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In the 8th Chapter of book Many-body Quantum Theory in Condensed Matter Physics by Henrik Bruus and Karsten Flensberg, they give an explicit form of retarded Green's function (GF) for the many-body system as $$ G^R(\textbf{r}\sigma t,\textbf{r}'\sigma' t') = -i\theta(t-t')\langle [\Psi_\sigma(\textbf{r} t),\Psi_{\sigma'}^\dagger(\textbf{r}'t')]_{B,F}\rangle \tag{8.28} $$ where $\theta(t-t')$ is step function, $\langle \cdot \rangle$ show thermal average, $[\cdot]_{B,F}$ show commutation (anti-commutation) relation for Bosons (Fermions), and $\Psi$ is field operator.

Also, the retarded GF (also called propagator) for single-particle wavefunction is given by $$ G^R(\textbf{r}t,\textbf{r}' t') = -i\theta(t-t')\langle \textbf{r}|e^{-iH(t-t')}|\textbf{r} \rangle. \tag{8.22} $$ They argue that for non-interacting particles, both equations are identical.

How are the two equations equal to each other for non-interacting particles? How can I prove that they are identical?

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We can change the basis of field operators according to $(1.69)$ of Bruus book. The relation between field operators and single-particle wavefunction $\phi(r)$ is $$ \Psi (r)=\sum_\mu \phi_\mu (r)a_\mu \quad ; \quad \Psi^+ (r)=\sum_\mu a_\mu^+ \phi_\mu^* (r)\tag{1.69} $$ here $\mu$ is the single-particle basis. Assume that this basis set diagonalizes single-particle hamiltonian, $H|\mu\rangle =\epsilon_\mu |\mu\rangle$. Ignore spin $\sigma$ degree of freedom in $(8.28)$ because $(8.22)$ is also defined without the spin degree of freedom. Now, put the above transformation in $(8.28)$ $$ G^R(\textbf{r} t,\textbf{r}' t') = -i\theta(t-t')\langle [\Psi(\textbf{r} t),\Psi^\dagger(\textbf{r}'t')]_{B,F}\rangle\\ = -i\theta(t-t')\left\langle \left[\sum_\mu \phi_\mu (r)a_\mu(t) ,\sum_\nu a_\nu^+(t')\phi_\nu^* (r')\right]_{B,F}\right\rangle\\ = -i\theta(t-t')\sum_{\mu\nu} \phi_\mu (r) \left\langle \left[ a_\mu(t) ,a_\nu^+(t')\right]_{B,F}\right\rangle\phi_\nu^* (r') $$ If we have a non-interacting system, the time dependence of operators is given by $(5.24)$, $a_\mu(t)=e^{-i\epsilon_\mu}a_\mu$ $$ = -i\theta(t-t')\sum_{\mu\nu} \phi_\mu (r) e^{-i\epsilon_\mu t}e^{i\epsilon_\nu t'} \left\langle \left[ a_\mu ,a_\nu^+ \right]_{B,F}\right\rangle\phi_\nu^* (r')\\ = -i\theta(t-t')\sum_{\mu} \phi_\mu (r) e^{-i\epsilon_\mu(t-t')} \phi_\mu^* (r') $$ by definition, we have $\phi_\mu(r)=\langle r|\mu\rangle$ $$ = -i\theta(t-t')\sum_{\mu} \langle r|\mu\rangle e^{-i\epsilon_\mu(t-t')} \langle \mu|r'\rangle \\ = -i\theta(t-t')\sum_{\mu} \langle r|e^{-iH(t-t')}|\mu\rangle \langle \mu|r'\rangle $$ use completeness identity $\sum_\mu |\mu\rangle \langle \mu|=1$ $$ = -i\theta(t-t') \langle r|e^{-iH(t-t')}|r'\rangle $$

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I have been sitting with the same problem now. I hope that by commenting it will be taken up again. So far the only way I seem to find that they are similar is by considering the Eigen basis. If you write the single particle propagator in the eigenbasis (eq: 8.23 in Bruus&Flensberg) you get: $$ G(nt,n't') = -i\theta(t-t')\langle n|e^{-iH(t-t')}|n'\rangle , $$ and since $H$ is diagonal in $n$ we get: $$ G(nt,n't') = -i\theta(t-t')\delta_{n,n'} e^{-iE_n(t-t')} .$$

Now for the many body Green's functions we have that for a diagonal Hamiltonian $H=\sum_n E_n c_n^\dagger c_n$ we can write the time-dependence as $c_n(t) = c_n e^{-iE_nt}$. We can now work out the commutator (for bosons and fermions): $$[c_n(t),c_{n'}(t')^\dagger] = e^{-iE_nt}e^{iE_{n'}t'}[c_n,c_{n'}^\dagger] = e^{-iE_n(t-t')}\delta_{n,n'}.$$

We now write the many-body GF in the eigenbasis instead of the $r$ basis (eq: 8.32 in Bruus&Flensberg). But since the commutator is a constant this just reduces to the equation from before:

\begin{align} G(nt,n't') &= -i\theta(t-t')\langle [c_n(t),c_{n'}(t')^\dagger] \rangle \\ &= -i\theta(t-t')e^{-iE_n(t-t')}\delta_{n,n'} \langle 1 \rangle \\ &= -i\theta(t-t')e^{-iE_n(t-t')}\delta_{n,n'}, \end{align} at all temperatures since the single-particle commutator simply gives a number. We then see that the single particle propagator and the many-body retarded Green's function are the same for non-interacting systems. Showing it directly in the $r$ basis, I couldn't do though but hopefully someone can help here.

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    $\begingroup$ Hey, interesting approach, diagonalization always makes things much more straightforward. I have posted an answer which uses the $|r\rangle$ basis. Let me know if you have any confusion. $\endgroup$ Nov 25, 2022 at 11:51

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