5 Strong Interactions

In this section we will develop the theory of the strong interactions, QCD. The major difference between QED and QCD is that the gluons are self-interacting because they also carry colour charge (unlike the charge-neutral photon). Experimental evidence suggest that each quark comes in $N_{c}=3$ different colours.

5.1 The QCD Lagrangian

In QCD we have two types of particles that carry colour

•

Six types of quarks (up, charm, top with charge $+2/3e$ and down, strange, bottom with charge $-1/3e$ ) with spin $1/2$ . For each flavour, there can be $N_{c}=3$ of these.
•

A massless, spin $1$ gluon of which there are $N_{c}^{2}-1=8$ versions.

The QCD Lagrangian for a single quark with mass $m$ is

\displaystyle\mathcal{L}_{\rm QCD}=-\frac{1}{4}(F^{a})^{\mu\nu}(F^{a})_{\mu\nu% }+\bar{\psi}_{i}(i\gamma\cdot D_{ij}-m\delta_{ij})\psi_{j}\,.

(151)

The indices $a=1,...,N_{c}^{2}-1$ , and $i,j=1,...,N_{c}$ are indices of this new colour charge and summed over. The covariant derivative and field-strength tensors are now

\displaystyle D_{ij}^{\mu}=\partial^{\mu}\delta_{ij}+ig_{s}t_{ij}^{a}A^{a\mu}% \,,\qquad(F^{a})_{\mu\nu}=\partial_{\mu}A^{a}_{\nu}-\partial_{\nu}A^{a}_{\mu}+% g_{s}f^{abc}A_{\mu}^{b}A_{\nu}^{c}\,.

(152)

The $t^{a}$ are $3\times 3$ matrices in colour space which fulfil a commutation relation

\displaystyle[t^{a},t^{b}]=if^{abc}t^{c}\,.

(153)

This is similar to the angular moment $[J_{i},J_{j}]=i\varepsilon_{ijk}J_{k}$ . In place of the anti-symmetric tensor we now have the structure constants $f^{abc}$ that also appear in $(F^{a})_{\mu\nu}$ which is also anti-symmetric under changes of indices but has a larger range of indices.

Just as the $J_{i}$ generate the rotation group ${\rm SU}(2)$ , the $t^{a}$ generate the colour symmetry group ${\rm SU}(3)$ . The $t^{a}$ are usually represented using Gell-Mann matrices $\lambda^{a}=2t^{a}$

\displaystyle\begin{split}\lambda^{1}=\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&0\end{pmatrix}\,,\quad\lambda^{2}=\begin{pmatrix}0&-i&0\\ i&0&0\\ 0&0&0\end{pmatrix}\,,\quad\lambda^{3}=\begin{pmatrix}1&0&0\\ 0&-1&0\\ 0&0&0\end{pmatrix}\,,\\ \lambda^{4}=\begin{pmatrix}0&0&1\\ 0&0&0\\ 1&0&0\end{pmatrix}\,,\quad\lambda^{5}=\begin{pmatrix}0&0&-i\\ 0&0&0\\ i&0&0\end{pmatrix}\,,\quad\lambda^{6}=\begin{pmatrix}0&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix}\,,\\ \lambda^{7}=\begin{pmatrix}0&0&0\\ 0&0&-i\\ 0&i&0\end{pmatrix}\,,\quad\quad\lambda^{8}=\frac{1}{\sqrt{3}}\begin{pmatrix}1&% 0&0\\ 0&1&0\\ 0&0&-2\end{pmatrix}\,.\end{split}

(154)

In practice, we are not interested in calculating one particular colour component and instead work with sums over all colours which ultimately leads to traces over the $t^{a}$ -matrices. We will see explicit examples of this in the sections that follow and here just collect some useful identities


