Solution to Exercise 17

We want to evaluate various Green's functions for a system of free fermions described by the Hamiltonian

$${\mathcal H}_0= \sum_{{\bf p}} \epsilon _{{\bf p}} a^{\dagger}_{{\bf p}} a_{{\bf p}}.$$

We will understand operator averages by

$$\langle {\mathcal O} \rangle = \frac{{\rm Tr} \left( e^{-\l... ...ight) }{{\rm Tr} \left(e^{-\lambda {\mathcal H}_0} \right)}.$$

Then, from quantum statistical mechanics we can show that

$$\langle a^{\dagger}_{{\bf p}} a_{{\bf p}} \rangle = n_{{\bf p}} = \frac{1}{e^{\lambda \epsilon _{{\bf p}}} + 1}.$$ (1)

In the following we will use the Fourier transform of the field operators, which are defined as

$$\psi ({\bf x}, t) = \frac{1}{\sqrt{V}} \sum_{{\bf k}} e^{i {\bf k}\cdot {\bf x}} a_{{\bf k}} (t)$$ (2)

$$\psi^{\dagger}({\bf x}, t) = \frac{1}{\sqrt{V}} \sum_{{\bf k}} e^{-i {\bf k}\cdot {\bf x}} a^{\dagger}_{{\bf k}} (t).$$ (3)

Here $V$ is the volume of the system. For a non-interacting system, the time evolution of the operators $a^{\dagger}_{{\bf k}}$ and $a_{{\bf k}}$ is very simple. We find that

$$a_{{\bf k}}(t) = e^{i {\mathcal H}_0t}a_{{\bf k}}e^{-i {\mathcal H}_0t}$$

$$= a_{{\bf k}} + \left( it \right) \left[ {\mathcal H}_0, a_{{\bf k}... ...ft[ {\mathcal H}_0, \left[ {\mathcal H}_0, a_{{\bf k}} \right] \right] + \cdots$$

$$= e^{- i \epsilon _{{\bf k}} t} a_{{\bf k}}.$$ (4)

Similarly we can show that

$$a^{\dagger}_{{\bf k}}(t) = e^{i \epsilon _{{\bf k}} t} a^{\dagger}_{{\bf k}}.$$ (5)

1. The time ordered Green's function is defined as $G^t(1,2)= -i \langle T ( \psi ( {\bf x}_1,t_1) \psi^{\dagger}({\bf x}_2,t_2)) \rangle$. Since the system is translation invariant in space and time variables, the Green's function is a function of $({\bf x}_1 - {\bf x}_2)$ and $(t_1 - t_2)$ only. So, without any loss of generality we can put ${\bf x}_2=0$ and $t_2 =0$. We can write
   

   $$G^t({\bf x},t) = -i \theta (t) \langle \psi ({\bf x},t) \psi^{\dagger}\rangle + i \theta (-t) \langle \psi^{\dagger} \psi ({\bf x},t)$$

   $$= \frac{1}{V} \sum_{{\bf k}_1,{\bf k}_2} e^{i {\bf k}_1 \cdot {\bf ... ... i \theta (-t) \langle a^{\dagger}_{{\bf k}_2} a_{{\bf k}_1}(t) \rangle \right)$$

   $$= \frac{1}{V} \sum_{{\bf k}_1,{\bf k}_2} e^{i {\bf k}_1 \cdot {\bf ... ...e + i \theta (-t) \langle a^{\dagger}_{{\bf k}_2} a_{{\bf k}_1} \rangle \right)$$

   $$= \frac{1}{V} \sum_{{\bf k}} e^{i {\bf k}\cdot {\bf x}} e^{-i \epsi... ...} t} \left( -i \theta (t) (1-n_{{\bf k}} ) + i \theta (-t) n_{{\bf k}} \right).$$ (6)

   
   
   The Fourier transform of $G^t({\bf x},t)$ into momentum space gives
   

   $$G^t({\bf k},t) = \int d {\bf x}e^{-i {\bf k}\cdot {\bf x}} G^t({\bf x},t)$$

   $$= -i \theta (t) e^{-i \epsilon _{{\bf k}} t} (1-n_{{\bf k}} ) + i \theta (-t) e^{-i \epsilon _{{\bf k}} t} n_{{\bf k}}.$$ (7)

   
   
   In the above the first term is for propagation of particle excitation in forward time, and the second term denotes propagation of hole excitation in backward time. Next we make use of the identities
   

   $$\int_{\infty}^{\infty} \frac{d \omega }{2 \pi} \frac{e^{-i ... ... k}} + i \eta} = -i \theta (t) e^{-i \epsilon _{{\bf k}} t},$$ (8)

   
   
   and
   

   $$\int_{\infty}^{\infty} \frac{d \omega }{2 \pi} \frac{e^{-i ... ... k}} - i \eta} = i \theta (-t) e^{-i \epsilon _{{\bf k}} t},$$ (9)

   
   
   to write
   
   $$G^t({\bf k},t) = \int_{\infty}^{\infty} \frac{d \omega }{2 ... ...ga - \epsilon _{{\bf k}} - i \eta} \right\} e^{-i \omega t}.$$
   
   
   Thus,
   

   $$G^t({\bf k},\omega ) = \frac{1 - n_{{\bf k}}}{\omega - \eps... ... \frac{n_{{\bf k}}} {\omega - \epsilon _{{\bf k}} - i \eta}.$$ (10)

   
   

