The Hubbard model in the atomic limit is given by,

$${\mathcal H}= -\mu \sum_{\sigma } f^{\dagger}_{\sigma}f_{\si... ...row } f_{\uparrow } f^{\dagger}_{\downarrow } f_{\downarrow }.$$ (1)

The system has four possible states, namely $\vert {\,\rangle }$ (no particle), $\vert \uparrow {\,\rangle }$ (up electron), $\vert \downarrow {\,\rangle }$ (down electron), and $\vert \uparrow \downarrow {\,\rangle }$ (doubly occopied). The grand canonical partition function for the system is given by,

$$Z = {\rm Tr} e^{- \beta {\mathcal H}}$$

$$= 1 + 2 e^{\beta \mu } + e^{\beta (2 \mu - U)}.$$ (2)

1. To calculate the retarded Green's function and the spectral function we will first calculate the temperature Green's function and then take appropriate analytic continuation.

   The temperature Green's function is defined as,
   

   $$g_{\sigma }(\tau ) = - {\langle\, }T_{\tau } f_{\sigma}(\tau ) f^{\dagger}_{\sigma}(0) {\,\rangle }$$

   $$= - \frac{1}{Z} {\rm Tr} \left( e^{-\beta {\mathcal H}} T_{\tau } f_{\sigma}(\tau ) f^{\dagger}_{\sigma}(0) \right).$$

   
   
   The Fourier transform in terms of fermionic Matsubara frequency $\omega _n = (2n+1) \pi/\beta$, is
   

   $$g_{\sigma } (i \omega _n) = \int_{0}^{\beta } d \tau g_{\sigma }(\tau ) e^{i \omega _n \tau }$$

   $$= - \frac{1}{Z} \int_{0}^{\beta } d \tau e^{i \omega _n \tau } \sum... ...thcal H}} f_{\sigma}(\tau ) f^{\dagger}_{\sigma}(0) \right\vert n {\,\rangle }.$$ (3)

   
   
   Since,
   

   $$\sum_n {\langle\, }n \left\vert e^{-\beta {\mathcal H}} f_{\sigma}(\tau ) f^{\dagger}_{\sigma}(0) \right\vert n {\,\rangle } = \sum_n {\langle\, }n \left\vert e^{-\beta {\mathcal H}} e^{\tau {... ...{\sigma}e^{- \tau {\mathcal H}} f^{\dagger}_{\sigma} \right\vert n {\,\rangle }$$

   $$= {\langle\, }0 \left\vert e^{-\beta {\mathcal H}} e^{\tau {\mathca... ...- \tau {\mathcal H}} f^{\dagger}_{\sigma}\right\vert \bar{\sigma } {\,\rangle }$$

   $$= e^{\tau \mu } + e^{\beta \mu + \tau (\mu - U)},$$ (4)

   
   
   we get from equations (3) and (4),
   

   $$g_{\sigma } (i \omega _n) = \frac{1}{Z} \left\{ \frac{e^{\be... ... \mu - U)} + e^{\beta \mu }}{ i \omega _n + \mu - U} \right\}.$$ (5)

   
   
   Then, after the analytic continuation $i \omega _n \rightarrow \omega + i\eta$, we have,
   

   $$G^R_{\sigma } (\omega ) = \frac{1}{Z} \left\{ \frac{e^{\beta... ... - U)} + e^{\beta \mu }}{ \omega + \mu - U + i\eta } \right\}.$$ (6)

   
   

   The spectral function is given by,
   

   $$A_{\sigma }(\omega ) = - \frac{1}{\pi} {\rm Im} G^R_{\sigma } (\omega )$$

   $$= \left(\frac{e^{\beta \mu } +1}{Z} \right) \delta ( \omega + \mu )... ...c{ e^{\beta (2 \mu - U)} + e^{\beta \mu }}{Z} \right) \delta (\omega + \mu -U).$$ (7)

   
   
   In the non-interacting case the spectral function has a single pole with weight one, i.e, $A_{\sigma }(\omega ) = \delta ( \omega + \mu )$. The effect of interaction is to produce two poles with finite energy gap $U$. The total spectral weight, which is always one, is shared between the two poles.
2. The Hubbard operators are defined as
   

   $$X_{\uparrow 0} \equiv f^{\dagger}_{\uparrow } (1-n_{\downarrow }) = \vert \uparrow {\,\rangle }{\langle\, }0 \vert$$

   $$X_{d \downarrow } \equiv f^{\dagger}_{\uparrow } n_{\downarrow } = \vert \uparrow \downarrow {\,\rangle }{\langle\, }\downarrow \vert.$$ (8)

   
   
   Their conjugate operators are repectively,
   

   $$X_{0 \uparrow } \equiv f_{\uparrow } (1-n_{\downarrow }) = \vert 0 {\,\rangle }{\langle\, }\uparrow \vert$$

   $$X_{\downarrow d} \equiv f_{\uparrow } n_{\downarrow } = \vert \downarrow {\,\rangle }{\langle\, }\uparrow \downarrow \vert.$$ (9)

   
   

   Now, it is easy to see that
   

   $$f^{\dagger}_{\uparrow } = X_{\uparrow 0} +X_{d \downarrow } ... ... \uparrow \downarrow {\,\rangle }{\langle\, }\downarrow \vert.$$ (10)

   
   

3. Similarly, we have,
   

   $$f^{\dagger}_{\downarrow }=X_{\downarrow 0} +X_{d \uparrow }=... ...rt \uparrow \downarrow {\,\rangle }{\langle\, }\uparrow \vert.$$ (11)

   
   