$\displaystyle{\rm tr}\big{(}t^{a}\big{)}$	$\displaystyle=0\,,$	(155a)
$\displaystyle{\rm tr}\big{(}t^{a}t^{b}\big{)}$	$\displaystyle=\frac{1}{2}\delta^{ab}\,,$	(155b)
$\displaystyle\sum_{a}t_{ij}^{a}t_{jk}^{a}$	$\displaystyle=C_{F}\delta_{ik}\,,$	(155c)
$\displaystyle\sum_{ab}f^{abc}f^{abd}$	$\displaystyle=C_{A}\delta^{cd}\,,$	(155d)

where $C_{F}$ and $C_{A}$ are the Casmir operators of the fundamental representation, i.e. the quarks, and the adjoint representation, i.e. the gluons, respectively. They evaluate to

\displaystyle C_{F}=\frac{N_{c}^{2}-1}{2N_{c}}\qquad\text{and}\qquad C_{A}=N_{% c}\,.

(156)

Note that, the label of the index matters. We have used $i$ , $j$ , $k$ for the fundamental representation and $a$ , $b$ , $c$ , $d$ for the adjoint. This is particularly important when calculating the trace of the identity matrix $\delta$

	$\displaystyle\delta^{aa}=N_{c}^{2}-1=8\,,$		(157)
	$\displaystyle\delta_{ii}=N_{c}=3\,.$		(158)

As in QED the QCD Lagrangian follows from a gauge symmetry and requires a gauge fixing term which we will discuss later.

5.2 Feynman rules

The Feynman rules of QCD are similar to those of QED (73) with a few simple modifications.

The propagators get a $\delta$ function in the correct representation (dropping spinor indices)

	For each internal fermion	$\displaystyle=\delta_{kj}\frac{i}{\gamma\cdot p-m+i\epsilon}\,,$	(159a)
	For each internal gluon	$\displaystyle=\delta^{ab}\frac{-ig^{\mu\nu}}{p^{2}+i\epsilon}\,.$	(159b)
For the quark-quark-gluon vertex, we also just add a $t$ matrix
	For each $q\bar{q}g$ vertex	$\displaystyle=-ig_{s}\gamma^{\mu}\,t_{kj}^{a}\,.$	(159c)
However, we now get new vertices from the $f^{abc}A_{\mu}^{b}A_{\nu}^{c}$ term in $(F^{a})_{\mu\nu}$ .
	For each $ggg$ vertex	$\displaystyle=-g_{s}f^{abc}\Big{(}g^{\mu\nu}(p-q)^{\rho}$
	$\displaystyle\qquad+g^{\nu\rho}(q-r)^{\mu}$
	$\displaystyle\qquad+g^{\rho\mu}(r-p)^{\nu}\Big{)}\,,$		(159d)
	For each $gggg$ vertex	$\displaystyle=-ig_{s}^{2}\Big{(}f^{eab}f^{ecd}(g^{\mu\rho}g^{\nu\sigma}-g^{\mu% \sigma}g^{\nu\rho})$
	$\displaystyle\qquad+f^{eac}f^{ebd}(g^{\mu\nu}g^{\rho\sigma}-g^{\mu\sigma}g^{% \nu\rho})$
	$\displaystyle\qquad+f^{ead}f^{ebc}(g^{\mu\nu}g^{\rho\sigma}-g^{\mu\rho}g^{\nu% \sigma})\Big{)}$		(159e)
The rules for the external particles remain unchanged since we treat these indices as ‘open’
	For each external gluon	$\displaystyle=\epsilon^{*}_{\mu}(p)\qquad\quad\ \text{(final)}\,,$	(159f)
	For each external gluon	$\displaystyle=\epsilon_{\mu}(p)\qquad\quad\ \text{(initial)}\,,$	(159g)
	For each external quark	$\displaystyle=\bar{u}_{s}(p)\qquad\quad\ \text{(final)}\,,$	(159h)
	For each external quark	$\displaystyle=u_{s}(p)\qquad\quad\ \text{(initial)}\,,$	(159i)
	For each external antiquark	$\displaystyle=v_{s}(p)\qquad\quad\ \text{(final)}\,,$	(159j)
	For each external antiquark	$\displaystyle=\bar{v}_{s}(p)\qquad\quad\ \text{(initial)}\,.$	(159k)

