For all methods, the expressions for the one- and two-body density matrices can be tested by generating the explicit expression for the CC-lagrangian
$$L_{\text{CC}} = \langle \Phi_0 | (1+\Lambda)e^{-T}He^{T} | \Phi0 \rangle,$$
and compare it to the value computed by taking the trace over the density matrices and Hamiltonian matrix elements
$$L{\text{CC}} = \sum_{pq} h^p_q \gamma^qp + \frac{1}{4} \sum{pqrs} u^{pq}_{rs} \Gamma^{rs}_{pq}.$$
Additionally, at convergence, we should have that $L{\text{CC}} = E{\text{CC}}$ where $E{\text{CC}}$ is the CC-energy: $$E{\text{CC}} = \langle \Phi_0 | e^{-T}He^{T} | \Phi_0\rangle.$$
For all methods, the expressions for the one- and two-body density matrices can be tested by generating the explicit expression for the CC-lagrangian $$L_{\text{CC}} = \langle \Phi_0 | (1+\Lambda)e^{-T}He^{T} | \Phi0 \rangle,$$ and compare it to the value computed by taking the trace over the density matrices and Hamiltonian matrix elements $$L{\text{CC}} = \sum_{pq} h^p_q \gamma^qp + \frac{1}{4} \sum{pqrs} u^{pq}_{rs} \Gamma^{rs}_{pq}.$$
Additionally, at convergence, we should have that $L{\text{CC}} = E{\text{CC}}$ where $E{\text{CC}}$ is the CC-energy: $$E{\text{CC}} = \langle \Phi_0 | e^{-T}He^{T} | \Phi_0\rangle.$$