7 Higher orders

So far we have only calculated observables at the first order in perturbation theory. This is usually fine to get a rough idea of the cross section but especially in QCD where $\alpha_{s}\sim 0.1$ , higher-order corrections can be very large (up to 100%). If we want to do any kind of precision measurement, either at the LHC or elsewhere like the anomalous magnetic moments from the abstract, we need more precision. Luckily, we have the tools to do this by simply drawing more complicated Feynman diagrams.

Consider for example the first correction to the $\phi$ propagator in $\phi^{4}$ (cf. (194))

	$\displaystyle\langle{\Omega}\|T\{\phi(x)\phi(y)\}\|{\Omega}\rangle$	$\displaystyle=\includegraphics{bmlimages/notes-151.svg}\bml@image@depth{151}% \bmlDescription{A lone propagator}+\includegraphics{bmlimages/notes-152.svg}% \bml@image@depth{152}\bmlDescription{A propagator with a tadpole}+\mathcal{O}(% \lambda^{2})$ + +𝒪⁢(λ2)\displaystyle=\includegraphics{bmlimages/notes-151.svg}\bml@image@depth{151}% \bmlDescription{A lone propagator}+\includegraphics{bmlimages/notes-152.svg}% \bml@image@depth{152}\bmlDescription{A propagator with a tadpole}+\mathcal{O}(% \lambda^{2})		(408)
		$\displaystyle=\frac{1}{p^{2}-m^{2}+i\epsilon}+\frac{1}{p^{2}-m^{2}+i\epsilon}% \frac{\lambda}{2}\int\frac{{\rm d}^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}-m^{2}+i% \epsilon}\frac{1}{p^{2}-m^{2}+i\epsilon}+\mathcal{O}(\lambda^{2})\,.$		(409)

Let us focus on the integral. To calculate it, we should first try to replace the integral over the Minkowskian momentum $k$ into a Euclidean one. We currently have $k^{2}=(k^{0})^{2}-\vec{k}^{2}$ . If we could replace $k^{0}\to ik^{4}$ , we have instead

	$\displaystyle k^{2}$	$\displaystyle=(k^{0})^{2}-\vec{k}^{2}\to-(k^{4})^{2}-\vec{k}^{2}\equiv-k_{E}^{% 2}\,,$		(410)
	$\displaystyle{\rm d}^{4}k$	$\displaystyle={\rm d}k^{0}{\rm d}^{3}\vec{k}\to i{\rm d}^{4}k_{E}\,.$		(411)

$k_{E}$ is now a normal Euclidean vector that integrate normally. By looking at the integration counter shown in Figure 6, we can see that this does not change the integral. The poles of the $k^{0}$ integration are at $k^{0}=\pm\sqrt{\vec{k}^{2}+m^{2}}\mp i\epsilon$ in the top-left and lower-right quadrant. This means the integral over the whole contour vanishes

\displaystyle 0=\oint{\rm d}k^{0}=\Bigg{(}\int_{-\infty}^{\infty}+\int_{-i% \infty}^{i\infty}+\text{arcs}\Bigg{)}\,.

(412)

Assuming the arcs vanish, we can write

\displaystyle\int\frac{{\rm d}^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}-m^{2}+i\epsilon% }=-i\int\frac{{\rm d}^{4}k_{E}}{(2\pi)^{4}}\frac{1}{k_{E}^{2}+m^{2}}\,.

(413)

This procedure is called Wick rotation.

The complex k^0 plane with poles at -\sqrt{k^2-m^2}+i ϵand +\sqrt{k^2-m^2}-i ϵ. The integration contour follows the real axis until it arcs to +i∞, follows the imaginary axis down, and arcs back to -∞where it closes — Figure 6: The contour of the Wick rotation

Before we start calculating, let us just look at the integral. If $k\gg m$ , the integral will scale like $k_{E}^{4}/k_{E}^{2}\sim k_{E}^{2}\to\infty$ , i.e. the integral is divergent! What does this mean for the whole concept of QFT? If higher-order terms can diverge, the $\lambda$ suppression does not really matter and we loose all predictive power.

Before we continue, we should introduce one more item of terminology. We already know how to classify by connected vs. disconnected and by amputated vs. non-amputated. Now we introduce one-particle irreducible (1PI) diagrams that cannot by split into two diagrams by cutting a single line. For an example of the three categorisation, see Table 7.

category	example	counter-example
connected
amputated
1PI

Figure 7: Some example diagrams that are connected or unconnect, amputated or non-amputated, 1PI or non-1PI.

7.1 Regularisation

To fix this problem, we first need to make it manifest. This means we need to introduce some way of parametrising the problem through a process called regularisation that we have encountered before. In the following section we will concurrently develop two regularisation techniques: cut-off regularisation and dimensional regularisation (dimreg). The former is conceptually easier to understand but very difficult to implement in practice. The latter may sound a bit more abstract and esoteric but is how almost all modern calculations are carried out.

