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Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

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Gamma fun. is defined for \(\Gamma(x)\), \(x > 0\).

for \(\Gamma\!\left(\dfrac{1-d}{2}\right)\) for \(\dfrac{1-d}{2} < 0\) we get poles.

\(\Gamma_m = (m-1)! = \displaystyle\int_{0}^{\infty} t^{m-1}e^{-t}\,dt\) \(\hookrightarrow\) has poles when \(m < 0\). Similarly \(\Gamma\!\left(\frac{1-d}{2}\right)\) has poles for \(\frac{1-d}{2} < 1\) \(\Rightarrow\) \(1-d < 2\) \(\Rightarrow\) \(d > -1\)

\(\uparrow\) we are extending domain of \(d\) from \(+\)ve integer to real numbers, by analytic continuation.

Expand \(\phi_d\) in powers of \(1-d \equiv \epsilon\)

\[ \text{if}\quad \frac{1-d}{2} = -\text{ve integer} \] \[ 1-d = -\text{ve even integer} \] \[ \Rightarrow\quad d = +\text{ve odd integer} \]

Note that we can't add quantities of different dimensions.

So \(\exp\left(\text{dimensionful quantity}\right)\) = does not make sense.

\(\mu^{d-1}\), \(x^{d-1}\) if expanded seperately then above problem arise.

So, we combine

\[ \left(\mu x\right)^{d-1} \;=\; \left(\mu x\right)^{-\epsilon} \] \[ =\; 1 - \epsilon\ln\mu x + O(\epsilon^2) \qquad {\text{> should be minus!}} \]

\(x^n = e^{\ln x^n}\); \((\mu x)^{-\epsilon} = e^{\ln(\mu x)^{-\epsilon}} = 1 + \ln(\mu x)^{-\epsilon} + \frac{\left(\ln(\mu x)^{-\epsilon}\right)^2}{2!} + ... = 1 - \epsilon\ln\mu x + O(\epsilon)^2\)

\[ \pi^{\frac{d-1}{2}} \;=\; 1 - \frac{\epsilon}{2}\ln\pi + O(\epsilon^2) \] \[ \Gamma\!\left(\frac{1-d}{2}\right) \;=\; \Gamma\!\left(-\epsilon/2\right) \;=\; \frac{-2}{\epsilon}\,\Gamma\!\left(1 - \epsilon/2\right) \qquad {\enclose{circle}{\text{?}}} \]

\(\Gamma(z+1) = z\,\Gamma(z)\); \(\Gamma(-\epsilon/2) = \left(-\frac{\epsilon}{2}\right)^{-1}\Gamma\!\left(\left(-\frac{\epsilon}{2}\right)+1\right) = \left(-\frac{2}{\epsilon}\right)\Gamma\!\left(1-\frac{\epsilon}{2}\right)\)

\[ =\; \frac{-2}{\epsilon}\left(1 + \frac{\epsilon\gamma}{2} + O(\epsilon^2) + \cdots\right) \]

So,

\[ \phi_d(x) \;=\; \frac{-\lambda}{2\pi\epsilon_0}\left[\frac{1}{\epsilon} + \underbrace{\frac{1}{2}\ln\pi + \frac{\gamma}{2} + \ln\mu x}_{\ln\mu' x} + O(\epsilon)\right] \]

\(\downarrow\) Can be dropped for \(d=1\) \(\Rightarrow\) \(_\epsilon = 0_\).

define

\[ \phi^{d}_{d=1}(x) \;=\; \frac{-\lambda}{2\pi\epsilon_0}\left(\ln\mu' x\right) \] \[ \phi^{R}(x) \;=\; \frac{A}{\epsilon} + \left(\frac{-\lambda}{2\pi\epsilon_0}\right)\ln\mu' x \] \[ \phi^{R}(y) \;=\; \frac{A}{\epsilon} + \left(\frac{-\lambda}{2\pi\epsilon_0}\right)\ln\mu' x' \] \[ \phi^{R}(x) - \phi^{R}(x') \;=\; \frac{-\lambda}{2\pi\epsilon_0}\,\ln\left(\frac{x}{x'}\right) \qquad {\longleftarrow \text{It is potential difference that are finite!}} \]

We have found this answer from basic defn of potential \(V(r) = \int \frac{k\,dq}{r}\)

For 2d charge confn.

