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  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %% Indranil's LaTeX macros
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \def\ek{\epsilon_{k}}
  \def\xk{\xi_k} 
  \def\la{{\langle\, }}
  \def\ra{{\,\rangle }}
  \def\p{\,|\,}
  \def\infintd3k{ \int_{-\infty}^\infty d^3r\,}

  % Equation environments
  \def\bea{\begin{eqnarray}}
  \def\eea{\end{eqnarray}}
  \def\ben{\begin{equation}}
  \def\een{\end{equation}}
  \def\benu{\begin{enumerate}}
  \def\enu{\end{enumerate}}

  
  
   \begin{document}

  
   
   \begin{center}
   Solution to Problem Set 1
   \end{center}
	


   For a system of non-interacting fermions described by the Hamiltonian,
   \[
   \mathcal{H}_0 = \sum_k \ek a^{\dagger}_k a_k
   \]
   we want to calculate Green's functions with respect to a state 
   $ \p \psi \ra = \prod_i a^{\dagger}_k \p 0 \ra $ , that is characterized
   by a distribution function 
   \[
   n_k = \la \psi \p a^{\dagger}_k a_k \p \psi \ra
   \] 
   The Fourier transform of a field operator 
   $ \psi(x,t) $ is defined as,
   \[
   \psi(x,t) = \int \frac{d^3k}{(2\pi)^3} e^{ikx} \psi(k,t)
   \]
   and the time evolution of the Heisenberg operator $\psi(k,t)$ is,
   \bea
   \psi(k,t) & = & e^{i\mathcal{H}_0t} a_k e^{-i\mathcal{H}_ot} \nonumber \\
             & = & e^{-i\ek t} a_k \nonumber
   \eea
   Note that time evolution giving just a phase factor is a simplification
   only for a noninteracting system (all the fuss about interacting
   systems is due to this unfortunate fact) 
   \begin{enumerate}
   \item  $G^>(1,2)$ is defined as
    $-i\la\psi(x_1,t_1)\psi^{\dagger}(x_2,t_2)\ra$. Performing the Fourier
    transform and the time evolution as defined above, we get,
    \[
    G^>(1,2) = -i \int \frac{d^3k_1}{(2\pi)^3} \int\frac{d^3k_2}{(2\pi)^3} 
    e^{ik_1x_1 -ik_2x_2} e^{-i\epsilon_{k_1}t_1 + i\epsilon_{k_2}t_2}
    \la a_{k_1}a^{\dagger}_{k_2} \ra
    \]       
    From the anticommutation of the second quantized operators we get,
    \[
    \la a_{k_1}a^{\dagger}_{k_2} \ra = (2\pi)^3 \delta(k_1 - k_2)(1 - n_{k_1})
    \] 
    so the above expression simplifies to 
    \[
    G^>(1,2) = -i \int \frac{d^3k}{(2\pi)^3}e^{ik(x_1 -x_2)}
    e^{-i\ek(t_1 - t_2)}(1 - n_k)  
    \]
    Since the system is translationally invariant, as expected,
    $G^>$ is a function of $(x_1 - x_2)$ and $(t_1 - t_2)$. Let  
    $(x_1- x_2) \equiv x$ and $(t_1 - t_2) \equiv t$. Then the Fourier
    transform of $G^>(x,t)$ is 
    \ben
    G^>(k,\omega)) = 2\pi i \delta (\omega - \ek) (n_k - 1)
    \een
    
    \item Evaluation of $ G^<(1,2)$ is similar and we get 
    \bea
    G^<(1,2) & = & -i\la\psi^{\dagger}(x_2,t_2)\psi(x_1,t_1)\ra \nonumber \\ 
    & = &  i \int \frac{d^3k}{(2\pi)^3} e^{ikx - i\ek t} n_k \nonumber
    \eea   
    whose Fourier transform is 
    \ben
    G^<(k,\omega)) = 2\pi i \delta (\omega - \ek) n_k  
    \een
    
    \item The time ordered Green's function is defined as 
     \bea
     G^t(1,2) &=& -i\la T(\psi(x_1,t_1)\psi^{\dagger}(x_2,t_2))\ra
     \nonumber \\ 
     &=& \theta(t_1 - t_2)G^>(1,2) + \theta(t_2 - t_1)G^<(1,2) \nonumber
     \eea
     or
     \[
     G^t(x,t) = -i \int \frac{d^3k}{(2\pi)^3} e^{ikx - i\ek t}\theta(t)
      (1 - n_k) + i \int \frac{d^3k}{(2\pi)^3} e^{ikx - i\ek t}\theta(-t) n_k
     \]   
     Since
     \[
     \int_{- \infty}^{\infty} \frac{d \omega}{2 \pi} 
     \frac{e^{-i\omega t}}{\omega - \ek + i \eta} = -i e^{-i \ek t} \theta(t)
     \]
     and 
     \[
     \int_{- \infty}^{\infty} \frac{d \omega}{2 \pi} 
     \frac{e^{-i\omega t}}{\omega - \ek - i \eta} = i e^{-i \ek t}\theta (-t)
     \]
     the Fourier transform of the time ordered Green's function is
     \ben
     G^t(k, \omega) = \frac{1 - n_k}{\omega - \ek + i \eta} + 
     \frac{n_k}{\omega - \ek - i \eta}
     \een
     The first term correspond to particle propagation in positive (future)
     time while the second term describes hole propagation in negative
     (past) time.
       \item The anti time ordered Green's function, defined as 
     \bea
     G^{\tilde{t}}(1,2) &=& -i\la \tilde{T}(\psi(x_1,t_1)
     \psi^{\dagger}(x_2,t_2))\ra
     \nonumber \\ 
     &=& \theta(t_2 - t_1)G^>(1,2) + \theta(t_1 - t_2)G^<(1,2) \nonumber
     \eea
     is evaluated similarly, and it's Fourier transform is 
     \ben
     G^{\tilde{t}}(k, \omega) = - \left[\frac{1 - n_k}{\omega - \ek - i \eta} + 
     \frac{n_k}{\omega - \ek + i \eta}\right]
     \een
     