A few comments are now in order:

•

The indices $a$ , $b$ , $c$ , $d$ , and $e$ are always in the adjoint representation (upper indices in $t$ , $f$ , and $\delta$ ). This means that they refer to the gluons. The lower indices $j$ and $k$ are in the fundamental representation (lower indices in $t$ and $\delta$ ) and refer to quarks.
•

The Dirac matrix $\gamma^{\mu}$ still appears. Even though we have dropped the explicit spinor indices, these are still present. However, $\gamma^{\mu}$ and $t^{a}$ act on different spaces and therefore commute.
•

This means that now have four kinds of indices: spinor indices, Lorentz indices, and two types of colour indices. Make sure to not mix them up!
•

The new vertices arise when multiplying out the $f^{abc}$ term in the field-strength tensor. In principle, this also exists in QED but there structure constant $f^{abc}=0$ so we do not usually write it.

5.3 Gauge invariance

Just as in QED, the Lagrangian is invariant under a gauge transformation

\displaystyle\begin{split}\psi_{i}(x)&\to\big{(}\delta_{ij}-ig_{s}\theta^{a}(x% )t_{ij}^{a}\big{)}\psi_{j}(x)\,,\\ A_{\mu}^{a}(x)&\to A_{\mu}^{a}(x)+D_{\mu}^{ab}\theta^{b}(x)\,,\end{split}

(160)

Here we have introduced another form of the covariant derivative. Unlike in (152) where we had it in the fundamental representation, here it is in the adjoint representation

\displaystyle D_{\mu}^{ab}=\partial_{\mu}\delta^{ab}+ig_{s}A_{\mu}^{c}\ (T^{c}% )^{ab}\,.

(161)

The $T^{c}$ is again a generator of the ${\rm SU}(3)$ colour algebra but for a different representation. In fact we have already encountered the matrices for this

\displaystyle(T^{c})^{ab}=if^{acb}=-if^{abc}\,.

(162)

These need to fulfil the commutation relations of the algebra (153)

\displaystyle[T^{a},T^{b}]=if^{abc}T^{c}\,,

(163)

which is equivalent to the Jacobi identity

\displaystyle f^{abc}f^{cde}+f^{bdc}f^{cae}+f^{dac}f^{cbe}=0\,.

(164)

As we have seen in (160) the gluon transforms in the adjoint representation while the quark transforms in the fundamental representation. Notice that the gauge transformation for $A_{\mu}^{a}$ involves the strong coupling $g_{s}$

\displaystyle A_{\mu}^{a}\to A_{\mu}^{a}+\partial_{\mu}\theta^{a}+g_{s}f^{abc}% \theta^{b}A^{c}_{\mu}=A_{\mu}^{a}+\partial_{\mu}^{a}\theta^{a}+\mathcal{O}(g_{% s})\,,

(165)

and only at lowest order in $g_{s}$ does it reduce to the analogous transformation for QED (63). The gauge fixing works exactly like in QED, cf. (68)

\displaystyle\mathcal{L}_{\rm gf}=-\frac{1}{2\xi}(\partial_{\mu}A_{a}^{\mu})^{% 2}\,,

(166)

5.3.1 Compton scattering in QCD

Let us calculate the equivalent of Compton scattering in QCD, namely

\displaystyle q(p_{1})\bar{q}(p_{2})\to g(p_{3})g(p_{4})\,.

