I am dealing with the anomalous dimension appearing in the n-point vertex green functions for a change of scale. I am following Ramond book, pages 188-189, Chapter 4, Section 5.

  1. Could someone give me a hand-waving explanation on how to go from eq 6.28 to eq. 6.31?

$$\left[-s \frac{\partial}{\partial s}+\beta(\lambda) \frac{\partial}{\partial \lambda}+\left(\gamma_{m}(\lambda)-1\right) m \frac{\partial}{\partial m}+d_{n}-n \gamma_{d}(\lambda)\right] \tilde{\Gamma}^{(n)}(s p ; m, \lambda, \mu)=0$$


$$\tilde{\Gamma}^{(n)}(s p ; m, \lambda, \mu)=s^{d_{n}} \tilde{\Gamma}^{(n)}(p ; \bar{m}(s), \bar{\lambda}(s), \mu) \exp \left\{-n \int_{1}^{s} \frac{d s^{\prime}}{s^{\prime}} \gamma_{d}\left(\bar{\lambda}\left(s^{\prime}\right)\right)\right\} $$

and why he defines

$$\begin{array}{l} s \frac{\partial \lambda(s)}{\partial s}=\beta(\bar{\lambda}(s)) \quad \bar{\lambda}(s=1)=\lambda \\ s \frac{\partial m(s)}{\partial s}=m(s)\left(\gamma_{m}(\bar{\lambda}(s))-1\right) \quad \bar{m}(s=1)=m \end{array}$$

  1. Also, the point of the result is that there is an additional scaling of the vertex functions besides the naive one. Is that correct?

  2. Does this scaling come from the non-trivial scaling of $m$ and $\lambda$?


1 Answer 1


Let us first remark that the RG equation that you quoted,

\begin{align} \left[-p \frac{\partial}{\partial p}+\beta(\lambda) \frac{\partial}{\partial \lambda}+\left(\gamma_{m}(\lambda)-1\right) m \frac{\partial}{\partial m}+d_{n}-n \gamma_{d}(\lambda)\right] \tilde{\Gamma}^{(n)}(p ; m, \lambda, \mu)=0, \end{align}

is a linear partial differential equation for $\tilde{\Gamma}^{(n)}$ in three parameters, $p$, $m$, and $\lambda$. To be explicit, let us remind ourselves that $\mu$ in the argument of $\tilde{\Gamma}^{(n)}$ is our reminder that our theory has been fixed by an RG condition at energy scale $\mu$ and we should not interpret it as a parameter of the vertex function. Now, mathematicians will immediately tell us that the solution to such a linear PDE can be given to us by the method of characteristics. The crux of the method is this: the linear PDE in question can be rewritten as an inner product:

\begin{align} (\,-p,\beta(\lambda),\gamma_m(\lambda)-1,[-d_n+n\gamma_d(\lambda)]\tilde{\Gamma}^{(n)}(p,\lambda,m)\,)\cdot (\,\partial_p\tilde{\Gamma}^{(n)},\partial_\lambda\tilde{\Gamma}^{(n)},\partial_m\tilde{\Gamma}^{(n)},-1\,) = 0. \end{align}

The second vector, $(\partial_p\tilde{\Gamma}^{(n)}, \partial_\lambda\tilde{\Gamma}^{(n)}, \partial_m\tilde{\Gamma}^{(n)},-1)$, describes a normal vector to the graph of $\tilde{\Gamma}^{(n)}$. Here I use the notation \begin{align} \textrm{Graph}(\tilde{\Gamma}^{(n)}) \equiv \{(p,\lambda,m,\tilde{\Gamma}^{(n)}(p,\lambda,m))\}_{p,\lambda,m\in\mathbb{R}} \subseteq \mathbb{R}^4. \end{align} In this geometrical way the RG equation tells you explicitly: for any choice of $p,\lambda,m$, the 4-vector $(-p,\beta(\lambda),m(\gamma_m(\lambda)-1),-d_n+n\gamma_d(\lambda))$ lies in the tangent space to the graph. We should now imagine that on the graph of $\tilde{\Gamma}^{(n)}$ we can draw a vector field given by these 4-vectors.

