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I am reading the Schwartz's Quantum field theory, p.269~p.272 ( 14.6 Fermionic path integral ) and some question arises.

In section 14.6, Fermionic path integral, p.272, $(14.100)$, he states that

$$ \int d\bar{\theta}_1d\theta_1 \cdots d\bar{\theta}_n d\theta_n e^{-\bar{\theta}_i A_{ij} \theta_{j} + \bar{\eta}_i \theta_{i}+ \bar{\theta}_i \eta_i} = e^{\bar{\vec{\eta}} A^{-1} \vec{\eta}} \int d\bar{\vec{\theta}}d \vec{\theta} e^{-(\bar{\vec{\theta}} - \bar{\vec{\eta}} A^{-1})A( \vec{\theta}-A^{-1}\vec{\eta})}= \operatorname{det}(A) e^{\bar{\vec{\eta}} A^{-1}\vec{\eta}} \tag{14.100}$$ where $\theta_i$ are grassmann numbers ( C.f. His book p.269 ) and $\bar{\theta}_i$ are defined in p.271. And $\eta_i$ and $\bar{\eta}_i$ are external currents.

Q. Why $$\int d\bar{\vec{\theta}}d \vec{\theta} e^{-(\bar{\vec{\theta}} - \bar{\vec{\eta}} A^{-1})A( \vec{\theta}-A^{-1}\vec{\eta})} = \operatorname{det}(A)~ ?$$ I don't understand this calculation at all.

In his book, p.271, $(14.98)$, he deduced that $$ \int d\bar{\theta}_1d\theta_1 \cdots d\bar{\theta}_n d\theta_n e^{-\bar{\theta}_i A_{ij} \theta_{j} } = \operatorname{det}(A). \tag{14.98}$$ Can we use this? How? Or by similar argument for derivation of the (14.98)? Perhaps, I think that we may try to do the calculation by change of variable but I don't know how to do it.

Can anyone help?

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    $\begingroup$ When integrating over all values, constant shifts do nothing. The integral that you don't know how to do, is reduced to the (14.98) that had been done before. Let $\vec\phi=\vec\theta-A^{-1}\vec\eta$, which implies $\mathrm d\vec\phi=\mathrm d\vec\theta$, and then you do the same for the other side. $\endgroup$ Oct 19, 2023 at 4:54
  • $\begingroup$ Aha. I got feeling. The external currents are constants ( vectors ) ? And $\bar{\vec{\phi}} = \bar{\vec{\theta}} - \bar{\vec{\eta}} A^{-1}$ ? $\endgroup$
    – Plantation
    Oct 19, 2023 at 5:06

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We can diagonalize $A$ and then perform a change of Grassmann integration variables so that only one terms remains: $$ -\bar{\theta}_i A_{ij} \theta_j \rightarrow -\bar{\theta}_i U_{ia}^{\dagger}D_{ab}U_{bj} \theta_j \rightarrow -\bar{\theta}_a' D_{ab} \theta_b' = -\sum_n \bar{\theta}_n' \lambda_n \theta_n' , $$ where $\lambda_n$ represents the eigenvalues of $A$. The only term in the expansion of the exponential function that contributes to the Grassmann integral is the term consisting of the product of all the Grassmann variables: $$ \exp(-\bar{\theta}_i A_{ij} \theta_j) = \prod_n \bar{\theta}_n' \lambda_n \theta_n . $$ The Grassmann integration now removes all the Grassmann variables, leaving only the product of the eigenvalues, which is the determinate $$ \prod_n \lambda_n = \det(A). $$

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  • $\begingroup$ Perhaps, did you provide proof of the $ \int d\bar{\theta}_1d\theta_1 \cdots d\bar{\theta}_n d\theta_n e^{-\bar{\theta}_i A_{ij} \theta_{j} } = \operatorname{det}(A)$ ? I asked why $\int d\bar{\vec{\theta}}d \vec{\theta} e^{-(\bar{\vec{\theta}} - \bar{\vec{\eta}} A^{-1})A( \vec{\theta}-A^{-1}\vec{\eta})} = \operatorname{det}(A)~ $. What relationship exists between your caclculation and my question? An issue that makes me annoying is the involved objects $\bar{\vec{\eta}}$ (and $\vec{\eta}$) (external currents). Perhaps can you provided explanation more step by step in detail? $\endgroup$
    – Plantation
    Oct 19, 2023 at 4:30
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    $\begingroup$ That is just a shift in the integration variables. You can perform a redefinition of the integration variables to get rid of the shift and end up with the same result. $\endgroup$ Oct 20, 2023 at 2:30
  • $\begingroup$ O.K. I willl think about it myselt. Thank you.~ $\endgroup$
    – Plantation
    Oct 20, 2023 at 11:38

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