Quantum field theory 1, lecture 27 (2024)

Higgs decay into photons

A Higgs particle can also decay into photons and this is in fact howit was discovered. How is this possible? If we try to write down adiagram in the theory introduced above we realize that there is no treediagram. However, there are loop diagrams!

Consider the diagrams

These terms arise from the expansion of the partition function if thefermion propagator appears \(3\) timesand there are \(2\) fermion-photon andone fermion-scalar vertices.

Signs in fermion loops

Schematically, the vertices are derivatives \[\left[(-e\gamma^\mu) \left(\frac{1}{i}\frac{\delta}{\delta J^\mu}\right)\left(i \frac{\delta}{\delta\eta}\right) \left(\frac{1}{i} \frac{\delta}{\delta\bar{\eta}}\right)\right] \quad\quad\quad \text{or} \quad\quad\quad\left[(-ig) \left(\frac{1}{i}\frac{\delta}{\deltaJ}\right)\left(i\frac{\delta}{\delta\eta}\right)\left(\frac{1}{i}\frac{\delta}{\bar{\eta}}\right)\right]\]and they act here on a chain like \[\left[(i\bar{\eta})\left(\frac{1}{i}S\right)(i\eta)\right]\left[(i\bar{\eta})\left(\frac{1}{i}S\right)(i\eta)\right]\left[(i\bar{\eta})\left(\frac{1}{i}S\right)(i\eta)\right].\] Note that the derivative with respect to\(\bar{\eta}\) can be commuted throughthe square brackets and acts on \(\bar{\eta}\) from the left. Factors \(1/i\) and \(i\) cancel. The derivative with respect to\(\eta\) receives an additional minussign from commuting and this cancels against \(i^2\). In this way the vertices can connectthe elements of the chain. However, for a closed loop also the beginningand end of the chain must be connected. To make this work, one can firstbring the \((i\eta)\) from the end ofthe chain to its beginning. This leads to one additional minus sign fromanti-commuting Grassmann fields. This shows that closedfermion lines have one more minus sign.

Position spacerepresentation

In position space and including sources, the first diagram is

Momentum spacerepresentation for first diagram

If one translates this now to momentum space and considers theamputated diagram for an S-matrix element, one finds that momentumconservation constrains momenta only up to one free integration momentumor loop momentum. In fact, more generally, there is one integrationmomentum for every closed loop. The first diagram is then (taking afactor \((-1)\) from the closed fermionloop, \(-ig\) from the Yukawa vertexand \(i^3\) from the LSZ reduction intoaccount)

\[\begin{split} &g e^2\;\epsilon^{*}_\mu(q_1) \epsilon^{*}_\nu(q_2) \int_l\frac{1}{[l+q_1)^2 + m^2 -i\epsilon][l^2 + m^2 -i\epsilon][l-q_2)^2 +m^2 + i\epsilon]}\\ & \times \text{tr} \left\{\left[-i(\slashed{l}+\slashed{q}_1)+m \right] \gamma^\mu\left[-i\slashed{l}+m \right] \gamma^nu\left[-i(\slashed{l}-\slashed{q}_2)+m \right] \right\} \end{split}\] Here we use here the abbreviation \[\int_l = \int \frac{d^4l}{(2\pi)^4}.\]

