Schwartz's Quantum field theory, (14.100) Fermionic path integral

AI Thread Summary
The discussion focuses on the derivation of the integral in Schwartz's Quantum Field Theory, specifically the equation involving the fermionic path integral. The key question is how to show that the integral of the exponential expression equals the determinant of matrix A. Participants suggest expanding the exponential into a power series due to the nilpotent nature of Grassmann numbers and recognizing the integral as a multivariable Berezin integral. The surviving terms from this expansion relate to the definition of the determinant through signed permutations of matrix entries. Further clarification and detailed steps are requested to understand the involvement of external currents in the integral.
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Homework Statement
$$\int d\bar{\vec{\theta}}d \vec{\theta} e^{-(\bar{\vec{\theta}} - \bar{\vec{\eta}} A^{-1})A( \vec{\theta}-A^{-1}\vec{\eta})} = \operatorname{det}(A)$$
Relevant Equations
$$ \int d\bar{\theta}_1d\theta_1 \cdots d\bar{\theta}_n d\theta_n e^{-\bar{\theta}_i A_{ij} \theta_{j} } = \operatorname{det}(A) \tag{14.98}$$
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)## ?

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 deduction of the (14.98)?
 
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I've seen the derivation but it has been a while.
  • The first step is to expand the exponential into its power series which will only have two terms since the Grassmann "numbers" are nilpotent.
e^{-\overline{\theta}_i A_{ij} \theta_j} = 1-\overline{\theta}_i A_{ij} \theta_j
(... searching for references...)
  • Next step is to recognize this as a multivariable: Berezin Integral.
  • You will then find the surviving terms are the definition of the determinant in terms of sums of signed permutations of products of entries. (Remembering that the Grassmann variables anti-commute.)
There are additional details but that's the big picture as I recall.
 
jambaugh said:
I've seen the derivation but it has been a while.
  • The first step is to expand the exponential into its power series which will only have two terms since the Grassmann "numbers" are nilpotent.
e^{-\overline{\theta}_i A_{ij} \theta_j} = 1-\overline{\theta}_i A_{ij} \theta_j
(... searching for references...)
  • Next step is to recognize this as a multivariable: Berezin Integral.
  • You will then find the surviving terms are the definition of the determinant in terms of sums of signed permutations of products of entries. (Remembering that the Grassmann variables anti-commute.)
There are additional details but that's the big picture as I recall.
O.K. Again.. How can we perform this integral : ##\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)## ? 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?
 
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