     \item The retarded Green's function, defined as
     \bea
     G^R(1,2) &=& -i \theta(t_1 - t_2) \la \{\psi(x_1,t_1) ,\psi^{\dagger}
     (x_2,t_2)\} \ra \nonumber \\
     &=& \theta(t_1 - t_2) \left[ G^>(1,2) - G^<(1,2) \right] \nonumber \\
     &=&  -i \int \frac{d^3k}{(2\pi)^3} e^{ikx - i\ek t}\theta(t) \nonumber
     \eea 
     and it's Fourier transform is 
     \ben
     G^R(k, \omega) = \frac{1}{\omega - \ek + i \eta}
     \een
     
     \item Similarly the advanced Green's function, defined as
      \[
      G^A(1,2) = i \theta(t_2 - t_1) \la \{\psi(x_1,t_1) ,\psi^{\dagger}
     (x_2,t_2)\} \ra
      \]
      has the Fourier transform
     \ben
     G^A(k, \omega) = \frac{1}{\omega - \ek - i \eta}
     \een
     The retarded and the advanced Green's functions do not carry any 
     information about the distribution function.
     \item \bea
     G^K(1,2) &=& -i \la [\psi(x_1,t_1) ,\psi^{\dagger}
     (x_2,t_2)] \ra \nonumber \\
     &=& G^>(1,2) + G^<(1,2) \nonumber
     \eea   
     and in the Fourier space this has the form
     \ben
     G^K(k,\omega) = 2\pi i \delta (\omega - \ek) (2n_k - 1)
     \een
     
     \item The Matsubara Green's function is defined as $g(\tau,x) =
     -\la T_{\tau} \psi(\tau,x) \psi^{\dagger}(0,0) \ra $, where $\tau$
     is imaginary time, and $T_{\tau}$ orders imaginary time. The (imaginary)
     time development of $\psi(\tau,k)$ is given by
     \bea
     \psi(\tau,k) &=& e^{\tau K_0} a_k e^{- \tau K_0} \nonumber \\
     &=& e^{- \tau \xk} a_k \nonumber 
     \eea
     where $ K_0 = \mathcal{H}_0 - \mu N$, and $\xk = \ek - \mu$. Then,
     \bea
     g(\tau,x) &=& - \int \frac{d^3k}{(2\pi)^3} e^{ikx} e^{- \tau \xk}
     \{ \theta(\tau) \la a_k a^{\dagger}_k \ra -  \theta(- \tau)
     \la a^{\dagger_k} a_k \ra \}  \nonumber  \\
     &=& - \int \frac{d^3k}{(2\pi)^3} e^{ikx} e^{- \tau \xk} \{ \theta(\tau)
     - n_F(\xk) \}  \nonumber
     \eea
     where $n_F(\xk) = \frac{1}{e^{\beta \xk} + 1}$, is the fermi function.
     For a fermionic system, $g(\tau)$ satisfies antiperiodic boundary
     condition, so that, $g(\tau + \beta) = - g(\tau)$, where $\beta$ is 
     inverse temperature in energy units. Therefore the Fourier transform of 
      $g(\tau)$ (in imaginary frequency) is defined as 
     \[
     g(i\omega_n, k) = \int_0^{\beta}d\tau g(\tau, k) e^{i \tau \omega_n}
     \]
     where $\omega_n = \frac{(2n + 1)\pi}{\beta}$, are the fermionic Matsubara
     frequencies. The result of the Fourier transform is,
     \ben
     g(i\omega_n, k) = \frac{1}{i\omega_n - \xk}
     \een
     $g(i\omega_n, k)$ has no explicit dependence on the distribution function
     $n_F(\xk)$. But while carrying out the discrete sum over fermionic 
     frequencies the distribution function reappears, since the discrete 
     frequencies are at the poles of the fermi distribution function. Analytic
     continuation to real frequencies $(i\omega_n \longrightarrow \omega 
     + \pm i\eta)$ give the retarded and advanced Green's functions 
     respectively, which we have already evaluated. Note that the
     corresponding expressions match.
     \end{enumerate}

\end{document}


   



