Impurity tensor of the pseudoscalar condensate in the
                     GN model




                          1
       Consider the single-flavor Gross-Neveu(GN) model on the two dimensional space-time
lattice, the action S looks like:


               1 X X  µδν,2
                                 ψ̄(n) (rI − γν ) ψ(n + ν̂) + e−µδν,2 ψ̄(n + ν̂) (rI + γν ) ψ(n)
                                                                                                 
              S=−            e
               2 n∈Λ ν=1,2
                     2
                  X                 g2 X h             2                   2 i
       + (m + 2r)      ψ̄(n)ψ(n) −          ψ̄(n)ψ(n) + ψ̄(n)iγ5 ψ(n)                              (1)
                   n
                                    2   n

       ψ = (ψ1 , ψ2 ) denotes the two-component Grassmann-valued field. The m is the mass
parameter, the r is the Wilson parameter which is set as 1, and gσ2 = gπ2 = g 2 denote
the strength of the four-Fermi interaction. I is the 2 × 2 identity matrix. In the chiral
representaion, the gamma matrices are set as γ1 ≡ σ̂x , γ2 ≡ σ̂y and γ5 ≡ σ̂z . The Grassmann
tensor of model (1) generally takes the form:


                                                                                                                                              0     0       0   0
                                                                                                                                             j i         j i
                                                       X
 T(η1 ξ1 )(η2 ξ2 )(ξ̄1 η̄1 )(ξ̄2 η̄2 ) =                                      T(i1 j1 )(i2 j2 )(i0 j 0 )(i0 j 0 ) (η1i1 ξ1j1 )(η2i2 ξ2j2 )(ξ¯11 η̄11 )(ξ¯22 η̄22 ) (2)
                                                                                                1 1       2 2
                                                             0 0     0 0
                                           (i1 j1 )(i2 j2 )(i1 j1 )(i2 j2 )


       Now we try to derive the concrete form of T(i1 j1 )(i2 j2 )(i0 j 0 )(i0 j 0 ) based on the general pro-
                                                                                                                1 1     2 2
                                                                       ν                ν
cedure. First, the hopping matrices X and Y are determined and decomposed as :

                                                                                                              !
                                          1                1                                1       −1                        1       1       1†
                             X ν=1     = − (rI − σ̂x ) = −                                                        ≡ −U X σ X V X                                    (3)
                                          2                2                                −1 1
                                                                                                                !
                                          1                   eµ                                1         i                       2       2    2†
                             X ν=2     = − eµ (rI − σ̂y ) = −                                                       ≡ −U X σ X V X                                  (4)
                                          2                   2  −i 1
                                                                  !
                                          1                1 1 1           1   1    1
                             Y ν=1     = − (rI + σ̂x ) = −           ≡ −U Y σ Y V Y †                                                                               (5)
                                          2                2 1 1
                                                                        !
                                          1 −µ               e−µ   1 −i            2  2  2
                             Y ν=2     = − e (rI + σ̂y ) = −               ≡ −U Y σ Y V Y †                                                                         (6)
                                          2                   2    i 1

       The matrices from (3) to (6) are:

                                                                !                                          !                                            !
                              X1               1        −1                    X1            1/2 0                       X1†               1 −1
                          U        =                                    σ          =                                V         =
                                               −1 −1                                        0         0                                   1 1


                                                                                       2
                                                         !                                              !                                      !
                   X2                   1        i                   X2              eµ /2 0                      X2†                1 i
               U            =                                    σ        =                                  V              =
                                    −i −1                                            0          0                                  1 −i
                                        !                                                       !                                     !
                    1               1 1                            1             1/2 0                           1†              1  1
               UY =                                          σY =                                           VY        =
                                    1 −1                                         0          0                                    −1 1
                                                     !                                                  !                                      !
                    2               1 −i                           2             e−µ /2 0                             2†             1 −i
               UY =                                          σY =                                            VY            =
                                    i −1                                         0              0                                    1 i
    Notice only the (1, 1) component of all the singular matrices σ is non-zero. We substi-
tude the decompositions above into (1), and introduce new fields {χ̄, χ} through performing
unitary transformations to the original fields {ψ̄, ψ}, we arrive at the following exponential
terms appeared in e−S :

