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Next: Bibliography Up: Quantum Chemistry Lecture Notes

Derivation of the Hamiltonian in terms of
Unitary Group Generators

C. David Sherrill
School of Chemistry and Biochemistry
Georgia Institute of Technology

March 1994

We begin with the second-quantized form of the one- and two-electron operators (see Szabo and Ostlund [1], p. 95),

\begin{displaymath}{\cal O}_1 = \sum_{ij}^{2K} \langle i \vert h \vert j \rangle a_{i}^{\dagger} a_{j}
\end{displaymath} (1)

\begin{displaymath}{\cal O}_2 = \frac{1}{2} \sum_{ijkl}^{2K} \langle ij \vert kl \rangle a_{i}^{\dagger}
a_{j}^{\dagger} a_{l} a_{k}
\end{displaymath} (2)

where the sums run over all spin orbitals $\{ \chi_{i} \}$. Thus the Hamiltonian is

\begin{displaymath}\hat{H} = \sum_{pq}^{2K} a_{p}^{\dagger} a_{q} [p\vert h\vert...
...{2K} a_{p}^{\dagger} a_{r}^{\dagger} a_{s} a_{q} [pq\vert rs]
\end{displaymath} (3)

Now integrate over spin, assuming that spatial orbitals are constrained to be identical for $\alpha$ and $\beta$ spins. A sum over all 2K spin orbitals can be split up into two sums, one over K orbitals with $\alpha$spin, and one over K orbitals with $\beta$ spin. Symbolically, this is

\begin{displaymath}\sum_{a}^{2K} = \sum_{a}^{K} + \sum_{\bar{a}}^{K}
\end{displaymath} (4)

The one-electron part of the Hamiltonian becomes

\begin{displaymath}\hat{H}_{\rm one} = \sum_{pq}^{K}
[p \vert h\vert q] a_{p \...
...\bar{p}\vert h\vert\bar{q}] a_{p \beta }^{\dagger} a_{q \beta}
\end{displaymath} (5)

After integrating over spin, this becomes

\begin{displaymath}\hat{H}_{\rm one} =
\sum_{pq}^{K} (p\vert h\vert q)
... a_{q \alpha}
+ a_{p \beta }^{\dagger} a_{q \beta}
\end{displaymath} (6)

The two-electron term can be expanded similarly to give

\begin{displaymath}\hat{H}_{\rm two} =
\frac{1}{2} \sum_{pqrs}^{K} (pq\vert rs...} a_{r \beta }^{\dagger} a_{s \beta } a_{q \beta }
\end{displaymath} (7)

Now we make use of the anticommutation relation

\begin{displaymath}\{ a_{j}, a_{i} \} = a_{j} a_{i} + a_{i} a_{j} = 0
\end{displaymath} (8)

and we swap the order of $a_{s \alpha}$ and $a_{q \alpha}$, introducing a minus sign. This yields

\begin{displaymath}\hat{H}_{\rm two} =
- \frac{1}{2} \sum_{pqrs}^{K} (pq\vert r...} a_{r \beta }^{\dagger} a_{q \beta } a_{s \beta }
\end{displaymath} (9)

Now we use the anticommutation relation between a creation and an annihilation operator, which is

\begin{displaymath}\{ a_{i}, a_{j}^{\dagger} \} = a_{i} a_{j}^{\dagger} + a_{j}^{\dagger} a_{i} =
\end{displaymath} (10)

This relation allows us to swap the aq and $a_{r}^{\dagger}$ in each term, to give
$\displaystyle \hat{H}_{\rm two}$ = $\displaystyle \frac{1}{2} \sum_{pqrs}^{K} (pq\vert rs) \left[
a_{p \alpha}^{\da...
...eta }
- \delta_{q \alpha, r \beta } a_{p \alpha}^{\dagger} a_{s \beta }
  + $\displaystyle \left.
a_{p \beta }^{\dagger} a_{q \beta } a_{r \alpha}^{\dagger}...
...beta }
- \delta_{q \beta, r \beta } a_{p \beta }^{\dagger} a_{s \beta }
\right]$ (11)

Now we observe that $\delta_{q \alpha, r \alpha}$ and $\delta_{q \beta, r \beta}$ can both be written $\delta_{qr}$, and also that $\delta_{q \alpha, r \beta }$ and $\delta_{q \beta, r \alpha}$ are both 0. This simplifies our equation to
$\displaystyle \hat{H}_{\rm two}$ = $\displaystyle \frac{1}{2} \sum_{pqrs}^{K} (pq\vert rs) \left[
a_{p \alpha}^{\da...
...a_{p \beta }^{\dagger} a_{q \beta } a_{r \beta }^{\dagger} a_{s \beta }
  - $\displaystyle \left.
\delta_{qr} a_{p \alpha}^{\dagger} a_{s \alpha}
- \delta_{qr} a_{p \beta }^{\dagger} a_{s \beta }
\right]$ (12)

Now we introduce the unitary group generators, which we write as [2]

\begin{displaymath}\hat{E}_{ij} = a_{i \alpha}^{\dagger} a_{j \alpha} +
a_{i \beta}^{\dagger} a_{j \beta}
\end{displaymath} (13)

and the Hamiltonian becomes

\begin{displaymath}\hat{H} =
\sum_{pq}^{K} (p\vert h\vert q) \hat{E}_{pq}
+ \...
... \hat{E}_{pq} \hat{E}_{rs} - \delta_{qr} \hat{E}_{ps}
\end{displaymath} (14)

This is the Hamiltonian in terms of the unitary group generators [3].

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Next: Bibliography Up: Quantum Chemistry Lecture Notes
C. David Sherrill