Momentum spacerepresentation for second diagram

For the second diagram we can write

Analytic continuationand Dirac traces

The Feynman \(i\epsilon\) termsallow to perform a Wick rotation to Euclidean space \(l^0 = i\tilde{l}^0_E\) so that \(l^2\) is then positive. First, in the Diractrace we have terms with up to 5 gamma matrices. However, only traces ofan even number of gamma matrices are non-zero. With a bit of algebra onefinds for the Dirac trace \[\begin{split} & \text{tr} \left\{ \left[-i (\slashed{l}+\slashed{q}_1)+m\right] \gamma^\mu \left[-i\slashed{l}+m \right] \gamma^\nu\left[-i(\slashed{l}-\slashed{q}_2)+m \right] \right\}\\ & = -m\;\text{tr}\left\{(\slashed{l}+\slashed{q}_1)\gamma^\mu\slashed{l}\gamma^\nu+ (\slashed{l}+\slashed{q}_1)\gamma^\mu\gamma^\nu(\slashed{l}-\slashed{q}_2)+\gamma^\mu\slashed{l}\gamma^\nu(\slashed{l}-\slashed{q}_2)\right\}+m^3\;\text{tr} \left\{\gamma^\mu\gamma^\nu \right\}\\ &=-4m {\Big [} (l+q_1)^\mu l^\nu + (l+q_1)^\nu l^\mu-(l+q_1) \cdot l\;\eta^{\mu\nu}\\ & \quad\quad\quad + (l+q_1)^\mu (l-q_2)^\nu + (l+q_1) \cdot(l-q_2) \eta^{\mu\nu} -(l+q_1)^\nu(l-q_2)^\mu\\ &\quad\quad\quad + l^\mu (l-q_2)^\nu + (l-q_2)^\mu l^\nu-\eta^{\mu\nu} l \cdot (l-q_2) {\Big ]} + 4 \eta^{\mu\nu} m^3\\ &= -4m \left[4 l^\mu l^\nu -l^2 \eta^{\mu\nu} + 2q^\mu_1l^\nu -2q^\nu_2 l^\mu -q^\mu_1 q^\nu_2 + q^\nu_1 q^\mu_2 -(q_1 \cdotq_2) \eta^{\mu\nu} \right] + 4\eta^{\mu\nu} m^3. \end{split}\]

Feynman parameters

Let us now consider the denominator. One can introduce so-calledFeynman parameters to write \[\begin{split} & \frac{1}{[(l+q_1)^2 + m^2][l^2+m^2][(l-q_2)^2+m^2]} \\ &= 2! \int_0^1 du_1 \cdots du_3 \; \delta(u_1+u_2+u_3 -1)\frac{1}{ \left[u_1[(l+q_1)^2+m^2]+ u_2[l^2+m^2] + u_3[(l-q_2)^2+m^2]\right]^3}\\ &= 2\int_0^1 du_1\cdots du_3\frac{\delta(u_1+u_2+u_3-1)}{\left[l^2 +2l(u_1 q_1 -u_3 q_2)+ u_1 q_1^2+u_3q_2^2+m^2 \right]^3}. \end{split}\] We have used here the identity (will be provenin the exercise classes) \[\frac{1}{p_1\cdots p_n} = (n-1)! \int_0^1 du_1 \ldots du_n \frac{\delta(u_1+ \ldots+ u_n -1)}{ \left[u_1 A_1 + \ldots + u_n A_n \right]^n}.\] In anext step one commutes the integral over \(u_1\ldots u_3\) with the integral over \(l\).

Shifting momenta

It is useful to change integration variables according to \[\begin{split}& l+ u_1 q_1 -u_3q_2 \to k, \\& l= k-u_1 q_1 + u_3 q_2.\end{split}\] Collecting terms we find for the first diagram\[g e^2 \; \epsilon^{*}_\mu(q_1)\;\epsilon^{*}(q_2)\; 2\int_0^1 du_1 \cdots du_3 \; \delta(u_1+u_2+u_3 -1)\int \frac{d^4 k}{(2\pi)^4} \frac{A^{\mu\nu}}{\left[k^2 + u_1q_1^2 +u_3q_2^2 -(u_1 q_1 -u_3q_2)^2 + m^2 \right]^3},\] where thenumerator contains the combination \[\begin{split}A^{\mu\nu} &= -4m {\Big [} 4k^\mu k^\nu -k^2 \eta^{\mu\nu}+\text{terms linear in } k\\& \quad\quad + 4(u_1q_1 -u_3q_2)^\mu (u_1q_1 -u_3q_2)^\nu -(u_1q_1-u_3q_2)^2 \eta^{\mu\nu}\\& \quad\quad - q_1^\mu q_2^\nu+ q_1^\nu q_2^\mu -(q_1 \cdotq_2)\eta^{\mu\nu} -\eta^{\mu\nu} m^2 {\Big ]}.\end{split}%\nonumber\] The integral over \(k\) is now symmetric around the origin.Accordingly, there is no contribution from linear terms in \(k\) \[\int\frac{d^d k}{(2\pi)^d} k^\mu f(k^2) = 0.\] Some simplificationsare also possible due to \(\epsilon^{*}_\mu(q_1) q_1^\mu =\epsilon^{*}_\nu(q_2) q_2^\nu =0\) and \(q_1^2 = q_2^2 = 0\), and we can replace\[\begin{split}A^{\mu\nu} &= -4m {\Big [} 4k^\mu k^\nu -k^2 \eta^{\mu\nu} + (1-4u_1 u_3) q_2^\mu q_1^\nu + (2 u_1 u_3 -1) q_1 \cdot q_2 \eta^{\mu\nu}-\eta^{\mu\nu} m^2 {\Big ]}.\end{split}%\nonumber\]

