fang Dec 5 '16 at 13:44
  • $\begingroup$ @ChenFang The fact that your theory is defined on the lattice just introduces the UV cut-off $\Lambda_{UV}=\frac{1}{a_{lattice}}$. It doesn't change the conclusion that the bare correlators don't know of any chosen scale $\mu$. $\endgroup$ – Andrey Feldman Dec 5 '16 at 13:53
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    $\begingroup$ This answer addresses your question. Essentially, the RG motion involved is infinitesimal, so $dt=-d\Lambda /\Lambda$. $\endgroup$ – Cosmas Zachos Aug 18 '19 at 0:34

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