Consider bosonic field operators in the Heisenberg picture: \begin{align} \Psi(x)=\int \frac{d^{3}p}{(2\pi)^{3}}e^{-ip\cdot x}a_{\bf{p}}\\ \Psi^{\dagger}(x)=\int \frac{d^{3}p}{(2\pi)^{3}}e^{+ip\cdot x}a^{\dagger}_{\bf{p}}, \end{align}
where the dot product $p\cdot x$ is Minkowskian with the signature $(+,-,-,-)$. The retarded Green's function is defined (see for example: Introduction to Many-body quantum theory in condensed matter physics by Henrik Bruus and Karsten Flensberg, Eq. 8.28) as: \begin{equation} G_{R}(x,x^{\prime})=\theta(t-t^{\prime})\langle 0|[\Psi(x),\Psi^{\dagger}(x^{\prime})]|0 \rangle. \end{equation} The first term in the commutator gives the greater Green's function and the second gives the lesser: \begin{align} G^{>}(x,x^{\prime})=\langle 0|\Psi(x)\Psi^{\dagger}(x^{\prime})|0 \rangle\\ G^{<}(x,x^{\prime})=\langle 0|\Psi^{\dagger}(x^{\prime})\Psi(x)|0 \rangle. \end{align} The first one is nonzero since the contribution $a_{\bf{k}}a^{\dagger}_{\bf{p}}$ gives a delta function. The second term has the order $a^{\dagger}_{\bf{k}}a_{\bf{p}}$ and gives zero. Shouldn't it vanish?