(167)

Up to the crossing, this is equivalent to what we have calculated in Section 4.4 except that we have now an additional diagram

\displaystyle\mathcal{M}

\displaystyle=\begin{gathered}\includegraphics{bmlimages/notes-27.svg}% \bml@image@depth{27}\bmlDescription{ An quark comes in with momentum p1 and emits a gluon with momentum p3 going % out. The quark line continues downwards where it joins with a anti-quark % coming in (momentum p2) that has emitted a gluon with momentum p4. }\end{gathered}\quad+\quad\begin{gathered}\includegraphics{bmlimages/notes-28.% svg}\bml@image@depth{28}\bmlDescription{ An quark comes in with momentum p1 and emits a gluon with momentum p4 going % out. The quark line continues downwards where it joins with a anti-quark % coming in (momentum p2) that has emitted a gluon with momentum p3. The % outgoing gluons are visibly swapped. }\end{gathered}\quad+\quad\begin{gathered}\includegraphics{bmlimages/notes-29.% svg}\bml@image@depth{29}\bmlDescription{ An quark and an anti-quark come in with momenta p1 and p2 and annihilate into % a gluon with momenta p1+p2. The gluon continues and turns into two gluons with% momenta p3 and p4. }\end{gathered}\,.

An quark comes in with momentum p1 and emits a gluon with momentum p3 going out. The quark line continues downwards where it joins with a anti-quark coming in (momentum p2) that has emitted a gluon with momentum p4.

An quark comes in with momentum p1 and emits a gluon with momentum p4 going out. The quark line continues downwards where it joins with a anti-quark coming in (momentum p2) that has emitted a gluon with momentum p3. The outgoing gluons are visibly swapped.

An quark and an anti-quark come in with momenta p1 and p2 and annihilate into a gluon with momenta p1+p2. The gluon continues and turns into two gluons with momenta p3 and p4.

.\displaystyle=\begin{gathered}\includegraphics{bmlimages/notes-27.svg}% \bml@image@depth{27}\bmlDescription{ An quark comes in with momentum p1 and emits a gluon with momentum p3 going % out. The quark line continues downwards where it joins with a anti-quark % coming in (momentum p2) that has emitted a gluon with momentum p4. }\end{gathered}\quad+\quad\begin{gathered}\includegraphics{bmlimages/notes-28.% svg}\bml@image@depth{28}\bmlDescription{ An quark comes in with momentum p1 and emits a gluon with momentum p4 going % out. The quark line continues downwards where it joins with a anti-quark % coming in (momentum p2) that has emitted a gluon with momentum p3. The % outgoing gluons are visibly swapped. }\end{gathered}\quad+\quad\begin{gathered}\includegraphics{bmlimages/notes-29.% svg}\bml@image@depth{29}\bmlDescription{ An quark and an anti-quark come in with momenta p1 and p2 and annihilate into % a gluon with momenta p1+p2. The gluon continues and turns into two gluons with% momenta p3 and p4. }\end{gathered}\,.

(171)

The first two diagrams evaluate just like in (141) but with colour matrices

\displaystyle\mathcal{M}\Big{|}_{\rm Abelian}=-ig_{s}^{2}\epsilon^{*\mu}(p_{3}% )\,\epsilon^{*\nu}(p_{4})\,\,\bar{v}(p_{2})\Bigg{(}\gamma_{\nu}t^{b}\frac{% \gamma\cdot(p_{1}-p_{3})}{(p_{1}-p_{3})^{2}}\gamma_{\mu}t^{a}\gamma_{\mu}t^{a}% \frac{\gamma\cdot(p_{1}-p_{4})}{(p_{1}-p_{4})^{2}}\gamma_{\nu}t^{b}\Bigg{)}u(p% _{1})\,.

(172)

If we just this Abelian part of $\mathcal{M}$ and replaced $\epsilon(p)$ with $p$ to verify the Ward identity (97), i.e. testing whether the replacement $\epsilon_{\mu}\to\epsilon_{\mu}+\lambda p$ is valid, we would find

\displaystyle p_{3}^{\mu}\ \mathcal{M}_{\mu\nu}\Big{|}_{\rm Abelian}=-ig_{s}^{% 2}\Big{[}t^{a},t^{b}\Big{]}\bar{v}(p_{2})\gamma_{\nu}u(p_{1})\neq 0\,.

