How to use Passarino-Veltman reduction for integral containing two tensors?

Question

I am trying to use Passarino-Veltman reduction to solve the following integral: $$ \int \frac{d^{D}k}{(2\pi)^{D}} \frac{k^{\mu}k^{\nu}}{k^{2}(k-q)^{2}} $$ However every ansatz I try, the integral goes to zero. For example $g^{\mu\nu}$ gives $k^{2}$ in the numerator which cancels with the $k^{2}$ in the denominator, making the integral equal to 0. Which ansatz is the best for an integral of this form?

Answer 1

You forgot that you can even have tensors of the form $q^\mu q^\nu$.

Let us first compute the simplest tensor integral \begin{equation} \mathcal{B}^\mu(p) = \int\frac{d^d{\ell}}{(2\pi)^d}\frac{\ell^\mu}{\ell^2(\ell+p)^2}, \end{equation} where $p^2\neq 0$ and we employ dimensional regularization. We can easily see that the only relevant $4$-vector on which the integral can depend is $p^\mu$, therefore we can write \begin{equation} \mathcal{B}^\mu(p) = B_{11}p^\mu. \end{equation} In order to find the coefficient, we can just project onto $p^\mu$ and get back to a scalar integral \begin{equation} p_\mu \mathcal{B}^\mu(p) = B_{11}p^2 = \int\frac{d^d{\ell}}{(2\pi)^d}\frac{p\cdot \ell}{\ell^2(\ell+p)^2}. \end{equation} Since $(\ell+p)^2 = p^2+\ell^2+2p\cdot\ell$ we have that \begin{equation} p\cdot\ell = \frac{1}{2}[(\ell+p)^2-\ell^2-p^2]. \end{equation} Using this in the integral we have \begin{equation} p^2 B_{11} = \frac{1}{2}\int\frac{d^d{\ell}}{(2\pi)^d}[\frac{1}{\ell^2}-\frac{1}{(\ell+p)^2}-\frac{p^2}{\ell^2(\ell+p)^2}], \end{equation} but this are just scalar one-loop integrals therefore \begin{equation} p^2 B_{11} = -\frac{p^2}{2}B_0(p^2) \implies B_{11} = -\frac{B_0(p^2)}{2}. \end{equation} This gives us the final result \begin{equation} \mathcal{B}^\mu(p^2) = -\frac{B_0(p^2)}{2}p^\mu. \end{equation}

Now we would like to compute now the $2$-tensor two point function, which is a tensor integral of the form \begin{equation} \mathcal{B}^{\mu\nu}(p) = \int\frac{d^d{\ell}}{(2\pi)^d}\frac{\ell^\mu \ell^\nu}{\ell^2(\ell+p)^2}. \end{equation} The only $2$-tensors we can construct are $p^\mu p^\nu$ and $g^{\mu\nu}$, therefore \begin{equation} \mathcal{B}^{\mu\nu}(p) = B_{21}p^\mu p^\nu + B_{22}g^{\mu\nu}. \end{equation} By projecting onto the two tensors \begin{align} &p_\mu \mathcal{B}^{\mu\nu}(p) = p^\nu(p^2 B_{21}+B_{22}) = \frac{1}{2}\int\frac{d^d{\ell}}{(2\pi)^d}\frac{\ell^\nu}{\ell^2(\ell+p)^2}[(\ell+p)^2-\ell^2-p^2],\tag{1}\\ &g_{\mu\nu} \mathcal{B}^{\mu\nu}(p) = p^2 B_{21}+d B_{22} = \int\frac{d^d{\ell}}{(2\pi)^d}\frac{\ell^2}{\ell^2(\ell+p)^2} = 0.\tag{2} \end{align} From $(2)$ one obtains that \begin{equation} B_{22} = -\frac{p^2}{d}B_{21}, \end{equation} while from $(1)$ \begin{equation} p^\nu(p^2-\frac{p^2}{d})B_{21} = -p^\nu \frac{p^2 B_{11}}{2}, \end{equation} which gives \begin{equation} B_{21} = \frac{d}{d-1}\frac{B_{11}}{2} = \frac{d}{d-1}\frac{B_0(p^2)}{4} \end{equation} and consequently \begin{equation} B_{21} = -\frac{p^2}{d-1}\frac{B_0(p^2)}{4}. \end{equation} Therefore \begin{equation} \mathcal{B}^{\mu\nu}(p) = \frac{1}{d-1}[\frac{d}{4}B_0(p^2)p^\mu p^\nu-\frac{p^2}{4}B_0(p^2)g^{\mu\nu}]. \end{equation}

Here \begin{equation} \begin{split} B_0(p^2) &= \int\frac{d^d\ell}{(2\pi)^d}\frac{1}{[\ell^2+i\epsilon][(\ell+p)^2+i\eta]}\\ &=\frac{i\Gamma(\frac{4-d}{2})}{(4\pi)^{d/2}}\int_0^1d{x}d{y}\delta(x+y-1)\frac{1}{(-xyp^2-i\eta)^{\frac{4-d}{2}}}\\ &=\frac{i\Gamma(\frac{4-d}{2})}{(4\pi)^{d/2}}\int_0^1d{x}(x(1-x)(-p^2-i\eta))^{-\frac{4-d}{2}}\\ &=\frac{i\Gamma(\frac{4-d}{2})}{(4\pi)^{d/2}}(-p^2-i\eta)^{\frac{4-d}{2}}\frac{\Gamma^2(\frac{d-2}{2})}{\Gamma(d-2)}. \label{eq:scalar_one_loop} \end{split} \end{equation}