• #PDF
• #Dimensional Regularisation --- (D R)
• #Volume of the unit sphere (working)
• [[#Continuing to \(w=2\)]]

Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

Previous: Lecture 8

Next: Lecture 10--11

Mid Term Exam --- 18th or 25th Feb. / 4 hours / 100 points.

We were talking about a graph that is \(\Lambda\)-sensitive has no \(\Lambda\)-sensitive subgraphs is called a primitively divergent graphs.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\node[align=left] at (-5.0,0.0) {Only\\ primitve\\ graphs};
\draw (-4.0,0.6) .. controls (-4.0,-0.4) and (-4.2,0.9) .. (-4.2,0.0) .. controls (-4.2,-0.9) and (-4.0,0.4) .. (-4.0,-0.6);
\draw (-3.4,0.0) -- (-1.7,0.0);
\draw (-2.55,0.26) circle (0.26);
\node[scale=0.85] at (-2.55,-0.45) {Tadpole};
\draw (-1.0,0.0) -- (0.9,0.0);
\draw (-0.05,0.0) ellipse (0.40 and 0.24);
\node[scale=0.85] at (-0.05,-0.6) {sunrise / saturn / sunset};
\draw (1.8,0.6) .. controls (2.5,0.05) and (3.0,0.05) .. (3.7,0.6);
\draw (1.8,-0.6) .. controls (2.5,-0.05) and (3.0,-0.05) .. (3.7,-0.6);
\draw (2.35,0.24) .. controls (2.65,-0.16) .. (3.1,-0.28);
\draw (2.35,-0.24) .. controls (2.65,0.16) .. (3.1,0.28);
\end{tikzpicture}

\(\left(\text{in } \lambda\phi^4 \text{ interaction},\ d=4\right)\) (every other diagram is built out of \(\Gamma^{(2)}\) \& \(\Gamma^{(4)}\))

(margin working) \(\displaystyle\int_{0}^{\Lambda}\frac{q^3\,dq}{q^2+\mu^2}\ \sim\ \int_{0}^{\Lambda}\frac{q^3\,dq}{q^2}\ \sim\ \int_{0}^{\Lambda} q\,dq\ \sim\ \Lambda^2\)

let's look at graph one by one.

1. Tadpole :

   \usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
   \begin{tikzpicture}[scale=0.9,>=Stealth]
   \draw (-0.9,0) -- (0.9,0);
   \draw (0,0.25) circle (0.25);
   \end{tikzpicture}
   

   \[ \int\frac{d^4\tilde{q}}{\left(q^2+\mu^2\right)}\ \sim\ \Lambda^2 \qquad \text{(quadratic UV--divergence)} \]

2)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (-1.1,0.65) .. controls (-0.45,0.08) and (0.15,0.08) .. (0.8,0.65);
\draw (-1.1,-0.65) .. controls (-0.45,-0.08) and (0.15,-0.08) .. (0.8,-0.65);
\draw (-0.6,0.25) .. controls (-0.25,-0.15) .. (0.25,-0.28);
\draw (-0.6,-0.25) .. controls (-0.25,0.15) .. (0.25,0.28);
\node[scale=0.7] at (-1.3,0.8) {$p_1$};
\node[scale=0.7] at (-1.3,-0.8) {$p_2$};
\node[scale=0.7] at (-0.15,0.62) {$q_1$};
\node[scale=0.7] at (-0.15,-0.55) {$q_2$};
\end{tikzpicture}

\[ \frac{\lambda^2}{2}\int\frac{d^4\tilde{q}_1\ d^4\tilde{q}_2}{\left(q_1^2+\mu^2\right)\left(q_2^2+\mu^2\right)}\ \delta\left(q_1+q_2-p\right) \] \[ \sim\ \frac{\lambda^2}{2}\int\frac{d^4\tilde{q}}{\left(q_1^2+\mu^2\right)\left(\left(p-q_1\right)^2+\mu^2\right)} \] \[ \longrightarrow\ \frac{\lambda^2}{2}\int\frac{d^4\tilde{q}}{q^4}\ \sim\ \lambda^2\ln\Lambda \]

If \(\mu^2=0\); \& \(\left(p_1+p_2\right)^2 = 0\), then there is also divergence for \(q\to 0\) (IR divergence)

Typical integrals ---

\[ I_4(k) \;=\; \int d^4\ell\ F(k,\ell) \]

For large \(\ell\) \(\longrightarrow\) \(\ell^{-2}\) for \(\Gamma^{(2)}\) , \(\ell^{-4}\) for \(\Gamma^{(4)}\)

For \(d=2\) ; \(\displaystyle\int\frac{d^2\ell}{\ell^4}\) in convergent in uv.

