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),

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

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

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

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

$$\sum_{a}^{2K} = \sum_{a}^{K} + \sum_{\bar{a}}^{K}$$ (4)

The one-electron part of the Hamiltonian becomes

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

After integrating over spin, this becomes

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

The two-electron term can be expanded similarly to give

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

Now we make use of the anticommutation relation

$$\{ a_{j}, a_{i} \} = a_{j} a_{i} + a_{i} a_{j} = 0$$ (8)

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

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

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

$$\{ a_{i}, a_{j}^{\dagger} \} = a_{i} a_{j}^{\dagger} + a_{j}^{\dagger} a_{i} = \delta_{ij}$$ (10)

This relation allows us to swap the a_q and $a_{r}^{\dagger}$ in each term, to give

$$\hat{H}_{\rm two} \frac{1}{2} \sum_{pqrs}^{K} (pq\vert rs) \left[ a_{p \alpha}^{\da... ...eta } - \delta_{q \alpha, r \beta } a_{p \alpha}^{\dagger} a_{s \beta } \right.$$ =

$$\left. a_{p \beta }^{\dagger} a_{q \beta } a_{r \alpha}^{\dagger}... ...beta } - \delta_{q \beta, r \beta } a_{p \beta }^{\dagger} a_{s \beta } \right]$$ + (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

$$\hat{H}_{\rm two} \frac{1}{2} \sum_{pqrs}^{K} (pq\vert rs) \left[ a_{p \alpha}^{\da... ...a_{p \beta }^{\dagger} a_{q \beta } a_{r \beta }^{\dagger} a_{s \beta } \right.$$ =

$$\left. \delta_{qr} a_{p \alpha}^{\dagger} a_{s \alpha} - \delta_{qr} a_{p \beta }^{\dagger} a_{s \beta } \right]$$ - (12)

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

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

and the Hamiltonian becomes

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

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