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$0000$. After using the mapping in the 3rd Equation, it is the same for the labeling of the spatial orb in the paper. The first two (from the right) means 2 0-th spatial orbs are occupied, and this can only be the case if both spin-up and down are occupied. Using the labeling of the spin-orbital, it would be 10 (or 10). Similarly for the next two indices. Thus we have 0110 in the spin-orbital basis (why it is not 1010, I am not sure, and maybe this is due to convention). By the symmetry in Eq. 4, we have also 1001.

$1111$. The argument is essentially the same as above, except that now we are dealing with 1st spatial orb, with the spin-up and down labeled as 23. Thus we have 3223 and 2332for the spin-orb in the paper.

$0000$. After using the mapping in the 3rd Equation, it is the same for the labeling of the spatial orb in the paper. The first two (from the right) means 2 0-th spatial orbs are occupied, and this can only be the case if both spin-up and down are occupied. Using the labeling of the spin-orbital, it would be 10 (or 01). Similarly for the next two indices. Thus we have 0110 in the spin-orbital basis (why it is not 1010, I am not sure, and maybe this is due to convention). By the symmetry in Eq. 4, we have also 1001.

$1111$. The argument is essentially the same as above, except that now we are dealing with 1st spatial orb, with the spin-up and down labeled as 23. Thus we have 3223 and 2332 for the spin-orb in the paper.

$ \int dx_1dx_2\chi_i^*(x_1)\chi_k^*(x_2)\chi_j(x_1)\chi_l(x_2) $

$ \int dx_1dx_2\chi_i^*(x_1)\chi_j^*(x_2)\chi_k(x_2)\chi_l(x_1) $

$ \text{ijkl in FCIDUMP file = iklj in the paper, for the spatial orb} $

$ ijkl = jilk = lkji $

$1100$. This is essentially the same as above, where we realize that it is 1001 for the spatial orb in the paper. With the same logic, we have 2002, 3003, 2112, and 3113. These can essentially be obtained with the symmetry property in Eq. 4

$$ \tag{1} \int dx_1dx_2\chi_i^*(x_1)\chi_k^*(x_2)\chi_j(x_1)\chi_l(x_2) $$

$$ \tag{2} \int dx_1dx_2\chi_i^*(x_1)\chi_j^*(x_2)\chi_k(x_2)\chi_l(x_1) $$

ijkl in FCIDUMP file = iklj in the paper, for the spatial orb

$$ \tag{3} ijkl = jilk = lkji $$

$1100$. This is essentially the same as above, where we realize that it is 1001 for the spatial orb in the paper. With the same logic, we have 2002, 3003, 2112, and 3113. These can essentially be obtained with the symmetry property in Eq. 4.

while the indices in the paper, it is (I thought I put the figures, but sorry it is not here. See below for the definition in the paper)

where the second equality is due to reality requirement of the integral. With these, we are ready to understand the integrals in FCIDUMP file one by one. First, we need to subtract all the indices by 1, so that it matches with those in the paper. So we have

6.74493103326006093745E-01 0 0 0 0

6.63472044860555665302E-01 0 0 1 1

6.63472044860555665302E-01 1 1 0 0

6.97397949820408036281E-01 1 1 1 1

1.81287535812332034624E-01 1 0 1 0

-1.25247730398215462166E+00 0 0 -1 -1

-4.75934461144127241017E-01 1 1 -1 -1

7.43077168397780152276E-01 -1 -1 -1 -1

0 0 0 0. After using the mapping in the 3rd Equation, it is the same for the labeling of the spatial orb in the paper. The first two (from the right) means 2 0-th spatial orbs are occupied, and this can only be the case if both spin-up and down are occupied. Using the labeling of the spin-orbital, it would be 10 (or 10). Similarly for the next two indices. Thus we have 0110 in the spin-orbital basis (why it is not 1010, I am not sure, and maybe this is due to convention). By the symmetry in Eq. 4, we have also 1001.

1 1 1 1. The argument is essentially the same as above, except that now we are dealing with 1st spatial orb, with the spin-up and down labeled as 23. Thus we have 3223 and 2332for the spin-orb in the paper.

0 0 1 1. Now with the mapping in Eq. 3, it is in fact 0110 for the spatial orb in the paper. The first two indices 10 means the 0th and 1st spatial orbs are occupied, and they could be either spin up and down. Thus we have $2\times2=4$ options: 20, 30, 21, 31 for the first two indices of the spin-orbs. Thus put it together, we have 0220, 0330, 1221, 1331 for the spin orbs. Again, I am not sure why we don't have 2020, maybe due to convention.

1 1 0 0. This is essentially the same as above, where we realize that it is 1001 for the spatial orb in the paper. With the same logic, we have 2002, 3003, 2112, and 3113. These can essentially be obtained with the symmetry property in Eq. 4

1 0 1 0. Ok, I am stuck here for now... Will update this after I figure it out.

while the indices in the paper, it is

where the second equality is due to reality requirement of the integral. With these, we are ready to understand the integrals in FCIDUMP file one by one. First, we need to subtract all the indices by 1, so that it matches with those in the paper. So we have

 6.74493103326006093745E-01 0 0 0 0 
 6.63472044860555665302E-01 0 0 1 1     
 6.63472044860555665302E-01 1 1 0 0  
 6.97397949820408036281E-01 1 1 1 1     
 1.81287535812332034624E-01 1 0 1 0     
-1.25247730398215462166E+00 0 0 -1 -1
-4.75934461144127241017E-01 1 1 -1 -1
 7.43077168397780152276E-01 -1 -1 -1 -1

$0000$. After using the mapping in the 3rd Equation, it is the same for the labeling of the spatial orb in the paper. The first two (from the right) means 2 0-th spatial orbs are occupied, and this can only be the case if both spin-up and down are occupied. Using the labeling of the spin-orbital, it would be 10 (or 10). Similarly for the next two indices. Thus we have 0110 in the spin-orbital basis (why it is not 1010, I am not sure, and maybe this is due to convention). By the symmetry in Eq. 4, we have also 1001.

$1111$. The argument is essentially the same as above, except that now we are dealing with 1st spatial orb, with the spin-up and down labeled as 23. Thus we have 3223 and 2332for the spin-orb in the paper.

$0011$. Now with the mapping in Eq. 3, it is in fact 0110 for the spatial orb in the paper. The first two indices 10 means the 0th and 1st spatial orbs are occupied, and they could be either spin up and down. Thus we have $2\times2=4$ options: 20, 30, 21, 31 for the first two indices of the spin-orbs. Thus put it together, we have 0220, 0330, 1221, 1331 for the spin orbs. Again, I am not sure why we don't have 2020, maybe due to convention.

$1100$. This is essentially the same as above, where we realize that it is 1001 for the spatial orb in the paper. With the same logic, we have 2002, 3003, 2112, and 3113. These can essentially be obtained with the symmetry property in Eq. 4

$1010$. Ok, I am stuck here for now... Will update this after I figure it out.