                                            X 1 χ̄X 1 (n)χX 1 (n+1̂)              X 2 χ̄X 2 (n)χX 2 (n+2̂)
                                        eσ1       1       1                   eσ1       1       1                                                   (7)
                                               1    1      Y1                        2    2      Y2
                                            σ1Y χ̄Y
                                                  1 (n+1̂)χ1 (n)                  σ1Y χ̄Y
                                                                                        1 (n+2̂)χ1 (n)
                                        e                                     e                                                                     (8)

    We introduce a pair of auxiliary bond Grassmann numbers to each hopping term:

                        1       1            1
                                                             Z                       √      1       1
                                                                                                                      √      1         1
                   σ1X χ̄X     X
                         1 (n)χ1 (n+1̂)                                                  σ1X χ̄X
                                                                                               1 (n)η1 (n)                σ1X η̄1 (n)χX
                                                                                                                                      1 (n+1̂)
                e                                        =                       e                               e                                  (9)
                                                                 η̄1 (n)η1 (n)
                        2       2            2
                                                             Z                       √      2       2
                                                                                                                      √      2         2
                   σ1X χ̄X     X
                         1 (n)χ1 (n+2̂)                                                  σ1X χ̄X
                                                                                               1 (n)η2 (n)                σ1X η̄2 (n)χX
                                                                                                                                      1 (n+2̂)
                e                                        =                       e                               e                                 (10)
                                                                 η̄2 (n)η2 (n)
                        1       1                1
                                                             Z                    √        1    1
                                                                                                                           √     1         1
                   σ1Y χ̄Y        Y
                         1 (n+1̂)χ1 (n)                                              σ1Y χ̄Y
                                                                                           1 (n+1̂)ξ̄1 (n) −                   σ1Y ξ1 (n)χY
                                                                                                                                          1 (n)
                e                                        =                    e                                       e                            (11)
                                                             ξ̄1 (n)ξ1 (n)
                        2       2                2
                                                             Z                    √        2    2
                                                                                                                           √     2         2
                   σ1Y χ̄Y        Y
                         1 (n+2̂)χ1 (n)                                              σ1Y χ̄Y
                                                                                           1 (n+2̂)ξ̄2 (n) −                   σ1Y ξ2 (n)χY
                                                                                                                                          1 (n)
                e                                        =                    e                                       e                            (12)
                                                             ξ̄2 (n)ξ2 (n)

    We select terms containg original fields on the n-th site and neglect the n-th labels
hereafter. Expand the (9) and substitude the matrix elements of the matrices:

                                        √                              √ 1 1
                                                                              X ψ̄ +U X 1 ψ̄
                                                                                             
                                                 1   1
                                            σ1X χ̄X                      σ1X U11            2 η1 (n)
                                    e             1 η1 (n)         = e            1  21
                                                                                                                                                   (13)
                                                                       √
                                                                         2 i1          i
                                                                   = (     ) ψ̄1 − ψ̄2 1 η1i1 (n)
                                                                        2

                             √ 1                    √ 1            
                                                                        1       1
                                                                                    
                                              1       σ X η̄ (n−1̂) V †X ψ +V †X ψ
                               X            X               1             1       2
                            e σ1 η̄1 (n−1̂)χ1 = e
                                                       1             11      12
                                                                                                                                                   (14)
                                                    √
                                                      2 i01 i01                     0
                                                = (     ) η̄1 (n − 1̂) (ψ1 − ψ2 )i1
                                                     2

                                                                              3
Expand the (10):

                        √                              √
                                                                                X 2 ψ̄ +U X 2 ψ̄
                                                                                                
                                 2      2                              2
                            σ1X χ̄X                          σ1X               U11              2 η2 (n)
                    e             1 η2 (n)       = e                                  1  21
                                                                                                                        (15)
                                                       √
                                                            2 i2 µ i2           i
                                                 = (         ) (e 2 ) ψ̄1 − iψ̄2 2 η2i2 (n)
                                                           2


              √ 2                    √ 2            
                                                         2       2
                                                                     
                               2       σ X η̄ (n−2̂) V †X ψ +V †X ψ
                X            X               2             1       2
             e σ1 η̄2 (n−2̂)χ1 = e
                                        1             11      12
                                                                                                                        (16)
                                     √
                                       2 i02 µ i02 i02                     0
                                 = (     ) (e 2 ) η̄2 (n − 2̂) (ψ1 + iψ2 )i2
                                      2
Expand the (11):

