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Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

Previous: Lecture 9

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\(\phi^4\) theory renormalisation ---

\[ S \;=\; \int d^{2w}x\left(\frac{1}{2}\left(\partial\phi\right)^2 - \frac{m^2\varphi^2}{2} - \frac{\lambda}{4!}\,\phi^4\left(\mu^2\right)^{2-w}\right) \]

Here \(m\) is bare mass in place of \(\mu\).

\(\mu\) is introduced to make action dimensionless. \& \(\lambda\) dimensionless in each `\(2w\)' dim.

(margin working)

\[ \int d^{2w}x\left(-m^2\varphi^2 - \frac{\lambda}{4!}\varphi^4\left(\mu^2\right)^n\right) \qquad [\varphi] \;=\; \frac{2w-2}{2} \;=\; w-1 \]
  1. \(2 + 2[\varphi] - 2w = 0\)
  2. \(-2w + 4[\varphi] + n(2) = 0\) \[ -2w + 4(w-1) + 2n = 0\ ;\qquad 2n \;=\; 2w - 4(w-1)\ ;\qquad 2n \;=\; -2w + 4\ ;\qquad \underline{n \;=\; 2-w} \]

Minor changes in Feynman Rules :

  1. Scalar product of \(k_1\) \& \(k_2\) is summed over \(2w\) components. \[ k_1\cdot k_2 \;=\; k_1^0 k_2^0 - \left(\vec{k}_1\cdot\vec{k}_2\right)_{\text{in }(2w-1)\text{ dim}} \]
  2. vertex \(-\lambda\ \longrightarrow\ -\lambda\left(\mu^2\right)^{2-w}\)

let's look at tadpole diagram ---

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\[ T \;=\; \frac{1}{2}\,(-\lambda)\left(\mu^2\right)^{2-w}\int\frac{d^{2w}\ell}{(2\pi)^{w}}\ \frac{1}{\left(\ell^2+m^2\right)} \hspace{2em}\text{---(1)} \]

(\(\tfrac{1}{2}\) : from where ?)

(Digression) Integrals over arbitrary dimensions ---

\[ I_N \;=\; \int d^N\ell\ F(\ell) \]

goto polar coordinates --- \(\left(\ell_1, \ell_2 \ldots \ell_N\right)\ \downarrow\ \left(L, \phi, \theta_1, \theta_2, \ldots \theta_{N-2}\right)\)

\[ L^2 \;=\; \ell_\mu \ell_\mu \qquad (\text{Euclidean space}) \] \[ d^N\ell \;=\; L^{N-1}\,dL\ \left(d\phi\right)\left(\sin\theta_1\,d\theta_1\right)\left(\sin^2\theta_2\,d\theta_2\right)\cdots\left(\sin^{N-2}\theta_{N-2}\,d\theta_{N-2}\right) \] \[ 0\le L<\infty\,,\qquad 0\le \phi<2\pi\,,\qquad 0\le \theta_i<\pi \]

Standard Formulae ---

\[ \int_{0}^{\pi/2}\left(\sin t\right)^{2x-1}\left(\cos t\right)^{2y-1} dt \;=\; \frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)} \qquad \begin{aligned}&\text{Re }x>0\\ &\text{Re }y>0\end{aligned} \]

\(\left\{A(2)\ \text{of Ramond QFT}\right\}\)

For \(y=\frac{1}{2}\)

\[ \int_{0}^{\pi}\sin^k t\ dt \;=\; 2\int_{0}^{\pi/2}\sin^k t\ dt \;=\; \frac{\sqrt{\pi}\ \Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(\frac{k+2}{2}\right)} \]

So,

\[ I_N \;=\; \frac{\pi^{N/2}}{\Gamma(N/2)}\int_{0}^{\infty} dx\ x^{\frac{N-2}{2}}\ F(x) \qquad \left(x = L^2\right)\hspace{2em}\text{---(2)} \] \[ \left\{\begin{aligned} &\text{comparing (1) \& (2)} \qquad \frac{d^{2w}\ell}{(2\pi)^{w}} \;=\; dx\\ &\hspace{8.5em}\frac{1}{\ell^2+m^2} \;=\; x^{\frac{N-2}{2}}\,F(x) \end{aligned}\right. \]

