# Quantum field theory 1, lecture 26

Feynman parameters. Let us now consider the denominator. One can introduce so-called Feynman parameters to write \begin{equation*} \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} \end{equation*} We have used here the identity (will be proven in the second part of the course QFT 2) \begin{equation*} \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}. \end{equation*} 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{equation*} \begin{split} & l+ u_1 q_1 -u_3q_2 \to k, \\ & l= k-u_1 q_1 + u_3 q_2. \end{split} \end{equation*} Collecting terms we find for the first diagram \begin{equation*}\begin{split} (-1) 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) \\ \times \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}, \end{split}\end{equation*} where \begin{equation*} \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 } -\eta ^{\mu \nu } m^2{\Big ]}. \end{split} \end{equation*} The integral over $$k$$ is now symmetric around the origin.

A further cancelation. There is no contribution from linear terms in $$k$$ and also the quadratic terms cancels. In fact, one can prove that \begin{equation*} \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. \end{equation*} We will develop the techniques to prove this in QFT2.

Result so far. Taking this as well as $$\epsilon ^{*}_\mu (q_1) q_1^\mu = \epsilon ^{*}_\nu (q_2) q_2^\nu =0$$ and $$q_1^2 = q_2^2 = 0$$ into account leads to \begin{equation*} A^{\mu \nu } = -4m \left [1-4 u_1 u_2 \right ] \left [q_1^\mu q_2^\nu -(q_1 \cdot q_2) \eta ^{\mu \nu }\right ]. \end{equation*} 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{equation*} \begin{split} i\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} \end{equation*}

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 \begin{equation*} \int \frac{d^4k}{(2\pi )^4} \frac{1}{[k^2 + m^2]^3} = i \frac{1}{(4\pi )^2} \frac{1}{2m^2}. \end{equation*} 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{equation*} \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} \end{equation*} Collecting terms we find \begin{equation*} 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 ]. \end{equation*}

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 \begin{equation*} \sum _{\text{polarizations}} |\mathcal{T}|^2 = \sum _{\text{polarizations}} \epsilon ^{*}_\mu (q) \epsilon _\nu (q) \mathcal{M}^\mu (q) \mathcal{M}^{\nu *}(q). \end{equation*} We have extended here the polarization vector of a photon from the amplitude by decomposing \begin{equation*} \mathcal{T} = \epsilon ^{*}_\mu (q) \mathcal{M}^\mu (q). \end{equation*} Let us choose without loss of generality $$q^\mu = (E,0,0,E)$$ and use the polarization vector introduced previously, \begin{equation*} \epsilon _\mu ^{(1)} =\left (0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0\right ), \end{equation*} \begin{equation*} \epsilon _\mu ^{(2)} =\left (0,\frac{1}{\sqrt{2}},\frac{i}{\sqrt{2}},0 \right ), \end{equation*} such that \begin{equation*} \epsilon _\mu ^{*(1)}\;\epsilon _\nu ^{(1)} + \epsilon _\mu ^{*(2)}\;\epsilon _\nu ^{(2)} = \begin{pmatrix} 0 & & & \\ &1 & & \\ & &1 & \\ & & &0 \end{pmatrix}. \end{equation*} This would give \begin{equation*} \sum _{j=1}^2 \epsilon _\mu ^{*(j)}\;\epsilon _\nu ^{(j)}\;\mathcal{M}^\mu \;\mathcal{M}^{*\nu } = |\mathcal{M}^1|^2 + |\mathcal{M}^2|^2. \end{equation*}

Ward identity. To simplify this one can use an identity we will prove later, \begin{equation*} q_\mu \mathcal{M}^\mu (q) = 0. \end{equation*} This is in fact a consequence of gauge symmetry known as Ward identity. For the above choice of $$q^\mu$$ it follows \begin{equation*} -\mathcal{M}^0 + \mathcal{M}^3 = 0 \end{equation*} Accordingly, one can add $$0= -|\mathcal{M}^0|^2 + |\mathcal{M}^3|^2$$ to the spin sum \begin{equation*} \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 }. \end{equation*} In this sense we can use for external photons \begin{equation*} \sum _{j=1}^2 \epsilon _\mu ^{*(j)}\;\epsilon _\nu ^{(j)} \to \eta _{\mu \nu }. \end{equation*}

Squared amplitude. With this we can now calculate the sums over final state photon polarizations \begin{equation*} \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} \end{equation*} 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 $$|\vec{q}_1|=M/2$$, \begin{equation*} \frac{d\Gamma }{d\Omega } = \frac{|\vec{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. \end{equation*} 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 \begin{equation*} \Gamma = \frac{g^2 \alpha ^2}{144 \pi ^3 m^2}\;M^3 \end{equation*} Note that because of $$g=m/v$$ this is in fact independant of the heavy fermion mass $$m$$.

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