# Current conservation beta function

1. May 18, 2015

### WannabeNewton

In section 19.5 of Peskin it is stated that if a scale transformation $\varphi \rightarrow e^{-\sigma}\varphi(xe^{-\sigma})$ is a symmetry of a theory then there is a current $D^{\mu} = \Theta^{\mu\nu}x_{\nu}$ (here $\Theta^{\mu\nu}$ is the Bellifante energy-momentum tensor) with $\partial_{\mu}D^{\mu} = \Theta^{\mu}{}{}_{\mu}$.

Furthermore if one considers loop corrections, so that the coupling $g$ acquires an RG flow, then this will no longer be a symmetry of the theory and under $g\rightarrow \sigma\beta(g)$ the current satisfies $\partial_{\mu}D^{\mu} = \sum_i \beta(g_i)\partial_{g_i} \mathcal{L}$ as an operator equation, where $\mathcal{L}$ is the Lagrangian and $\beta(g) = \Lambda \frac{d g}{d\Lambda}$ is the usual beta function. However this claim is not proven and I have been trying for hours to prove this relation starting from the partition function $\mathcal{Z} = \int \mathcal{D}\varphi e^{S}$ and trying a calculation similar to the proof of the Ward identity, with no luck. I suspect my issue is I do not know how exactly the beta function is defined in terms of $\mathcal{Z}$ as opposed to its definition in terms of the RG flow.

Does anyone know of or have a reference for a calculation proving this claim? Thanks in advance!

2. May 19, 2015

### vanhees71

Look for "trace anomaly" (particularly in QCD). A very nice treatment can be found in the marvelous book

Coleman, S.: Aspects of Symmetry, Cambridge University Press, 1985

The qualitative reason for the anomaly is simple. To begin with you need a theory where there are no dimensionful coupling constants, which implies that there are no mass terms in the Lagrangian. The most simple example is massless $\phi^4$ theory in 1+3 dimensions, because there the one coupling constant is dimensionless.

Now, since you have massless particles, you are plagued with both UV and IR (low momenta and collinear singularities). To get rid of the UV divergences you thus cannot renormalize the self-energy "on shell", i.e., at $p^2=0$, because the cut in the complex $s$ plane of the self-energy diagram (seen as an analytic function in the $s$ plane) starts precisely there. Thus you have to choose a renormalization scale, where you subtract the corresponding logarithmic divergence, i.e., you introduce an energy-momentum scale into the game.

One can show that there's no way to avoid this breaking of scale invariance. E.g., if you try dimensional regularization, you have to introduce a dimensionful coupling constant or, and that's the way a scale enters in this renormalization scale, write $\lambda \rightarrow \lambda \mu^{2 \epsilon}$ (working in $d=4-2 \epsilon$ dimensions). That's how the scale is usually defined in QCD, where the modified minimal-subtraction scheme (which is one choice of a mass-independent renormalization scheme, which you must use if there are massless particles in the theory, particularly in gauge theories with massless gauge bosons) is used. That's how $\Lambda_{\text{QCD}}$ enters the game and that's how the trace anomaly occurs in QCD. The trace anomaly is responsible for around 98% of the mass of the matter surrounding us. The current-quark masses due to the Higgs boson makes only about 2%.

This aspect is nicely described in

Donoghue, J. F., Golowich, E., Holstein, B. R.: Dynamics of the Standard Model, Cambridge University press, 1992

3. May 20, 2015

### samalkhaiat

I have never bothered myself with that book. I am not sure what method they are using and it is not clear to me whether your confusion is about calculating the variation or showing its relation to the beta function. And, since the rigorous proof of the final result is rather lengthy [*], I will sketch the derivation for you.

I will start with the regularized generating functional (to avoid UV divergences) $$W_{\Lambda}[J] = \int [ \mathcal{D} \Phi ]_{\Lambda} \ e^{ i ( S[\Phi] + \langle J \Phi \rangle )} ,$$ where $$S[\Phi] = \int d x \sum_{k=1}^{n} \mathcal{L}_{k} ( \Phi , \partial \Phi ) ,$$ $$\langle J \Phi \rangle = \int d x J_{i} (x) \Phi_{i} (x) ,$$ and $[\mathcal{D} \Phi]_{\Lambda}$ indicates that we are integrating over fields with frequency smaller than some UV cutoff $\Lambda$. Assuming that the theory is renormalizable in the usual sense, we introduce the appropriate cutoff-dependent counterterms and let $\Lambda \to \infty$. So, we define the renormalized generating functional by $$W_{r}[J] = \lim_{\Lambda \to \infty} \int [ \mathcal{D} \Phi_{r}]_{\Lambda} \exp \left[ i \left( \int d x \sum_{k} Z_{k} ( \frac{\Lambda}{\mu}) \mathcal{L}_{k} ( \Phi_{r} , \partial \Phi_{r} ) + \langle J \Phi_{r} \rangle \right) \right] ,$$ or $$W_{r}[J] = \lim_{\Lambda \to \infty} \int [ \mathcal{D} \Phi_{r}]_{\Lambda} \exp \left[ i \left( \int d x \mathcal{L}_{eff} ( \Phi_{r} , \partial \Phi_{r} ; \frac{\Lambda}{\mu} ) + \langle J \Phi_{r} \rangle \right) \right] ,$$ where $Z_{k}$ are the renormalization constants. Clearly this (renormalization) procedure introduces an anomaly in ,otherwise, scale invariant theory. Now, consider the scale transformations $$\delta_{\alpha} x = - x , \ \ \ \delta_{\alpha} \Phi = \Delta \Phi ,$$ and $$\delta_{\alpha} \Lambda = \Lambda .$$ This gives $$\delta Z_{n} = Z_{n} ( \frac{e^{\alpha} \Lambda}{\mu} ) - Z_{n} (\frac{\Lambda}{\mu}) ,$$ or, to first order $$\delta_{\alpha} Z_{n} = \frac{d Z_{n}}{d \log \Lambda} .$$ Assuming that $S[\Phi]$ is invariant, we find that $$\delta_{\alpha} \mathcal{L}_{eff} = \sum_{k} \frac{d Z_{k}}{d \log \Lambda} \mathcal{L}_{k} = \theta^{\nu}_{\nu} .$$ As an example, consider pure Y-M theories. They are conformal invariant at the classical level (contain no dimensionful parameters) and, owing to gauge invariance, they can be renormalized with only one $Z$: since $Z_{1} = Z_{3}$ and then the bare coupling $$g_{0} = Z_{3}^{- 1/2} g_{r} , \ \ \ (R)$$ leads to $$\mathcal{L}_{YM}^{(r)} = \mathcal{L}_{YM} ( A_{0} ) = Z_{3} \mathcal{L}_{YM} ( A_{(r)} ) .$$ Thus, in this case, we have the following trace anomaly $$\delta_{\alpha} \mathcal{L}_{YM} = \frac{d Z_{3}}{d \log \Lambda} \mathcal{L}_{YM} .$$ Using eq(R) together with the definition of beta function $$\beta ( g_{r} ) = \frac{d g_{r}}{ d \log \mu} ,$$ we find $$\frac{d Z_{3}}{d \log \Lambda} = - 2 Z_{3}\frac{\beta ( g_{r})}{g_{r}} .$$ Thus our final result is $$\partial_{\sigma} D^{\sigma} = \theta^{\mu}_{\mu} = - 2 \frac{\beta (g_{r})}{g_{r}} \mathcal{L}^{(r)}_{YM} .$$

Sam

[*]

Adler, S. L. et al (1977), Phys. Rev. D15, 1712

Nielson, N. K. (1977), Nucl. Phys. B120, 212