For \(F = (x+a)^{-A}\) ; \(A = 2,3,\ldots\)

\[ \int_{0}^{\infty} dx\ \frac{x^{\frac{N-2}{2}}}{\left(x+a^2\right)^A} \;=\; \left(a^2\right)^{-A+\frac{N}{2}}\int_{0}^{\infty} dy\ y^{\frac{N-2}{2}}\left(1+y\right)^{-A} \]

Recall,

\[ B\left(\frac{N}{2},\,A-\frac{N}{2}\right) \;=\; \frac{\Gamma(N/2)\,\Gamma\left(A-N/2\right)}{\Gamma(A)} \;=\; \int_{0}^{\infty} dy\ y^{\frac{N}{2}-1}\left(1+y\right)^{-A} \]

\(\left\{\text{Re}\left(N/2\right)>0\ ;\ \text{Re}\left(A-N/2\right)>0\right\}\)

\[ \int\frac{d^N\ell}{\left(\ell^2+a^2\right)^A} \;=\; \frac{\pi^{N/2}\,\Gamma\left(A-N/2\right)}{\Gamma(A)}\ \frac{1}{\left(a^2\right)^{A-N/2}} \]

We take this formula valid to complex values of \(N\) by analytic continuation.

Another integral we need --- (we are doing in euclidean signature)

\[ \int d^N\ell\ \frac{\ell_\mu \ell_\nu}{\left(\ell^2+2p\cdot\ell+a^2\right)^A} \;=\; \frac{\pi^{N/2}}{\Gamma(A)\left(a^2-p^2\right)^{A-\frac{N}{2}}} \left[\Gamma\left(A-\frac{N}{2}\right)p_\mu p_\nu + \frac{\delta_{\mu\nu}}{2}\Gamma\left(A-1-\frac{N}{2}\right)\left(a^2-p^2\right)\right] \]

\(\downarrow\) (\(p\) is an) Arbitrary vector.

(margin note) Such Formula arrives in

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Considering \(A\) can take complex values.

\[ T \;=\; \frac{1}{2}\,(-\lambda)\left(\mu^2\right)^{2-w}\int\frac{d^{2w}\ell}{\left(2\pi\right)^{2w}}\ \frac{1}{\left(\ell^2+m^2\right)} \] \[ =\; \frac{-\lambda m^2}{2\left(4\pi\right)^2}\left(\frac{4\pi^2\mu^2}{m^2}\right)^{2-w}\Gamma\left(1-w\right) \]

expand it (1) ; \(\searrow\) pole at \(w=1\).

(margin working)

\[ \Gamma(m) \;=\; (m-1)! \;=\; \int_{0}^{\infty} t^{m-1}e^{-t}\,dt \] \[ \Gamma(1-w) \;=\; \left(1-w-1\right)! \;=\; \int_{0}^{\infty} t^{-w}e^{-t}\,dt \] \[ =\; \int_{0}^{\infty} t^{-w}\left(1 - t + \frac{t^2}{2!} - \frac{t^3}{3!} + \cdots\right)dt \] \[ =\; \int_{0}^{\infty} dt\left(t^{-w} - t^{1-w} + \frac{t^{2-w}}{2!} + \cdots\right) \;=\; \left(\frac{t^{1-w}}{-w+1} - \frac{t^{1-w}}{2-w} + \frac{t^{2-w}}{(3-w)2!} + \cdots\right)\Bigg|_{0}^{\infty} \]

Expand about \(\underline{2-w}\ (\equiv \epsilon)\)

\[ T \;=\; \frac{-\lambda m^2}{32\pi^2}\left[1 + (2-w)\ln\left(\frac{4\pi\mu^2}{m^2}\right) + \cdots\right] \left[\frac{-1}{2-w} - \psi(2) + \cdots\right] \]

pole term from lowest series \(\downarrow\)

\[ =\; \frac{\lambda m^2}{32\pi^2}\left(\frac{1}{2-w} + \psi(2) - \underbrace{\ln\left(\frac{m^2}{4\pi^2\mu^2}\right) + O(2-w)}_{\text{finite part}}\right) \]

Next look at \(\Gamma^{(4)}\)

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\[ =\; \frac{+1}{2}\,(-\lambda)^2\left(\mu^2\right)^{4-2w}\int\frac{d^{2w}\ell}{(2\pi)^{2w}}\ \frac{1}{\left(\ell^2+m^2\right)}\ \frac{1}{\left(\ell-p\right)^2+m^2} \]

