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values λ K are complex and can turn to zero for certain fields φ$_{r}$ ( τ ). 

Suppose that the modulus of an eigenvalue λ becomes very small or can even turn to zero and let us show that, nevertheless, the corresponding eigenvalue E remains fi- nite for a finite γ . 

Let v$_{r,r}$ ′ ( τ ) be the eigenfunction corresponding to the eigenvalue λ in the first equation in (2.46). Another func- tion ¯ v ∗ r,r ′ ( τ ) should correspond to the eigenvalue λ ∗ in the second equation in (2.47). 

The small values of | λ | correspond to small values of the eigenvalues E of Eq. (2.48) of a state with an eigenvector S$_{r,r}$ ′ ( τ ). At γ = 0 and λ = 0 the solution of Eq. (2.48) is very simple. One obtains immediately E = 0, while the solution S$_{r,r}$ ′ ( τ ) can be written in the form 

with the vectors S a r,r ′ ( τ ) and S b r,r ′ ( τ ) given by 

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and arbitrary coefficients c$_{1}$ and c$_{2}$ . This means that the state with E = 0 is degenerate. 

The case of small λ and γ can be considered using quantum mechanical perturbation theory. As the state with E = 0 is degenerate we seek for the solution S$_{r,r}$ ′ ( τ ) writing it in the form of Eq. (A1). Following the standard scheme other states are neglected in this expansion. 

Substituting Eq. (A1) into Eq. (2.48) we multiply both sides of the equation first by S a ∗ r,r ′ ( τ ) , then by S b ∗ r,r ′ ( τ ), sum over r, r ′ and integrate over τ . This gives us a system of two equations for the coefficients c$_{1}$ and c$_{2}$ , 

where 

Non-trivial solutions of Eq. (A3) exist provided the de- terminant equals to zero. This condition gives us the eigenvalues E 

Eq. (A5) demonstrates explicitly that the eigenvalues cannot turn to zero even in the situation when λ = 0. This is a well known effect of level repulsion. The ab- sence of the zero eigenvalues E justifies our method of the regularization. 

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## Appendix B: Checking the regularization for a static Hubbard-Stratonovich field 

In this Appendix we demonstrate that using the spec- tral expansion, Eqs. (2.52, 2.54), we come in the case of static HS fields back to Eq. (2.21). This can serve as a check of the transformations we have carried out in or- der to derive Eqs. (2.52, 2.54) and can help the reader to visualize our regularization scheme. 

Assuming that the HS field φ$_{r}$ does not depend on time we write time independent solutions S K r,r ′ of Eq. (2.48) in the form 

where a and b are coefficients and the functions v k r are the eigenfunctions of the operator ˆ h$_{r}$, Eq. (2.24) 

In the static case, only time independent solutions con- tribute into the exponent in Eq. (2.54) and for such states K = { k, k $^{′}$} . Substituting Eq. (B1) into Eq. (2.48) we come to a 

system of two linear equations 

Solution of these equations leads to the eigenvalue 

Substituting Eq. (2.52) into Eq. (2.54) we have to calcu- late the sum over the eigenstates S K r,r ′ and thus find the solutions for a and b from Eq. (B3) using the eigenvalues E $^{K}$, Eq. (B4). The function Z [ φ ] can be written as 

<img src='content_image/393097.jpg'>

The subscripts 1 and 2 in Eq. (B5) correspond to two different components of the vector S K r,r ′ , Eq. (B1). 

One can easily obtain from Eqs. (B3, B4) that the product ab entering Eq. (B5) takes the form 

and we come to the following form of the function Z [ φ ] 

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