Additionally to $\langle{\Omega}|T\{\phi(x)\phi(y)\}|{\Omega}\rangle$ , we will also calculate the four-point function

\displaystyle\langle{p_{4}p_{3}}|T|{p_{2}p_{1}}\rangle\Big{|}_{1L,p=0}=% \includegraphics{bmlimages/notes-161.svg}\bml@image@depth{161}\bmlDescription{% Two particles coming in, scattering into a bubble-shaped loop, scattering % again and leaving}+\text{permutations}=(-i)(-i\lambda)^{2}\frac{3}{2}\int\frac% {{\rm d}^{4}k}{(2\pi)^{4}}\frac{i^{2}}{(k^{2}-m^{2}+i\epsilon)^{2}}

+permutations=(−i)⁢(−i⁢λ)2⁢32⁢∫d4⁢k(2⁢π)4⁢i2(k2−m2+i⁢ϵ)2\displaystyle\langle{p_{4}p_{3}}|T|{p_{2}p_{1}}\rangle\Big{|}_{1L,p=0}=% \includegraphics{bmlimages/notes-161.svg}\bml@image@depth{161}\bmlDescription{% Two particles coming in, scattering into a bubble-shaped loop, scattering % again and leaving}+\text{permutations}=(-i)(-i\lambda)^{2}\frac{3}{2}\int\frac% {{\rm d}^{4}k}{(2\pi)^{4}}\frac{i^{2}}{(k^{2}-m^{2}+i\epsilon)^{2}}

(414)

7.1.1 Cut-off regularisation

Since our problem is due to $k$ being very large, let us just truncate the integral at some large value $\Lambda$ . We can now simply write

\displaystyle\langle{\Omega}|T\{\phi(x)\phi(y)\}|{\Omega}\rangle\Big{|}_{1L}=% \frac{-\lambda}{32\pi^{4}}\int_{\Lambda}\frac{1}{k_{E}^{2}+m^{2}}=\frac{-% \lambda}{16\pi^{2}}\int_{0}^{\Lambda}{\rm d}k\frac{k^{3}}{k^{2}+m^{2}}=\frac{-% \lambda}{32\pi^{2}}\bigg{(}\Lambda^{2}-m^{2}\log\frac{m^{2}+\Lambda^{2}}{m^{2}% }\bigg{)}\,.

(415)

Expanding this in $\Lambda^{2}/m^{2}$ , we find

\displaystyle\langle{\Omega}|T\{\phi(x)\phi(y)\}|{\Omega}\rangle\Big{|}_{1L}=-% \frac{m^{2}\lambda}{32\pi^{2}}\bigg{(}\frac{\Lambda^{2}}{m^{2}}-\log\frac{% \Lambda^{2}}{m^{2}}\bigg{)}+\mathcal{O}\big{(}(\Lambda/m)^{0}\big{)}\,.

(416)

This does not solve the problem but makes it explicit enough that we can talk about it.

Suggested Exercise

Calculate the other process to show that

\displaystyle\langle{p_{4}p_{3}}|T|{p_{2}p_{1}}\rangle\Big{|}_{1L,p=0}=\frac{3% \lambda^{2}}{32\pi^{2}}\log\frac{\Lambda^{2}}{m^{2}}

(417)

7.1.2 Dimensional regularisation

A big downside of cut-off regularisation is that it breaks Lorentz invariance until we set $\Lambda\to\infty$ . Combined with the fact that it makes integrals more complicated, it is no surprise that it is rarely used in practical calculations. Instead, we shift the spacetime dimension away from four, i.e. we work in $d=4-2\epsilon$ dimensions. This can be formalised but what matters for us is that it regulates the divergences. We write

\displaystyle\langle{\Omega}|T\{\phi(x)\phi(y)\}|{\Omega}\rangle\Big{|}_{1L}=% \lambda\int\frac{{\rm d}^{d}k}{(2\pi)^{d}}\frac{i}{k^{2}-m^{2}}=-\lambda\int% \frac{{\rm d}^{d}k_{E}}{(2\pi)^{d}}\frac{1}{k_{E}^{2}+m^{2}}=-\lambda\int\frac% {{\rm d}\Omega^{(d)}}{(2\pi)^{d}}{\rm d}k\frac{k^{d-1}}{k^{2}+m^{2}}\,.

(418)

The $d$ -dimensional spherical integral can be solved as

\displaystyle\int{\rm d}\Omega^{(d)}=\frac{2\pi^{d/2}}{\Gamma(d/2)}\,.

(419)

Proof

Consider the following trick that uses the normalisation of the Gaussian distribution

	$\displaystyle(\sqrt{\pi})^{d}$	$\displaystyle=\Bigg{(}\int{\rm d}x\ {\rm e}^{-x}\Bigg{)}^{d}=\int{\rm d}^{d}x% \ {\rm e}^{-\vec{x}^{2}}=\int{\rm d}\Omega^{(d)}\int_{0}^{\infty}{\rm d}x\ x^{% d-1}{\rm e}^{-x^{2}}$
		$\displaystyle=\int{\rm d}\Omega^{(d)}\frac{1}{2}\int_{0}^{\infty}{\rm d}y\ y^{% d/2-1}{\rm e}^{-y}=\int{\rm d}\Omega^{(d)}\frac{1}{2}\Gamma(d/2)\,,$		(420)

with $y=x^{2}$ .