4. The anticommutator
   

   $$\left\{ X_{\uparrow 0}, X_{0 \uparrow } \right\} = \left\{ f^{\dagger}_{\uparrow }(1- n_{\downarrow }), (1- n_{\downarrow })f_{\uparrow } \right\}$$

   $$= (1- n_{\downarrow })^2 \left\{ f^{\dagger}_{\uparrow },f_{\uparrow } \right\}$$

   $$= (1- n_{\downarrow }),$$ (12)

   
   
   since $(1- n_{\downarrow })$ is either zero or one. Thus,
   

   $${\langle\, }\{ X_{\uparrow 0}, X_{0 \uparrow }\}{\,\rangle }... ...\, }n_{\downarrow } {\,\rangle } =\frac{e^{\beta \mu } +1}{Z},$$ (13)

   
   
   is the average occupation of a $\downarrow$-hole. We also note, from equation (7), that this is also the spectral weight for the pole corresponding to single occupancy.

   Similarly, the anticommutator $\left\{ X_{d \downarrow }, X_{\downarrow d} \right\} = n_{\downarrow }$. Thus,
   

   $${\langle\, }\{X_{d \downarrow }, X_{\downarrow d} \} {\,\ran... ...\,\rangle } =\frac{e^{\beta (2 \mu - U)} + e^{\beta \mu }}{Z},$$ (14)

   
   
   is the average occupation of a $\downarrow$-electron. Again from equation (7) we note that this is the spectral weight for the pole corresponding to double occupancy.

5. To evaluate the commutator $[{\mathcal H}, X_{0\uparrow }]$, it is enough to consider the state $\vert \uparrow {\,\rangle }$, since all other states give zero when acted on by $X_{0\uparrow }$. We have $[{\mathcal H}, X_{0\uparrow }] \vert \uparrow {\,\rangle }= \mu \vert 0 {\,\rangle }= \mu X_{0\uparrow } \vert \uparrow {\,\rangle }$. Thus, $[{\mathcal H}, X_{0\uparrow }] = \mu X_{0\uparrow }$, and
   

   $${\langle\, }\left\{ X_{\uparrow 0}, \left[ {\mathcal H}, X_{... ...\mu \left(1 - {\langle\, }n_{\downarrow } {\,\rangle }\right).$$ (15)

   
   
   Comparing with equation (7) we find that the energy coefficient in the above expression gives the pole for the singly occupied state.

   Similarly, we can show that $[{\mathcal H}, X_{\downarrow d}] = (\mu - U) X_{\downarrow d}$, and
   

   $${\langle\, }\left\{ X_{d \downarrow }, \left[ {\mathcal H}, ... ...eft( \mu - U \right) {\langle\, }n_{\downarrow } {\,\rangle }.$$ (16)

   
   
   Here the energy coefficient gives the pole for the doubly occupied state.

6. Comment

   To put parts (1), (4) and (5) into proper perspective we will re-evaluate the retarded Green's function using the Hubbard operators. From definition,
   

   $$G^R_{\sigma } (t) = -i \theta (t) {\langle\, }\left\{ f_{\sigma}(t), f^{\dagger}_{\sigma}(0) \right\} {\,\rangle }$$

   $$= -i \theta (t) {\langle\, }\left\{ X_{0 \sigma }(t) + X_{\bar{\sigma}d}(t), X_{\sigma 0}(0) + X_{d \bar{\sigma}} (0) \right\} {\,\rangle }$$

   $$= -i \theta (t) \left[ {\langle\, }\left\{X_{0 \sigma }(t), X_{\sig... ...eft\{ X_{\bar{\sigma}d}(t), X_{d \bar{\sigma}}(0) \right\} {\,\rangle }\right].$$

   
   
   Here $\bar{\sigma}$ is the spin opposite to $\sigma$. In the last line above, we have used the fact that cross-terms are zero. Now, from our discussion in part (5), we know that $[{\mathcal H}, X_{0\sigma }] = \mu X_{0\sigma }$ and that $[{\mathcal H}, X_{\bar{\sigma}d}] = (\mu - U) X_{\bar{\sigma}d}$. Then, $X_{0 \sigma }(t) = e^{i \mu t}X_{0 \sigma }(0)$, and $X_{\bar{\sigma}d}(t)= e^{i(\mu - U)t}X_{\bar{\sigma}d}(0)$. We can write,
   
   $$G^R_{\sigma } (t)= -i \theta (t) \left[ e^{i \mu t} {\langle... ...ar{\sigma}d}, X_{d \bar{\sigma}}\right\} {\,\rangle } \right].$$
   
   
   From our discussion in part (4) we know that ${\langle\, }\{ X_{\sigma 0}, X_{0 \sigma }\}{\,\rangle } = 1 - {\langle\, }n_{\bar{\sigma}} {\,\rangle }$ and that ${\langle\, }\{ X_{\bar{\sigma}d}, X_{d \bar{\sigma}} \} {\,\rangle } ={\langle\, }n_{\bar{\sigma}} {\,\rangle }$. So, finally we have,
   

   $$G^R_{\sigma } (t) = -i \theta (t) \left\{ (1 - {\langle\, }n... ...ngle\, }n_{\bar{\sigma}} {\,\rangle }e^{i(\mu - U)t} \right\}.$$ (17)

   
   
   Taking the Fourier transform of the above equation we get,
   

   $$G^R_{\sigma } (\omega ) = \frac{1 - {\langle\, }n_{\bar{\sig... ...\, } n_{\bar{\sigma}} {\,\rangle }}{\omega + \mu -U + i\eta }.$$ (18)

   
   
   This is nothing but equation (6) in a more compact form.