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\node[scale=0.8] at (-0.25,0.16) {$0$};
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\node[right,scale=0.85] at (3.2,0.20) {$V(z=0) = 0$};
\node[scale=0.9] at (1.9,-0.95) {$E = \dfrac{\sigma}{2\epsilon_0}$};
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\[ \displaystyle Z[J] \;=\; e^{\frac{1}{2}\langle JGJ\rangle}\left[e^{\langle JG\delta\rangle}\ \mathcal{F}[\varphi]\right]_{\varphi=0} \]

Where,

\[ \langle JGJ\rangle \;=\; \int d^4x\,d^4y\ J(x)\,G(x,y)\,J(y) \qquad {\text{Check}\ \varphi^4 d^4x\,?} \] \[ \mathcal{F}[\varphi] \;=\; N\,e^{\frac{1}{2}\langle \delta G\delta\rangle}\ e^{-\frac{i\lambda}{4!}\int \varphi^4 dx} \] \[ Z_0[J] \;=\; e^{\frac{1}{2}\langle JGJ\rangle} \]

So,

\[ Z[J] \;=\; N\left[e^{\frac{i\lambda}{4!}\int d^4x \left(\frac{\delta}{\delta J(x)}\right)^4}\right] Z_0[J] \]

Gaussian functional integral\ \(\downarrow\)\ Greens fun are functional integrals\ \(\downarrow\)\ \(Z_0[J]\) as a functional integral\ \(\downarrow\)\ \(Z\) as a functional integral.

\[ Z[J] \;=\; N\int \left[d\varphi\right]\ e^{iS - \epsilon\int\varphi^2 + J\varphi} \] \[ S \;\to\; S_0 - \frac{\lambda}{4!}\int d^4x\ \varphi^4(x) \qquad \left\{\begin{aligned} S &= S_M\\ S_E &= -i S_M \end{aligned}\right. \]

The gaussian integral is of type

\[ \int dx\ e^{i\alpha x^2 - \epsilon x^2} \;=\; \sqrt{\frac{\pi}{\epsilon - i\alpha}} \]

We can equally work with euclidean action.

\[ t \to -i\tau \ ,\qquad S_E = -i S_M \qquad \text{here,}\ \ \tau = \text{imag}\ ,\ t = \text{real} \] \[ Z[J] \;\sim\; \int \left[d\varphi_E\right] e^{-S_E} \] \[ S_E \;=\; \int d^4x\ \left(\frac{\left(\partial\varphi\right)^2}{2} + \frac{\mu^2\varphi^2}{2} + \frac{\lambda\varphi^4}{4}\right) \]

• loop integrals involve integrals over arbitrary large momenta.
• Theory (i.e. our \(\mathcal{L}\)) is not valid at extremely high energies / momenta. (We impose a cut off on momenta, then integrals will be dependent on momentum cut off (say \(\Lambda\)). (known as \(\Lambda\)-sensitive integrals for a given diagram.) \(\Rightarrow\) Then loop integrals become finite but depend on \(\Lambda\).

This is like cutting infinite wire to make it wire of finite length \(\ell\).

where, \(\ell \to \infty\) is approximation of real world.

\[ \ell\left[\ \underline{V \sim \ln r}\right. \qquad \longrightarrow\ r' \quad \left(V(r') \neq \sim \ln(r) \sim \frac{1}{r}\right) \]

Transition from \(\ln(r)\ \longrightarrow\ \frac{1}{r}\).

• Scattering amplitudes depend on parameters of \(\mathcal{L}\) (like \(\lambda_0, \mu_0 ...\)) \(\hookleftarrow\) Bare Coupling const. & the actual measured couplings \((\lambda, \mu ...)\) are functions of \(\lambda_0, \mu_0\) and \(\Lambda\) (momentum cut off). \[ \lambda = f\left(\lambda_0, \mu_0, \Lambda\right)\ ;\qquad \mu = g\left(\lambda_0, \mu_0, \Lambda\right) \]

What ?\(\left\{\right.\) The \(\Lambda\)-dependence of calculated results can be absorbed between \(\mu_0, \lambda_0\) and \(\mu, \lambda ...\)

\(\longrightarrow\) let us see this for \(\varphi^4\) theory.