We now have

\displaystyle\langle{\Omega}|T\{\phi(x)\phi(y)\}|{\Omega}\rangle\Big{|}_{1L}=-% \frac{\lambda}{(4\pi)^{d/2}}\Gamma\Big{(}1-\tfrac{d}{2}\Big{)}(m^{2})^{d/2-1}\,.

(421)

Since this effectively changes the dimension of the coupling or the action, it is customary to add a factor $\mu^{2\epsilon}$

\displaystyle\langle{\Omega}|T\{\phi(x)\phi(y)\}|{\Omega}\rangle\Big{|}_{1L}=-% m^{2}\Big{(}\frac{\mu^{2}}{m^{2}}\Big{)}^{\epsilon}\frac{\lambda}{(4\pi)^{2-% \epsilon}}\Gamma(\epsilon-1)=\frac{m^{2}\lambda}{16\pi^{2}}\frac{1}{\epsilon}+% \mathcal{O}(\epsilon^{0})\,.

(422)

Once again, this does not solve our problem but it makes it manifest as a pole in $1/\epsilon$ .

Suggested Exercise

Show that in dimreg

\displaystyle\langle{p_{4}p_{3}}|T|{p_{2}p_{1}}\rangle\Big{|}_{1L,p=0}=\frac{-% 3\lambda^{2}}{(4\pi)^{2-\epsilon}}\frac{\Gamma(2-\epsilon)\Gamma(\epsilon-1)}{% \Gamma(1-\epsilon)}\Big{(}\frac{\mu^{2}}{m^{2}}\Big{)}^{\epsilon}=\frac{3% \lambda^{2}}{16\pi^{2}}\frac{1}{\epsilon}+\mathcal{O}(\epsilon^{0})\,.

(423)

7.2 Renormalisation

Now that we have the divergences explicit, we can think about fixing them. So far we have just assumed that our semi-classical construction of the fields $\phi$ was a good one. But in reality, there is no physical interpretation in the parameters of the Lagrangian, be they $\phi$ , $m$ , or $\lambda$ . The only thing that is physical are $\mathcal{S}$ matrix elements and the location of the pole of the propagator (which we called mass before). We have now found $\mathcal{S}$ matrix elements that made no sense whatsoever. Is it therefore maybe possible that our choice of parameters in $\mathcal{L}$ were bad?

Since we would have to measure these parameters by studying $\mathcal{S}$ matrix elements, we can not really predict $\phi\phi\to\phi\phi$ scattering since we do not yet know $\lambda$ . Would it therefore be possible to first measure $\phi\phi\to\phi\phi$ , calculate $\lambda$ and then measure for example $\phi\phi\to\phi\phi\phi\phi$ as a prediction? The parameter $\lambda$ in the Lagrangian is meaningless, the only thing that matters are relations between observables; $\lambda$ is just a convenient intermediary.

Let us therefore add labels to our old Lagrangian to indicate that these quantities were a first guess, called bare quantities

\displaystyle\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi_{0})^{2}-\frac{1}{2}m_% {0}^{2}\phi_{0}^{2}-\frac{\lambda_{0}}{4!}\phi_{0}^{4}\,.

(424)

When we calculated $\mathcal{S}$ matrix elements with these bare quantities, we would them to depend on the regulator. To cancel this dependency, the bare parameters need to depend on the regulator, denoted by $\mathcal{R}$ , themselves, i.e.

\displaystyle\mathcal{L}_{0}(\mathcal{R})=\frac{1}{2}(\partial_{\mu}\phi_{0}(% \mathcal{R}))^{2}-\frac{1}{2}m_{0}(\mathcal{R})^{2}\phi_{0}(\mathcal{R})^{2}-% \frac{\lambda_{0}(\mathcal{R})}{4!}\phi_{0}(\mathcal{R})^{4}\,.

(425)

This means that the bare coupling $\lambda_{0}$ , mass $m_{0}$ , and field $\phi_{0}$ are meaningless, so let us relate them to meaningful quantities

\displaystyle\phi_{0}(\mathcal{R})=Z_{\phi}^{1/2}(\mathcal{R})\ \phi\,,\quad m% _{0}(\mathcal{R})=Z_{m}(\mathcal{R})\ m\,,\quad\lambda_{0}(\mathcal{R})=Z_{% \lambda}(\mathcal{R})\ \lambda\,.

(426)

We want the renormalised quantities to be physical, i.e. not depend on $\mathcal{R}$ which means that the $Z$ factors also need to depend on the regulator. These quantities are called renomalisation constants and more specifically, $Z_{\phi}$ is the field strength renormalisation, $Z_{m}$ is the mass renormalisation, and $Z_{\lambda}$ is the coupling renormalisation.