Dimensional regularization

In manipulating the remaining integral over \(k\) we need to be careful because a firstanalysis based on power counting suggests that the integral of the termsquadratic in \(k\) in the numeratormight not converge (in the numerator we have \(d^4 k \times k^2\) and in the denominator\((k^2+A)^3\) which allows alogarithmic divergence in the UV regime). We need to first introducesome regularization and then do the analysis carefully. An often usedmethod is to extend the integrals from four spacetime dimensions to\(d\) spacetime dimensions, where \(d\) can actually be taken a complex number,see below. For \(d\) slightly differentfrom \(4\) one finds convergentexpressions and one can take the limit \(d\to4\) at the end. Let us now proceed using this dimensionalregularization.

For the quadratic terms we use the identity \[\int \frac{d^d k}{(2\pi)^d} k^\mu k^\nu f(k^2) =\frac{1}{d} \eta^{\mu\nu} \int \frac{d^dk}{(2\pi)^d} k^2 f(k^2).\] The overall structure follows fromLorentz invariance and the prefactor from taking the trace on bothsides.

With this, all remaining integrals are of the form \[\begin{split}\int \frac{d^d p}{(2\pi)^d} \frac{p^{2a}}{(p^2+A)^b} =\frac{\Omega_d}{(2\pi)^d} \int_0^\infty dp \frac{p^{2a+d-1}}{(p^2+A)^b}.\end{split}\] Here we have used the surface of the unit sphere in\(d\) dimensions \(\Omega_d\). To derive a formula for thelatter we perform a \(d\)-dimensionalGaussian integral twice, first in cartesian then in polar coordinates,\[\int d^d x \, e^{-x^2} = \pi^{d/2} =\Omega_d \int_0^\infty d x \, x^{d-1} e^{-x^2} = \frac{\Omega_d}{2}\int_0^\infty dt \, t^{\frac{d}{2}-1} e^{-t} = \frac{\Omega_d }{2}\Gamma(d/2).\] This yields the formula \[\Omega_d=\frac{2\pi^{d/2}}{\Gamma(d/2)}.\]We used here the Euler Gamma function \(\Gamma(z)\) with the notable properties\(\Gamma(z+1) = z \Gamma(z)\) forRe\((z)>0\) and \[\Gamma(n+1) = n!, \quad\quad\quad \Gamma(n+1/2) =\frac{(2n)!}{n! 4^n} \sqrt{\pi},\] for non-negative integers\(n\). Indeed one finds with this theknown special cases \[\Omega_1 = 2,\quad\quad\quad \Omega_2 = 2\pi, \quad\quad\quad \Omega_3 = 4\pi,\quad\quad\quad \Omega_4 = 2\pi^2.\] The remaining integral overthe magnitude of momentum can actually also be evaluated in terms ofGamma functions, \[\int_0^\inftydp \frac{p^{2a+d-1}}{(p^2+A)^b} =\frac{\Gamma(b-a-d/2 )\Gamma(a+d/2)}{2\Gamma(b)}A^{-(b-a-d/2)}.\] Taken together, this leads to the useful result\[\begin{split} \int \frac{d^d p}{(2\pi)^d} \frac{p^{2a}}{(p^2+A)^b} = \frac{\Gamma(b-a-d/2 )\Gamma(a+d/2)}{(4\pi)^{d/2} \Gamma(b)\Gamma(d/2)} A^{-(b-a-d/2)}.\end{split}\label{eq:ScalarIntegralsDimReg}\] In this context,the following properties of the gamma function is very useful forinteger \(n\geq 0\) and small \(x\), \[\Gamma(-n+x) = \frac{(-1)^n}{n!} \left[\frac{1}{x} - \gamma + \sum_{k=1}^n \frac{1}{k} + {\calO}(x)\right],\] where \(\gamma \approx0.5772\) is the Euler-Mascheroni constant. In particular \(\Gamma(z)\) has a simple pole at theorigin.

