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• #Mass Renormalisation --
• [[#Renormalisation of \(\lambda\) :--]]

Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

Previous: Lecture 6

Next: Lecture 8

Reading Assignment -- 2 Read Srednicki chapters 12--20. and solve all problems.

\(\Gamma^{(2)}(k)\) depends on \(\Lambda\) only then \(\exists\) relation b/w \(\mu\) & \(m\).

\[ \int\frac{d^4\tilde{q}}{\alpha^2 q^2 + \mu^2} \;=\; \frac{V\!\left(S^{d-1}\right)}{(2\pi)^d}\int_{0}^{\Lambda}\frac{q^{d-1}\,dq}{\alpha^2 q^2 + \mu^2}\ ; \qquad \text{where}\ \left\{ d^d\tilde{q} = \frac{d^d q}{(2\pi)^d}\right. \]

For \(d=4\) ;

\[ \frac{V\!\left(S^{3}\right)}{(2\pi)^4}\int_{0}^{\Lambda}\frac{q^{3}\,dq}{\alpha^2 q^2 + \mu^2} \;=\; \left(\alpha^2\Lambda^2 + \mu^2\ln\mu^2 - \underbrace{\mu^2\ln\left(\mu^2 + \alpha^2\Lambda^2\right)}_{\textstyle \downarrow}\right) \]

Subleading term.

(margin working)

\[ \begin{aligned} \int\frac{dq\ q^3}{\alpha^2 q^2 + \mu^2} &= \frac{1}{\alpha^2}\int_{0}^{\Lambda}\frac{dq\left(\alpha^2 q^2 + \mu^2 - \mu^2\right)q}{\alpha^2 q^2 + \mu^2} &&\left.\begin{aligned}\alpha^2 q^2 + \mu^2 &= t\\ 2\alpha^2 q\,dq &= dt\end{aligned}\right.\\ &= \frac{1}{2\alpha^4}\int\frac{dt\,\left(t - \mu^2\right)}{t}\\ &= \frac{1}{2\alpha^4}\int dt\left(1 - \frac{\mu^2}{t}\right) = \frac{1}{2\alpha^4}\left(t - \mu^2\ln t\right)\\ &= \frac{1}{2\alpha^4}\left(\alpha^2 q^2 + \mu^2 - \mu^2\ln\left(\alpha^2 q^2 + \mu^2\right)\right)\Big|_{0}^{\Lambda}\\ &= \frac{1}{2\alpha^4}\left(\alpha^2\Lambda^2 + \mu^2 - \mu^2\ln\left(\alpha^2\Lambda^2+\mu^2\right) - \left(\mu^2 - \mu^2\ln\mu^2\right)\right)\\ &{= \frac{1}{2\alpha^4}\left(\alpha^2\Lambda^2 + \cancel{\mu^2} - \mu^2\ln\left(\frac{\alpha^2\Lambda^2+\mu^2}{\mu^2}\right)\right)}\\ &{= \frac{1}{2\alpha^4}\left(\alpha^2\Lambda^2 - \mu^2\ln\left(\frac{\alpha^2\Lambda^2}{\mu^2} + 1\right)\right)} \end{aligned} \]

From eqn (1) of 6th lecture --

\[ m^2 \;=\; \mu^2 + \frac{\lambda}{2}\left(\alpha^2\Lambda^2 + \mu^2\ln\left(\frac{\mu^2}{\mu^2 + \alpha^2\Lambda^2}\right)\right) \] \[ \text{This gives}\qquad \mu^2 \;=\; \mu^2\left(\Lambda;\, m^2\right) \]

look at 2 loops --

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\begin{tikzpicture}[scale=1.0,>=Stealth,baseline=(current bounding box.center)]
\node at (0,0) {$\Gamma^{(2)} \;=\; \left(\text{tree level} + 1\ \text{loop}\right)\ +$};
% double tadpole
\begin{scope}[xshift=3.9cm,yshift=-0.1cm]
  \draw[->] (-0.95,0) -- (0.05,0);
  \draw (0,0) -- (0.95,0);
  \draw (0,0.34) circle (0.34);
  \draw (0,1.02) circle (0.34);
  \node[scale=0.8] at (0.60,1.15) {$q_2$};
  \node[scale=0.8] at (0.62,0.42) {$q_1$};
\end{scope}
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\node at (-1.3,0) {$+$};
% sunset / bubble
\draw[->] (-0.85,0) -- (0.05,0);
\draw (0,0) -- (0.35,0);
\draw (0.35,0) .. controls (0.75,0.60) and (1.55,0.60) .. (1.95,0);
\draw (0.35,0) .. controls (0.75,-0.60) and (1.55,-0.60) .. (1.95,0);
\draw[->] (1.95,0) -- (2.75,0);
\node[scale=0.8] at (-0.45,0.24) {$k$};
\node[scale=0.8] at (1.15,0.55) {$q_1$};
\node[scale=0.8] at (1.35,-0.55) {$q_2$};
\draw[->] (0.95,0.44) -- (1.25,0.44);
\end{tikzpicture}

