Prof. Sachindeo Vaidya (CHEP, IISc) | PDF
Previous: Lecture 12
Next: A Short Break from Renormalisation
The idea is to express bare parameters as a laurent series in the renormalized parameters
\[
\lambda_0 \;=\; \mu^{2\epsilon}\left(a_0\left(\lambda,\frac{m}{\mu},\epsilon\right)
+ \sum_{k=1}^{\infty}\frac{a_k\left(\lambda, m/\mu\right)}{\epsilon^k}\right)
\]
\[
m_0^2 \;=\; m^2\left(b_0\left(\lambda,\frac{m}{\mu},\epsilon\right)
+ \sum_{k=1}^{\infty}\frac{b_k\left(\lambda, m/\mu\right)}{\epsilon^k}\right)
\]
\[
Z_\varphi \;=\; C_0\left(\lambda,\frac{m}{\mu},\epsilon\right)
+ \sum_{k=1}^{\infty}\frac{C_k\left(\lambda, m/\mu\right)}{\epsilon^k}
\]
Where,
\[
a_0\left(\lambda,\frac{m}{\mu},\epsilon\right) \;=\; \lambda\left(1 + \frac{3}{2}\hat{\lambda}G_1\right)
+ O(\lambda^2)
\]
and
\[
b_0\left(\lambda,\frac{m}{\mu},\epsilon\right) \;=\; 1 + \frac{1}{2}\left(\hat{\lambda}F_1
+ \hat{\lambda}^2 F_2 + \hat{\lambda}^2 H_2\right) + O(\lambda^3)
\]
and
\[
C_0\left(\lambda,\frac{m}{\mu},\epsilon\right) \;=\; 1 - \hat{\lambda}^2 H_2\left(m/\mu,\epsilon\right)
+ O(\lambda^3)
\]
\[
a_1\left(\lambda, m/\mu\right) \;=\; \frac{3}{2}\frac{\lambda^2}{16\pi^2} + O(\lambda^3)
\]
\[
b_1\left(\lambda, m/\mu\right) \;=\; \frac{1}{2}\left(\hat{\lambda}
+ \frac{\hat{\lambda}^2}{4}\left(F_1 + 3G_1 + 1\right) + \frac{\hat{\lambda}^2}{24}\right) + O(\lambda^5)
\]
\[
b_2\left(\lambda, m/\mu\right) \;=\; \frac{1}{2}\hat{\lambda}^2 + O(\lambda^3)
\]
\[
C_1\left(\lambda, m/\mu\right) \;=\; -\frac{\hat{\lambda}^2}{24} + O(\lambda^3)
\]
----------------- \(\times\) -----------------
\(\phi^6\) theory in 4d, do as before: renormalize \(m_0^2, \lambda_4, Z_\varphi, \lambda_6\)
Now look at 8 pt. function. (it turns out this is not finite ! at loop level (not at tree level).)
\[
V \;\sim\; \frac{m^2\varphi^2}{2!} + \lambda_4\frac{\varphi^4}{4!} + \lambda_6\frac{\varphi^6}{6!}
\]
\(\downarrow\)
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we need \(\dfrac{\lambda_8}{8!}\varphi^8\) when we add counterterm to get rid of infinities in
But in \(\varphi^4\) theory 4pt, 6pt, 8pt\(\ldots\) fun. are all finite once we have renormalized
So For \(\varphi^6\) theory we need infinite no. of coupling to be added in \(\mathcal{L}\) to
\[
V \;=\; \sum \lambda_{2n}\varphi^{2n}\qquad\text{to be added in }\mathcal{L}_0.
\]
Renormalizable theory requires to make finite couplings renormalized.
Non renormalized theory requires renormalization of infinite couplings. (we need infinite
measurements). Not a good theory (has weak predictive power).
\underline{For low momenta it is good at providing prediction.}
Remember we wrote down ---
\[
\widetilde{\Gamma}_0^{(n)}\left(p_1,p_2\ldots p_n;\lambda_0,m_0,\epsilon\right)
\;=\; Z_\varphi^{-n/2}\ \widetilde{\Gamma}^{(n)}\left(p_1,p_2\ldots p_n; m,\lambda,\mu,\epsilon\right)
\]
\(\downarrow\) Bare \(n\)-pt fun. \(\downarrow\) renormalized \(n\)-pt fun.
Now, vary \(\widetilde{\Gamma}^{(n)}\) w.r.t. physical mass `\(m\)', this gives us
----------------- \(\times\) -----------------
We were looking at renor. of \(\varphi^4\) theory (to 2 loops). We introduced arbitary scale \(\mu\).
We arrived at analytic fun. in counterterms \(\left(F_1, F_2 \ldots G_1, H_2 \ldots\right)\)
These fun. are the finite part of the counterterms.
\[
\mathcal{L}_{\text{ren}} \;=\; \mathcal{L} + \mathcal{L}_{ct}.
\]
The finite part of \(\mathcal{L}_{ct}\) can be fixed by defining the parameters appearing in
physical mass \(=\) zero of 2 pt fun (2pt fun. with zero momenta)
\[
\left(\eta = (+1,1,1,1)\ \textcolor{red}{\text{check it !}}\right)
\]
\[
\widetilde{\Gamma}^{(2)}\left(p, m_A\right) \;=\; p^2 + m_A^2 \qquad\left(\text{mean }\vec{p}^{\,2}=0\right)
\]
\[
\widetilde{\Gamma}^{(4)}\left(p_1 p_2 p_3 p_4\right) \;=\; -\mu^{2\epsilon}\lambda_A
\qquad\text{at } p_i = 0.
