I generated an 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 pictures for their definitions and results which are taken from here). 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.

enter image description here

I have checked following references:

  • This one, 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!

  • 1
    $\begingroup$ +1. I believe they are literally just the spin orbital indices. Might be able to answer in more detail after the Raptors game. $\endgroup$ – Nike Dattani Aug 6 at 1:28
  • $\begingroup$ @nikedattani, Thank you for editing the question. I am in fact not sure if the output is spatial overlap integral or spin overlap integral. I feel it is the latter, but I was not able to reproduce the table attached. Any help will be appreciated. $\endgroup$ – fagd Aug 6 at 3:18
  • $\begingroup$ The orbitals have to be spin orbitals, because there's 4 of them. If they were spatial orbitals there would be only 2 of them. $\endgroup$ – Nike Dattani Aug 6 at 3:20

"I was able to guess most of results, except the ones with value 0.663472."

I am able to confirm that your integrals are correct even for the 0.663472 case, because by calculating them myself in MOLPRO (a completely different program from what you used), I get the same indices of 2 2 1 1:

 &FCI NORB=  2,NELEC=  2,MS2= 0, ORBSYM=1,5,ISYM=0,
  0.6744931033260081E+00   1   1   1   1  
  0.6634720448605567E+00   2   2   1   1   <---------------
  0.6973979494693358E+00   2   2   2   2  
  0.1812875358123322E+00   2   1   2   1  
 -0.1252477303982147E+01   1   1   0   0  
 -0.4759344611440753E+00   2   2   0   0  
  0.7137758743754461E+00   0   0   0   0

Hence the reason why the indices are different in the paper you're looking at, would be due to symmetry considerations.

Now that I've confirmed that your indices are indeed correct, I would like you to work out how the known symmetry relations for integrals lead from our indices to those shown in the paper (if I do that part for you, then it would be like solving a "homework question"). The most important way I could help you was to give you the confidence that your indices are indeed correct.

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  • $\begingroup$ Sure. I would be happy to do that, and I have spent the whole day for that. The symmetry I know is <ij|kl> = <ji|lk>. And for the indices 1111, the first two has to be for spin up and down, and similarly for the latter two. That means it should corresponds to 0110 (It is not 1001 for convention I guess), where 0, 1 are for spin up and down in the 1st spin-orb. This is the same as 1001 using the symmetry above. Similarly 2222 would be 2332 and 3223. The 1100 and 2200 is easy. And the 1122, 2211, 2121 I am not quite sure. $\endgroup$ – fagd Aug 6 at 3:24
  • $\begingroup$ Don't forget that $\langle ij | kl \rangle = \langle kj | il \rangle$, so 1221 in the paper is the same as 2211 in our output files :) $\endgroup$ – Nike Dattani Aug 6 at 3:25
  • $\begingroup$ How did you arrive at βŸ¨π‘–π‘—|π‘˜π‘™βŸ©=βŸ¨π‘˜π‘—|π‘–π‘™βŸ©? If I use the definition of βŸ¨π‘–π‘—|π‘˜π‘™βŸ© in theochem.github.io/horton/2.0.2/user_hamiltonian_io.html, I can only have βŸ¨π‘–π‘—|π‘˜π‘™βŸ©=βŸ¨π‘—i|𝑙k⟩, if I keep x_{1,2} not touched. Please let me know if this is just a simple algebra. I am doing the checking now. $\endgroup$ – fagd Aug 6 at 3:35
  • $\begingroup$ You also have to enforce that if the integral is real-valued (meaning it has no imaginary parts), then the integral equals its complex conjugate. $\endgroup$ – Nike Dattani Aug 6 at 3:40
  • 1
    $\begingroup$ interesting, let me think abt it... $\endgroup$ – fagd Aug 6 at 3:56

I would like to post a partial answer that I have for now.

For the indices ijkl in the FCIDUMP file, it corresponds to the integral (see https://theochem.github.io/horton/2.0.2/user_hamiltonian_io.html)

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

while the indices in the paper, it is

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

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

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

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.

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  • $\begingroup$ Could you help me to resolve the very last case? Also there are some questions why certain cases are not included. Your help will be very important for my subsequent research. Thank you! $\endgroup$ – fagd Aug 7 at 2:59
  • $\begingroup$ So the 3003 is fine now and you're wondering about an integral with a different value? $\endgroup$ – Nike Dattani Aug 7 at 12:37
  • $\begingroup$ Yes, the one with 1.81287. Thanks for the help! $\endgroup$ – fagd Aug 7 at 16:01
  • $\begingroup$ Seems the last digit should be 8 not 7. Anyway, what is the reason why you are so curious about matching up these integrals? This job has been carefully done by several generations of quantum chemists, and now we have tools (like the ones you used and I used here) to generate the integrals for us (and they all agree with each other). Your quantum computing project seems to be getting delayed by this distraction of trying to figure out the inner workings of quantum chemistry software that print only 7 integrals rather than all 256 of them. Real molecules (not H2/STO3G) will be 90GB of integrals $\endgroup$ – Nike Dattani Aug 7 at 16:09
  • $\begingroup$ No, I am not interested in how these integrals are calculated, but I will need to use them as the coefficients in the second quantized Hamiltonian. At the end of the day, if anyone wants to use these results, they will need to know which is which, isn't it? And I feel this should be straightforward for anyone who has been in this field long enough. So could you please tell me the answer if you know it? $\endgroup$ – fagd Aug 7 at 16:39

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