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Matrix Elements of the Hamiltonian and One- and Two-Electron Integrals

Now that we have a much more compact notation for the electronic Hamiltonian, we need to discuss how to evaluate matrix elements of this Hamiltonian in a basis of $N$-electron Slater determinants $\{ \vert I \rangle \}$. A matrix element between Slater determinants $\vert I \rangle $ and $\vert J \rangle $ will be written $\langle I \vert {\hat H} \vert J \rangle $, where we have dropped the ``el'' subscript on ${\hat H}$ because we will discuss the electronic Hamiltonian exclusively from this point. Because the Hamiltonian, like most operators in quantum mechanics, is a Hermitian operator, $\langle I \vert {\hat H} \vert J \rangle =
\langle J \vert {\hat H} \vert I \rangle ^*$.

It would seem that computing these matrix elements would be extremely tedious, because both $\vert I \rangle $ and $\vert J \rangle $ are Slater determinants which expand into $N!$ different products of $N$ orbitals, giving a total of $(N!)^2$ different terms! However, all but $N!$ of these terms go to zero because the orbitals are orthonormal. If we were computing a simple overlap like $\langle I \vert I \rangle$, all of the remaining $N!$ terms are identical and cancel the $1/N!$ in the denominator from the Slater determinant normalization factor.

If $\vert I \rangle $ and $\vert J \rangle $ differ, then any term which places electron $i$ in some orbital $\chi_p(i)$ in $\vert I \rangle $ and a different orbital $\chi_q(i)$ in $\vert J \rangle $ must also go to zero unless the integration over electron $i$ is not just a simple overlap integral like $\int \chi_p^*(i)
\chi_q(i) d{\mathbf x}_i$, but involves some operator $h(i)$ or $v(i,j)$. Since a particular operator $h(i)$ can only affect one coordinate $i$, all the other spin orbitals for other electrons $j$ must be identical, or the integration will go to zero (orbital orthonormality). Hence, $h(i)$ allows, at most, only one spin orbital to be different in $\vert I \rangle $ and $\vert J \rangle $ for $\langle I \vert h(i) \vert J \rangle $ to be nonzero. Integration over the other electron coordinates $i \neq j$ will give factors of one, resulting in a single integral over a single set of electron coordinates ${\mathbf x}_i$ for electron $i$, which is called a one-electron integral,

\langle p \vert h(1) \vert q \rangle = \langle p \vert h \ve...
...hi_p^*({\mathbf x}_1)
h({\mathbf x}_1) \chi_q({\mathbf x}_1),
\end{displaymath} (35)

where $\chi_p$ and $\chi_q$ are the two orbitals which is allowed to be different.

Likewise, the operator $v(i,j)$ allows up to two orbitals to be different in Slater determinants $\vert I \rangle $ and $\vert J \rangle $ before matrix element $\langle I \vert {\hat H} \vert J \rangle $ goes to zero. Integration over other electron coordinates gives factors of one, leading to an integral over two electronic coordinates only, a two-electron integral,

\langle pq \vert v(1,2) \vert rs \rangle = \langle pq \vert ...
\chi_r({\mathbf x}_1) \chi_s({\mathbf x}_2).
\end{displaymath} (36)

There are other ways to write this integral. The form above, which places the complex conjugate terms on the left, is called ``physicist's notation,'' and is usually written in a bra-ket form.

There is a very simple set of rules, called Slater's rules, which explain how to write matrix elements $\langle I \vert {\hat H} \vert J \rangle $ in terms of these one- and two-electron integrals. Most of quantum chemistry is derived in terms of these quantities.

next up previous
Next: Bibliography Up: intro_estruc Previous: Simplified Notation for the
David Sherrill 2003-08-07