                        √                                    √
                                                                                     Y 1 ψ̄ +U Y 1 ψ̄
                                                                                                     
                                    1   1                                      1
                            σ1Y χ̄Y                                σ1Y              U11              2 ξ̄1 (n−1̂)
                    e             1 ξ̄1 (n−1̂)    = e                                      1  21
                                                                                                                        (17)
                                                             √
                                                                    2 j10       j 0 j 0
                                                  = (                ) ψ̄1 + ψ̄2 1 ξ¯11 (n − 1̂)
                                                                   2


                            √                                  √          1
                                                                                 
                                                                                    †Y 1       †Y 1
                                                                                                       
                                   1
                                σ1Y ξ1 (n)χY
                                             1             −           σ1Y ξ1 (n) V11    ψ1 +V12    ψ2
                    e−                     1     = e                                                                    (18)
                                                                                   √
                                                                                     2 j1 j1
                                                 = (−1)j1 (                           ) ξ1 (n) (ψ1 + ψ2 )j1
                                                                                    2
Expand the (12):

                √                                  √
                                                                        Y 2 ψ̄ +U Y 2 ψ̄
                                                                                        
                        2       2                              2
                σ1Y χ̄Y                                σ1Y             U11              2 ξ̄2 (n−2̂)
            e         1 ξ̄2 (n−2̂)           = e                              1  21
                                                                                                                        (19)
                                                   √
                                                       2 j20 − µ j20       j 0 j 0
                                             = (        ) (e 2 ) ψ̄1 + iψ̄2 2 ξ¯22 (n − 2̂)
                                                      2


                √                                     √      2
                                                                    
                                                                       †Y 2       †Y 2
                                                                                          
                       2
                    σ1Y ξ2 (n)χY
                                 2                −       σ1Y ξ2 (n) V11    ψ1 +V12    ψ2
           e−                  1            = e                                                                         (20)
                                                     √
                                                      2 j2 − µ j2 j2
                                                  j2
                                            = (−1) (   ) (e 2 ) ξ2 (n) (ψ1 − iψ2 )j2
                                                     2
Expand the rest terms in (1):



            e−(m+2r)ψ̄ψ = 1 − (m + 2r)ψ̄ψ + (m + 2r)2 ψ̄1 ψ1 ψ̄2 ψ2
                                                                                                                    
                                                                                                                        (21)

                                                                       4
                              g 2h                2                2
                                                                       i
                             e2 ( ) ( 5 )
                                 ψ̄ψ + ψ̄iγ ψ
                                                                           = 1 + 2g 2 ψ̄1 ψ1 ψ̄2 ψ2                                (22)

    Starting from (22) to (13), we move all the terms containing original fields ψ1 , ψ2 , ψ̄1
                                                                                                                           0   0
and ψ̄2 to the left side. This process would generate a sign factor of (−1)j2 +j1 +i2 +i1 , which
                                                                                                              0   0
can be incorporated with that in (18) and (20), leading to (−1)i2 +i1 . We then adjust the
order of the original and new Grassmann fields at the same time while maintaining the mirror
symmetry within them(thus no additional sign factor appears), and arrives at:



                     1 − (m + 2r)ψ̄ψ + (m + 2r)2 ψ̄1 ψ1 ψ̄2 ψ2 1 + 2g 2 ψ̄1 ψ1 ψ̄2 ψ2
                                                                                     
                                                                                                                                   (23)
                               j 0        j 0         i          i
               ×     ψ̄1 + iψ̄2 2 ψ̄1 + ψ̄2 1 ψ̄1 − iψ̄2 2 ψ̄1 − ψ̄2 1
                                 0                         0
               × (ψ1 + iψ2 )i2 (ψ1 − ψ2 )i1 (ψ1 − iψ2 )j2 (ψ1 + ψ2 )j1
                                   i      0     i              0             j            j           0                0
               × ξ1j1 (n)ξ2j2 (n)η̄11 (n − 1̂)η̄22 (n − 2̂)η1i1 (n)η2i2 (n)ξ¯11 (n − 1̂)ξ¯22 (n − 2̂)