When expressing the bare Lagrangian using renormalised objects, we find

\displaystyle\mathcal{L}_{0}=\frac{1}{2}Z_{\phi}(\partial_{\mu}\phi)^{2}-Z_{% \phi}Z_{m}^{2}\frac{1}{2}m^{2}\phi^{2}-Z_{\phi}^{2}Z_{\lambda}\frac{\lambda}{4% !}\phi^{4}\,.

(427)

Expanding the $Z_{i}=1+\lambda\delta Z_{i}+\mathcal{O}(\lambda^{2})$ , we can rearrange this to be

\displaystyle\begin{split}\mathcal{L}_{0}=\frac{1}{2}(\partial_{\mu}\phi)^{2}&% -\frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4}\\ &+\lambda\ \delta Z_{\phi}\frac{1}{2}(\partial_{\mu}\phi)^{2}-\lambda(\delta Z% _{\phi}+2\delta Z_{m})\frac{1}{2}m^{2}\phi^{2}-\lambda(\delta Z_{\lambda}+2% \delta Z_{\phi})\frac{\lambda}{4!}\phi^{4}+\mathcal{O}(\lambda^{2})\,.\end{split}

(428)

The first line of this is just the same as before and we have the same Feynman rules to use for our one-loop calculation. The new terms essentially give rise to new Feynman rules that are $\mathcal{O}(\lambda)$ . The $\delta Z_{i}$ are usually referred to as counterterms and therefore the resulting vertices are called counterterm vertices. We therefore complement our set of Feynman rules by

		$\displaystyle=i\lambda\Big{(}p^{2}\delta Z_{\phi}-m^{2}(\delta Z_{\phi}+2% \delta Z_{m})\Big{)}\,,$		(429)
		$\displaystyle=-i\lambda^{2}\Big{(}2\delta Z_{\phi}+\delta Z_{\lambda}\Big{)}\,.$		(430)

When adding these to our calculations, we need to be careful and expand to the same order in $\lambda$ for each term.

7.2.1 Cut-off regularisation

We have

	$\displaystyle\langle{\Omega}\|T\{\phi(x)\phi(y)\}\|{\Omega}\rangle\Big{\|}_{1L+CT}$	$\displaystyle=-\frac{m^{2}\lambda}{32\pi^{2}}\bigg{(}\frac{\Lambda^{2}}{m^{2}}% -\log\frac{\Lambda^{2}}{m^{2}}\bigg{)}+\lambda\Big{(}p^{2}\delta Z_{\phi}-m^{2% }(\delta Z_{\phi}+2\delta Z_{m})\Big{)}+\mathcal{O}\big{(}(\Lambda/m)^{0}\big{% )}\,,$		(431)
	$\displaystyle\langle{p_{4}p_{3}}\|T\|{p_{2}p_{1}}\rangle\Big{\|}_{1L+CT,p=0}$	$\displaystyle=\frac{3\lambda^{2}}{32\pi^{2}}\log\frac{\Lambda^{2}}{m^{2}}-% \lambda^{2}\Big{(}2\delta Z_{\phi}+\delta Z_{\lambda}\Big{)}+\mathcal{O}\big{(% }(\Lambda/m)^{0}\big{)}\,.$		(432)

We can now all but read of the counterterms. For $\langle{\Omega}|T\{\phi(x)\phi(y)\}|{\Omega}\rangle$ , we want this to hold regardless of what $m^{2}$ and $p^{2}$ are, so just collect coefficients. The only solution is


$\displaystyle\delta Z_{\phi}$	$\displaystyle=0\,,$	(433a)
$\displaystyle\delta Z_{m}$	$\displaystyle=-\frac{1}{64\pi^{2}}\Big{(}\frac{\Lambda^{2}}{m^{2}}-\log\frac{% \Lambda^{2}}{m^{2}}\Big{)}\,,$	(433b)
$\displaystyle\delta Z_{\lambda}$	$\displaystyle=\frac{3}{32\pi^{2}}\log\frac{\Lambda^{2}}{m^{2}}\,.$	(433c)

Note that this choice was not unique. We could have added more or less finite terms as long as we remove the $\Lambda$ dependence. The choice we made is called the renormalisation scheme and it is possible to convert between different schemes. For example, we could have chosen $Z_{\lambda}$ such that the one-loop four-point function is not just finite but zero for $p=0$ .

7.2.2 Dimensional regularisation

We can do the same here

	$\displaystyle\langle{\Omega}\|T\{\phi(x)\phi(y)\}\|{\Omega}\rangle\Big{\|}_{1L+CT}$	$\displaystyle=\frac{m^{2}\lambda}{16\pi^{2}}\frac{1}{\epsilon}+\lambda\Big{(}p% ^{2}\delta Z_{\phi}-m^{2}(\delta Z_{\phi}+2\delta Z_{m})\Big{)}+\mathcal{O}(% \epsilon^{0})\,,$		(434)
	$\displaystyle\langle{p_{4}p_{3}}\|T\|{p_{2}p_{1}}\rangle\Big{\|}_{1L+CT,p=0}$	$\displaystyle=\frac{3\lambda^{2}}{16\pi^{2}}\frac{1}{\epsilon}-\lambda^{2}\Big% {(}2\delta Z_{\phi}+\delta Z_{\lambda}\Big{)}+\mathcal{O}(\epsilon^{0})\,,$		(435)

and find


$\displaystyle\delta Z_{\phi}$	$\displaystyle=0\,,$	(436a)
$\displaystyle\delta Z_{m}$	$\displaystyle=\frac{1}{32\pi^{2}}\frac{1}{\epsilon}\,,$	(436b)
$\displaystyle\delta Z_{\lambda}$	$\displaystyle=\frac{3}{16\pi^{2}}\frac{1}{\epsilon}\,.$	(436c)

This scheme is famous enough to have its own name, minimal subtraction (MS).