\(\left(P = p_1+p_2\right)\)

Feynman parameterization ---

\[ \frac{1}{D_1^{a_1}D_2^{a_2}\cdots D_K^{a_K}} \;=\; \frac{\Gamma\left(a_1+a_2+\cdots a_K\right)}{\Gamma(a_1)\Gamma(a_2)\cdots\Gamma(a_K)}\int_{0}^{1}\!\!\cdots\!\int_{0}^{1} dx_1\,dx_2\cdots dx_K \] \[ \times\ \frac{\delta\left(1-\left(x_1+x_2+\cdots x_K\right)\right)}{\left(D_1x_1+D_2x_2+\cdots D_Kx_K\right)^{\left(a_1+a_2+\cdots a_K\right)}}\left(x_1^{a_1-1}x_2^{a_2-1}\cdots x_K^{a_K-1}\right) \] \[ \frac{1}{\left(\ell^2+m^2\right)\left(\left(\ell+p\right)^2+m^2\right)} \;=\; \int_{0}^{1} dx\ \frac{1}{\left[\ell^2+m^2+2\,\ell\cdot p\,(1-x) + p^2(1-x)\right]^2} \]

\(\downarrow\ (\ell-p)^2\) (?)

\[ D_r \;=\; \left(\ell'\right)^2 + m^2 + p^2 x(1-x)\ ;\qquad d^{2w}\ell' \;=\; d^{2w}\ell \]

So, the bubble is

\[ \frac{\lambda^2}{2}\left(\mu^2\right)^{4-2w}\int_{0}^{1} dx\int\frac{d^{2w}\ell}{(2\pi)^{2w}}\ \frac{1}{\left(\ell^2+m^2+p^2x(1-x)\right)^2} \]

use,

\[ \int\frac{d^{2w}\ell}{(2\pi)^{2w}}\ \frac{1}{\left(\ell^2+M^2+2\ell\cdot p\right)^A} \;=\; \frac{\Gamma(A-w)}{\left(4\pi\right)^{w}\Gamma(A)}\ \frac{1}{\left(M^2-p^2\right)^{A-w}} \] \[ \text{(bubble)} \;=\; \frac{\lambda^2}{2}\left(\mu^2\right)^{4-2w}\int_{0}^{1} dx\ \frac{\Gamma(2-w)}{\left(4\pi\right)^{w}}\ \frac{1}{\left(M^2+p^2x(1-x)\right)^{2-w}} \]

Expand in powers of \(\epsilon\) \((=2-w)\) :

\[ \text{(bubble)} \;=\; \left(\mu^2\right)^{2-w}\frac{\lambda}{32\pi^2}\left(\frac{1}{2-w} + \psi(1) - \int_{0}^{1} dx\ \ln\left(\frac{M^2+p^2x(1-x)}{4\pi\mu^2}\right)\right) \]

The finite part depends upon \(\mu^2\) \& \(p\) (ext. momentum)

We know

\[ \int_{0}^{1} dx\ \ln\left(1 + \frac{4x}{a}(1-x)\right) \;=\; -2 + \sqrt{1+a}\ \ln\left(\frac{\sqrt{1+a}+1}{\sqrt{1+a}-1}\right) \]

Using above,

\[ \text{(bubble)} \;=\; \left(\mu^2\right)^{2-w}\left(\frac{\lambda}{32\pi^2}\right) \left[\frac{1}{2-w} + \psi(1) + 2 + \ln\frac{4\pi\mu^2}{m^2}\right. \] \[ \left. -\ \left(1+\frac{4m^2}{p^2}\right)^{1/2}\ln\left(\frac{\sqrt{1+\frac{4m^2}{p^2}}+1}{\sqrt{1+\frac{4m^2}{p^2}}-1}\right)\right] \qquad (\ln 4\pi m) \]

2-loop diagrams ---

(To be done!)