Suggest the following math. trick / procedure ---

Consider, \(\displaystyle I(w,k) \;=\; \int \left(d^{2w}\ell\right) F(\ell,k)\) , \(\left(2w = \text{dimension}\right)\)

Treat `\(w\)' as complex variable. (We are extending \(\int d^d q\) integral to complex plane, via analytic continuation) \(\overline{n+1} = n!\)

We evaluate this integral in a region (of \(w\)-space), where there are no singularities. We then invent \(I'\) (new integral) which is convergent over bigger region / domain. This process is called analytic continuation in complex analysis.

\[ \int d^{2w}\ell\ \boxed{F(\ell,k)} \qquad \text{converges for } w<2 \qquad\quad \left(\sim \frac{1}{\ell^2}\cdot\frac{1}{\ell^2}\right) \]

For our divergent integrals, this region is \(w<2\). We write :

\[ \int d^{2w}\ell \;=\; \underbrace{\int d^4\ell}_{\text{radial}}\ \underbrace{\int d^{2w-4}\ell}_{\text{angular}} \] \[ I \;=\; \int d^4\ell\ \underbrace{\int d\Omega_{2w-5}}_{\text{easy}}\ \int_{0}^{\infty} dL\ \frac{L^{2w-5}}{\left(L^2+\ell^2+\mu^2\right)} \] \[ \int_{-\infty}^{\infty} dx_1 \cdots \int_{-\infty}^{\infty} dx_N\ e^{-a\left(x_1^2+x_2^2+\cdots+x_N^2\right)} \;=\; \left(\frac{\pi}{a}\right)^{N/2} \]

Changing to polar coordinates

\[ \underbrace{\int d\Omega_{N-1}}_{V\left(S^{N-1}\right)}\int r^{N-1}\,dr\ e^{-ar^2} \qquad \left(\textstyle\sum x_i^2 = r^2\right) \]

\(\downarrow\) Volume of sphere of \((N-1)\) dim.

\[ \Rightarrow\quad V\left(S^{N-1}\right)\int r^{N-1}e^{-ar^2}\,dr \qquad \left.\begin{aligned} ar^2 &= t\\ ar\,dr &= dt/2\end{aligned}\right. \] \[ V\left(S^{N-1}\right)\int \left(\frac{t}{a}\right)^{\frac{N-1}{2}} e^{-t}\ \frac{dt}{2a\left(\frac{t}{a}\right)^{1/2}} \] \[ \frac{1}{2a}V\left(S^{N-1}\right)\int\left(\frac{t}{a}\right)^{\frac{N-2}{2}} e^{-t}\,dt \;=\; \left(\frac{\pi}{a}\right)^{N/2} \] \[ \frac{1}{2a^{N/2}}V\left(S^{N-1}\right)\int t^{\frac{N}{2}-1}e^{-t}\,dt \;=\; \frac{1}{2a^{N/2}}V\left(S^{N-1}\right)\Gamma\left(N/2\right) \;=\; \left(\frac{\pi}{a}\right)^{N/2} \] \[ \boxed{\ V\left(S^{N-1}\right) \;=\; \frac{2}{\Gamma(N/2)}\ \pi^{N/2}\ } \]

check: For \(N=3\)

\[ V\left(S^2\right) = \frac{2\pi^{3/2}}{\Gamma(3/2)} = \frac{2\pi^{3/2}}{\frac{1}{2}\Gamma(1/2)} \qquad \left(\Gamma(1/2)=\sqrt{\pi}\right) \] \[ =\; 4\,\pi^{3/2-\frac{1}{2}} \;=\; 4\pi \]

So,

\[ I \;=\; \frac{2\pi^{w-2}}{\Gamma(w-2)}\int d^4\ell\ \int_{0}^{\infty}\frac{dL\ L^{2w-5}}{\left(L^2+\ell^2+\mu^2\right)} \qquad \left(\begin{aligned}2w-6&\ge 0\\ w&\ge 3\end{aligned}\right. \]

uv divergent for \(w\ge 3\) ; IR divergence for \(w\le 1\) (check by \(\mu=0\)).