2. The anti-time ordered Green's function defined as $G^{\tilde{t}}({\bf x},t) = -i \langle \tilde{T} \psi ({\bf x},t) \psi^{\dagger}\rangle$, can be calculated along similar lines. We get
   

   $$G^{\tilde{t}}({\bf k},t) = i \theta (t) e^{-i \epsilon _{{\... ...i \theta (-t) e^{-i \epsilon _{{\bf k}} t} (1- n_{{\bf k}}).$$ (11)

   
   
   The Fourier transform into frequency space gives,
   

   $$G^{\tilde{t}}({\bf k},\omega ) - \left\{ \frac{n_{{\bf k}}}... ...n_{{\bf k}}}{\omega - \epsilon _{{\bf k}} - i \eta} \right\}.$$ (12)

   
   

3. The correlation function $G^>({\bf x},t) = -i \langle \psi ({\bf x},t) \psi^{\dagger}\rangle$ has Fourier transforms
   

   $$G^>({\bf k},t) = -i \left(1 - n_{{\bf k}} \right) e^{-i \epsilon _{{\bf k}} t}$$ (13)

   
   
   and
   

   $$G^>({\bf k},\omega ) = 2 \pi i \delta (\omega - \epsilon _{{\bf k}}) ( n_{{\bf k}} -1 ).$$ (14)

   
   

4. For $G^<({\bf x},t) = i \langle \psi^{\dagger}\psi ({\bf x},t) \rangle$ we get
   

   $$G^<({\bf k},t) = i n_{{\bf k}} e^{-i \epsilon _{{\bf k}} t},$$ (15)

   
   
   and
   

   $$G^<({\bf k},\omega ) = 2 \pi i n_{{\bf k}} \delta (\omega - \epsilon _{{\bf k}}).$$ (16)

   
   

5. 
   

   $$G^K({\bf x},t) = -i \langle \left[ \psi ({\bf x},t), \psi^{\dagger}\right] \rangle$$

   $$= G^>({\bf x},t) + G^<({\bf x},t).$$ (17)

   
   
   The Fourier transform is
   

   $$G^K({\bf k},\omega ) = 2 \pi i \delta (\omega - \epsilon _{{\bf k}}) ( 2n_{{\bf k}} -1 ).$$ (18)

   
   

6. 
   

   $$G^R({\bf x},t) = -i \theta (t) \langle \left\{ \psi ({\bf x},t), \psi^{\dagger}\right\} \rangle$$

   $$= \theta (t) \left\{ G^>({\bf x},t) - G^<({\bf x},t) \right\}.$$ (19)

   
   
   The Fourier transforms are
   

   $$G^R({\bf k},t) = -i \theta (t) e^{-i \epsilon _{{\bf k}} t},$$ (20)

   
   
   and
   

   $$G^R({\bf k},\omega ) = \frac{1}{\omega - \epsilon _{{\bf k}} + i \eta}.$$ (21)

   
   

7. 
   

   $$G^A({\bf k},t) = i \theta (-t) e^{-i \epsilon _{{\bf k}} t}.$$ (22)

   
   
   
   

   $$G^A({\bf k},\omega ) = \frac{1}{\omega - \epsilon _{{\bf k}} - i \eta}.$$ (23)

   
   
   It is important to note that in the above calculations we have not used the explicit form of the distribution function $n_{{\bf k}}$. The form of the distribution function depends on the density operator, and so the above results are true for arbitrary density operators. Also note that, unlike the other functions, $G^R({\bf k},\omega )$ and $G^A({\bf k},\omega )$ have information only about the excitation spectrum, but they have no information about the distribution function $n_{{\bf k}}$.
8. The Matsubara Green's function is given by
   

   $$g(\tau ,{\bf x}) = - \langle T_{\tau } \psi ({\bf x},\tau ) \psi^{\dagger}\rangle$$

   $$= - \frac{1}{V} \sum_{{\bf k}} \langle T_{\tau } a_{{\bf k}} (\tau ) a^{\dagger}_{{\bf k}} \rangle e^{i {\bf k} \cdot {\bf x}}.$$ (24)

   
   
   The evolution in imaginary time is given by
   

   $$a_{{\bf k}} (\tau ) = e^{\tau {\mathcal H}_0} a_{{\bf k}} e... ... {\mathcal H}_0} = e^{-\tau \epsilon _{{\bf k}}}a_{{\bf k}}.$$ (25)

   
   
   Then,
   

   $$g(\tau ,{\bf k}) = e^{-\tau \epsilon _{{\bf k}}} \left( n_{{\bf k}} - \theta (\tau ) \right).$$ (26)

   
   
   The principal region where $g(\tau )$ is defined is $- \lambda < \tau < \lambda$. Over this region, it can be seen explicitly from the above equation that $g(\tau )$ is antiperiodic, i.e., $g(\tau <0) = - g(\tau + \lambda )$. It is important to note that this property depends on the explicit form of the distribution function. Due to the antiperiodicity, the Fourier transform to imaginary frequencies,
   

   $$g(i \omega _n, {\bf k}) = \int_0^{\tau } d \tau g(\tau , {\bf k}) e^{i \tau \omega _n}$$ (27)

   
   
   is nonzero only for $\omega _n = (2n+1) \pi /\lambda$. These are fermionic Matsubara frequencies. We get,
   

   $$g(i \omega _n, {\bf k}) = \frac{(n_{{\bf k}} - 1)}{i \omega _n - \epsilon _{{\bf k}} } \left( e^{(i \omega _n - \epsilon _{{\bf k}} ) \lambda } - 1 \right)$$

   $$= \frac{1}{i \omega _n - \epsilon _{{\bf k}}},$$ (28)

   
   
   because $e^{i \omega _n \lambda } = -1.$