How to interpret the fci dump file

I generated a fcidump file, which is copied below

6.74493103326006093745E-01 1 1 1 1
6.63472044860555665302E-01 1 1 2 2
6.63472044860555665302E-01 2 2 1 1
6.97397949820408036281E-01 2 2 2 2
1.81287535812332034624E-01 2 1 2 1
-1.25247730398215462166E+00 1 1 0 0
-4.75934461144127241017E-01 2 2 0 0
7.43077168397780152276E-01 0 0 0 0

I am not really sure what the indices mean. In particular, I would like to link it to the following definition of the one and two electron integrals (see attached picture 1). It is not obvious to me how to match the indices. The result I am trying to obtain is in the second attached figure. I was able to guess most of results, except the ones with value 0.663472.

I have checked following references, which is not really helpful yet. I also found this link describing the fcidump file explicitly, it is close to the integrals I want, but not quite exactly.

Any help is really appreciated. Thanks!

The indices in what you linked are spin orbital indices, and the indices in the FCIDUMP file are spatial orbital indices. So for example, indices 0 and 1 in your screenshot correspond to the alpha and beta versions of orbital 1 in the FCIDUMP file.

Yes, this part I understood. So for example, for 1111 in the fcidump file, it would be 0110 in the paper where 0,1 are for spin up and down in the 1st spin-orb, and this is equal to 1001 by symmetry. Similarly, 2222 in the fcidump file corresponds to 2332 in the paper. But how about 1122? I don’t understand it. From the logic above, it should be sth like 1032, but this does not seems to be correct. Could you help me on this? Thanks!

You’re also missing differences in indexing. In this paper’s convention, the outermost indices are for the same electron. In FCIDUMP convention (see Sec. of the page you linked), the outermost indices are for different electrons.

This issue seems to be only tangentially related to Psi, so I’m going to be less responsive.

Thank you jmisiewicz, that is very helpful. With that observation, I now realize that the ijkl index in the FCIDUMP convention is really iklj in the paper. With that, for 2211 in the FCIDUMP file, it is really 2112 for the spatial orb in the paper. Then to switch to spin-orb, we have 0220, 0330, 1331 and 1221, which is nice. But I still have problems for 2121 in the FCIDUMP file, which is 2211 for the spatial orb in the paper. First, how to decide if this corresponds to 2310 or 3210? I assume the rest can be obtained via symmetry? Also I don’t see how to get 2002 in the very last row in the table.

Thank you very much for the help! I will write a complete post once this is sorted out.

I don’t understand where you’re stuck.

For all of these paper integrals, perform the spin-integration (end here if the integral is zero), convert from paper’s spinorbital indices to FCIDUMP’s spatialorbital indices, move the order of the indices to the FCIDUMP table’s ordering, and look up the relevant integral in the FCIDUMP table.

I would like to post a partial answer that I have for now. (sorry I am not sure how to type in equation here, this is copied from my answer in here

For the indices ijkl in the FCIDUMP file, it corresponds to the integral (see

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

while the indices in the paper, it is

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

I have ignored some other factors which are not relevant here.
We see that there are two differences, one is the order of the indices, and the other is the $x_{1,2}$ arguments. Thus we have the following mapping

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

Further, with the second equation above, we have the following symmetry property

ijkl = jilk = lkji

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

Next, the very last row is the nuclear repulsion energy, which is not of our interest here. The 2nd and 3rd to last rows are the 1-body integral, which are pretty easy to understand. So let’s focus on the 2-body integral.

$0000$. 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.

$0011$. 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.