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\[ =\; \frac{\lambda^2}{4}\left(\mu^2\right)^{4-2w}\int\frac{d^{2w}\ell}{(2\pi)^{2w}}\ \frac{1}{\left(\ell^2+m^2\right)} \int\frac{d^{2w}q}{\left(q^2+m^2\right)\left(q^2+m^2\right)} \]

Subleading part \(\downarrow\)

\[ =\; \frac{\lambda^2 m^2}{1024\,\pi^4}\left(\frac{1}{\left(2-w\right)^2} + \frac{1}{\left(2-w\right)}\left(2\ln\frac{4\pi^2\mu^2}{m^2} + \psi(2) + \psi(1)\right)\right. \] \[ \left.\hspace{2em} +\ 2\ln^2\frac{4\pi\mu^2}{m^2} + 2\ln\frac{4\pi\mu^2}{m^2}\left(\psi(2)+\psi(1)\right)\right. \] \[ \left.\hspace{2em} +\ \frac{1}{2}\left\{\left(\psi(2)+\psi(1)\right)^2 + \frac{2\pi^2}{3} - \psi'(2) - \psi'(1) + O(2-w)\right\}\right) \]

The other 2-loop diagram ---

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\[ \Sigma(p) \;=\; \frac{\lambda^2}{6}\left(\mu^2\right)^{4-2w}\int\left\{\frac{d^{2w}\hat{\ell}}{(2\pi)^{2w}}\ \frac{d^{2w}\check{q}}{(2\pi)^{2w}}\ \frac{1}{\left(\ell^2+m^2\right)}\ \frac{1}{\left(q^2+m^2\right)}\ \frac{1}{\left(\left(P-\ell+q\right)^2+m^2\right)}\right\} \] \[ \text{use}\qquad \frac{1}{4w}\left(\frac{\partial \ell_\mu}{\partial \ell_\mu} + \frac{\partial q_\mu}{\partial q_\mu}\right) \;=\; 1 \qquad \left(\downarrow 2w\ ,\ \downarrow 2w\right) \]

(margin working)

\[ \Sigma(p) \;=\; \frac{\lambda^2}{6}\left(\mu^2\right)^{4-2w}\int\left\{\frac{d^{2w}\hat{\ell}}{(2\pi)^{2w}}\ \frac{d^{2w}\check{q}}{(2\pi)^{2w}} \cdot\left(\ell_\mu\frac{\partial}{\partial \ell_\mu} + q_\mu\frac{\partial}{\partial q_\mu}\right)\right. \] \[ \left.\times\ \frac{1}{\left(\ell^2+m^2\right)}\ \frac{1}{\left(q^2+m^2\right)}\ \frac{1}{\left(\left(p+q-\ell\right)^2+m^2\right)}\right. \]

Integrate by parts

\[ \Sigma(p) \;=\; \frac{1}{4w}\ \frac{\lambda^2}{6}\left(\mu^2\right)^{4-2w}\int\frac{d^{2w}\ell}{(2\pi)^{2w}}\int\frac{d^{2w}q}{(2\pi)^{2w}} \] \[ \times\ \frac{3m^2 + p\cdot\left(p+q-\ell\right)}{\left(q^2+m^2\right)\left(\ell^2+m^2\right)\left[\left(q+p-\ell\right)^2+m^2\right]^2} \] \[ =\; \frac{1}{2w-3}\ \frac{\lambda^2}{6}\left(\mu^2\right)^{4-2w}\left[3m^2 K(p) + p^{\mu}K_\mu(p)\right],\ \text{where} \] \[ K(p) \;=\; \int\frac{d^{2w}\ell}{(2\pi)^{2w}}\int\frac{d^{2w}q}{(2\pi)^{2w}}\ \frac{1}{\left(q^2+m^2\right)\left(\ell^2+m^2\right)\left(\left(q+p-\ell\right)^2+m^2\right)} \] \[ K_\mu(p) \;=\; \int\frac{d^{2w}\ell}{(2\pi)^{2w}}\int\frac{d^{2w}q}{(2\pi)^{2w}}\ \frac{\left(p+q-\ell\right)_\mu}{\left(q^2+m^2\right)\left(\ell^2+m^2\right)\left(\left(q+\ell-p\right)^2+m^2\right)} \]

