# Problem using spin-restricted form of the second-quantized nonrelativistic Hamiltonian

I have a problem that confuses me a lot. The two-electron part of the electronic nonrelativistic Hamiltonian can be written $$\frac{1}{2}\sum_{pqrs} (pq|rs) [a^\dagger_{p\alpha}a^\dagger_{r\alpha}a_{s\alpha}a_{q\alpha} + a^\dagger_{p\alpha}a^\dagger_{r\beta}a_{s\beta}a_{q\alpha} + a^\dagger_{p\beta}a^\dagger_{r\alpha}a_{s\alpha}a_{q\beta} + a^\dagger_{p\beta}a^\dagger_{r\beta}a_{s\beta}a_{q\beta}],$$ where the two-electron integral is defined via $$(pq|rs) = \iint p^*(\mathbf r_1) q(\mathbf r_1) \frac{1}{r_{12}} r^*(\mathbf r_2) s(\mathbf r_2) d\mathbf r_1 d\mathbf r_2.$$ I will assume that the single particle wavefunctions (orbitals) are a complete set of eigenfunctions of an effective one-electron Hamiltonian (for example the Fock operator of Hartree-Fock theory) for the hydrogen molecule, written as the product of an orbital part and an eigenfunction of $S_z$ ($\alpha$ or $\beta$).

Then the ground state of the hydrogen molecule can approximately be written as $$|\Psi_0\rangle = a^\dagger_{1\alpha}a^\dagger_{1\beta}|\mathrm{vac}\rangle,$$ where 1 is the index of the lowest-energy orbital. So far for the introduction. My problem arises when I consider an alternative form of the two-electron part of the Hamiltonian, $$a^\dagger_{p\alpha}a^\dagger_{r\alpha}a_{s\alpha}a_{q\alpha} + a^\dagger_{p\alpha}a^\dagger_{r\beta}a_{s\beta}a_{q\alpha} + a^\dagger_{p\beta}a^\dagger_{r\alpha}a_{s\alpha}a_{q\beta} + a^\dagger_{p\beta}a^\dagger_{r\beta}a_{s\beta}a_{q\beta} = E_{pq}E_{rs} - \delta_{qr}E_{ps},$$ where the so-called singlet excitation operators are defined via $$E_{pq} = a^\dagger_{p\alpha}a_{q\alpha} + a^\dagger_{p\beta}a_{q\beta}.$$ When I calculate the expectation value of the two-electron part in the molecule's ground state with the first formula, I get the result $(11|11)$, which I know is correct. For the form using singlet excitation operators, however, I obtain $$\langle \Psi_0 | E_{pq} E_{rs} | \Psi_0\rangle = 2\delta_{qr} \delta_{s1} \delta_{p1}$$ and $$\langle \Psi_0 | E_{ps} | \Psi_0\rangle = 2 \delta_{s1} \delta_{p1}.$$ When combining these two parts, one sees that they exactly cancel, meaning that the total expectation value would vanish. Can anyone see where the error is? I have obviously become blind for it, I have calculated it several times forwards and backwards and always get to the same results.

Your problem is in the term $\langle \Psi_0| E_{pq}E_{rs}|\Psi_0\rangle$:
$$\langle \Psi_0| E_{pq}E_{rs}|\Psi_0\rangle = \langle 1\alpha 1\beta \;| \left( a_{p\alpha}^\dagger a_{q\alpha} + a_{p\beta}^\dagger a_{q\beta}\right) \left( a_{r\alpha}^\dagger a_{s\alpha} + a_{r\beta}^\dagger a_{s\beta}\right)|\; 1\alpha 1\beta \rangle =\\ = \delta_{s1} \langle 1\alpha 1\beta \;| \left( a_{p\alpha}^\dagger a_{q\alpha} + a_{p\beta}^\dagger a_{q\beta}\right) \left( |\;r\alpha\; 1\beta \rangle + |\;1\alpha \;r\beta\rangle \right) = \\ = \delta_{p1} \delta_{s1} \left( \langle q\alpha \;1\beta \;| + \langle 1\alpha \;q\beta \;|\right)\left( |\;r\alpha\; 1\beta \rangle + |\;1\alpha \;r\beta\rangle \right) = \\ = 2 \;\delta_{p1} \delta_{s1} \left( \delta_{qr} + \delta_{q1}\delta_{r1}\right)$$ etc, etc.
• I always unconsciously imagined q to be an excited orbital when seeing terms like $|q_\alpha 1_\beta\rangle$ and thought "then it must be orthogonal to the ket where the beta electron is excited". I hope this will no happen to me again. Thank you very much! – LLang Dec 9 '15 at 7:38