• #PDF
• [[#\(\Gamma^{(2)}\) to two loops ---]]
• [[#Counterterms in \(\mathcal{L}\) ---]]
• #Bare fields & bare couplings ---
• #(margin working)
• #Rescaling the momenta

Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

Previous: Lecture 10--11

Next: Lecture 13

Similarly we can study \(\Gamma^{(2)}\) to 2 loops :

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There are 2 new diagrams---

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\[ =\; \frac{m^2}{4}\,\hat{\lambda}^2\left(\frac{1}{\epsilon^2} \;+\; \frac{1}{\epsilon} \left(\psi(1) + F_1 - \ln\hat{m}^2\right) + \cdots\right) \]

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\[ =\; \frac{3m^2}{4}\,\hat{\lambda}^2\left(\frac{1}{\epsilon^2} \;+\; \frac{1}{\epsilon} \left(\psi(2) + G_1 - \ln\hat{m}^2\right) + \cdots\right) \]

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\[ =\; -\frac{m^2}{4}\,\hat{\lambda}^2\left(\frac{1}{\epsilon^2} \;+\; \frac{1}{\epsilon} \left(\psi(2) + \psi(1) - 2\ln\hat{m}^2\right) + \cdots\right) \]

Adding up we get---

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\[ =\; \frac{-\hat{\lambda}^2}{24\,\epsilon}\,p^2 \;+\; \frac{m^2\lambda^2}{2} \left(\frac{1}{\epsilon^2} \;+\; \frac{1}{\epsilon}\left(F_1 + 3G_1 + 1\right) + \cdots \;+\; O(\lambda^3)\right) \]

diverges as \(\epsilon\to 0\)

\[ \ln\hat{m}^2 \;=\; \ln\frac{m^2}{4\pi\mu^2}\qquad\text{has disappeared.} \qquad \textcolor{red}{\text{Check !}} \]

The new mass counterterm feynman rule become,

\[ -\frac{m^2}{2}\left(\frac{\hat{\lambda}^2}{\epsilon^2} \;+\; \frac{1}{\epsilon} \left(\hat{\lambda} + \frac{\hat{\lambda}^2}{4}\left(F_1 + 3G_1 + 1\right) + \hat{\lambda}^2 F_2 + \hat{\lambda}F_1\right)\right) \]

\(\downarrow\)
arbitrary \(F^n\) & \(\epsilon\,\ln\hat{m}^2\) (\(\to\) finite as \(\epsilon\to 0\))

Counterterm in \(\mathcal{L}\) is

\[ \frac{m^2}{4}\,\varphi^2\left(\frac{\hat{\lambda}^2}{\epsilon} \;+\; \frac{1}{\epsilon} \left(\hat{\lambda} + \frac{\hat{\lambda}^2}{4}\left(F_1 + 3G_1 + 1\right) + \hat{\lambda}^2 F_2 + \hat{\lambda}F_1\right)\right) \]

The other infinity \(\left(\dfrac{-\hat{\lambda}}{24\epsilon}p^2\right)\) is taken care by additional
counterterm of form

\[ \frac{1}{2}\left(\partial_\mu\varphi\right)\left(\partial^\mu\varphi\right) \left(\left(\frac{-\hat{\lambda}^2}{24\,\epsilon}\right) + \hat{\lambda}^2 H_2\left(\epsilon,m^2\right)\right) \]

\(\downarrow\)
How to fix this pieces

does this piece has important physics information ?

We have eliminated UV divergence upto \(O(\lambda^2)\) (i.e. upto 2 loops)