Using Feynman parameterization ---

\[ K(p) \;=\; \int d^{2w}\tilde{\ell}\ d^{2w}\tilde{q}\ \frac{1}{\left(q^2+m^2\right)}\int_{0}^{1} dx\ \frac{1}{\left(\underbrace{\ell^2+m^2+\left(p+q\right)^2 x(1-x)}_{M^2}\right)^2} \]

do the \(\ell\)-integral first

\[ \int d^{2w}\ell\ \frac{1}{\left(\ell^2+M^2+2\ell\cdot p\right)^A} \;=\; \frac{\Gamma(A-w)}{\left(4\pi\right)^{w}\Gamma(A)}\ \frac{1}{\left(M^2-p^2\right)^{A-w}} \]

Comparing \(A=2\) \(\left(\Gamma(2) = 1! = 1\right)\)

\[ K(p) \;=\; \frac{\Gamma(2-w)}{\left(4\pi\right)^{w}}\int_{0}^{1} dx\int d^{2w}\check{q}\ \frac{1}{\left(q^2+m^2\right)^2}\ \frac{1}{\left[m^2+\left(p+q\right)^2 x(1-x)\right]^{2-w}} \]

use Feynman parameterization once more to deal with

\[ \frac{1}{\left(q^2+m^2\right)^2}\ \frac{1}{\left(\left(p+q\right)^2 x(1-x)+m^2\right)^{2-w}} \] \[ K(p) \;=\; \frac{\Gamma(2-w)}{\left(4\pi\right)^{w}}\int_{0}^{1} dx\ \left(x(1-x)\right)^{w-2}\int_{0}^{1} dy\ y^{1-w}(1-y)\cdot\int\frac{d^{2w}q}{(2\pi)^{w}} \left(q^2 + p^2 y(1-y)\right. \] \[ \left. +\ m^2\left(1-y+\frac{y}{x(1-x)}\right)\right)^{w-4} \]

do the \(q\) integral first,

\[ K(p) \;=\; \frac{\Gamma(4-2w)}{\left(4\pi\right)^{2w}}\int_{0}^{1} dx\ \left(x(1-x)\right)^{w-2}\int_{0}^{1} dy\ y^{1-w}(1-y) \] \[ \times\ \left(p^2y(1-y) + m^2\left(1-y+\frac{y}{x(1-x)}\right)\right)^{-2\epsilon} \]

Expand around \(\epsilon=0\) \((w=2)\).

We have to isolate poles from finite part.

\[ K(p) \;=\; \frac{\Gamma(2\epsilon)}{\left(4\pi\right)^{4-2\epsilon}}\int_{0}^{1} dx\ \left(x(1-x)\right)^{-\epsilon}\int_{0}^{1} dy\ y^{-1+\epsilon}(1-y) \] \[ \times\ \left[p^2y(1-y) + m^2\left(1-y+\frac{y}{x(1-x)}\right)\right]^{-2\epsilon} \] \[ =\;\text{use}\quad y^{-1+\epsilon} = \frac{1}{\epsilon}\frac{d}{dy}y^{\epsilon}\ ,\quad \text{integrate by parts} \] \[ K(p) \;=\; \frac{\Gamma(2\epsilon)}{\left(4\pi\right)^{4-2\epsilon}}\ \frac{1}{\epsilon}\int dx\ \left(x(1-x)\right)^{1-\epsilon}\int_{0}^{1} dy\ y^{\epsilon}\times\left\{1 + 2\epsilon(1-y)\frac{d}{dy}\right. \] \[ \times\ \ln\left[p^2y(1-y) + m^2\left(1-y+\frac{y}{x(1-x)}\right)\right] \] \[ \times\ p^2y(1-y) + m^2\left(1-y+\frac{y}{x(1-x)}\right)^{-2\epsilon} \]

Expand the integral about \(\epsilon=0\)

\[ K(p) \;=\; \frac{\Gamma(2\epsilon)}{\left(4\pi\right)^{4-2\epsilon}}\ \frac{1}{\epsilon}\left(1 + \epsilon - 2\epsilon\ln m^2 + O(\epsilon^2)\right) \]

Now, Evaluate \(K_\mu(p)\) similarly \(\left\{\text{Mathew's \& walker metn}\right\}\)

Since we are working in Euclidean metric. \(\left(K_\mu = K^\mu\right)\).

\[ p^{\mu}K_\mu(p) \;=\; \frac{p^2\,\Gamma(2\epsilon)}{\left(4\pi\right)^{4-2w}}\int_{0}^{1} dx\ \left(x(1-x)\right)^{1-\epsilon}\int_{0}^{1} dy\ y^{\epsilon}(1-y) \] \[ \times\ \left[p^2y(1-y) + m^2\left(1-y+\frac{y}{x(1-x)}\right)\right]^{-2\epsilon} \]