(173)

The non-zero commutator makes these diagrams alone not gauge-invariant. Adding the final diagram gives a contribution which exactly cancels this (try this!) but yields another term proportional to $p_{4}^{\nu}$ . This vanishes when we remember the whole expression is contracted with $\epsilon^{*\nu}(p_{4})$ , and so gauge invariance is only obeyed once we project onto physical polarisations. This was not necessary in QED.

This problem is due to the sum over polarisations (102) where we did the replacement

\displaystyle\sum_{\alpha=1}^{2}\epsilon_{\mu}^{\alpha}\epsilon_{\nu}^{*\alpha% }\to-g_{\mu\nu}\,.

(102)

Although the right-hand summed all polarisations and not only the physical transverse ones, in actual calculations the unphysical longitudinal gluon polarisations automatically cancelled. This is no longer the case in QCD, where one has to sum strictly over physical polarisations. However, this can make calculations more cumbersome, so it might still be useful to sum over all polarisations, and to cancel in some way the unphysical degrees of freedom. How this cancellation is performed depends on the gauge. In covariant gauges, like the Feynman gauge, this is done by introducing extra fields, called the ghost fields. The alternative is to use the so-called physical gauges, that ensure that that only physical degrees of freedom propagate on shell.

5.3.2 Unitarity

Before we can understand this, we need to return to $ee\to\gamma\gamma$ which we have calculated in Section 4.4. It is possible to relate the amplitude for $ee\to\gamma\gamma$ to the process $ee\to ee$ using unitarity. One can show that the imaginary part of forward scattering of $ee\to ee$ is related to the amplitude square of $ee\to\gamma\gamma$ , i.e.

\displaystyle 2\Im\Bigg{(}\ \begin{gathered}\includegraphics{bmlimages/notes-3% 0.svg}\bml@image@depth{30}\bmlDescription{ A fermion coming in, emitting a photon, going down and emitting another photon% before meeting up with the other incoming anti-fermion. The two photons % travel horizontally and then meet up with another fermion line that is % outgoing. The diagram is vaguely box-shaped. }\end{gathered}\ +\ \begin{gathered}\includegraphics{bmlimages/notes-31.svg}% \bml@image@depth{31}\bmlDescription{ A fermion coming in, emitting a photon, going down and emitting another photon% before meeting up with the other incoming anti-fermion. The two photons cross% over without interacting and then meet up with another fermion line that is % outgoing. The diagram is shaped liked a twisted box. }\end{gathered}\ +\genfrac{}{}{0.0pt}{0}{\text{diags w/o}}{\text{imaginary}}% \Bigg{)}=\int{\rm d}\Phi\sum\Bigg{|}\ \begin{gathered}\includegraphics{% bmlimages/notes-32.svg}\bml@image@depth{32}\bmlDescription{ An quark comes in and emits a photon going out. The quark line continues % downwards where it joins with a anti-quark coming in that has emitted a photon% . }\end{gathered}+\begin{gathered}\includegraphics{bmlimages/notes-33.svg}% \bml@image@depth{33}\bmlDescription{ An quark comes in and emits a photon going out. The quark line continues % downwards where it joins with a anti-quark coming in that has emitted a photon% . The outgoing photons are visibly swapped. }\end{gathered}\ \Bigg{|}^{2}\,.

A fermion coming in, emitting a photon, going down and emitting another photon before meeting up with the other incoming anti-fermion. The two photons travel horizontally and then meet up with another fermion line that is outgoing. The diagram is vaguely box-shaped.