\[ \mathcal{L}^{\text{ren}} \;=\; \underset{\text{\small Bare}}{\mathcal{L}} + \mathcal{L}_{ct} \qquad\qquad \mathcal{L}_{ct} = \mathcal{L}\ \text{(due to additional counterterms).} \] \[ \mathcal{L}^{\text{ren}} \;=\; \frac{1}{2}\partial_\mu\varphi\,\partial^\mu\varphi - \frac{m^2\varphi^2}{2} - \frac{\lambda}{4!}\mu^{2\epsilon}\varphi^4\ ; \qquad \left\{\epsilon = 2-w\right. \] \[ \mathcal{L}_{ct} \;=\; \frac{1}{2}A\,\partial_\mu\varphi\,\partial^\mu\varphi + \frac{1}{2}m^2 B\,\varphi^2 + \frac{\lambda}{4!}C\,\mu^{2\epsilon}\varphi^4 \hspace{2em}\text{---(3)} \] \[ \mathcal{L}_{\text{Bare}} \;=\; \frac{1}{2}\partial_\mu\varphi_0\,\partial^\mu\varphi_0 - \frac{m_0^2\phi_0^2}{2} - \frac{\lambda_0}{4!}\mu^{2\epsilon}\varphi_0^4 \hspace{2em}\text{---(4)} \]

\(\mathcal{L}_{\text{ren}}\) is exactly like `\(\mathcal{L}\)' (except in \(\mathcal{L}\) we write
\(m\to m_{\text{bare}}\), \(\lambda\to\lambda_{\text{bare}}\)).

\(A, B, C\) are specially choosen so that the green's fun. are finite as \(\epsilon\to 0\)

\[ \text{Notation :}\qquad \left.\begin{aligned} A &\longrightarrow Z_\varphi - 1\\ B &\longrightarrow Z_m\\ C &\longrightarrow Z_\lambda \end{aligned}\right\}\ \text{used in some books} \]

We redefine Bare fields & bare coupling to write \(\mathcal{L}_{\text{ren}}\) in terms of bare
fields & coupling

Write,

\[ \mathcal{L}_{\text{ren}} \;=\; \frac{1}{2}\partial_\mu\varphi_0\,\partial^\mu\varphi_0 - \frac{1}{2}m_0^2\varphi_0^2 - \frac{\lambda_0}{4!}\varphi_0^4 \hspace{2em}\longleftarrow\ \text{eq}^\text{n}\ \text{(3)} + \text{(4)} \]

where,

\[ \varphi_0 \;=\; (1+A)^{1/2}\varphi \;\equiv\; Z_\varphi^{1/2}\,\varphi \] \[ m_0 \;=\; m^2\left(\frac{1-B}{1+A}\right) \;=\; m^2(1-B)\,Z_\varphi^{-1} \] \[ \lambda_0 \;=\; \lambda\,\mu^{2\epsilon}\left(\frac{1-C}{(1+A)^2}\right) \;=\; \lambda\,\mu^{2\epsilon}(1-C)\,Z_\varphi^{-2} \]

\(\varphi_0, \lambda_0, m_0\) are bare field, bare coupling & bare mass which diverges as
\(\epsilon\to 0\), the renormalized ones do not.

Source \(J\) is also rescaled to \(J_0 = Z_\varphi^{-1/2}J\),

\(\mathcal{L}_{\text{ren}}\) gives us the Green's function, with \(m\) & \(\lambda\) replaced by
\(m_0\) & \(\lambda_0\).

\[ \widetilde{\Gamma}_0^{(n)}\left(p_1,\ldots p_n;\lambda_0,m_0,\epsilon\right) \;=\; Z_\varphi^{-n/2}\ \widetilde{\Gamma}^{(n)}\left(p_1,p_2,\ldots p_n; m,\lambda,\mu,\epsilon\right) \]

\(\uparrow\) compute with bare quantities. (\(\mu\) = mass scale to make \(\lambda\)
dimensionless in each dimension.)

\(\widetilde{\Gamma}^{(n)}\) are finite as \(\epsilon\to 0\).