Start with,

\[ \mathcal{L}_M \;=\; \frac{1}{2}\alpha^2\left(\partial\varphi\right)^2 - \frac{\mu^2\varphi^2}{2} - \frac{\lambda}{4}\varphi^4 - J\varphi \] \[ \mathcal{L}_E \;=\; \frac{1}{2}\alpha^2\left(\partial\varphi\right)^2 + \frac{\mu^2\varphi^2}{2} + \frac{\lambda}{4}\varphi^4 + J\varphi \]

So,

\[ Z[J] \;=\; \int\left[d\varphi\right]\ e^{-S_E + J\varphi} \qquad \text{(Meaning of source term ?)} \]

\(\downarrow\) generates correlation functions / greens fun. which gives scattering amplitudes.

Correlation functions

\[ G^{(n)}\left(x_1, x_2 \cdots x_n;\, J\right) \;=\; \frac{1}{Z[J]}\ \frac{\delta}{\delta J(x_1)}\ \frac{\delta}{\delta J(x_2)}\ \cdots\ \frac{\delta}{\delta J(x_n)}\ Z[J] \]

\(\longleftarrow\) includes disconnected diagram as well

Connected correlation functions. (How ?) \(\to\) will derive tomorrow!

\[ G_c^{(n)}\left(x_1, x_2 \cdots x_n;\, J\right) \;=\; \frac{\delta}{\delta J(x_1)}\ \cdots\ \frac{\delta}{\delta J(x_n)}\ \ln Z[J] \]

(Srednicki S.24)

Quantity of interest is 2-point vertex function

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\begin{tikzpicture}[scale=1.0,>=Stealth,baseline=(current bounding box.center)]
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% inverse propagator
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\[ =\; \left(\alpha^2 k^2 + \mu^2\right) + \frac{\lambda}{2}\int^{\Lambda}\frac{d^4q}{\left(\alpha^2 q^2 + \mu^2\right)(2\pi)^4}\ +\ O(\lambda^2) \qquad -\ \enclose{circle}{\text{1}} \]

\(\Lambda \leftarrow\) cut off

We demand that, \(\Gamma^{(2)}(k=0)\) be equal to \(\left(\text{physical mass}\right)^2\).

\[ \left.\Gamma^{(2)}(k)\right|_{k=0} \;=\; m^2 \qquad\qquad \left(\tilde{q} = \frac{q}{2\pi}\right) \] \[ \Gamma^{(2)}(0) \;=\; m^2 \;=\; \alpha^2(0) + \mu^2 + \frac{\lambda}{2}\int_{0}^{\Lambda}\frac{d^4\tilde{q}}{\left(\alpha^2 q^2 + \mu^2\right)} + O(\lambda^2) \] \[ \begin{aligned} \Gamma^{(2)}(k) &= \Gamma^{(2)}(k) + 0\\ &= \Gamma^{(2)}(k) - \Gamma^{(2)}(0) + m^2\\ &= \alpha^2 k^2 + \mu^2 + \frac{\lambda}{2}\int_{0}^{\Lambda}\cancel{\frac{d^4 q}{(2\pi)^4}}\ \frac{1}{\left(\alpha^2 q^2+\mu^2\right)} + O(\lambda^2)\\ &\quad - \left(0 + \mu^2 + \frac{\lambda}{2}\int_{0}^{\Lambda}\cancel{\frac{d^4\tilde{q}}{\left(\alpha^2 q^2 + \mu^2\right)}} + O(\lambda^2)\right)\\ &\quad + m^2 \end{aligned} \] \[ \boxed{\ \Gamma^{(2)}(k) \;=\; \alpha^2 k^2 + m^2 + O(\lambda^2)\ } \]