There is one more modification we would like to make. Consider the renormalised result expanded to the finite term

\displaystyle\langle{p_{4}p_{3}}|T|{p_{2}p_{1}}\rangle\Big{|}_{1L+CT,p=0}=-% \frac{3\lambda^{2}}{16\pi^{2}}\Big{(}\underbrace{\gamma_{E}-\log(4\pi)}-\log% \frac{\mu^{2}}{m^{2}}\Big{)}\,,

(437)

with Euler’s constant (not be confused with ${\rm e}$

\displaystyle\gamma_{E}=\frac{{\rm d}}{{\rm d}x}\Gamma(1-x)\Big{|}_{x=0}=0.577% 216..\,.

(438)

This and the $\log(4\pi)$ are artefacts of our calculation and not physical. They are therefore almost universally removed by modifying the renormalisation constants to be


$\displaystyle\delta Z_{\phi}$	$\displaystyle=0\,,$	(439a)
$\displaystyle\delta Z_{m}$	$\displaystyle=\frac{1}{32\pi^{2}}\frac{1}{\epsilon}(4\pi)^{\epsilon}{\rm e}^{-% \gamma_{E}\epsilon}\,,$	(439b)
$\displaystyle\delta Z_{\lambda}$	$\displaystyle=\frac{3}{16\pi^{2}}\frac{1}{\epsilon}(4\pi)^{\epsilon}{\rm e}^{-% \gamma_{E}\epsilon}\,.$	(439c)

This scheme is now called modified minimal subtraction ( $\overline{\rm MS}$ ).

7.3 Calculation of $\phi\phi\to\phi\phi$ at non-zero momentum

We have used the $p=0$ case to fix the coupling but we can still calculate the one-loop corrections to $\phi\phi\to\phi\phi$ scattering. To do this, we write down the full diagram, including momentum dependence

	$\displaystyle\langle{p_{4}p_{3}}\|T\|{p_{2}p_{1}}\rangle\Big{\|}_{1L}$	$\displaystyle=$		(440)
		$\displaystyle=\frac{-i\lambda^{2}}{2}\underbrace{\mu^{2\epsilon}\int\frac{{\rm d% }^{d}k}{(2\pi)^{d}}\frac{1}{k^{2}-m^{2}}\frac{1}{(k+p_{1}+p_{2})^{2}-m^{2}}}_{% \mathcal{I}}+(p_{2}\leftrightarrow p_{3})+(p_{2}\leftrightarrow p_{4})\,.$		(441)

To solve this loop integral we employ a trick called Feynman parametrisation

\displaystyle\frac{1}{AB}=\int_{0}^{\infty}{\rm d}x{\rm d}y\frac{\delta(\cdots% )}{(xA+yB)^{2}}\,.

(442)

The delta function can either be $\delta(1-x-y)$ or $\delta(1-x)$ . We will choose the former. We can now write with $Q=p_{1}+p_{2}$

	$\displaystyle\mathcal{I}$	$\displaystyle=\mu^{2\epsilon}\int{\rm d}x{\rm d}y\frac{{\rm d}^{d}k}{(2\pi)^{d% }}\frac{\delta(1-x-y)}{\Big{(}x\big{[}k^{2}-m^{2}\big{]}+y\big{[}(k+Q)^{2}-m^{% 2}\big{]}\Big{)}^{2}}$		(443)
		$\displaystyle=\mu^{2\epsilon}\int_{0}^{1}{\rm d}y\int\frac{{\rm d}^{d}k}{(2\pi% )^{d}}\frac{1}{\Big{(}(k+Qy)^{2}-m^{2}+(1-y)yQ^{2}\Big{)}^{2}}\,.$		(444)

In the last step, we have completed the square in the denominator and can now shift $k\to k+Qy$ to have once again

\displaystyle\mathcal{I}

\displaystyle=\mu^{2\epsilon}\int_{0}^{1}{\rm d}y\int\frac{{\rm d}^{d}k}{(2\pi% )^{d}}\frac{1}{\Big{(}k^{2}-m^{2}+(1-y)yQ^{2}\Big{)}^{2}}=\mu^{2\epsilon}\frac% {\Gamma(\epsilon)}{(2\pi)^{d}}\int_{0}^{1}{\rm d}y\ \big{(}m^{2}+Q^{2}(y-1)y% \big{)}^{-\epsilon}\,.

