# Gaussian integrals with gamma matrices in their exponents

I should evaluate Gaussian integrals in the 1+1 Minkowski space, which read

$$I_{1}= \int d^{2}k \, {\rm Tr}\big[ \gamma^{5} \gamma^{\eta} \gamma^{\kappa} e^{\alpha k^{\mu}k_{\mu} + \beta \gamma^{\mu} \gamma^{\nu} M^{* \sigma}_{\mu} M^{\iota}_{\nu} k_{\sigma} k_{\iota}} \big],\\ I_{2}= \int d^{2}k \, {\rm Tr}\big[ \gamma^{5} e^{\alpha k^{\mu}k_{\mu} + \beta \gamma^{\mu} \gamma^{\nu} M^{* \kappa}_{\mu} M^{\iota}_{\nu} k_{\kappa} k_{\iota}} \big],\\ I_{3}= \int d^{2}k \, {\rm Tr}\big[ \gamma^{5} e^{\alpha k^{\mu}k_{\mu} + \beta \gamma^{\mu} \gamma^{\nu}\gamma^{5} M^{* \kappa}_{\mu} M^{\iota}_{\nu} k_{\kappa} k_{\iota}} \big],$$ where $$\alpha$$ and $$\beta$$ are real constants and the diagonal matrix $$M$$ has nonzero complex elements~($$M_{\mu}^{\mu}=v_{\mu}$$). Using the Euclidean convention $$g^{\mu \nu} =- \delta^{\mu \nu}$$ at $$\beta=0$$ results in $$I_{1}= \int d^{2}k \, {\rm Tr}\big[ \gamma^{5} \gamma^{\eta} \gamma^{\kappa} e^{-\alpha k_{\mu}k_{\mu} } \big] =-2 \frac{\pi}{\alpha} \varepsilon^{\eta \kappa} , \\ I_{2}=0,\\ I_{3}=0,$$ where I have used $${\rm Tr}[\gamma^{5}]=0$$, and $${\rm Tr}[\gamma^{5} \gamma^{\eta} \gamma^{\kappa}] = -2 \varepsilon^{\eta \kappa}$$. How can I calculate $$I_{1}$$, $$I_{2}$$, and $$I_{3}$$ with nonzero $$\beta$$?

EDIT: A simplified question is how the following integral should be calculated $$I_{0}= \int d^{2}k \, e^{ \gamma^{\mu} \gamma^{\nu} M^{* \sigma}_{\mu} M^{\iota}_{\nu} k_{\sigma} k_{\iota}} .$$

• There's a problem of indices in your exponent, $\mu$ and $\nu$ appear three times in a product. Commented Feb 27, 2021 at 11:33
• @JeanbaptisteRoux Corrected! Commented Feb 27, 2021 at 15:59
• Have you tried calculating $\int dxe^{-Ax^2}$ where $A$ is some matrix? If you can find what properties $A$ must have, the generalization should be about as straightforward as the usual generalization from the 1d Gaussian integral to more complicated versions. I think it should be pretty lax...might need some weak form of invertibility like a pseudoinverse. Commented Feb 27, 2021 at 20:38
• In this way, the matrix $A$ comprises $MM^{*}$ and $\gamma$ matrices. This Gaussian integral after the integration then generates a factor with the determinate of A in the denominator. This seems very weird to me. I think I should treat gamma matrices differently. Commented Mar 6, 2021 at 23:39

You need to expand out the exponential using identites such a $$(v^\mu \gamma_\mu)^2 = |v|^2{\mathbb I}$$. It may well help introduce a new variable $$\omega$$ and us a Hubbard Stratanovich transform to replace $$(v^\mu \gamma_\mu)(v^{*\nu} \gamma_\nu)$$ by something like $$i\omega (v^\mu \gamma_\mu) + \frac 12 \omega^2$$ (signs not checked) befoe expanding.