    Now we arrange the new Grassmann fields into the desired order(move from the under-
lined position to the blue-star position sequentially):


                                 i0                   i0                                        j0                j0
             (∗)ξ1j1 (n)ξ2j2 (n)η̄11 (n − 1̂)η̄22 (n − 2̂)η1i1 (n)η2i2 (n)ξ¯11 (n − 1̂)ξ¯22 (n − 2̂)
                                             i0            i0                   j0           j0
              η1i1 (n)ξ1j1 (n) (∗)ξ2j2 (n)η̄11 (n − 1̂)η̄22 (n − 2̂)η2i2 (n)ξ¯11 (n − 1̂)ξ¯22 (n − 2̂)
                              
        →
                                                        i0            i0           j0           j0
              η1i1 (n)ξ1j1 (n) η2i2 (n)ξ2j2 (n) (∗)η̄11 (n − 1̂)η̄22 (n − 2̂)ξ¯11 (n − 1̂)ξ¯22 (n − 2̂)
                                                
        →
                                                   j0            i0
                                                                                
                                                                                        i0            j0
              η1i1 (n)ξ1j1 (n) η2i2 (n)ξ2j2 (n) ξ¯11 (n − 1̂)η̄11 (n − 1̂) (∗)η̄22 (n − 2̂)ξ¯22 (n − 2̂)
                              
        →
                                                   j0            i0
                                                                                 0
                                                                                      j            i0
                                                                                                         
              η1i1 (n)ξ1j1 (n) η2i2 (n)ξ2j2 (n) ξ¯11 (n − 1̂)η̄11 (n − 1̂) ξ¯22 (n − 2̂)η̄22 (n − 2̂)
                              
        →

    The accumulated sign factors in the above process are:



                       (−1)i1 ×(i2 +i1 +j2 +j1 )+i2 ×(i2 +i1 +j2 )+j1 ×(i2 +i1 )+j2 ×i2
                                      0       0                             0       0   0   0    0        0   0
                                                                                                                                   (24)
                                                                                                R
    Now we would like to perform the Grassmann integral                                              dψ1 dψ̄1 dψ2 dψ̄2 to integrate the
original Grassmann fields on the n-th site. (23) contain the following contributions:



        2g 2 + (m + 2r)2 ψ̄1 ψ1 ψ̄2 ψ2 → 2g 2 + (m + 2r)2 δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0
                                                        
                                                                                                                                   (25)

                                                                                5
                                                      0                    0        0
 −(m + 2r)ψ̄1 ψ1 → −(−1)i1 +i2 +i1 +j2 × (i)i2 +i2 +j2 +j2 × (m + 2r)δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1 (26)




                               −(m + 2r)ψ̄2 ψ2 → −(m + 2r)δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1                                    (27)




                                                     1 → −Āi1 ,i2 ,j10 ,j20 Aj1 ,j2 ,i01 ,i02                                         (28)

      The minus sign in (28) comes from adjusting the Grassmann integral measure from
R                       R
    dψ1 dψ̄1 dψ2 dψ̄2 to dψ2 dψ1 dψ̄2 dψ̄1 . The Āi1 ,i2 ,j10 ,j20 is given by the following Grassmann
integral:

                                     Z
                                                                   j20                 j10                  i2                i1
              Āi1 ,i2 ,j10 ,j20 =       dψ̄2 dψ̄1 ψ̄1 + iψ̄2              ψ̄1 + ψ̄2             ψ̄1 − iψ̄2         ψ̄1 − ψ̄2          (29)

       Similarly, the Aj1 ,j2 ,i01 ,i02 is given by the following Grassmann integral:

                                     Z
                                                                       0                  0
               Aj1 ,j2 ,i01 ,i02 =       dψ2 dψ1 (ψ1 + iψ2 )i2 (ψ1 − ψ2 )i1 (ψ1 − iψ2 )j2 (ψ1 + ψ2 )j1                                 (30)

       The non-zero tensor elements of Āi1 ,i2 ,j10 ,j20 and Aj1 ,j2 ,i01 ,i02 are listed:



      Ā1,1,0,0 = −1 + i, Ā1,0,1,0 = −2, Ā1,0,0,1 = −1 − i, Ā0,1,1,0 = −1 − i, Ā0,1,0,1 = −2i, Ā0,0,1,1 = 1 − i
      A1,1,0,0 = 1 + i, A1,0,1,0 = 2, A1,0,0,1 = 1 − i, A0,1,1,0 = 1 − i, A0,1,0,1 = −2I, A0,0,1,1 = −1 − i

       Finally, we gather all the information below and obtain the coefficient tensor T(i1 j1 )(i2 j2 )(i0 j 0 )(i0 j 0 ) :
                                                                                                                                         1 1   2 2




                                                                                                                         µ
      T(i1 j1 )(i2 j2 )(i0 j 0 )(i0 j 0 ) = (−1)i1 ×(i2 +i1 +j2 +j1 )+i2 ×(i2 +i1 +j2 )+j1 ×(i2 +i1 )+j2 ×i2 +i1 +i2 × e 2 (i2 +i2 −j2 −j2 )
                                                      0   0                 0   0        0    0   0    0   0   0   0             0       0

                         1 1      2 2
        √ !i1 +j1 +i2 +j2 +i01 +j10 +i02 +j20
            2
                                                    × 2g 2 + (m + 2r)2 δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0
                                                                                 
    ×
          2
                           0                     0          0
    − (−1)i1 +i2 +i1 +j2 × (+i)i2 +i2 +j2 +j2 × (m + 2r)δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1
                                                                                               
    − (m + 2r)δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1 − Āi1 ,i2 ,j10 ,j20 Aj1 ,j2 ,i01 ,i02


                                                                               6
     The expectation value of the pseudscalar condensate hψ̄iγ5 ψi:


                                                 i ψ̄1 ψ1 − ψ̄2 ψ2 e−S
                                                                  
                                                  ψ̄iγ5 ψe−S
                                              Z                        Z
                 hψ̄iγ5 ψi =                                 =                    (31)
                                                      Z    Z
                                                                  
    Therefore, we only need to add the term i ψ̄1 ψ1 − ψ̄2 ψ2 to (23), and perform the
                  R
Grassmann integral dψ1 dψ̄1 dψ2 dψ̄2 . The contributions are:



                                      iψ̄1 ψ1 → −i(m + 2r)δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0                                      (32)



                                          0       0                        0
                  iψ̄1 ψ1 → i(+i)j2 +i2 +i2 +j2 (−1)i2 +i1 +i1 +j2 δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1                             (33)




                                      −iψ̄2 ψ2 → i(m + 2r)δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0                                      (34)




                                      −iψ̄2 ψ2 → −iδi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1                                             (35)

     Therefore, the impurity tensor corresponds to the pseudoscalar condensate hψ̄iγ5 ψi can
be obtained as:


                                                                                                                                   µ
                                               i1 ×(i02 +i01 +j2 +j1 )+i2 ×(i02 +i01 +j2 )+j10 ×(i02 +i01 )+j20 ×i02 +i01 +i02       i +i02 −j2 −j20 )
                                                                                                                                   2( 2
             ψ
   T(iψ̄iγj 5)(i         0 0     0 0 = (−1)                                                                                    × e
        1 1      2 j2 )(i1 j1 )(i2 j2 )
       √ !i1 +j1 +i2 +j2 +i01 +j10 +i02 +j20 h
           2                                                      0       0                      0
 ×                                                 × i(+i)j2 +i2 +i2 +j2 (−1)i2 +i1 +i1 +j2 δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1
         2
                                              
 − iδi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1




                                                                      7
    The pseudoscalar condensate hψ̄iγ5 ψi can be also computed through finite difference
methods by adding the following term in the action (1):

                                                         X
                                              Sh = −h              ψ̄(n)iγ5 ψ(n)                                             (36)
                                                           n

    Similarly, it would contribute a term eih(ψ̄1 ψ1 −ψ̄2 ψ2 ) to (23), which would expand like:


                         eih(ψ̄1 ψ1 −ψ̄2 ψ2 ) = 1 + ih ψ̄1 ψ1 − ψ̄2 ψ2 + h2 ψ̄1 ψ1 ψ̄2 ψ2
                                                                      
                                                                                                                             (37)

    The additional contributions from (37) are:


