Symmetry Restrictions on the CI Space

In this section we discuss some general considerations concerning which N-electron basis functions should be included in the CI space (given that, in the general case, there are too many for us to include all of them). Certainly if we can find a class of N-electron functions which rigorously have zero Hamiltonian matrix elements with the desired CI wavefunction, then none of these basis functions will contribute at all to our approximate wavefunction and they should not be included in the CI space; the Hamiltonian will be block diagonal, and all the functions in this class will contribute to the wrong block. We will now prove the following: consider an operator $\hat{A}$ which commutes with $\hat{H}$. If

$${\hat A} \vert \Phi_1 \rangle = a_1 \vert \Phi_1 \rangle$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred1 Â | _1 = a_1 | _1 )

and

$${\hat A} \vert \Phi_2 \rangle = a_2 \vert \Phi_2 \rangle, a_1 \neq a_2$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred2 Â | _2 = a_2 | _2 , a_1 a_2 )

then

$$\langle \Phi_1 \vert \hat{H} \vert \Phi_2 \rangle = 0$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred3 _1 | H | _2 = 0 )

First we show that $\hat{H} \vert \Phi_2 \rangle$ is an eigenfunction of $\hat{A}$with eigenvalue a₂. Define

$$\hat{H} \vert \Phi_2 \rangle = \vert \Phi_2' \rangle$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred4 H | _2 = | _2' )

Now apply $\hat{A}$ to $\vert \Phi_2' \rangle$

 

$${tex2html_deferred}{{tex2html_deferred}4.{tex2html_deferred}5}\hat{A} \left[ \hat{H} \vert \Phi_2 \rangle \right] \hat{A} \hat{H} \vert \Phi_2 \rangle$$ = (tex2html_deferredtex2html_deferred4.tex2html_deferred5Â [ H | _2 ] & = & Â H | _2 )

$$\hat{H} \hat{A} \vert \Phi_2 \rangle$$ =

$$\hat{H} a_2 \vert \Phi_2 \rangle$$ =

$$a_2 \left[ \hat{H} \vert \Phi_2 \rangle \right]$$ =

Where we have used the given that $\hat{A} \hat{H} = \hat{H} \hat{A}$. We may now write

$$\hat{A} \vert \Phi_2' \rangle = a_2 \vert \Phi_2' \rangle$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred6 Â | _2' = a_2 | _2' )

Now consider again equation (4.1). If we take the adjoint of this equation we obtain

$$\langle \Phi_1 \vert \hat{A}^{\dag } = \langle \Phi_1 \vert a_1^{*}$$

Now use the fact that $\hat{A} = \hat{A}^{\dag }$ (we assumed $\hat{A}$ was Hermitian) and that the eigenvalues of a Hermitian operator are real. This yields

$$\langle \Phi_1 \vert \hat{A} = \langle \Phi_1 \vert a_1$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred8 _1 | Â = _1 | a_1 )

Multiply on the right by $\vert \Phi_2' \rangle$

$$\langle \Phi_1 \vert \hat{A} \vert \Phi_2' \rangle = a_1 \langle \Phi_1 \vert \Phi_2' \rangle$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred9 _1 | Â | _2' = a_1 _1 | _2' )

Now multiply equation (4.6) on the left by $\langle \Phi_1 \vert$ to obtain

$$\langle \Phi_1 \vert \hat{A} \vert \Phi_2' \rangle = a_2 \langle \Phi_1 \vert \Phi_2' \rangle$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred10 _1 | Â | _2' = a_2 _1 | _2' )

If we subtract equation (4.10) from equation (4.9) we arrive at

$$(a_1 - a_2) \langle \Phi_1 \vert \Phi_2' \rangle = 0$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred11 (a_1 - a_2) _1 | _2' = 0 )

Since we assumed $a_1 \neq a_2$, then $\langle \Phi_1 \vert \Phi_2' \rangle = 0$. Recalling the definition of $\vert \Phi_2' \rangle$, we have

$$\langle \Phi_1 \vert \hat{H} \vert \Phi_2 \rangle = 0$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred12 _1 | H | _2 = 0 )

which was to be proven. Thus if our desired wavefunction is an eigenfunction of some Hermitian operator that commutes with the Hamiltonian, our CI space need not include those N-electron functions which are eigenfunctions of this operator with different eigenvalues. As an example, consider the spin angular momentum operator $\hat{S}^2$. If we want to solve for a state $\vert \Psi \rangle$ of spin S, then we know

$$\hat{S}^2 \vert \Psi \rangle = S(S+1) \vert \Psi \rangle$$ (tex2html_deferredtex2html_deferred4.tex2html_deferred13S^2 | = S(S+1) | )

and any basis function of a different spin can be excluded from the CI. Slater determinants are generally not eigenfunctions of $\hat{S}^2$. However, we can take linear combinations of Slater determinants so that we do have eigenfunctions of $\hat{S}^2$; such functions are generally called configuration state functions, or CSF's. The advantage of using CSF's is that we can throw out all functions with the wrong eigenvalue S--they contribute to another, noninteracting block of the Hamiltonian matrix. This reasoning also applies to symmetry operations of point groups, such that we can throw away any N-electron basis functions (whether determinants or CSF's) which have the wrong irreducible representation. We can also restrict the basis functions according to their eigenvalues with respect to the operator S_z. For a triplet state, we can perform the calculation using basis functions which have M_s = -1, 0, or 1. If we were to include basis functions of all these values of M_s, we would obtain a triply-degenerate answer--as one should expect!³