Are the bare parameters of a renormalizable field theory infinitesimal or infinite? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-15T18:24:15Z https://physics.stackexchange.com/feeds/question/224858 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/224858 8 Are the bare parameters of a renormalizable field theory infinitesimal or infinite? guillefix https://physics.stackexchange.com/users/29621 2015-12-18T18:52:48Z 2016-03-25T06:46:56Z <p>I think this should be an easy question.</p> <p>Several sources I've read say that the bare parameters in a quantum field theory are "infinite" so that the renormalized values are "finite". However, in renormalization group flow, relevant parameters are amplified infinitely as the cut-off is pushed to infinity. So I think the bare parameters (which are the parameters at the cut-off scale right?) should go to zero so that they are finite at low energy scales. Is this right?</p> https://physics.stackexchange.com/questions/224858/-/225306#225306 1 Answer by Abdelmalek Abdesselam for Are the bare parameters of a renormalizable field theory infinitesimal or infinite? Abdelmalek Abdesselam https://physics.stackexchange.com/users/7266 2015-12-21T18:49:34Z 2016-02-24T00:09:36Z <p>Take $\phi^4$ theory in 3D for example. Let $L&gt;1$ be some fixed rescaling ratio for each RG transformation. The dimensionless $\phi^4$ coupling $g$ evolves according to $$g\rightarrow g'=L^{3-4[\phi]}g- Ag^2+\cdots$$ where the scaling dimension $[\phi]$ is $\frac{1}{2}$. So indeed the (relevant) coupling gets amplified by a factor of $L$ each time one shrinks space by $L$. But if one sets the UV cut-off scale at $x\sim L^r$ for $r$ a very negative integer, then the bare (dimension-full) coupling is $g_{B,r}=L^{-r(3-4[\phi])}g_r$ in terms of the dimensionless coupling $g_r$ which is the one that evolves according to the RG equation above. For the construction of a typical massive theory one can take the bare coupling $g_{B,r}$ to be constant with respect to the UV cut-off $r$ that is taken to $-\infty$. That implies that the dimensionless coupling goes to zero as you said. However if you want to construct the CFT corresponding to the non-trivial infrared fixed point, then the dimensionless couplings are bounded away from zero and $\infty$, which means that the bare coupling $g_{B,r}\rightarrow\infty$ when $r\rightarrow -\infty$.</p> https://physics.stackexchange.com/questions/224858/-/239430#239430 2 Answer by Adam for Are the bare parameters of a renormalizable field theory infinitesimal or infinite? Adam https://physics.stackexchange.com/users/29944 2016-02-24T03:19:00Z 2016-02-24T03:19:00Z <p>The subtility in the OP's question is in the use of dimensionfull and dimensionless quantities. This is easier to understand if one does not work at the upper critical dimension ($d=4$ in most standard field theories), where the interaction has dimension (even using dimensionfull quantities). I will first comment on the case where the UV cut-off $\Lambda$ is finite, and then comment on the limit $\Lambda\to \infty$.</p> <p>To give an example, let us look at a massless $\phi^4$ theory in $d=5$, which would correspond to a trivial theory in a renormalization group sense, the infrared fixed point being the gaussian fixed point. This, however, does not mean that the renormalized parameter of the theory are all zero ! The most interesting quantities are low energy quantities, which correspond to the "renormalized parameters" of the OP, like the (renormalized) mass $m_R$ (here equal to zero) and the (renormalized) interaction $g_R$. Physically, $g_R$ corresponds to a scattering amplitude at zero 5-momentum $g_R$, and it is a priori finite (that is, unless the bare theory is non-interacting), as one could compute in a perturbative expansion in $g_B$, the bare value of the interaction. (Note that $g_R$ will depend on $\Lambda$, but that is not a problem, since $\Lambda$ is assumed finite.)</p> <p>Now, what do we expect in a renormalization group point of view? First, we need to use dimensionless quantities, and introducing $\tilde g= \mu g$, we find that after a short transient regime (corresponding in the dimensionfull units to the transition from $g(\mu=\Lambda)=g_B$ to $g(\mu\ll\Lambda)=g_R$), the renormalized dimensionless interaction flows to zero (i.e. to the gaussian fixed point) $\tilde g(\mu)\propto \mu$, while the dimensionfull interaction is constant.</p> <p>The OP was interested in the case of a relevant variable, which can be included easily in the example by assuming that the theory is massive, but with a very small (dimensionfull, renormalized) mass $m_R\ll \Lambda$. To study the RG flow, we need a dimensionless mass $\tilde m = m/\mu$, and we see that in the IR, $\tilde m$ is a relevant perturbation, that diverges as $\mu\to0$ (although, of course, the "real" mass stays finite and equal to $m_R$).</p> <p>Let me now comment briefly on the issue of the limit $\Lambda\to\infty$ and the "infinite" value of the bare parameter, which is in my view an artificial problem introduced by an old-school approach to field theory (and unfortunately still promoted in textbooks and pop-science). In the present case (as well as in $d=4$), we cannot take the limit $\Lambda\to\infty$ while keeping $g_R$ finite, as there is no UV fixed point to control the flow at high-energy, and the only consistent theory is the free theory $g_B=0$. I do not really want to go further into this subtle problem (and how one could work this out $d=3$), but a related discussion is given here : <a href="https://physics.stackexchange.com/questions/73403/why-do-we-expect-our-theories-to-be-independent-of-cutoffs/79053#79053">Why do we expect our theories to be independent of cutoffs?</a></p> https://physics.stackexchange.com/questions/224858/-/245442#245442 0 Answer by innisfree for Are the bare parameters of a renormalizable field theory infinitesimal or infinite? innisfree https://physics.stackexchange.com/users/23389 2016-03-25T06:39:31Z 2016-03-25T06:46:56Z <p>Bare parameters are the coefficients of the operators in a Lagrangian (without any counter-terms). They depend on the cut-off scale. </p> <p>The coefficients of relevant operators grow in the IR. That means, for example, the bare Higgs mass parameter at low energy (IR) is $$\mu_{IR}^2 \sim \mu_{UV}^2 + \text{const}\cdot\Lambda^2$$ This is much bigger than its high-energy (UV) value at the cut-off scale. The IR value is relevant for the physical mass of the Higgs.</p> <p>If we take the cut-off to infinity, it must be the case that $\mu_{UV}^2$ also diverges to cancel the cut-off in the above equation, and keep the Higgs mass light.</p> <p>So that's the answer. Because relevant operators grow in the IR, receiving divergent corrections, their UV values must diverge to cancel the divergent corrections. The corrections are additive, not multiplicative, which may be a source of confusion.</p>