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Why are Coupled-Cluster and MBPT Energies not Variational?

  Electron correlation methods other than CI may not be variational. For example, consider the coupled-cluster energy expression

 \begin{displaymath}
E = \frac{\langle \Phi_0 \vert e^{-\hat{T}} H e^{\hat{T}} \vert \Phi_0 \rangle}
{\langle \Phi_0 \vert \Phi_0 \rangle}
\end{displaymath} (tex2html_deferredtex2html_deferred3.tex2html_deferred14 E = _0 | e^-T H e^T | _0 _0 | _0 )

If the operator $e^{\hat{T}}$ is not trunctated, then we know that $e^{\hat{T}} \vert \Phi_0 \rangle = \vert \Psi_0 \rangle$. Generally, however, the operator is truncated. Let us define $e^{\hat{T'}} \vert \Phi_0 \rangle =
\vert \Theta_A \rangle$ for our truncated $\hat{T'}$. Now define $\langle \Phi_0 \vert e^{-\hat{T'}} = \langle \Theta_B \vert$. Note that in general $\vert \Theta_B \rangle \neq \vert \Theta_A \rangle$, which would have occured had we used $\langle \Phi_0 \vert(e^{\hat{T'}})^{\dag }$ on the left. Then the energy expression is

 \begin{displaymath}
E = \frac{\langle \Theta_B \vert H \vert \Theta_A \rangle}{\langle \Phi_0 \vert \Phi_0 \rangle}
\end{displaymath} (tex2html_deferredtex2html_deferred3.tex2html_deferred15 E = _B | H | _A _0 | _0 )

which, after expansion over the complete set of eigenvectors, becomes

 \begin{displaymath}
E = \frac{\sum_{ij} c_i^{*} d_j \langle \Psi_i \vert H \vert \Psi_j \rangle}
{\langle \Phi_0 \vert \Phi_0 \rangle}
\end{displaymath} (tex2html_deferredtex2html_deferred3.tex2html_deferred16 E = _ij c_i^* d_j _i | H | _j _0 | _0 )

This simplifies to

\begin{displaymath}E = \frac{\sum_i c_i^{*} d_i {\cal E}_i}{\langle \Phi_0 \vert \Phi_0 \rangle}
\end{displaymath} (tex2html_deferredtex2html_deferred3.tex2html_deferred17E = _i c_i^* d_i E_i_0 | _0 )

At this point we can go no farther, because the terms ci* di may be negative, in contrast to the situation in equation (3.12).

For completeness, we also show that MBPT energies are not variational. The nth order MBPT wavefunction may be written [12] as

\begin{displaymath}\vert \Theta_{\rm MBPT}^{(n)} \rangle = \vert \Phi_0 \rangle ...
...Phi_0 \vert)}
{E_0 - H_0} \right] ^k \vert \Phi_0 \rangle_{L}
\end{displaymath} (tex2html_deferredtex2html_deferred3.tex2html_deferred18| _MBPT^(n) = | _0 + _k=1^n [ V (1 - | _0 _0 |) E_0 - H_0 ] ^k | _0 _L )

where the sum is over ``linked diagrams'' only. The nth order energy is then given by

\begin{displaymath}E_{\rm MBPT}^{(n)} = \langle \Phi_0 \vert \hat{H} \vert \Theta_{\rm MBPT}^{(n-1)} \rangle
\end{displaymath} (tex2html_deferredtex2html_deferred3.tex2html_deferred19E_MBPT^(n) = _0 | H | _MBPT^(n-1) )

Since this integral is not symmetric, the energy is not variational. Only the first-order perturbation theory energy (which is also the Hartree-Fock energy) is variational, since it uses $\vert \Theta_{\rm MBPT}^{(0)} \rangle = \vert \Phi_0 \rangle$.    


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Next: Application of the Variational Up: The Variational Theorem Previous: Variational Theorem for the
C. David Sherrill
2000-04-18