How to prove zero classical classical Green's function contribution

Question

Much of the Keldysh formalism is based on the following identity $$G^{T}(t,t') + G^{\tilde{T}}(t,t') = G^{<}(t,t') + G^{>}(t,t'),$$ which using their defintions is equivalent to $$G^{\text{cl,cl}}(t,t') \equiv i\langle (\psi^+(t) - \psi^{-}(t))(\psi^+(t') - \psi^{-}(t')) \rangle = 0,$$ I can see how this holds for a Harmonic oscillator, but I do not see how this holds in general. Here $\psi^+$ is on the forward and $\psi^-$ on the backward parts of the closed time contour respectively. Obviously if one assumes that $\psi^+(t) = \psi^-(t)$, then this holds, but is the point not to treat them as independent?

In Kamenev continuous notation, we have for instance

$$\langle \psi^{\pm}(t) \psi^{\pm}(t') \rangle = \int D[\psi \bar{\psi}] \psi^\pm(t) \psi^\pm(t') e^{S[\bar{\psi},\psi]}$$ with the action $$S[\bar{\psi},\psi] = \int_{-\infty}^{\infty}d\tau \mathcal{L}[\bar{\psi}^+(\tau),\psi^+(\tau)] - \int_{-\infty}^{\infty}d\tau \mathcal{L}[\bar{\psi}^-(\tau),\psi^-(\tau)].$$ The minus sign is from flipping the backwards time integrals bounds.

Kamenev states (paraphrasing): The continuum notation creates the impression that the $\psi^+$ and $\psi^-$ fields are completely uncorrelated. In fact, they are connected due to the presence of the non-zero off-diagonal blocks in the discrete form of $\mathcal{L}$

It is this correlation that somehow still means that $G^{cl,cl} = 0$ even though $\psi^+(t) \neq \psi^-(t)$.

Answer 1

By definition: $$ G^T(t,t') = G^>(t,t')\theta(t-t') + G^<(t,t')\theta(t'-t),\\ G^{\tilde{T}}(t,t') = G^>(t,t')\theta(t'-t) + G^<(t,t')\theta(t-t'). $$ By summing these two equations we readily obtain: $$ G^T(t,t') + G^{\tilde{T}}(t,t') = G^>(t,t') + G^<(t,t'). $$ This equality is thus the consequence of the definitions of the Green's functions involved.

I sketch the derivation of the second identity without delving too much on the notation (apparently defined in Kamenev's review). $$ i\langle \left(\psi^+(t) -\psi^-(t)\right)\left(\psi^+(t') -\psi^-(t')\right)\rangle= \\ i\langle \psi^+(t)\psi^+(t')\rangle + i\langle \psi^-(t)\psi^-(t')\rangle - i\langle \psi^+(t)\psi^-(t')\rangle - i\langle \psi^+(t)\psi^-(t')\rangle=\\ G^T(t,t') + G^{\tilde{T}}(t,t') - G^>(t,t') - G^<(t,t') $$ This is zero due to the first equality.

The full set of relations can be found, e.g., in the review by Rammer&Smith.

Update
To add some relevant material that was discussed in the comments. Green's function
$$G(t,t')=-i\langle T_c\left[\psi(t)\bar{\psi}(t')\right]\rangle,$$ with times $t,t'$ taken on the Keldysh contour, is quivalent to four Green's functions with real time arguments, which are distinguished by whether each time lies on the forward ($C_+/C_1$) or backward ($C_-/C_2$) branch of the Keldysh contour. Specifically, $$G_{12}(t,t') = i\langle \psi^+(t)\bar{\psi}^-(t')\rangle,$$ since, according to the countour ordering, $t$ on the forward branch of the contour is "lesser" than $t'$ on the backward branch, $t<_ct'$. The crucial point here is that superscripts $\pm$ are only a book-keeping tool for ordering the operators. Once the order is established, they are no more needed, and the above Green's function can be written as $$G_{12}(t,t') = i\langle \psi(t)\bar{\psi}(t')\rangle = G^{<}(t,t'),$$ which is by definition the "lesser" Green's function.

Indeed, we can see that the operators taken at the same time $t$ at the two branches of the contour are the same, if we avoid taking the Keldysh limit of extending contour to $+\infty$ and stay within the Kadanoff-Baym framework of the contour running from $-\infty$ to arbitrarily chosen time and backward to $-\infty$. Deforming the contour in such a way that it turns back at time $t$, we see that $\psi^+(t)$ is identical with $\psi^-(t)$.

Similar manipulations with the contour are also possible when considering two-time Green's function - by deforming the contour one could place the two times either on the same or on the different branches.

Keldysh trick of extending the contour to $+\infty$ and seemingly cutting it into two separate branches makes the formalism less transparent, but significantly simplifies the calculations.