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I have to obtain a one-particle Green function for phonons at $T=0$ $$D^{0}(\mathbf{x},t)=\frac{1}{iV}\sum_{\mathbf{k}}\frac{\omega_\mathbf{k}}{2}\big(\theta(t)e^{i(\mathbf{kx}-\omega_{\mathbf{k}} t)}+\theta(-t)e^{-i(\bf{kx}-\omega_{\mathbf{k}} t)}\big)$$ So, I start from $$D^{0}(\mathbf{x},t;\mathbf{x'},t')=-i\langle T\varphi(\mathbf{x},t)\varphi(\mathbf{x'},t') \rangle_0$$ where $$\varphi(\mathbf{x},t)=\frac{1}{\sqrt{V}}\sum_{\mathbf{k}}(\frac{\omega_{\mathbf{k}}}{2})^{\frac{1}{2}}\big[b_{\mathbf{k}}e^{i(\mathbf{kx}-\omega_{\mathbf{k}} t)}+b^{\dagger}_{\mathbf{k}}e^{-i(\mathbf{kx}-\omega_{\mathbf{k}} t)}\big]$$

$$D^{0}(\mathbf{x},t;\mathbf{x'},t')=D^{0}(\mathbf{x}-\mathbf{x'},t-t';0,0)=D^{0}(\mathbf{x},t;0,0)=-i\theta(t)\langle\varphi(\mathbf{x},t)\varphi(0,0) \rangle_0 -i\theta(-t)\langle \varphi(0,0)\varphi(\mathbf{x},t) \rangle_0=\frac{1}{iV}\sum_{\mathbf{k},\mathbf{k'}}\frac{\sqrt{\omega_{\mathbf{k}}\omega_{\mathbf{k'}}}}{2}\langle\big[\theta(t)(b_{\mathbf{k}}e^{i(\mathbf{kx}-\omega_{\mathbf{k}} t)}+b^{\dagger}_{\mathbf{k}}e^{-i(\mathbf{kx}-\omega_{\mathbf{k}} t)})(b_{\mathbf{k'}}+b^{\dagger}_{\mathbf{k'}})+\theta(-t)(b_{\mathbf{k}}+b^{\dagger}_{\mathbf{k}})(b_{\mathbf{k}}e^{i(\mathbf{kx}-\omega_{\mathbf{k}} t)}+b^{\dagger}_{\mathbf{k'}}e^{-i(\mathbf{kx}-\omega_{\mathbf{k'}} t)})\big]\rangle_0$$ Here I use the bosonic commutation relations $$[b^{\dagger}_{\mathbf{k}},b_{\mathbf{k'}}]=\delta_{\mathbf{k}\mathbf{k'}},[b^{\dagger}_{\mathbf{k}},b^{\dagger}_{\mathbf{k'}}]=[b_{\mathbf{k}},b_{\mathbf{k'}}]=0$$ and fact that $\theta(t)+\theta(-t)=1$ and gettting $$D^{0}(\mathbf{x},t)=\frac{1}{iV}\sum_{\mathbf{k},\mathbf{k'}}\frac{\sqrt{\omega_\mathbf{k}\omega_\mathbf{k'}}}{2}\langle\big(\theta(t)\delta_{\mathbf{k}\mathbf{k'}}e^{+}+\theta(-t)\delta_{\mathbf{k}\mathbf{k'}}e^{-}+b_{\mathbf{k}}b_{\mathbf{k'}}e^{+}+b^{\dagger}_{\mathbf{k}}b^{\dagger}_{\mathbf{k'}}e^{-}+b^{\dagger}_{\mathbf{k'}}b_{\mathbf{k}}e^{+}+b^{\dagger}_{\mathbf{k}}b_{\mathbf{k'}}e^{-}\big)\rangle_0,$$ I use a short notation for the exponents $e^{i(\mathbf{kx}-\omega_{\mathbf{k}} t)}=e^{+},e^{-i(\mathbf{kx}-\omega_{\mathbf{k}} t)}=e^{-}$.

I don't know what to do with these terms: $$\langle b^{\dagger}_{\mathbf{k}}b_{\mathbf{k'}}\rangle_{0},\langle b^{\dagger}_{\mathbf{k'}}b_{\mathbf{k}}\rangle_{0},\langle b^{\dagger}_{\mathbf{k}}b^{\dagger}_{\mathbf{k'}}\rangle_{0},\langle b_{\mathbf{k}}b_{\mathbf{k'}}\rangle_{0}$$ Should they be equal to zero? If it so, why? In that case I can get right answer.

I know that in ground state should be no phonons, that means that $\langle b^{\dagger}_{\mathbf{k}}b_{\mathbf{k}}\rangle_{0}=0$

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Your last argument can be applied to all of them including $\langle 0 | b_k^\dagger b_{k'}^\dagger |0 \rangle = 0$, because $b_k |0\rangle = 0$ implies that $\langle 0 | b_k^\dagger = 0$.

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