(445)

This integral can be evaluated for example using Mathematica

	$\displaystyle\mathcal{I}$	$\displaystyle=\Big{(}\frac{\mu^{2}}{m^{2}}\Big{)}^{\epsilon}\frac{\Gamma(-1+% \epsilon)}{4(4\pi)^{d/2}}\frac{\beta^{2}-1}{\beta}\Bigg{(}{}_{2}F_{1}\biggl{[}% \genfrac{.}{.}{0.0pt}{}{1\mskip 8.0mu2-2\epsilon}{2-\epsilon};\frac{\beta-1}{2% \beta}\biggr{]}-{}_{2}F_{1}\biggl{[}\genfrac{.}{.}{0.0pt}{}{1\mskip 8.0mu2-2% \epsilon}{2-\epsilon};\frac{\beta+1}{2\beta}\biggr{]}\Bigg{)}$		(446)
		$\displaystyle=\frac{1}{16\pi^{2}}\Bigg{(}\frac{1}{\epsilon}-\gamma_{E}+\log(4% \pi)+2+\log\frac{\mu^{2}}{m^{2}}+\beta\log\frac{\beta-1}{\beta+1}+\mathcal{O}(% \epsilon)\Bigg{)}\,,$		(447)

where we have introduced the Gauss-hypergeometric function and $\beta=\sqrt{1-4m^{2}/Q^{2}}$ . Adding all diagrams, we have with $\beta_{s}$ , $\beta_{t}$ , and $\beta_{u}$ defined by their Mandelstam variables

\displaystyle\begin{split}\langle{p_{4}p_{3}}|T|{p_{2}p_{1}}\rangle\Big{|}_{1L% }=\frac{\lambda^{2}}{16\pi^{2}}\Bigg{(}\frac{3}{\epsilon}-3\gamma_{E}&+3\log(4% \pi)+6+3\log\frac{\mu^{2}}{m^{2}}\\ &+\beta_{s}\log\frac{\beta_{s}-1}{\beta_{s}+1}+\beta_{t}\log\frac{\beta_{t}-1}% {\beta_{t}+1}+\beta_{u}\log\frac{\beta_{u}-1}{\beta_{u}+1}+\mathcal{O}(% \epsilon)\Bigg{)}\,.\end{split}

(448)

Renormalising the coupling in the $\overline{\rm MS}$ scheme, we arrive at

\displaystyle\langle{p_{4}p_{3}}|T|{p_{2}p_{1}}\rangle\Big{|}_{1L}=\frac{% \lambda^{2}}{16\pi^{2}}\Bigg{(}6+3\log\frac{\mu^{2}}{m^{2}}+\beta_{s}\log\frac% {\beta_{s}-1}{\beta_{s}+1}+\beta_{t}\log\frac{\beta_{t}-1}{\beta_{t}+1}+\beta_% {u}\log\frac{\beta_{u}-1}{\beta_{u}+1}\Bigg{)}\,.

(449)

7.4 Renormalisibility

You may now wonder whether renormalisation is always possible. Can we always find finitely many $Z_{i}$ to fix all divergences of our theory to any order in perturbation theory? If so, the theory is predictive once all $n$ parameters of the Lagrangian have been fixed using $n$ measurements. Any such theory is called renormalisable and one can show that the Standard Model of particle physics (as well as QED and QCD separately) is renormalisable. However, certain theories like the Fermi description of the beta decay or the simplest quantum theory of gravity are not renormalisable.

7.4.1 For scalar theories

The first step of showing whether a theory is renormalisable is to consider the superficial degree of divergence of the diagrams it can generate. The singularities we need to remove are due to the large $k$ behaviour and we have seen above that each loop gives a factor of ${\rm d}^{4}k$ and each propagator a factor of $1/(k^{2}-m^{2})\sim k^{-2}$ Consider therefore a diagram at $L$ loop with $P$ internal lines. The degree of divergence $D$ is defined as the scaling of the integrand for large $k$ and for our scalar theory it is

\displaystyle D=4L-2P\,.

(450)

If $D<0$ , the resulting integral is superficially finite and we can ignore it for our discussion of renormalisation. If $D>0$ , the integral is definitely divergent and needs to be considered. The case of $D=0$ cannot be decided through power-counting and the integral actually needs to be computed (hence the superficial).

One can show that for any Feynman diagram with $V$ vertices (Euler’s formula)

\displaystyle L=P-V+1

(451)

You may have seen this written as the Euler characteristic $\chi=V-E+F=2$ for a polyhedron with $V$ vertices, $E$ edges and $F=L+1$ faces (those included by the loop and the outside).

Proof of Euler’s formula for graphs

Begin with a single vertex, i.e. $V=1$ , $P=0$ , $L=0$ . This satisfies Euler’s formula trivially. Using induction, we can now either

•

add a vertex by connecting it with a propagator/edge to the existing ones ( $V\to V+1$ , $P\to P+1$ ). (451) remains satisfied.
•

connect two existing vertices with a propagator/edge. This creates a new loop/face ( $P\to P+1$ , $L\to L+1$ ). (451) remains satisfied.

