to arrive at Φ GW xc = − 1 2 ∞ ∑ j =1 ∫ ∞ −∞ dω 2 πi · 1 j tr $^{{}$[ V ¯ P ( ω )] j $^{}}$. j = 1 term : − 1 2 ∫ ∞ −∞ dω 2 πi tr $^{{}$V ¯ P ( ω ) } = i 2 tr $^{{}$V ∫ ∞ −∞ dω 2 π ¯ P ( ω ) } = i 2 tr $^{{}$V ¯ P ( t = 0) $^{}}$. The j = 1 term in the series is Writing the trace explicitly the x basis and using Eq. (21) shows that the j = 1 term is just the Fock exchange energy E$_{X}$ j = 1 term : − 1 2 ∫ dx$^{∫}$ dx ′ ρ$_{0}$ ( x, x $^{′}$) ρ$_{0}$ ( x $^{′}$, x ) V ( x, x $^{′}$) = E$_{X}$ [ ρ$_{0}$ ] . (22) Φ GW xc = E$_{X}$ [ ρ$_{0}$ ] − 1 2 ∞ ∑ j =2 1 j ∫ ∞ −∞ dω 2 πi tr $^{{}$[ V ¯ P ( ω )] j $^{}}$. Thus, we have This has naturally separated E$_{X}$ from the correlation energy Φ$_{c}$, The remaining frequency integrals with j ≥ 2 correspond to j th order autocorrelations of ¯ P ( t ) evaluated at t = 0, so the infinitesimal time shifts actually do not matter. Specifically, the j th term has an ω integral Φ$_{c}$ = Φ$_{xc}$ − E$_{X}$ . ∫ ∞ −∞ dω 2 π tr $^{{}$[ V ¯ P ( ω )] j } = ∫ ∞ −∞ dω 2 π ∫ ∞ −∞ dt$_{1}$ · · ·$^{∫}$ ∞ −∞ dt$_{j}$ e iω ( t$_{1}$ + ··· + t$_{j}$ ) tr $^{{}$V ¯ P ( t$_{1}$ ) · · · V ¯ P ( t$_{j}$ ) } = ∫ ∞ −∞ dt$_{1}$ · · ·$^{∫}$ ∞ −∞ dt$_{j}$ δ ( t$_{1}$ + · · · + t$_{j}$ ) tr $^{{}$V ¯ P ( t$_{1}$ ) · · · V ¯ P ( t$_{j}$ ) } . Φ GW xc = E$_{X}$ [ ρ$_{0}$ ] − 1 2 ∞ ∑ j =2 1 j ∫ ∞ −∞ dω 2 πi tr $^{{}$[ V P ( ω )] j } = E$_{X}$ [ ρ$_{0}$ ] + 1 2 ∫ ∞ −∞ dω 2 πi tr $^{{}$V P ( ω ) } + 1 2 ∫ ∞ −∞ dω 2 πi tr $^{{}$ln ε ( ω ) } = E$_{X}$ [ ρ$_{0}$ ] − i 2 tr $^{{}$V P ( t = 0) } + 1 2 ∫ ∞ −∞ dω 2 πi tr $^{{}$ln ε ( ω ) $^{}}$. We integrate over the times so that the infinitesimal region about t = 0 where ¯ P ( t ) = P ( t ) is of zero measure. Hence, we can safely replace ¯ P by P for the j ≥ 2 terms to get In the second line, we added and subtracted the corresponding j = 1 term to the first line and summed up the series for the logarithm. In the third line, evaluating P ( t = 0) requires some care as plugging in t = 0 directly in Eq. (8) yields an indeterminate result. Rather, we take the limit t → 0 of P ( t ) in Eq. (8) to find P ( x, x $^{′}$, t → 0) = − i ∑ c ψ$_{c}$ ( x ) ψ$_{c}$ ( x $^{′}$)∗ ∑ v ψ$_{v}$ ( x $^{′}$) ψ$_{v}$ ( x ) ∗ 11