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I am learning quantum chemistry at the moment and I'm trying to understand the Hamiltonian generated by the OpenFermion package. I'm now stuck at understanding how openfermion calculates the coefficients in second quantization Hamiltonians.

Take the for Hydrogen molecule for example. My understanding is that the coefficients

  • $h_{00}$, represents electron 1 in 1s orbit with spin up.
  • $h_{11}$, represents electron 1 in 1s orbit with spin down.
  • $h_{22}$, represents electron 2 in 1s orbit with spin up.
  • $h_{33}$, represents electron 2 in 1s orbit with spin down.

As the wavefunction of 1s orbits takes the form

$$\psi_{1s}(r) = \frac{1}{\sqrt{\pi}}e^{-r}$$

The $h_{ii}$ should be

\begin{equation} 2\pi \int_0^{\infty}\int_0^{\pi}\psi_{1s}(r)\left(-\frac{1}{2}\nabla^2-\frac{1}{r}-\frac{1}{\sqrt{r^2+R^2-2r R\cos\theta}}\right)\psi_{1s}(r)r^2 \sin\theta d\theta dr = -1.07123 \end{equation} when $R = 0.74$ Bohr radius, which does not match with openfermion's \begin{align} h_{00} &= h_{11} = -1.2524635735648988\\ h_{22} &= h_{33} = -0.47594871522096416 \end{align}

Can anyone tell me why I am wrong?

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  • $\begingroup$ Is there spin down in $h_{22}$ ? Is it not spin up ? $\endgroup$
    – Maurice
    Commented Nov 17, 2022 at 17:10
  • $\begingroup$ yeah it should be spin up. I just fixed the typo $\endgroup$
    – QF2QP
    Commented Nov 17, 2022 at 17:14
  • $\begingroup$ Isn't this just basically considering two separate H atoms, not a hydrogen molecule? I'm quite inexperienced in this, but do we not have to consider the attraction of two electrons to both separate nuclei? Also, why is $h_{11}$ not equal to $h_{33}$ as both have the same spin and the assignment of which electron is electron 1 or 2 is arbitrary. $\endgroup$
    – M.L
    Commented Nov 17, 2022 at 19:56

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