Next: Bibliography
Up: Quantum Chemistry Lecture Notes
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}](img1.png) |
(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}](img2.png) |
(2) |
where the sums run over all spin orbitals
.
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}](img4.png) |
(3) |
Now integrate over spin, assuming that spatial orbitals are constrained
to be identical for
and
spins. A sum over all 2K spin
orbitals can be split up into two sums, one over K orbitals with
spin, and one over K orbitals with
spin. Symbolically, this is
![\begin{displaymath}\sum_{a}^{2K} = \sum_{a}^{K} + \sum_{\bar{a}}^{K}
\end{displaymath}](img7.png) |
(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}](img8.png) |
(5) |
After integrating over spin, this becomes
![\begin{displaymath}\hat{H}_{\rm one} =
\sum_{pq}^{K} (p\vert h\vert q)
\lbra...
... a_{q \alpha}
+ a_{p \beta }^{\dagger} a_{q \beta}
\rbrace}
\end{displaymath}](img9.png) |
(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...
...er} a_{r \beta }^{\dagger} a_{s \beta } a_{q \beta }
\rbrace}
\end{displaymath}](img10.png) |
(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}](img11.png) |
(8) |
and we swap the order of
and
,
introducing a minus sign. This yields
![\begin{displaymath}\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}
\end{displaymath}](img14.png) |
(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} =
\delta_{ij}
\end{displaymath}](img15.png) |
(10) |
This relation allows us to swap the aq and
in each term,
to give
Now we observe that
and
can both be written
,
and also that
and
are both 0.
This simplifies our equation to
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}](img27.png) |
(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}
\right)
\end{displaymath}](img28.png) |
(14) |
This is the Hamiltonian in terms of the unitary group generators
[3].
Next: Bibliography
Up: Quantum Chemistry Lecture Notes
C. David Sherrill
2000-04-18