Result so far

Using dimensional regularization as outlined above, it is now astraight-forward excercise to prove that \[\lim_{d \to 4} \int \frac{d^d k}{(2\pi)^d}\frac{4 k^\mu k^\nu -(k^2+A) \eta^{\mu\nu}}{(k^2+A)^3} = 0.\]Taking this into account leads to \[A^{\mu\nu} = -4m \left[1-4 u_1 u_2 \right]\left[q_1^\nu q_2^\mu -(q_1 \cdot q_2) \eta^{\mu\nu}\right].\]Note that this is symmetric with respect to \((q_1,\mu) \leftrightarrow (q_2,\nu),\) sowe can add the second diagram by multiplying with 2. We obtain \[\begin{split}\mathcal{T} = & -8 g e^2 m \, \epsilon^{*}_\mu(q_1)\epsilon^{*}_\nu(q_2)\left[q_1^\nu q_2^\mu -(q_1 \cdot q_2) \eta^{\mu\nu} \right] \\& \times 2 \int_0^1 du_1\cdots du_3\; \delta(u_1+u_2+u_3 -1) [1-4u_1u_3] \int \frac{d^4 k}{(2\pi)^4} \frac{1}{ \left[k^2 + 2u_1u_3 q_1 \cdotq_2+m^2 \right]^3 }\end{split}%\nonumber\]

Momentum integral

To evaluate the integral over \(k\)we note that in the rest frame of the decaying scalar boson \(p = q_1+ q_2 = (M,0,0,0)\) such that \(p^2 = 2q_1 \cdot q_2 = -M^2.\) If weconcentrate on fermions that are very heavy such that \(m \gg M\) we can expand in the term \(u_1 u_3 q_1 \cdot q_2\) in the integralover \(k\). One finds to lowest order\[\int \frac{d^4k}{(2\pi)^4} \frac{1}{[k^2 +m^2]^3} = i \frac{1}{(4\pi)^2} \frac{1}{2m^2}.\] This \(i\) is due to the Wick rotation \(k^0 = ik^0_E.\)

Integral over Feynmanparameters

Also the integral over Feynman parameters can now easily be performed\[\begin{split} & 2\int_0^1 du_1\ldotsdu_3\;\delta(u_1+u_2+u_3-1)[1-4u_1u_3]\\ &= 2\int_0^1 du_1 du_3 \, \theta(1-u_1-u_3) \left[1-4u_1u_3\right]\\ &=2 \int_0^1 du_1 \int_0^{1-u_1} du_3 \left[1-4u_1u_3\right]\\ &=2 \int_0^1 du_1 [(1-u_1) -4u_1 \tfrac{1}{2}(1-u_1)^2]\\ &= 2-3+\frac{8}{3}-1 = \frac{2}{3}. \end{split} %\nonumber\] Collecting terms we find \[i\mathcal{T} = i \frac{8ge^2}{3(4\pi)^2m}\;\epsilon^{*}_\mu(q_1)\;\epsilon^{*}_\nu(q_2) \left[ q_1^\nu q_2^\mu-(q_1 \cdot q_2)\eta^{\mu\nu} \right].\]

