I have now used the equation in the book chapter to get the energy:
E = < Hartree Fock | exp(-T) H exp(T) | Hartree Fock>
It does work for H2 not for H4 and larger molecules probably due to wrong signs or missing information how to convert cc amplitudes from MO to the proper spin orbitals.
I need to get the T operator in spin orbital picture (still in second quantization) using my numbering. For H2 this works easy as TIjAb 0 0 0 0 in spin orbital picture can only mean that
T = 0.1153404926 a_0 a_1 a_2^dagger a_3^dagger
With some trying I figured out there needs to be a minus sign.
Once I have the T operator in proper spin orbital numbering, I use the Jordan Wigner transformation to map fermions to distinguishable spin 1/2 particles. Hence all creation and annihilation operators transform to sums of tensor products of Pauli Matrices. I can therefore numerically build a matrix representation for T and H and exponentiate the matrix T and apply the energy formula. Yes this is exponentially costly but for small stuff I don’t mind. It gives me a fast and clean way to test if the T operator in second quantization using spin orbital numbering is correct. It is for H2 except the sign. All my hamiltonians are already transformed by Jordan Wigner and fixed. So any wrong sign in T will change the energy…
For H4 I have more difficult TIjAb so I am not sure I am converting them correctly to spin orbital operators, e.g.
TIjAb
0 1 0 1 -0.0468249907
This would mean a term a_0 a_3 a_4^dagger a_7^dagger -0.0468249907
Do I need to add symmetric terms manually? I.e. I can also expect a transition TIJAB and Tijab? Does it mean I need to add the terms
a_0 a_2 a_4^dagger a_6^dagger -0.0468249907
and
a_1 a_3 a_5^dagger a_7^dagger -0.0468249907
What about the coefficients ? Now I have three terms each with the same strength, do I need to adjust the coefficient by a factor?
H2 worked because a term
TIjAb 0 0 0 0
Can only mean that there is one down and one up electron involved, so I didn’t need to think about adding extra terms.