+diags w/oimaginary)=∫dΦ⁢∑|

An quark comes in and emits a photon going out. The quark line continues downwards where it joins with a anti-quark coming in that has emitted a photon.

|2.\displaystyle 2\Im\Bigg{(}\ \begin{gathered}\includegraphics{bmlimages/notes-3% 0.svg}\bml@image@depth{30}\bmlDescription{ A fermion coming in, emitting a photon, going down and emitting another photon% before meeting up with the other incoming anti-fermion. The two photons % travel horizontally and then meet up with another fermion line that is % outgoing. The diagram is vaguely box-shaped. }\end{gathered}\ +\ \begin{gathered}\includegraphics{bmlimages/notes-31.svg}% \bml@image@depth{31}\bmlDescription{ A fermion coming in, emitting a photon, going down and emitting another photon% before meeting up with the other incoming anti-fermion. The two photons cross% over without interacting and then meet up with another fermion line that is % outgoing. The diagram is shaped liked a twisted box. }\end{gathered}\ +\genfrac{}{}{0.0pt}{0}{\text{diags w/o}}{\text{imaginary}}% \Bigg{)}=\int{\rm d}\Phi\sum\Bigg{|}\ \begin{gathered}\includegraphics{% bmlimages/notes-32.svg}\bml@image@depth{32}\bmlDescription{ An quark comes in and emits a photon going out. The quark line continues % downwards where it joins with a anti-quark coming in that has emitted a photon% . }\end{gathered}+\begin{gathered}\includegraphics{bmlimages/notes-33.svg}% \bml@image@depth{33}\bmlDescription{ An quark comes in and emits a photon going out. The quark line continues % downwards where it joins with a anti-quark coming in that has emitted a photon% . The outgoing photons are visibly swapped. }\end{gathered}\ \Bigg{|}^{2}\,.

(178)

This result is known as the optical theorem and it can be extremely powerful. For our purposes, it is enough to note that the sum on the right-hand side goes onlt over physical polarisations.

The way this is done in practice is by cutting propagators, i.e. replacing $i/(p^{2}-m^{2}+i\epsilon)$ with $(2\pi)\Theta(p^{0})\delta(p^{2}-m^{2})$ . Everything on the right-hand side of the cut then gets complex-conjugated. Cuts that violate momentum conservation do not contribute. By drawing the cut as a dashed line, we can write the optical theorem as

\displaystyle 2\Im\Bigg{(}\ \begin{gathered}\includegraphics{bmlimages/notes-3% 4.svg}\bml@image@depth{34}\bmlDescription{ A fermion and an anti-fermion enter a blob. Two photons leave the blob before % entering another blob. A fermion and an anti-fermion leave the blob. }\end{gathered}\ \Bigg{)}={\ \begin{gathered}\includegraphics{bmlimages/notes-% 35.svg}\bml@image@depth{35}\bmlDescription{ A fermion and an anti-fermion enter a blob. Two photons leave the blob before % entering another blob. A fermion and an anti-fermion leave the blob. There is a dashed cut through the photons }\end{gathered}\ }\,,

A fermion and an anti-fermion enter a blob. Two photons leave the blob before entering another blob. A fermion and an anti-fermion leave the blob.

A fermion and an anti-fermion enter a blob. Two photons leave the blob before entering another blob. A fermion and an anti-fermion leave the blob.
There is a dashed cut through the photons

,\displaystyle 2\Im\Bigg{(}\ \begin{gathered}\includegraphics{bmlimages/notes-3% 4.svg}\bml@image@depth{34}\bmlDescription{ A fermion and an anti-fermion enter a blob. Two photons leave the blob before % entering another blob. A fermion and an anti-fermion leave the blob. }\end{gathered}\ \Bigg{)}={\ \begin{gathered}\includegraphics{bmlimages/notes-% 35.svg}\bml@image@depth{35}\bmlDescription{ A fermion and an anti-fermion enter a blob. Two photons leave the blob before % entering another blob. A fermion and an anti-fermion leave the blob. There is a dashed cut through the photons }\end{gathered}\ }\,,

(181)

where the blob indicates all possible ways to connect the photon, i.e. the two diagrams in $t$ and $u$ channel. In Feynman gauge we replace

\displaystyle-\frac{ig_{\mu\nu}}{k^{2}+i\epsilon}\to-g_{\mu\nu}\ (2\pi)\Theta(% k^{0})\delta(k^{2})\,.