\[ \begin{aligned} =\;& \mu^2 + \alpha^2 k^2 + \frac{1}{2}\int\frac{d^d\tilde{q}}{\alpha^2 q^2 + \mu^2} \;-\; \lambda^2\left(\frac{1}{4}\int\frac{d^d\tilde{q}_1\ d^d\tilde{q}_2}{\left(\alpha^2 q_1^2+\mu^2\right)\left(\alpha^2 q_2^2 + \mu^2\right)}\right.\\ &\left.\hspace{6em} +\ \frac{1}{6}\int\frac{d^d\tilde{q}_1\ d^d\tilde{q}_2}{\left(\alpha^2 q_1^2+\mu^2\right)\left(\alpha^2 q_2^2+\mu^2\right)\left(\alpha^2\left(k - Q\right)^2 + \mu^2\right)}\right)\ +\ O(\lambda^3) \end{aligned} \]

Again define \(\Gamma^{(2)}(k=0) = m^2\) and write

\[ \Gamma^{(2)}(k) \;=\; \Gamma^{(2)}(k) - \Gamma^{(2)}(0) + m^2 + O(\lambda^3) \] \[ \Gamma^{(2)}(k) \;=\; \alpha^2 k^2 + m^2 - \frac{\lambda^2}{6}\,\Delta A(k) \] \[ \Delta A(k) \;=\; A(k) - A(0) \qquad\qquad {Q = \tilde{q}_1 + \tilde{q}_2} \] \[ \Delta A(k) \;=\; \int\frac{d^d\tilde{q}_1\ d^d\tilde{q}_2}{\left(\alpha^2 q_1^2+\mu^2\right)\left(\alpha^2 q_2^2+\mu^2\right)} \left(\frac{1}{\alpha^2\left(k-Q\right)^2+\mu^2} - \frac{1}{\alpha^2 Q^2 + \mu^2}\right) \] \[ =\; \int\frac{d^d\tilde{q}_1\ d^d\tilde{q}_2}{\left(\alpha^2 q_1^2+\mu^2\right)\left(\alpha^2 q_2^2+\mu^2\right)} \left(\frac{\alpha^2 Q^2 - \alpha^2\left(k-Q\right)^2}{\left(\alpha^2\left(k-Q\right)^2+\mu^2\right)\left(\alpha^2 Q^2+\mu^2\right)}\right) \]

It is divergent for \(\left(2d - 4 + 1 - 4 = 2d - 7\right)\) ( ?)

\(\lambda\) should be small so that, it nullifies effect of large \(\Lambda\) when series expanded in \(\lambda\). Why so much small ?

Eliminate \(\mu\) in favour of \(m\) \(\longrightarrow\) replace \(\mu \leftrightarrow m\) in RHS; ( ?)

\[ \Gamma^{(2)}(k) \;=\; m^2 + \alpha^2 k^2 - \frac{\lambda^2}{6}\,\Delta A(k) \] \[ \mu^2 \;=\; m^2 - \frac{\lambda}{2}\int\frac{d^d\tilde{q}}{\alpha^2 q^2 + m^2} + \lambda^2\left[\int\frac{d^d\tilde{q}_1\ d^d\tilde{q}_2}{\left(\alpha_1^2 q_1^2+m^2\right)\left(\alpha_2^2 q_2^2+m^2\right)} + \frac{1}{6}A(0)\right] + O(\lambda^3) \]

We have one more relation available to re-define

\[ a^2 \;\equiv\; \left.\frac{d\Gamma^{(2)}(k)}{dk^2}\right|_{k=\lambda} \] \[ \Gamma^{(2)}(k) \;=\; \Gamma^{(2)}(k) - \left.k^2\frac{d\Gamma^{(2)}}{dk^2}\right|_{k} + k^2\frac{d\Gamma^{(2)}}{dk^2} \] \[ \begin{aligned} &=\; m^2 + \alpha^2 k^2 - \frac{\lambda^2}{6}\left(\Delta A(k)\right) - \left.k^2\frac{d\Gamma^{(2)}}{dk^2}\right|_{k}\\ &\hspace{10em} +\ k^2 a^2 + O(\lambda^2) \end{aligned} \] \[ =\; m^2 + k^2\left(a^2 - \frac{\lambda^2}{6}B(k,\lambda)\right) + O(\lambda^3) \] \[ \downarrow \qquad \frac{\Delta A(k)}{k^2} + \frac{6}{\lambda^2}\left(\frac{d\Gamma^{(2)}}{dk^2}\right)_{k} \qquad\qquad -\alpha^2 \]

So we know \(B\);

\[ \frac{d\Gamma^{(2)}}{dk^2} \;=\; [?] \;=\; -\alpha \qquad \left(\text{in free theory}\right) \]