Again expand around \(\epsilon=0\).

\[ p^{\mu}K_\mu(p) \;=\; \frac{p^2\,\Gamma(2\epsilon)}{\left(4\pi\right)^{4-2\epsilon}}\left(\frac{1}{2} + O(\epsilon)\right) \]

Putting All together :

\[ \Sigma(p) \;=\; \frac{-\lambda^2}{6\left(16\pi^2\right)^2}\left[\frac{3m^2}{2\epsilon^2} + \frac{3m^2}{\epsilon}\left(\frac{1}{2} + \psi(1) + \ln\frac{4\pi^2\mu^2}{m^2}\right)\right. \] \[ \left.\hspace{6em} +\ \frac{p^2}{4\epsilon} + \text{finite term}\right] \]

Renormalisation ---

poles in `\(\epsilon\)' can be eliminated order by order in \(\lambda\).

\(\left(\hat{\lambda} \equiv \dfrac{\lambda}{16\pi^2}\right)\) , \(\hat{m}^2 = \dfrac{m^2}{16\pi^2\mu^2}\)

Look at the tadpole digram ---

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\[ =\; m^2\,\frac{\hat{\lambda}}{2}\left(\frac{1}{\epsilon} + \psi(2) - \ln\hat{m}^2 + O(\epsilon)\right)\hspace{2em}\text{---(3)} \]

\(\nwarrow\) divergent as \(\epsilon\to 0\) ; finite \(F_1\left(\epsilon,m^2\right)\)

Original lagrangian

\[ \mathcal{L} \;=\; \frac{1}{2}\left(\partial\varphi\right)^2 - \frac{m^2\varphi^2}{2} - \frac{\lambda}{4}\varphi^4 \ \underbrace{-\ g\varphi^2}_{\mathcal{L}_{int}} \]

(\(\longleftarrow\) bare mass ; the counterterm vertex is drawn as a cross with coefficient \(g\))

(Choose coff \(g\) such that Counter term kills off divergence in the tadpole digram.)

Actually, Bare \(\mathcal{L}\) itself is not defined as \(\epsilon\to 0\) bare parameters depend on \(\epsilon\).

Add an extra term

\[ \frac{m^2}{4}\,\hat{\lambda}\left(\frac{1}{\epsilon} + F_1\left(\epsilon,m^2\right)\right)\varphi^2 \]

, where \(F_1\) is arbitrary (dimensionless) function, analytic as \(\epsilon \to 0\).

Puzzle :

Finite part is dependent on renormalization scheme, but we expect it to be same, wheather we do cut off renormalisation or DR. So how to reconcile ?

Our \(\mathcal{L}\) looks like :

\[ \mathcal{L} \;=\; \frac{1}{2}\left(\partial\varphi\right)^2 - \frac{m^2\varphi^2}{2} - \frac{\lambda}{4!}\varphi^4 + \underline{\frac{m^2\hat{\lambda}}{4}\left(\frac{1}{\epsilon} + F_1\right)\varphi^2} \]

\(\downarrow\) Tree level contribution

This \(\mathcal{L}\) has extra Feynman rule ---

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\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (-1.3,0) -- (1.3,0);
\node at (0,0.05) {$\times$};
\end{tikzpicture}
\[ =\; -\frac{1}{2}\,m^2\hat{\lambda}\left(\frac{1}{\epsilon}+F_1\right)\hspace{2em}\text{---(4)} \]

Lets compute \(\Gamma^{(2)}\) or inverse propagator ---

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\begin{tikzpicture}[scale=1.0,>=Stealth]
\node at (-4.6,0) {$\Gamma^{(2)}_{\text{new}} \;=\;$};
\draw (-3.7,0) -- (-2.6,0);
\node at (-2.25,0) {$+$};
\draw (-1.9,0) -- (-0.8,0);
\draw (-1.35,0.25) circle (0.25);
\node at (-0.45,0) {$+$};
\draw (-0.1,0) -- (1.0,0);
\node at (0.45,0.05) {$\times$};
\node at (1.8,0) {$+\ O(\lambda)$};
\end{tikzpicture}