(182)

Let us call the diagram on the left side of the cut $\mathcal{M}^{\mu\nu}$ . Then the Ward identity $k_{\mu}\mathcal{M}^{\mu\nu}=0$ , means we can obtain the contribution for a photon $k$ to the imaginary part as

\displaystyle\int\frac{{\rm d}^{4}k}{(2\pi)^{4}}\frac{-ig_{\mu\nu}}{k^{2}+i% \epsilon}\mathcal{M}^{\mu\nu}\to\int\frac{{\rm d}^{4}k}{(2\pi)^{4}}(-g_{\mu\nu% })\ (2\pi)\Theta(k^{0})\delta(k^{2})\mathcal{M}^{\mu\nu}=\int\frac{{\rm d}^{3}% k}{(2\pi)^{3}2k^{0}}\mathcal{M}^{\mu\nu}\sum_{\alpha=1}^{2}\epsilon_{\mu}^{% \alpha}(k)\epsilon_{\nu}^{*\alpha}(k)\,.

(183)

This verifies the explicit relation (102) and means that, in QED, we obtain (102) from unitarity.

In QCD, the fact that the Ward identity is partially broken in (172) implies that the amplitude for the process $q\bar{q}\to gg$ is not given by the imaginary part of the forward amplitude for $q\bar{q}\to q\bar{q}$ if we only consider gluons as internal particles. In fact, the cut forward amplitude contains the contribution of non-physical longitudinal polarisations, which do not contribute to the amplitude squared for $q\bar{q}\to gg$ . This would violate unitarity, so there has to be additional fields that are responsible for the cancellation of the contribution of non-physical polarisations in the imaginary part of the forward amplitude. These new fields are called ghosts.

5.3.3 Ghosts

Ghosts are scalar fields but follow Pauli’s exclusion principle and pick up a sign under swapping, just like fermions. Unlike fermions, however, they transform like gluons in the adjoint representation of ${\rm SU}(3)$ . The Feynman rules for the ghosts are


For each internal ghost	$\displaystyle=\delta^{ab}\frac{i}{p^{2}+i\epsilon}\,,$	(184a)
For each $c\bar{c}g$ vertex	$\displaystyle=g_{s}f^{abc}q^{\mu}\,.$	(184b)

Note that ghost only appear in internal lines, never as final states.

Consider now again the relation (181) but in QCD. We now know that there is an additional diagram with a ghost

\displaystyle{\ \begin{gathered}\includegraphics{bmlimages/notes-38.svg}% \bml@image@depth{38}\bmlDescription{ A fermion and an anti-fermion enter a blob. Two gluons leave the blob before % entering another blob. A fermion and an anti-fermion leave the blob. There is a dashed cut through the gluons }\end{gathered}\ }+\begin{gathered}\includegraphics{bmlimages/notes-39.svg}% \bml@image@depth{39}\bmlDescription{ An quark and an anti-quark come in with and annihilate into a gluon. The gluon% continues and turns into a ghost bubble before re-combining back into a gluon% . The gluon continues further and turns into two gluons. }\end{gathered}

A fermion and an anti-fermion enter a blob. Two gluons leave the blob before entering another blob. A fermion and an anti-fermion leave the blob.
There is a dashed cut through the gluons

An quark and an anti-quark come in with and annihilate into a gluon. The gluon continues and turns into a ghost bubble before re-combining back into a gluon. The gluon continues further and turns into two gluons.

