Prof. Dr. Stefan Floerchinger

Researcher and professor for quantum field theory.

Higgs decay into photons

Higgs decay into photons

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

Consider the diagrams

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

Signs in fermion loops

Schematically, the vertices are derivatives \[\left[(-e\gamma^\mu) \left(\frac{1}{i} \frac{\delta}{\delta J^\mu}\right)\left(i \frac{\delta}{\delta \eta}\right) \left(\frac{1}{i} \frac{\delta}{\delta \bar{\eta}}\right)\right] \quad\quad\quad \text{or} \quad\quad\quad \left[(-ig) \left(\frac{1}{i}\frac{\delta}{\delta J}\right)\left(i\frac{\delta}{\delta \eta}\right)\left(\frac{1}{i}\frac{\delta}{\bar{\eta}}\right)\right]\] 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 through the 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 minus sign from commuting and this cancels against \(i^2\). In this way the vertices can connect the elements of the chain. However, for a closed loop also the beginning and end of the chain must be connected. To make this work, one can first bring the \((i\eta)\) from the end of the chain to its beginning. This leads to one additional minus sign from anti-commuting Grassmann fields. This shows that closed fermion lines have one more minus sign.

Position space representation

In position space and including sources, the first diagram is

\[\begin{split} & (-1) (-ig) \int_{x,y,z} \text{tr}\left\{ \left[\frac{1}{i}S(x-y) \right](-e\gamma^\mu) \left[\frac{1}{i}S(y-z) \right] (-e\gamma^\nu) \left[\frac{1}{i}S(z-x) \right]\right\}\\ & \times \int_{u,v,w} \left[\frac{1}{i}\Delta_{\mu\alpha}(y-u) \right] (iJ^\alpha(u)\left[\frac{1}{i}\Delta_{\nu\beta}(z-v) \right] (iJ^\beta(v) \left[\frac{1}{i}\Delta(x-w) \right](iJ(w)) \end{split}\] The trace is for the Dirac matrix indices.

Momentum space representation for first diagram

If one translates this now to momentum space and considers the amputated diagram for an S-matrix element, one finds that momentum conservation constrains momenta only up to one free integration momentum or loop momentum. In fact, more generally, there is one integration momentum for every closed loop. The first diagram is then (taking a factor \((-1)\) from the closed fermion loop, \(-ig\) from the Yukawa vertex and \(i^3\) from the LSZ reduction into account)

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

Momentum space representation for second diagram

For the second diagram we can write

\[ge^2\;\epsilon^{*}_\mu(q_1) \;\epsilon^{*}_\nu(q_2) \int_l \ldots\] where the integrand is the same up to the interchange \(q_1 \leftrightarrow q_2\) and \(\mu \leftrightarrow \nu\). We can therefore concentrate on evaluating the first diagram.

Analytic continuation and Dirac traces

The Feynman \(i\epsilon\) terms allow 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 Dirac trace we have terms with up to 5 gamma matrices. However, only traces of an even number of gamma matrices are non-zero. With a bit of algebra one finds for the Dirac trace \[\begin{split} & \text{tr} \left\{ \left[-i (\slashed{l}+\slashed{q}_1)+m \right] \gamma^\mu \left[-i\slashed{l}+m \right] \gamma^\nu \left[-i(\slashed{l}-\slashed{q}_2)+m \right] \right\}\\ & = -m \;\text{tr}\left\{(\slashed{l}+\slashed{q}_1)\gamma^\mu\slashed{l}\gamma^\nu + (\slashed{l}+\slashed{q}_1) \gamma^\mu\gamma^\nu(\slashed{l}-\slashed{q}_2)+ \gamma^\mu\slashed{l}\gamma^\nu(\slashed{l}-\slashed{q}_2) \right\}+m^3\;\text{tr} \left\{\gamma^\mu\gamma^\nu \right\}\\ &=-4m {\Big [} (l+q_1)^\mu l^\nu + (l+q_1)^\nu l^\mu -(l+q_1) \cdot l\;\eta^{\mu\nu}\\ & \quad\quad\quad + (l+q_1)^\mu (l-q_2)^\nu + (l+q_1) \cdot (l-q_2) \eta^{\mu\nu} -(l+q_1)^\nu(l-q_2)^\mu\\ &\quad\quad\quad + l^\mu (l-q_2)^\nu + (l-q_2)^\mu l^\nu -\eta^{\mu\nu} l \cdot (l-q_2) {\Big ]} + 4 \eta^{\mu\nu} m^3\\ &= -4m \left[4 l^\mu l^\nu -l^2 \eta^{\mu\nu} + 2q^\mu_1 l^\nu -2q^\nu_2 l^\mu -q^\mu_1 q^\nu_2 + q^\nu_1 q^\mu_2 -(q_1 \cdot q_2) \eta^{\mu\nu} \right] + 4\eta^{\mu\nu} m^3. \end{split}\]

Feynman parameters

Let us now consider the denominator. One can introduce so-called Feynman 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 proven in 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 a next 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 the numerator contains the combination \[\begin{split} A^{\mu\nu} &= -4m {\Big [} 4k^\mu k^\nu -k^2 \eta^{\mu\nu}+ \text{terms linear in } k\\ & \quad\quad + 4(u_1q_1 -u_3q_2)^\mu (u_1q_1 -u_3q_2)^\nu -(u_1q_1 -u_3q_2)^2 \eta^{\mu\nu}\\ & \quad\quad - q_1^\mu q_2^\nu+ q_1^\nu q_2^\mu -(q_1 \cdot q_2)\eta^{\mu\nu} -\eta^{\mu\nu} m^2 {\Big ]}. \end{split} %\nonumber\] 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 simplifications are 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 \[\begin{split} A^{\mu\nu} &= -4m {\Big [} 4k^\mu k^\nu -k^2 \eta^{\mu\nu} + (1-4 u_1 u_3) q_2^\mu q_1^\nu + (2 u_1 u_3 -1) q_1 \cdot q_2 \eta^{\mu\nu} -\eta^{\mu\nu} m^2 {\Big ]}. \end{split} %\nonumber\]

