I recently went through Field Theory of Non-Equilibrium Systems by Alex Kamenev (2011) and I wanted to understand certain points in detail. My current problem is to understand how one can calculate expectation values in the continuum notation presented in this book.
Specifically, I do not understand completely the following equalities (page 27): $$\langle \hat F(t) \rangle \equiv \langle \hat b^\dagger(t) \hat b(t) + \hat b(t) \hat b^\dagger(t) \rangle = \langle \bar\phi^+(t) \phi^+(t) + \bar\phi^-(t) \phi^-(t) \rangle \tag{1}$$ $$ iG^K(t,t) = F(\omega_0) = 2n_B(\omega_0) +1 \tag{2}$$ With $\hat b$ a bosonic annihilation operator, $\phi$ its corresponding eigenvalue, $G^K$ the Keldysh Green's function and $n_B$ the Bose-Einstein distribution. The $+$ and $-$ superscripts indicate that the field is evaluated on the forward and backward branches of the Keldysh time contour respectively. In this chapter, one deals with a free bosonic particle.
1. Regarding equation $(1)$
I made several attempts but I do not see how to prove this without going back to the discrete notation. Here is my current reasoning. I inserted $\hat b^\dagger \hat b$ at time $t$ on the forward branch of the contour and $\hat b \hat b^\dagger$ at time $t$ on the backward branch. Following equation $(2.17)$ page 13, which in my case becomes $ \langle \phi_j| \hat b^\dagger \hat b e^{-i\omega_0\delta t \hat b^\dagger \hat b} |\phi_{j-1}\rangle = \langle \phi_j|\phi_{j-1}\rangle \bar \phi_j \phi_{j-1}e^{-i\omega_0\delta t \bar \phi_j \phi_{j-1}} e^{-i\omega \delta t}$, I get:
$$ \langle \hat b^\dagger(t) \hat b(t) + \hat b(t) \hat b^\dagger(t) \rangle = \langle \bar\phi^+_j \phi^+_{j-1} e^{-i\omega \delta t}+ \bar\phi^-_{j-1} \phi^-_j e^{i\omega \delta t} +1 \rangle \tag{3}$$
Going to the continuum notation (omitting that one field is taken a time step ahead), I would get:
$$ \langle \hat b^\dagger(t) \hat b(t) + \hat b(t) \hat b^\dagger(t) \rangle = \langle \bar\phi^+(t) \phi^+(t) + \bar\phi^-(t) \phi^-(t) +1 \rangle \tag{4}$$
I realize there is something I might have not understood about what one should do at equal time without resorting to the general case of different times. Also, it is not clear to me where the resolution of unity should be inserted. For instance one could insert it between $\hat b$ and $\hat b^\dagger$. In this case, using the commutation relation to transform $\hat b^\dagger \hat b$ into $\hat b \hat b^\dagger -1$ before inserting the resolution of unity, I would get:
$$ \langle \hat b^\dagger(t) \hat b(t) + \hat b(t) \hat b^\dagger(t) \rangle = \langle \phi^+_j \bar\phi^+_{j} + \phi^-_{j} \bar\phi^-_j -1 \rangle = \langle \phi^+(t) \bar\phi^+(t) + \phi^-(t) \bar\phi^-(t) -1 \rangle \tag{5}$$
Where the last equality comes from going to the continuum notation. Kamenev says that $\langle \bar\phi^+(t) \phi^+(t) + \bar\phi^-(t) \phi^-(t) \rangle$ cannot be found by inserting $\hat b^\dagger \hat b$ at time t on the forward and backward branch because one field is taken a time step ahead (as one can see in equation 3). So, I inserted the resolution of unity between $\hat b$ and $\hat b^\dagger$ but it will not work if I do the same between $\hat b^\dagger$ and $\hat b$. Has anyone any idea how this should be done please?
2. Regarding equation $(2)$
I know this equation directly comes from the expression for $iG^K(t,t')$ where $t \neq t'$. However, it seems that Kamenev does not discute the equal time case as he does for the advanced and retarded Green's functions. Following what is written in the book, when $t=t'$, one gets from equations $(2.35)$ and the discussion regarding the Heaviside function at $0$: $$ \langle\phi^+(t) \bar\phi^-(t)\rangle = iG^<(t,t) = n_B \tag{6}$$ $$ \langle\phi^-(t) \bar\phi^+(t)\rangle = iG^>(t,t) = n_B +1\tag{7}$$ $$ \langle\phi^+(t) \bar\phi^+(t)\rangle = iG^\mathbb T(t,t) = n_B +1\tag{8}$$ $$ \langle\phi^-(t) \bar\phi^-(t)\rangle = iG^\overset{\sim}{\mathbb T}(t,t) = n_B +1\tag{9}$$ Thus using the definition of $G^K$, one gets: $$ iG^K(t,t') \equiv \dfrac{1}{2}(iG^<(t,t)+iG^>(t,t)+iG^\mathbb T(t,t)+iG^\overset{\sim}{\mathbb T}(t,t)) = 2n_B+\dfrac{3}{2} \neq 2n_B+1\tag{10}$$ Moreover, from equations $(1)$, $(8)$ and $(9)$ I would expect that: $$ \langle \bar\phi^+(t) \phi^+(t) + \bar\phi^-(t) \phi^-(t) \rangle = 2n_B+2 \tag{11}$$ Where the left hand side is equal to $iG^K(t,t)$ as can be seen using the generating function (equation $(2.60)$). How can this issue be resolved?