\]
As long as, \(m_A^2 \neq 0\) (non-zero mass theory) above prescription is well defined.
This fixes \(F_1, G_1, H_2\) for
\[
\widetilde{\Gamma}^{(2)}(p) \;=\; p^2 + m^2\left(1 + \frac{\hat{\lambda}}{2}
\left(\psi(2) - \ln\hat{m}^2 - F_1\right)\right) + O(\lambda^2)
\]
\[
F_1^{(A)} \;=\; \psi(2) - \ln\hat{m}_A^2
\]
Similarly,
\[
G_1^{A} \;=\; \psi(1) - \ln\hat{m}_A^2 \ ;\qquad H_2^{A} \;=\; 0
\]
One can change the subtraction pt, as long as it does not mess up the analytic continuation to
As long as we perform this subtraction on Euclidean Green function (vertex fun), it will give us
\[
\left(p^0, 0\right) \;\longrightarrow\; \left(0, \vec{p}\right)
\]
time like space like
\[
\widetilde{\Gamma}^{(2)}\left(p, m_B\right) \;=\; p^2 + m_B^2 \qquad\text{at } p^2 = M^2
\]
\[
\widetilde{\Gamma}^{(4)}\left(\{p_j\}\right) \;=\; -\lambda\,\mu^{2\epsilon}
\qquad\text{at } p_i p_j = M^2\left(\delta_{ij} - \frac{1}{4}\right)
\]
\[
\left(s = t = u = M^2\right)
\]
\[
F_1^{B} \;=\; \psi(2) - \ln\hat{m}_B^2\ ,\qquad H_2^{B} = 0
\]
\[
G_1^{B} \;=\; \psi(1) - \ln\hat{m}_B^2 - \int_0^1 dx\,\ln\left(1 + \frac{M^2}{m_B^2}x(1-x)\right)
\]
In scheme B, the renormalisation eqn are not easy to solve, except deep in
We set \underline{all} the finite parts of counterterm to be zero (order by order in \(\lambda\)).
\[
0 \;=\; F_1^{C} \;=\; F_2^{C} \;=\; G_1^{C} \;=\; F_2^{C}
\]
The advantage of scheme `C' becomes clear when we look at \(a_0, b_0, C_0\ldots\)
\[
a_0\left(\lambda,\frac{m}{\mu},\epsilon\right) \;=\; \lambda\left(1 + \frac{3}{2}\hat{\lambda}G_1\right)
+ O(\lambda^2)\ ,
\]
\[
b_0(\quad) \;=\; (\qquad\qquad)
\]
All the coefficients are independent of \(m\).
This prescription is called \underline{mass -- independent renormalization}
\[
\lambda_0 \;=\; \mu^{2\epsilon}\left(\lambda + \sum_{k=1}^{\infty}\frac{a_k(\lambda)}{\epsilon^k}\right)
\qquad\text{etc.}
\]
differentiate w.r.t \(\mu\dfrac{\partial}{\partial\mu}\) at fixed \(\lambda_0\) we get,
\[
0 \;=\; 2\epsilon\left(\lambda + \sum_{k=1}^{\infty}\frac{a_k(\lambda)}{\epsilon^k}\right)
+ \mu\frac{\partial\lambda}{\partial\mu}\left(1 + \sum_{k=1}^{\infty}\frac{a_k'(\lambda)}{\epsilon^k}\right)
\]
here \(\lambda\) & \(\dfrac{\partial\lambda}{\partial\mu}\) are analytic at \(\epsilon = 0\). so we get---
\[
\mu\frac{\partial\lambda}{\partial\mu} \;=\; -2\epsilon\lambda - 2a_1(\lambda) + 2\lambda\,a_1'(\lambda)
\]
\[
\lim_{\epsilon\to 0}\ \mu\frac{\partial\lambda}{\partial\mu} \;=\; \beta(\lambda)
\;=\; -2\left(1 - \lambda\frac{\partial}{\partial\lambda}\right)a_1(\lambda)
\]
\[
\left(1 - \lambda\frac{\partial}{\partial\lambda}\right)a_{k+1}(\lambda)
\;=\; a_k'(\lambda)\left(1 - \lambda\frac{\partial}{\partial\lambda}\right)a_1(\lambda)
\]
\[
a_1(\lambda) \;=\; \frac{3}{2}\frac{\lambda^2}{16\pi^2} + O(\lambda^2)
\]
\[
\Rightarrow\qquad \mu\frac{\partial\lambda}{\partial\mu} \;=\; \frac{3\lambda^2}{16\pi^2} + O(\lambda^3)
\]
integrate, we get
\[
\int\frac{\partial\lambda}{\frac{3\lambda^2}{16\pi^2} + O(\lambda^3)} \;\simeq\; \int\frac{\partial\mu}{\mu}
\]
\[
\lambda(\mu) \;=\; \lambda_s\ \frac{1}{1 - \frac{3}{16}\lambda_s\ln\frac{\mu}{\mu_s}}
\ ;\qquad \lambda_s \equiv \lambda(\mu_s)
\]