\underline{There is no `\(\mu\)' dependence on RHS.} ! (as LHS does not.)

\[ \mu\frac{\partial}{\partial\mu}\widetilde{\Gamma}_0^{(n)}\left(p_1 p_2\ldots p_n;\lambda_0,m_0,\epsilon\right) \;=\; 0 \;=\; \mu\frac{\partial}{\partial\mu}\left(Z_\varphi^{-n/2}\, \widetilde{\Gamma}^{(n)}\left(\{p_i\}; m,\lambda,\mu,\epsilon\right)\right) \] \[ \left(\mu\frac{\partial}{\partial\mu} + \mu\frac{\partial\lambda}{\partial\mu}\frac{\partial}{\partial\lambda} + \mu\frac{\partial m}{\partial\mu}\frac{\partial}{\partial m} - n\left(\frac{\mu}{2}\frac{\partial}{\partial\mu}\ln Z_\varphi\right)\right)\widetilde{\Gamma}^{(n)} \;=\; 0 \]

where the underbraced pieces are, in order, dimensionless; \
\(\beta\left(\lambda,\frac{m}{\mu},\epsilon\right)\); \
\(\dfrac{\mu}{2m}\dfrac{\partial m^2}{\partial\mu} \equiv \gamma_m\); \
\(\gamma_d\left(\lambda,\frac{m}{\mu},\epsilon\right)\) \(\uparrow\) field \(\gamma\) function.

So all objects shall be dimensionless.

These functions \(\beta,\gamma_m\) & \(\gamma_d\) are dimensionless and analytic as \(\epsilon\to 0\).
(they depend only on \(\lambda\) & \(\frac{m}{\mu}\))

\(\widetilde{\Gamma}^{(n)}\) has \(n\) fields

\[ \text{So mass dim.\ of }\widetilde{\Gamma}^{(n)} \;=\; n\times\text{mass dim.\ of one field} \;=\; 4 - n - \epsilon(n-2) \] \[ = w - 2 + 1\ ;\qquad w-1 \;=\; 1-\epsilon\ ;\qquad \epsilon = 2-w \] \[ G^{(n)} \;=\; \left\langle \phi(x)\ldots\phi(x_n)\right\rangle \] \[ S \;=\; \int d^{2w}\left(-\frac{m^2\varphi^2}{2} - \frac{\lambda}{4!}\mu^{x}\varphi^4\right) \]

1. \(0 = -2w + 2 + 2[\varphi]\ \Rightarrow\ 2[\varphi] = 2w-2\ ;\ = w-1\)
2. \(0 = [\lambda]^{0} + x + 4[\varphi]\) \((\epsilon = 2-w)\) \[ x \;=\; -4(w-1) \;=\; -4w+4 \;=\; -4(2-\epsilon)+4 \;=\; -4-4\epsilon\ ! \] \[ \left[\Gamma^{(n)}\right] \;=\; n\cdot(w-1) \;=\; n\left((2-\epsilon)-1\right) \] \[ G^{(n)} \;=\; V(x_1 x_2 \ldots x_n)\, G^{(2)} G^{(2)} \ldots G^{(2)} \]

\[ \widetilde{\Gamma}^{(n)}\left(\{s\,p_i\}; m,\lambda,\mu,\epsilon\right) \]

rescaling \(p_i\) --- we find

\[ \left(\mu\frac{\partial}{\partial\mu} + s\frac{\partial}{\partial s} + m\frac{\partial}{\partial m}\right)\widetilde{\Gamma}^{(n)} \;=\; \left(4 - n + \epsilon(n-2)\right)\times \widetilde{\Gamma}^{(n)}\left(\{s p_i\}; m,\lambda,\mu,\epsilon\right) \]

with \(\mu\dfrac{\partial}{\partial\mu}\ \longrightarrow\ \dfrac{\partial}{\partial(\ln\mu)}\)

Eliminate \(\mu\dfrac{\partial}{\partial\mu}\) & get,

\[ \left(-s\frac{\partial}{\partial s} + \beta\left(\lambda,\frac{m}{\mu}\right)\frac{\partial}{\partial\lambda} + \left(\gamma_m\left(\lambda,\frac{m}{\mu}\right)-1\right)m\frac{\partial}{\partial m} - n\,\gamma_d\left(\lambda,\frac{m}{\mu}\right) + (4-n)\right) \] \[ \times\ \widetilde{\Gamma}^{(n)}\left(\{s p_i\}; m,\lambda,\mu\right) \;=\; 0 \]

\underline{Renormalization Group eqn or Gell mann -- Law eqn.}

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