Write \(\displaystyle L^{2w-5} \;=\; \frac{1}{w-2}\ \frac{d}{dL^2}\left(\left(L^2\right)^{w-2}\right)\)

\[ dL\ L^{2w-5} \;=\; L\,dL\ L^{2w-6} \;=\; \frac{1}{2}\,dL^2\left(L^2\right)^{w-3} \]

Integrate by parts over \(L^2\). \& discard the surface term.

\[ I \;=\; \underbrace{\frac{\pi^{w-2}}{(w-2)\,\Gamma(w-2)}}_{\Gamma(w-1)}\int d^4\ell \int dL^2\ \left(L^2\right)^{w-2}\left(-\frac{d}{dL^2}\right)\frac{1}{L^4+\ell^2+\mu^2} \]

Again, IR divergent for \(w\le 1\) ; uv '' '' \(w\ge 1\)

Do integral by parts once more ---

\[ I \;=\; \frac{\pi^{w-2}}{\Gamma(w)}\int d^4\ell\int dL^2\ \left(L^2\right)^{w-1}\left(-\frac{d}{dL^2}\right)^2\frac{1}{L^4+\ell^2+\mu^2} \]

well defined for \(\left(0<w<1\right)\)

We need to look for \(I'\) which extends the domain \& can be analytically continued to \(w=2\).

\[ \text{use}\quad 1 \;=\; \frac{1}{(5)}\left(\frac{\partial L}{\partial L} + \frac{\partial \ell_\mu}{\partial \ell_\mu}\right) \qquad \left\{\frac{\partial \ell_\mu}{\partial \ell_\mu} = 4\right\} \]

Integrate by parts ---

\[ I \;=\; -\frac{2\pi^{w-2}}{5\,\Gamma(w)}\int d^4\ell\int dL^2\left(\ell_\mu\frac{\partial}{\partial \ell_\mu} + 2L^2\frac{\partial}{\partial L^2} + 1\right) \cdot\left(\frac{\left(L^2\right)^{w-1}}{\left(L^2+\ell^2+\mu^2\right)^3}\right) \] \[ =\; \frac{-3\mu^2}{w-1}\ \frac{2\pi^{w-2}}{\Gamma(w)}\int d^4\ell\int_{0}^{\infty} dL^2\ \frac{\left(L^2\right)^{w-1}}{\left(L^2+\ell^2+\mu^2\right)^4} \]

Now, the only divergence is from the pole at \(w=2\).

Check,

\[ I \;=\; \int\frac{d^{2w}\ell}{\left(\ell^2+\mu^2\right)^6} \ \longrightarrow\ I \] \[ I \;=\; \frac{\pi^{w-2}}{\Gamma(w-1)}\int d^4\ell\int_{0}^{\infty} dL^2\ \left(L^2\right)^{w-2}\left(-\frac{d}{dL}\right)\left(\frac{1}{L^4+\ell^2+\mu^2}\right) \] \[ I(w=2) \;=\; \int d^4\ell\ \frac{(-1)}{\left(L^2+\mu^2+\mu^2\right)^6}\Bigg|_{0}^{\infty} \;=\; \int d^4\ell\ \frac{1}{\left(\ell^2+\mu^2\right)^6} \]

use,

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

\(\longleftarrow\) expand around \(w=2\). (use \(\epsilon = 2-w\))

\[ \Gamma\left(-n+\epsilon\right) \;=\; \frac{(-1)^n}{n!}\left(\frac{1}{\epsilon} + \psi(n+1) + \frac{\epsilon}{2}\left(\frac{\pi^2}{3} + \psi^2(n+1) - \psi'(n+1)\right) + O(\epsilon^2)\right) \] \[ \frac{d\,\psi(s)}{ds} \;=\; \frac{d}{ds}\ln\Gamma(s) \] \[ \psi(n+1) \;=\; 1 + \frac{1}{2} + \cdots\ \frac{1}{n} - \gamma \]

\(\searrow\) euler maclaurian const.

\[ \psi(1) \;=\; -\gamma \;=\; -0.577 \] \[ \psi'(n+1) \;=\; \frac{\pi^2}{6} - \sum_{1}^{n}\frac{1}{k^2}\ , \qquad \psi'(1) \;=\; \pi^2/6 \]

So,

\[ \lim_{w\to 2}\int\frac{d^{2w}\ell}{\ell^2+\mu^2} \;=\; -\pi^2\mu^2\left(\frac{1}{2-w} + \psi(2) + O(\epsilon)\right) \]

\(\nwarrow\) Blows up. \(\searrow\) quadratically divergent for \(w=2\)