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

[Plantation]

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)?

 

[jambaugh]

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.

 

[Plantation]

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?