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

$$E = \frac{\langle \Phi_0 \vert e^{-\hat{T}} H e^{\hat{T}} \vert \Phi_0 \rangle} {\langle \Phi_0 \vert \Phi_0 \rangle}$$ (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

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

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

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

This simplifies to

$$E = \frac{\sum_i c_i^{*} d_i {\cal E}_i}{\langle \Phi_0 \vert \Phi_0 \rangle}$$ (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 c_i^* d_i 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

$$\vert \Theta_{\rm MBPT}^{(n)} \rangle = \vert \Phi_0 \rangle ... ...Phi_0 \vert)} {E_0 - H_0} \right] ^k \vert \Phi_0 \rangle_{L}$$ (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

$$E_{\rm MBPT}^{(n)} = \langle \Phi_0 \vert \hat{H} \vert \Theta_{\rm MBPT}^{(n-1)} \rangle$$ (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 $\vert \Theta_{\rm MBPT}^{(0)} \rangle = \vert \Phi_0 \rangle$.