Eliminate \(\lambda\) in favour of \(g \equiv \Gamma^{(4)}(0,0,0)\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
% doodle: vertex with loop, and plain X
\draw (-2.5,0.55) -- (-1.15,-0.55);
\draw (-2.5,-0.55) -- (-1.15,0.55);
\draw (-1.35,0.0) circle (0.30);
\draw (-0.35,0.55) -- (0.55,-0.55);
\draw (-0.35,-0.55) -- (0.55,0.55);
\end{tikzpicture}

Proceed as earlier :--

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
% tree-level 4 point
\draw (-1.0,0.75) -- (0.0,-0.75);
\draw (-1.0,-0.75) -- (0.0,0.75);
\node[scale=0.8] at (-1.25,0.85) {$k_1$};
\node[scale=0.8] at (-1.25,-0.85) {$k_2$};
\node[scale=0.8] at (0.25,0.85) {$k_3$};
\node[scale=0.8] at (0.25,-0.85) {$k_4$};
\draw[->] (-0.72,0.33) -- (-0.55,0.08);
\draw[->] (-0.55,-0.08) -- (-0.35,-0.38);
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\node at (-2.6,0) {$+$};
% s-channel bubble drawn as crossing curves
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\draw (-1.9,-0.75) .. controls (-0.9,-0.1) and (-0.3,-0.1) .. (0.5,-0.75);
\draw (-1.25,0.32) .. controls (-0.85,-0.15) .. (-0.35,-0.32);
\draw (-1.25,-0.32) .. controls (-0.85,0.15) .. (-0.35,0.32);
\node[scale=0.8] at (-2.15,0.85) {$k_1$};
\node[scale=0.8] at (-2.15,-0.85) {$k_2$};
\node[scale=0.8] at (0.75,0.85) {$k_3$};
\node[scale=0.8] at (0.75,-0.85) {$k_4$};
\node at (1.5,0) {$+$};
% t-channel: legs meeting in a lens
\draw (2.2,0.75) .. controls (2.8,0.25) .. (3.05,0.0);
\draw (2.2,-0.75) .. controls (2.8,-0.25) .. (3.05,0.0);
\draw (3.05,0.0) .. controls (3.35,0.25) .. (3.9,0.75);
\draw (3.05,0.0) .. controls (3.35,-0.25) .. (3.9,-0.75);
\draw (2.85,0.15) circle (0.0);
\node[scale=0.8] at (1.95,0.85) {$k_1$};
\node[scale=0.8] at (1.95,-0.85) {$k_2$};
\node[scale=0.8] at (4.15,0.85) {$k_3$};
\node[scale=0.8] at (4.15,-0.85) {$k_4$};
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\node at (-1.6,0) {$+$};
\draw (-0.9,0.75) -- (0.5,-0.75);
\draw (-0.9,-0.75) -- (0.5,0.75);
\draw (-0.20,0.42) ellipse (0.16 and 0.30);
\end{tikzpicture}

\[ =\; \frac{\langle 0|\,T\left\{\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\ e^{i\int \mathcal{L}_{int}\,d^4y}\right\}|0\rangle} {\langle 0|T\left\{e^{i\int \mathcal{L}_{int}\,d^4y}\right\}|0\rangle} \]

disconnected diagrams cancels away because of \(\log[Z]\) term in path integral.

Path integrals shouldn't be worried about because they fall apart into 2 diagrams.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
% X with wiggly + bubble
\draw (-3.4,0.85) -- (-1.9,-0.85);
\draw (-3.4,-0.85) -- (-1.9,0.85);
\draw (-2.55,0.62) ellipse (0.16 and 0.28);
\draw[decorate,decoration={snake,amplitude=1.6pt,segment length=5pt}] (-2.70,0.34) -- (-2.05,0.02);
\node at (-1.1,0) {$\longrightarrow$};
\draw (-0.4,0.85) -- (1.1,-0.85);
\draw (-0.4,-0.85) -- (1.1,0.85);
\node at (1.7,0) {$+$};
\draw (2.2,0.0) -- (3.6,0.0);
\draw (2.95,0.20) ellipse (0.20 and 0.20);
\end{tikzpicture}

\[ \Gamma^{(2)}(k_1, k_2, k_3) \;=\; \lambda - \frac{\lambda^2}{2}\int\frac{d^d\tilde{q}}{\alpha^2 q^2+m^2} \left(\frac{1}{a^2\left(k-q\right)^2+m^2} + \frac{1}{a^2\left(k_1+k_3-q\right)^2+m^2}\right. \] \[ \left.+\ \frac{1}{a^2\left(k_1+k_3-q\right)^2+m^2}\right) \]

Add and substract \(g = \Gamma^{(4)}(0,0,0)\)

\[ \Gamma^{(4)}(k_1\,k_2\,k_3) \;=\; \Gamma^{(4)}(k_1, k_2, k_3) - \Gamma^{(4)}(0,0,0) \]