Commutation and Anticommutation relations in lattice QCD

Question

The article "Construction of a selfadjoint, strictly positive transfer matrix for euclidean lattice gauge theories" (Lüscher 1977), about lattice QCD, says the following:

> The fermion Hilbert space $\mathscr{H}_F$ is the Fock space built from an operator spinor field $\hat{\chi}_n$ which satisfies the usual canonical anticommutation relations:
> $$\{\hat{\chi}_{n,\alpha},\hat{\chi}^\dagger_{m,\beta} \}=\delta_{n,m} \delta_{\alpha,\beta}\qquad[...]\tag{11}$$
> The field operator $\hat{\psi}$ acts in $\mathscr{H}_F \otimes \mathscr{H}_G$.

(This is the hilbert space of fermions tensor product the hilbert space of the gauge fields.)

> It does _not_ have a canonical anticommutator, but:
> $$\{\hat{\psi}_{n,\alpha},\hat{\psi}^\dagger_{m,\beta} \}= B^{-1}_{n\alpha,m\beta}\qquad [...]\tag{12}$$
> The matrix $B_{n\alpha,m\beta}$ depends on the gauge field and is given by
> $$B_{n\alpha,m\beta} = \delta_{n,m}\delta_{\alpha,\beta} - K \sum_{j=1,2,3} U(n,j)_{\alpha,\beta}\delta_{n+\hat{j},m}+U(m,j)^{\dagger}_{\alpha,\beta}\delta_{m+\hat{j},n}\tag{13}$$

Then the article says in equation (14) that the relation between the field $\psi$ and the canonical field $\chi$ is
$$\psi_{n\alpha} = \sum_{m\beta} (B^{-1/2})_{n\alpha,m\beta} \chi_{m,\beta}$$
where $n$ and $m$ are the lattice sites, $U$ are the links variables (the gauge fields), $K$ is the Wilson hopping parameter, and the action is supposed to be the improved Wilson action:
$$S_F= \sum_{n} \biggl\{\bar{\psi}(n)\psi(n)-K \sum_{j=1,2,3}\bar{\psi}(n) U(n,j)(1-\gamma_j)\psi(n+\hat{j})+\bar{\psi}(n+\hat{j})U(n,j)^{\dagger}(1+\gamma_j)\psi(n) \biggr\}$$

How can these equations be justified? I have tried for days (without any success) to come up with an explanation nor I have found any demonstration from other sources for these assertions.
Especially on why the anticommutator of the fields is $B^{-1}$.

Comments

Could you add a reference?

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I edited the post