Now comes the crux! The integral curves that follow this vector field are given by equations:

\begin{align} \frac{\partial p(\ell)}{\partial \ell} &= -p(\ell)\\ \frac{\partial \lambda(\ell)}{\partial \ell} &= \beta(\lambda(\ell))\\ \frac{\partial m(\ell)}{\partial \ell} &= m(\ell)(\gamma_m(\lambda(\ell))-1)\\ \frac{\partial \tilde{\Gamma}^{(n)}(\ell;p(\ell),\lambda(\ell),m(\ell))}{\partial \ell} &= [-d_n+n\gamma_d(\lambda(\ell))]\tilde{\Gamma}^{(n)}(\ell;-p(\ell),\lambda(\ell),m(\ell))\\ \end{align}

The first equation is easy to solve; $p(\ell) = p e^{-\ell}$. (Quick aside: Ramond uses the notation to identifying \begin{align}s \equiv e^{\ell}\end{align} so that $p(s) \equiv p/s$.) In the last equation, or more precisely all four equations, the partial derivatives, $\partial/\partial\ell$ only act on the explicit $\ell$ dependence. The solution to the fourth equation is then easy to write down as well: \begin{align} \tilde{\Gamma}^{(n)}(\ell;p(\ell),\lambda(\ell),m(\ell)) = \exp\left(-d_n\ell + n\int_0^\ell \mathrm{d}\ell' \gamma_d(\lambda(\ell'))\right)\tilde{\Gamma}^{(n)}(0;p(0),\lambda(0),m(0)). \end{align}

Rewriting the last equation in terms of $s$, we arrive at \begin{align}\tilde{\Gamma}^{(n)}(p(s), m(s), \lambda(s)) =s^{-d_n}\exp \left\{n \int_{1}^{s} \frac{d s^{\prime}}{s^{\prime}} \gamma_{d}\left(\bar{\lambda}\left(s^{\prime}\right)\right) \right\} \tilde{\Gamma}^{(n)}(p,\lambda,m). \end{align} Rearranging, and applying the additional scaling $p\to sp$ we arrive at precisely the equation that you quoted from Ramond. \begin{align}\tilde{\Gamma}^{(n)}(sp ; m, \lambda, \mu)=s^{d_{n}} \tilde{\Gamma}^{(n)}(p ; \bar{m}(s), \bar{\lambda}(s), \mu) \exp \left\{-n \int_{1}^{s} \frac{d s^{\prime}}{s^{\prime}} \gamma_{d}\left(\bar{\lambda}\left(s^{\prime}\right)\right)\right\}. \tag{1} \end{align}

In summary, this is the answer to your first question, the mathematical reason for the appearance of the two differential equations describing the flow of $\lambda$ and $m$ comes from solving for $\tilde{\Gamma}$ along the vector field given by the RG equation.

However, there is an additional physical interpretation of the above solution and of the two differential equations that Ramond has written. Let us agree that a theory is well-defined only upon specifying the RG condition. Notice, then, that in equation (1) there are two different "theories": on the LHS, the theory is defined at $\mu$ with the parameters $(sp,m,\lambda)$; while the RHS defines a theory at $\mu$ with parameters $(p,\bar{m}(s),\bar{\lambda}(s))$. In this way the solution of the RG equation describes the relation between two theories defined by different RG conditions. The two differential equations describing the flow of $\lambda$ and $m$ precisely capture the relation between these theories.

I think, as you said, the utility of this solution to the RG equation is indeed to capture the scaling behaviour of the vertex functions. When you have a non-interacting (Gaussian) theory the anomalous dimensions will vanish because of the absence of any quantum corrections and the scaling of the vertex functions will go like $s^{d_n}$ which is the same as the tree-level scaling coming only from the engineering dimensions. However, with non-trivial interactions you see a non-trivial scaling of the entire vertex function.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.