Dimensional regularization

In manipulating the remaining integral over \(k\) we need to be careful because a first analysis based on power counting suggests that the integral of the terms quadratic in \(k\) in the numerator might not converge (in the numerator we have \(d^4 k \times k^2\) and in the denominator \((k^2+A)^3\) which allows a logarithmic divergence in the UV regime). We need to first introduce some regularization and then do the analysis carefully. An often used method 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 different from \(4\) one finds convergent expressions and one can take the limit \(d\to 4\) at the end. Let us now proceed using this dimensional regularization.

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

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 the latter we perform a \(d\)-dimensional Gaussian 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 \(\Gamma(z+1) = z \Gamma(z)\) for Re\((z)>0\) 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 the known 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 over the magnitude of momentum can actually also be evaluated in terms of Gamma functions, \[\int_0^\infty dp \frac{p^{2a+d-1}}{(p^2+A)^b} = \frac{\Gamma(b-a-d/2 )\Gamma(a+d/2)}{2\Gamma(b)} A^{-(b-a-d/2)}.\] 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 for integer \(n\geq 0\) and small \(x\), \[\Gamma(-n+x) = \frac{(-1)^n}{n!} \left[ \frac{1}{x} - \gamma + \sum_{k=1}^n \frac{1}{k} + {\cal O}(x)\right],\] where \(\gamma \approx 0.5772\) is the Euler-Mascheroni constant. In particular \(\Gamma(z)\) has a simple pole at the origin.

Result so far

Using dimensional regularization as outlined above, it is now a straight-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 \((q_1,\mu) \leftrightarrow (q_2,\nu),\) so we can add the second diagram by multiplying with 2. We obtain \[\begin{split} \mathcal{T} = & -8 g e^2 m \, \epsilon^{*}_\mu(q_1) \epsilon^{*}_\nu(q_2) \left[q_1^\nu q_2^\mu -(q_1 \cdot q_2) \eta^{\mu\nu} \right] \\ & \times 2 \int_0^1 du_1\cdots du_3\; \delta(u_1+u_2+u_3 -1) [1-4u_1 u_3] \int \frac{d^4 k}{(2\pi)^4} \frac{1}{ \left[k^2 + 2u_1u_3 q_1 \cdot q_2+m^2 \right]^3 } \end{split}%\nonumber\]

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 we concentrate 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 integral over \(k\). 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 Feynman parameters

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

Photon polarization sums and Ward identity

Before we continue we need to develop a method to perform the spin sums for photons. In the squared amplitude expressions like the following 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 here the 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 the polarization vector introduced previously, \[\epsilon_\mu^{(1)} =\left(0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0\right),\] \[\epsilon_\mu^{(2)} =\left(0,\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}},0 \right),\] such that \[\epsilon_\mu^{*(1)}\;\epsilon_\nu^{(1)} + \epsilon_\mu^{*(2)}\;\epsilon_\nu^{(2)} = \begin{pmatrix} 0 & & & \\ &1 & & \\ & &1 & \\ & & &0 \end{pmatrix}.\] 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 Ward identity, \[q_\mu \mathcal{M}^\mu(q) = 0.\] This is a direct consequence of gauge symmetry. For the above 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 external photons \[\sum_{j=1}^2 \epsilon_\mu^{*(j)}\;\epsilon_\nu^{(j)} \to \eta_{\mu\nu}.\]

Squared amplitude

With this we can now calculate the sums over final state photon polarizations \[\begin{split} \sum_{\text{pol.}}\;|\mathcal{T}|^2 = & \left(\frac{8ge^2}{3\;(4\pi)^2\,m}\right)^2 {\Big [}q_1^\nu q_2^\mu -(q_1 \cdot q_2)\eta^{\mu\nu} {\Big ]} {\Big [} q_1^\beta q_2^\alpha-(q_1 \cdot q_2)\eta^{\alpha\beta} {\Big ]} \\ & \times \sum_{\text{pol.}} \epsilon_\mu^{*}(q_1)\;\epsilon_\alpha(q_1)\;\sum_{\text{pol.}}\epsilon_\nu^{*}(q_2)\;\epsilon_\beta(q_2)\\ =& \left(\frac{8ge^2}{3\;(4\pi)^2\,m}\right)^2\;2(q_1 \cdot q_2)^2 =\frac{2g^2\alpha^2}{9\pi^2 m^2} M^4. \end{split} %\nonumber\] In the last step we have used that the momentum of the 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 also used that the photons are massless, \(q_1^2=q_2^2=0\). We also used the fine structure constant \(\alpha=e^2/(4\pi)\).

Decay rate

For the differential particle decay rate \(\varphi \to \gamma\gamma\) this gives in the 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, we integrate over solid angle \(\Omega = (1/2)4\pi\) where the factor \((1/2)\) is due to the fact that the photons in the final state are indistinguishable. The decay rate for \(\varphi \to \gamma\gamma\) through a heavy fermion 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 the heavy fermion mass \(m\).

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