Photon polarizationsums and Ward identity

Before we continue we need to develop a method to perform the spinsums for photons. In the squared amplitude expressions like thefollowing appear \[\sum_{\text{polarizations}} |\mathcal{T}|^2 =\sum_{\text{polarizations}} \epsilon^{*}_\mu(q) \epsilon_\nu(q)\mathcal{M}^\mu(q) \mathcal{M}^{\nu*}(q).\] We have extended herethe polarization vector of a photon from the amplitude by decomposing\[\mathcal{T} = \epsilon^{*}_\mu(q)\mathcal{M}^\mu(q).\] Let us choose without loss of generality\(q^\mu = (E,0,0,E)\) and use thepolarization vector introduced previously, \[\epsilon_\mu^{(1)}=\left(0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0\right),\] \[\epsilon_\mu^{(2)}=\left(0,\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}},0 \right),\] suchthat \[\epsilon_\mu^{*(1)}\;\epsilon_\nu^{(1)} +\epsilon_\mu^{*(2)}\;\epsilon_\nu^{(2)} =\begin{pmatrix}0 & & & \\ &1 & & \\ & &1 & \\ & & &0\end{pmatrix}.\] This would give \[\sum_{j=1}^2\epsilon_\mu^{*(j)}\;\epsilon_\nu^{(j)}\;\mathcal{M}^\mu\;\mathcal{M}^{*\nu}= |\mathcal{M}^1|^2 + |\mathcal{M}^2|^2.\]

Ward identity

To simplify this one can use an identity known as the Wardidentity, \[q_\mu \mathcal{M}^\mu(q) =0.\] This is a direct consequence of gauge symmetry. For theabove choice of \(q^\mu\) it follows\[-\mathcal{M}^0 + \mathcal{M}^3 =0.\] Accordingly, one can add \(0=-|\mathcal{M}^0|^2 + |\mathcal{M}^3|^2\) to the spin sum \[\sum_{j=1}^2\epsilon_\mu^{*(j)}\;\epsilon_\nu^{(j)}\;\mathcal{M}^\mu\;\mathcal{M}^{*\nu}= -|\mathcal{M}^0|^2 + |\mathcal{M}^1|^2 + |\mathcal{M}^2|^2 +|\mathcal{M}^3|^2 =\eta_{\mu\nu}\mathcal{M}^\mu\mathcal{M}^{*\nu}.\] In this sense we can use for externalphotons \[\sum_{j=1}^2\epsilon_\mu^{*(j)}\;\epsilon_\nu^{(j)} \to \eta_{\mu\nu}.\]

Squared amplitude

With this we can now calculate the sums over final state photonpolarizations \[\begin{split}\sum_{\text{pol.}}\;|\mathcal{T}|^2 = &\left(\frac{8ge^2}{3\;(4\pi)^2\,m}\right)^2 {\Big [}q_1^\nu q_2^\mu-(q_1 \cdot q_2)\eta^{\mu\nu} {\Big ]} {\Big [} q_1^\betaq_2^\alpha-(q_1 \cdot q_2)\eta^{\alpha\beta} {\Big ]} \\& \times \sum_{\text{pol.}}\epsilon_\mu^{*}(q_1)\;\epsilon_\alpha(q_1)\;\sum_{\text{pol.}}\epsilon_\nu^{*}(q_2)\;\epsilon_\beta(q_2)\\=& \left(\frac{8ge^2}{3\;(4\pi)^2\,m}\right)^2\;2(q_1 \cdot q_2)^2=\frac{2g^2\alpha^2}{9\pi^2 m^2} M^4.\end{split}%\nonumber\] In the last step we have used that the momentum ofthe incoming Higgs particle is \(p=q_1 +q_2\). The square is given by the rest mass, \(p^2=-M^2 = 2(q_1 \cdot q_2)\). Here we alsoused that the photons are massless, \(q_1^2=q_2^2=0\). We also used the finestructure constant \(\alpha=e^2/(4\pi)\).

Decay rate

For the differential particle decay rate \(\varphi \to \gamma\gamma\) this gives inthe rest frame of the Higgs particle with \(|\mathbf{q}_1|=M/2\), \[\frac{d\Gamma}{d\Omega} =\frac{|\mathbf{q}_1|}{32 \pi^2 M^2}\;\sum_\text{pol.}|\mathcal{T}|^2 =\frac{g^2 \alpha^2}{9 \times 32 \pi^4 m^2}M^3.\] Finally, weintegrate over solid angle \(\Omega =(1/2)4\pi\) where the factor \((1/2)\) is due to the fact that the photonsin the final state are indistinguishable. The decay rate for \(\varphi \to \gamma\gamma\) through a heavyfermion loop is finally \[\Gamma = \frac{g^2\alpha^2}{144 \pi^3 m^2}\;M^3\] Note that because of \(g=m/v\) this is in fact independant of theheavy fermion mass \(m\).