If we work in a $\phi^{n}$ theory, any vertex needs to be connected to $n$ different lines. These could either be one of the $P$ internal (in which case they are shared between two vertices) or $N$ external lines Mathematically,

\displaystyle 2P+N=nV

(452)

Therefore, we find for $D$

\displaystyle D=(n-4)V+4-N\,.

(453)

For $\phi^{3}$ and $\phi^{4}$ , we can easily see that the only divergent diagrams are $N\leq 4$ . Since there are only finitely many such diagrams we can always subtract the divergence with a counterterm.

Note that we have to go through the above discussion step-by-step, order-by-order. We can only consider $L=2$ once we have calculated all counterterms for $L=1$ etc. This is because of diagrams like

\displaystyle\mathcal{M}^{(2)}(\phi\phi\phi\to\phi\phi\phi)\supset\ % \includegraphics{bmlimages/notes-164.svg}\bml@image@depth{164}\bmlDescription{ A Feynman diagram of three \phi incoming and three \phi outgoing. The diagram consist of two loops. One loop is connected to all six \phi particles. The second loop is connected to the first loop. }\,.

A Feynman diagram of three ϕincoming and three ϕoutgoing.
The diagram consist of two loops.
One loop is connected to all six ϕparticles.
The second loop is connected to the first loop.

.\displaystyle\mathcal{M}^{(2)}(\phi\phi\phi\to\phi\phi\phi)\supset\ % \includegraphics{bmlimages/notes-164.svg}\bml@image@depth{164}\bmlDescription{ A Feynman diagram of three \phi incoming and three \phi outgoing. The diagram consist of two loops. One loop is connected to all six \phi particles. The second loop is connected to the first loop. }\,.

(454)

Superficially, this diagram ( $n=4$ , $L=2$ , $N=6$ , $V=4$ , $P=5$ ) should be finite with $D=-2$ . However, if we only consider the tadpole that is attached to the ‘main’ loop

\displaystyle\mathcal{M}^{(2)}(\phi\phi\phi\to\phi\phi\phi)\supset\ % \includegraphics{bmlimages/notes-165.svg}\bml@image@depth{165}\bmlDescription{ A part of the previous diagram focussing only on the top loop. }\,.

.\displaystyle\mathcal{M}^{(2)}(\phi\phi\phi\to\phi\phi\phi)\supset\ % \includegraphics{bmlimages/notes-165.svg}\bml@image@depth{165}\bmlDescription{ A part of the previous diagram focussing only on the top loop. }\,.

(455)

This sub-diagram ( $L=1$ , $N=2$ , $V=1$ , $P=1$ ), which we could have decided to calculate first, has $D=2$ and is clearly divergent. However, it also renormalised by the counterterm for $Z_{m}$ and $Z_{\phi}$ so that

\displaystyle\mathcal{M}^{(2)}(\phi\phi\phi\to\phi\phi\phi)\supset\ % \includegraphics{bmlimages/notes-166.svg}\bml@image@depth{166}\bmlDescription{ A Feynman diagram of three \phi incoming and three \phi outgoing. The diagram consist of two loops. One loop is connected to all six \phi particles. The second loop is connected to the first loop. }+\includegraphics{bmlimages/notes-167.svg}\bml@image@depth{167}% \bmlDescription{The diagram with the top loop replaced with a counterterm}=% \text{finite}\,.

=finite.\displaystyle\mathcal{M}^{(2)}(\phi\phi\phi\to\phi\phi\phi)\supset\ % \includegraphics{bmlimages/notes-166.svg}\bml@image@depth{166}\bmlDescription{ A Feynman diagram of three \phi incoming and three \phi outgoing. The diagram consist of two loops. One loop is connected to all six \phi particles. The second loop is connected to the first loop. }+\includegraphics{bmlimages/notes-167.svg}\bml@image@depth{167}% \bmlDescription{The diagram with the top loop replaced with a counterterm}=% \text{finite}\,.

(456)

This concept of sub-diagrams being renormalised order-by-order is crucial to the concept of renormalisibility. We can therefore confidently state that for $n=4$ the theory is renormalisable because $N=1$ or $N=3$ are excluded from symmetry.

For the $n=3$ case we already know that we only need to consider $N\leq 4$ and we can introduce counterterms $Z_{\phi}$ and $Z_{m}$ to handle the $N=2$ case (which can be divergent for $V=2$ since $D=2-V$ ). Crucially, at higher loops the number of vertices grows and therefore $D$ further decreases. This means that only finitely many diagrams are divergent.

The

N=1

case

For the $N=1$ case, we get these following divergent diagrams

\displaystyle\Delta=\includegraphics{bmlimages/notes-168.svg}\bml@image@depth{% 168}\bmlDescription{Feynman diagram with a loop that is connected to a single % external line}\,\,+\,\,\includegraphics{bmlimages/notes-169.svg}% \bml@image@depth{169}\bmlDescription{Feynman diagram with a loop that is % connected to a single external line}\,.