                           ihψ̄1 ψ1 → −i(m + 2r)hδi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0                                   (38)



                                          0                    0        0
            ihψ̄1 ψ1 → (−1)i1 +i2 +i1 +j2 × (i)i2 +i2 +j2 +j2 × ihδi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1                  (39)



                           −ihψ̄2 ψ2 → i(m + 2r)hδi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0                                   (40)



                                −ihψ̄2 ψ2 → −ihδi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1                                     (41)



                               h2 ψ̄1 ψ1 ψ̄2 ψ2 → h2 δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0                               (42)

    Therefore, the Grassmann tensor now looks like:


                                                                                                                      µ
   T(i1 j1 )(i2 j2 )(i0 j 0 )(i0 j 0 ) = (−1)i1 ×(i2 +i1 +j2 +j1 )+i2 ×(i2 +i1 +j2 )+j1 ×(i2 +i1 )+j2 ×i2 +i1 +i2 × e 2 (i2 +i2 −j2 −j2 )
                                                   0   0                 0   0        0    0   0    0   0   0   0             0       0

                      1 1      2 2
     √ !i1 +j1 +i2 +j2 +i01 +j10 +i02 +j20
         2
                                                 × 2g 2 + (m + 2r)2 + h2 δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0
                                                                                    
 =
       2
                     0                    0       0
 − (−1)i1 +i2 +i1 +j2 × (+i)i2 +i2 +j2 +j2 × (m + 2r − ih)δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1
                                                                                                 
 − (m + 2r + ih)δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1 − Āi1 ,i2 ,j10 ,j20 Aj1 ,j2 ,i01 ,i02


                                                                    8
    现在我们考虑一个更 general 的情况，也就是对存在 external field (36) 的模型使用杂
志点方法。存在 external field 展开后的 action 是：



                       1 − (m + 2r)ψ̄ψ + (m + 2r)2 ψ̄1 ψ1 ψ̄2 ψ2 1 + 2g 2 ψ̄1 ψ1 ψ̄2 ψ2
                                                                                       
                                                                                                                             (43)
                   × 1 + ih ψ̄1 ψ1 − ψ̄2 ψ2 + h2 ψ̄1 ψ1 ψ̄2 ψ2
                                                             
                               j 0        j 0           i        i
                   × ψ̄1 + iψ̄2 2 ψ̄1 + ψ̄2 1 ψ̄1 − iψ̄2 2 ψ̄1 − ψ̄2 1
                                         0                0
                   × (ψ1 + iψ2 )i2 (ψ1 − ψ2 )i1 (ψ1 − iψ2 )j2 (ψ1 + ψ2 )j1

    如果在 (43) 的基础上，再添加上杂志点的贡献：


                                                                         
                                                  i ψ̄1 ψ1 − ψ̄2 ψ2                                                          (44)

    则执行 Grassmann 积分后得到的非零项为：



                       iψ̄1 ψ1 → −i(m + 2r)δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0
                       iψ̄1 ψ1 → hδi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0
                                                      0         0       0
                       iψ̄1 ψ1 → i(−1)i2 +i1 +i1 +j2 (i)j2 +i2 +i2 +j2 δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1
                       −iψ̄2 ψ2 → i(m + 2r)δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0
                       −iψ̄2 ψ2 → hδi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0
                       −iψ̄2 ψ2 → −iδi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1

    最终的杂志点张量因而可以表示为：


                                                                                                                      µ
   T(iψ̄γj5 ψ)(i j )(i0 j 0 )(i0 j 0 ) = (−1)i1 ×(i2 +i1 +j2 +j1 )+i2 ×(i2 +i1 +j2 )+j1 ×(i2 +i1 )+j2 ×i2 +i1 +i2 × e 2 (i2 +i2 −j2 −j2 )
                                                   0   0                   0   0            0   0 0 0   0   0   0             0       0

        1 1     2 2   1 1      2 2
       √ !i1 +j1 +i2 +j2 +i01 +j10 +i02 +j20 h
           2                                                  0         0                     0
 ×                                               × i(+i)j2 +i2 +i2 +j2 (−1)i2 +i1 +i1 +j2 δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1
         2
                                                                                            
 − iδi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1 + 2hδi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0