\displaystyle{\ \begin{gathered}\includegraphics{bmlimages/notes-38.svg}% \bml@image@depth{38}\bmlDescription{ A fermion and an anti-fermion enter a blob. Two gluons leave the blob before % entering another blob. A fermion and an anti-fermion leave the blob. There is a dashed cut through the gluons }\end{gathered}\ }+\begin{gathered}\includegraphics{bmlimages/notes-39.svg}% \bml@image@depth{39}\bmlDescription{ An quark and an anti-quark come in with and annihilate into a gluon. The gluon% continues and turns into a ghost bubble before re-combining back into a gluon% . The gluon continues further and turns into two gluons. }\end{gathered}

\displaystyle=\int{\rm d}\Phi\sum\Bigg{|}\ \begin{gathered}\includegraphics{% bmlimages/notes-40.svg}\bml@image@depth{40}\bmlDescription{ An quark and an anti-quark come in and enter a blob. Out of the blob come two % gluon. }\end{gathered}\ \Bigg{|}^{2}

An quark and an anti-quark come in and enter a blob. Out of the blob come two gluon.

|2\displaystyle=\int{\rm d}\Phi\sum\Bigg{|}\ \begin{gathered}\includegraphics{% bmlimages/notes-40.svg}\bml@image@depth{40}\bmlDescription{ An quark and an anti-quark come in and enter a blob. Out of the blob come two % gluon. }\end{gathered}\ \Bigg{|}^{2}

(188)

The ghost-antighost loop contributes to the imaginary part of the forward amplitude with a factor $(-1)$ , just like a normal fermion loop, so as to cancel the contribution of the unphysical longitudinal gluon polarisations when summing over all diagrams. The resulting imaginary part equals the amplitude squared for the process $q\bar{q}\to gg$ , integrated over the gluon phase space and summed over physical gluon polarisations, as required by unitarity of QCD.

5.3.4 Physical gauges

Alternatively, we can impose a so-called physical gauge condition on the gluon fields to eliminate unphysical polarisations from the start. This eliminates the need for ghosts, which do not interact with gluons anymore, but complicates the gluon propagator. In place of the Lorent gauge condition $\partial^{\mu}A_{\mu}^{a}=0$ , we impose

\displaystyle A_{\mu}^{a}n^{\mu}=0

(189)

for some arbitrary reference vector $n^{\mu}$ . This is also sometimes called axial gauge. The gauge fixing Lagrangian is

\displaystyle\mathcal{L}_{\rm gf}=-\lim_{\xi\to 0}\frac{1}{2}\xi\Big{(}A_{\mu}% ^{a}n^{\mu}\Big{)}^{2}\,.

(190)

The gluon propagator changes to

For each internal gluon

A gluon of momentum p flowing from μ,a to ν,b

\displaystyle=\delta^{ab}\frac{i}{p^{2}+i\epsilon}\Bigg{(}-g^{\mu\nu}+\frac{p^% {\mu}n^{\nu}+p^{\nu}n^{\mu}}{p\cdot n}+n^{2}\frac{p^{\mu}p^{\nu}}{(p\cdot n)^{% 2}}\Bigg{)}\,.

(191)

And the polarisation sum becomes

\displaystyle\sum_{\alpha=1}^{2}\epsilon_{\mu}^{\alpha}\epsilon_{\nu}^{*\alpha% }\to-g_{\mu\nu}+\frac{k_{\mu}n_{\nu}+k_{\nu}n_{\mu}}{k\cdot n}+n^{2}\frac{k_{% \mu}k_{\nu}}{(k\cdot n)^{2}}\,.

(192)

Physical quantities cannot depend on the choice of $n$ as long as $n\neq k$ and one can check this explicitly.

A relevant example of a physical gauge is the light-cone gauge, in which $n^{2}=0$ . We would for example choose $k=(\omega,\vec{k})$ and $n=(1,-\vec{k}/\omega)$ so that the $n^{2}$ term drops out and we have an exact equality in (192) (rather than the “ $\to$ ” we have used previously).