.\displaystyle\Delta=\includegraphics{bmlimages/notes-168.svg}\bml@image@depth{% 168}\bmlDescription{Feynman diagram with a loop that is connected to a single % external line}\,\,+\,\,\includegraphics{bmlimages/notes-169.svg}% \bml@image@depth{169}\bmlDescription{Feynman diagram with a loop that is % connected to a single external line}\,.

(457)

Calculating the first term using cut-off regularisation, we find

\displaystyle\Delta=\frac{\lambda}{2}\int\frac{{\rm d}^{4}k}{(2\pi)^{4}}\frac{% 1}{k^{2}-m^{2}+i\epsilon}+\mathcal{O}(\lambda^{2})=\frac{\lambda}{2}\frac{i}{1% 6\pi^{2}}\Lambda^{2}\,.

(458)

There is no operator in our Lagrangian to fix this, meaning we have to introduce a new term $\mathcal{L}\supset Y\phi$ . The resulting Feynman rule is

\displaystyle\includegraphics{bmlimages/notes-170.svg}\bml@image@depth{170}% \bmlDescription{Feynman diagram with a blob that is connected to a single % external line}=-iY\,.

=−i⁢Y.\displaystyle\includegraphics{bmlimages/notes-170.svg}\bml@image@depth{170}% \bmlDescription{Feynman diagram with a blob that is connected to a single % external line}=-iY\,.

(459)

The renormalisation of this operator $Z_{Y}$ fixes the divergent diagrams but the operator also leads to a non-zero vev

\displaystyle\langle{0}|\phi(0)|{0}\rangle=iY+\Delta\,.

(460)

As long as the loop diagrams are also imaginary (which they are), the vev can be forced back to be zero, i.e. $\langle{0}|\phi(0)|{0}\rangle=0$ , while keeping $Y$ real.

For the $n=5$ case we have a problem. Because $D=-V+4-N$ , the number of counterterms we need, i.e. the number of distinctly divergent $N$ , grows as we increase the number of loops or vertices. This means that, as we go higher in the perturbative expansion, we need ever more counterterms, severely limiting the predictive power of our theory.

$n$	divergent $N$				$[\lambda]$
$n$	$L=1$	$L=2$	$L=3$	$L=4$	$[\lambda]$
3	1, 2	1	none	none	${\rm GeV}^{1}$
4	2, 4	2, 4	2, 4	2, 4	${\rm GeV}^{0}$
5	$\leq 6$	$\leq 7$	$\leq 8$	$\leq 9$	${\rm GeV}^{-1}$
6	$\leq 8$	$\leq 10$	$\leq 12$	$\leq 14$	${\rm GeV}^{-2}$

Figure 8: A list of

\phi^{n}

theories and the number of counterterms required at each loop order.

Base on the above discussion, we can define three types of theories (cf. also Figure 8).

•

super-renormalisable theories like $n=3$ where only finitely many diagrams are divergent (not counting divergent sub-diagrams).
•

renormalisable theories like $n=4$ where infinitely many diagrams are divergent (not counting divergent sub-diagrams) but the divergence can be remedied order-by-order using finitely many counterterms.
•

non-renormalisable theories like $n\leq 5$ where we need infinitely many counterterms.

Non-renormalisable theories were long-held to be useless because of their lack of predictive power. However, if we view these theories merely as describing the low-energy behaviour of some unknown theory we can still use them as an effective field theory (EFT) description.

We can view this description also in terms of the mass-dimension of the operator or coupling. As we have discussed in the very beginning of the course, the action is dimensionless, meaning that $\mathcal{L}$ has mass-dimension $[\mathcal{L}]={\rm GeV}^{4}$ . Since $[\partial]=[m]={\rm GeV}$ , $[\phi]={\rm GeV}$ as well. This means that the coupling of the renormalisable $\phi^{4}$ theory is $[\lambda]=1$ and for the $\phi^{3}$ theory $[\lambda]={\rm GeV}$ . For the non-renormalisable $\phi^{5}$ theory we have $[\lambda]={\rm GeV}^{-1}$ etc. Therefore, there is a direct mapping between the renormalisibility of an operator and its mass dimension.

7.4.2 For QED

For QED the arguments work exactly the same way except that the mass dimension due to $P$ changes. For fermions, we have $S\sim 1/k$ and for photons $D_{F}^{\mu\nu}\sim 1/k^{2}$ . Therefore,

\displaystyle D=4L-P_{e}-2P_{\gamma}=4-N_{\gamma}-\frac{3}{2}N_{e}\,,

(461)

for a diagram with $N_{\gamma}$ external photons and $N_{e}$ external electrons. Therefore, we deduce that there are up to seven divergent amplitudes (since $N_{e}$ needs to be even), as shown in Figure 9.

$n_{e}$	$N_{\gamma}$	$D$
0	0	4	unobservable vacuum shift
0	1	3	vanishes because of $\langle j^{\mu}\rangle$
0	2	2	$Z_{A}$
0	3	1	zero by Furry theorem
0	4	0	actually finite
2	0	1	$Z_{\psi}$
2	1	0	$Z_{e}$