                                                                9
                     2
        再考虑 h ψ̄iγ5 ψ i = 2hψ̄1 ψ1 ψ̄2 ψ2 i。 将其添加到 (23) 式中并执行 Grassmann 积分
    dψ1 dψ̄1 dψ2 dψ̄2 ，其贡献只包含：
R



                                                   2 × δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0                                   (45)

        因而相应的杂志点张量为：


          (ψ̄iγ5 ψ)2                                                                                                           µ
                                            = (−1)i1 ×(i2 +i1 +j2 +j1 )+i2 ×(i2 +i1 +j2 )+j1 ×(i2 +i1 )+j2 ×i2 +i1 +i2 × e 2 (i2 +i2 −j2 −j2 )
                                                            0   0                   0    0         0   0   0   0   0   0   0        0      0
        T(i                0 0     0 0
           1 j1 )(i2 j2 )(i1 j1 )(i2 j2 )
          √ !i1 +j1 +i2 +j2 +i01 +j10 +i02 +j20
           2
    ×                                           × 2δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0
          2




                                                                          10
   考虑 hψ̄ψi = ψ̄1 ψ1 +ψ̄2 ψ2 ，这个情况和计算 pseudoscalar condensate 差不多。Grassmann
积分后存在的贡献包括：



                                           ψ̄1 ψ1 → −(m + 2r)δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0                               (46)



                                            0          0                     0
                   ψ̄1 ψ1 → (+i)j2 +i2 +i2 +j2 (−1)i2 +i1 +i1 +j2 δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1                          (47)




                                           ψ̄2 ψ2 → −(m + 2r)δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0                               (48)




                                                ψ̄2 ψ2 → δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1                                   (49)



                                                                                                                                 µ
                                         = (−1)i1 ×(i2 +i1 +j2 +j1 )+i2 ×(i2 +i1 +j2 )+j1 ×(i2 +i1 )+j2 ×i2 +i1 +i2 × e 2 (i2 +i2 −j2 −j2 )
                                                            0   0                  0   0             0   0   0   0   0   0   0       0      0
     T(iψ̄ψj            0 0     0 0
         1 1 )(i2 j2 )(i1 j1 )(i2 j2 )
    √ !i1 +j1 +i2 +j2 +i01 +j10 +i02 +j20 h
       2                                               0      0                      0
×                                              × (+i)j2 +i2 +i2 +j2 (−1)i2 +i1 +i1 +j2 δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1
      2
                                                                                                    
+ δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1 − 2 × (m + 2r)δi1 +i2 +j10 +j20 ,0 δj1 +j2 +i01 +i02 ,0




                                                                          11
    如果这个作用量 S 包含 lattice anisotropy term:


        1 X X  δν,2 µδν,2
                              ψ̄(n) (rI − γν ) ψ(n + ν̂) + γ δν,2 e−µδν,2 ψ̄(n + ν̂) (rI + γν ) ψ(n)
                                                                                                    
     S=−              γ e
        2 n∈Λ ν=1,2
              2
           X                g2 X h              2                    2 i
+ (m + 2r)      ψ̄(n)ψ(n) −           ψ̄(n)ψ(n) + ψ̄(n)iγ5 ψ(n)                                  (50)
            n
                            2 n

    那相应的 pseudoscalar 期望值就会变成：


                                                                                                                         µ
                                          = (−1)i1 ×(i2 +i1 +j2 +j1 )+i2 ×(i2 +i1 +j2 )+j1 ×(i2 +i1 )+j2 ×i2 +i1 +i2 × e 2 (i2 +i2 −j2 −j2 )
                                                        0   0                   0 0       0   0   0    0   0   0   0             0          0
               ψ
     T(iψ̄iγj 5)(i           0 0     0 0
         1 1         2 j2 )(i1 j1 )(i2 j2 )
                                          √ !i1 +j1 +i2 +j2 +i01 +j10 +i02 +j20 h
     √           0             0           2                                            0       0                    0
×        γ i2 +i2 +j2 +j2 ×                                                    × i(+i)j2 +i2 +i2 +j2 (−1)i2 +i1 +i1 +j2 δi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1
                                          2
                                                
− iδi1 +i2 +j10 +j20 ,1 δj1 +j2 +i01 +i02 ,1




                                                                        12