eq (3) \(\downarrow\)

\[ =\; p^2 + m^2 + \frac{m^2\hat{\lambda}}{2}\left[\frac{1}{\epsilon} + \psi(2) - \ln\hat{m}^2 + O(\epsilon)\right] \] \[ -\ \frac{1}{2}m^2\hat{\lambda}\left(\frac{1}{\epsilon}+F_1\right) + O\left(\hat{\lambda}\right) \] \[ =\; p^2 + m^2 + \frac{m^2\hat{\lambda}}{2}\left(\psi(2) - \ln\hat{m}^2 - F_1\right) + O\left(\hat{\lambda}\right) \] \[ =\; p^2 + m^2\left(1 + \frac{\hat{\lambda}}{2}\left(\psi(2) - \ln\hat{m}^2 - F_1\right)\right) + O\left(\hat{\lambda}\right) \]

\(\longrightarrow\) means upto one loop.

2 pt. Fun. is now finite (upto order \(\hat{\lambda}\))

The extra term is called counterterm \(\left(\frac{m^2}{4}\hat{\lambda}\left(\frac{1}{\epsilon}+F_1\right)\right)\)

look at \(O\left(\lambda^2\right)\)

The 1 P I 4 pt. function ---

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\node at (-0.2,0) {$+$};
\draw (0.3,0.8) .. controls (0.9,0.3) .. (1.15,0.0);
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\node[align=center,scale=0.85] at (-3.9,-1.6) {$\downarrow$\\ Not 1PI (?)};
\end{tikzpicture}
\[ \Gamma^{(4)}\left(p_1,p_2,p_3,p_4\right) \;=\; -\mu^{2\epsilon}\lambda\left(1 - \frac{3\hat{\lambda}}{2}\left(\frac{1}{\epsilon} + \psi(1) + 2 + \ln\hat{m}^2\right.\right. \] \[ \left.\left. -\ \frac{1}{3}A\left(s,t,u\right) + O(\epsilon)\right)\right) + O\left(\lambda^3\right) \] \[ A\left(s,t,u\right) \;=\; \sum_{z=s,t,u}\left(1+\frac{4m^2}{z}\right)^{1/2} \ln\left(\frac{\left(1+\frac{4m^2}{z}\right)^{1/2}+1}{\left(1+\frac{4m^2}{z}\right)^{1/2}-1}\right) \]

\(s,t,u\) are mendelstam variables

\[ \begin{aligned} s &= \left(p_1+p_2\right)^2 = \left(p_1+p_2\right)\cdot\left(p_1+p_2\right)\\ t &= \left(p_1+p_3\right)^2\\ u &= \left(p_1+p_4\right)^2 \end{aligned} \]

\(\left(\Gamma^{(4)}\ \text{diverges as } \frac{1}{\epsilon}\right)\)

So we add a counterterm to \(\mathcal{L}\) of the form ---

\[ \text{Counterterm} \;=\; \frac{1}{4}\,\mu^{2\epsilon}\lambda\ \frac{3\hat{\lambda}}{2}\left(\frac{1}{\epsilon} + G_1\left(\epsilon,m^2\right)\right)\varphi^4 \]

\(\downarrow\) arbitrary but analytic as \(\epsilon\to 0\).

One more Feynman rule gets to be added due to counterterm.

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\[ =\; -\frac{3}{2}\,\mu^{2\epsilon}\lambda\,\hat{\lambda}\left(\frac{1}{\epsilon}+G_1\right) \] 
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\begin{tikzpicture}[scale=0.9,>=Stealth]
\node at (-6.4,0) {$\Gamma^{4}_{(\text{new})} \;=\;$};
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\[ \Gamma^{(4)}_{\text{new}}\left(p_1,p_2,p_3,p_4\right) \;=\; -\mu^{2\epsilon}\lambda\left(1 - \frac{3}{2}\hat{\lambda}\left(-G_1 + \psi(1) + 2 + \ln\hat{m}^2\right.\right. \] \[ \left.\left. -\ \frac{1}{3}A\left(s,t,u\right)\right) + O(\epsilon)\right) + O\left(\lambda^3\right). \]

Now divergence is gone!

We need only to renormalize \(\Gamma^{(2)}\) \& \(\Gamma^{(4)}\) then all 8 point \& 16 point functions

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