Truncation of the Exponential Ansatz

Recall that the exponentiated operator may be expanded in a power series as

$$e^{\hat{T}} = 1 + \hat{T} + \frac{\hat{T}^2}{2!} + \frac{\hat{T}^3}{3!} + \ldots .$$ (3.5)

Inserting this into the energy expression Eq. [3.3] we obtain

$$\langle \Phi_0 \vert\hat{H}(1 + \hat{T} + \frac{\hat{T}^2}{2!} + \frac{\hat{T}^3}{3!} + \ldots)\vert \Phi_0 \rangle = E,$$ (3.6)

which becomes, after distributing terms,

$$\langle \Phi_0 \vert\hat{H}\vert \Phi_0 \rangle + \langle \Ph... ...t\hat{H}\frac{\hat{T}^3}{3!}\vert \Phi_0 \rangle + \ldots = E.$$ (3.7)

Note that $\hat{H}$ is at most a two-particle operator and that $\hat{T}$ is at least a one-particle excitation operator. Then, assuming that the reference wavefunction is a single determinant constructed from a set of one-electron functions, Slater's rules[82] state that matrix elements of the Hamiltonian between determinants that differ by more than two orbitals are zero. Thus, the fourth term on the left-hand side of the above equation contains, at the least, threefold excitations, and, as a result, that matrix element (and all higher-order elements) necessarily vanish. The energy equation then simplifies to

$$\langle \Phi_0 \vert \hat{H} \vert \Phi_0 \rangle + \langle \... ...hi_0 \vert\hat{H}\frac{\hat{T}^2}{2!}\vert \Phi_0 \rangle = E.$$ (3.8)

This is the natural truncation of the coupled cluster energy equation; an analogous phenomenon occurs for the amplitude equation (Eq. [3.4]). This truncation depends only on the form of $\hat{H}$ and not on that of $\hat{T}$ or on the number of electrons. Equation [3.8] is correct even if $\hat{T}$ is truncated to a particular excitation level.