f-sum rule

The retarded current-current correlation function is defined by

$$\Pi_{jj}^R (\omega ) = \langle \langle j,j \rangle \rangle (... ...\rangle \right\vert ^2 \frac{1}{\omega + E_n - E_m + i \eta }.$$ (1)

The corresponding spectral function is

$$\rho_{jj} (\omega ) = - \frac{1}{\pi} {\rm Im} \Pi_{jj}^R (\omega )$$

$$= \frac{1}{Z} \sum_{nm} \left( e^{- \beta E_n} - e^{- \beta E_m} \r... ...ert \langle n \vert j \vert m \rangle \right\vert^2 \delta (\omega + E_n - E_m)$$

$$= \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \omega t} \langle [ j(t) , j] \rangle .$$ (2)

In terms of the spectral function, the real part of the conductivity (which is what enters the f-sum rule) is given by,

$$\sigma (\omega ) = - \frac{1}{V} \frac{1}{\omega } {\rm Im} \Pi_{jj}^R (\omega ) = \frac{\pi}{V} \rho_{jj} (\omega ).$$ (3)

We define the polarization operator $P$ such that $\frac{\partial P}{\partial t} = j$. Then,

$$\rho_{jj} (\omega ) = \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \omega t} \langle [ \frac{\partial P}{\partial t}, j ] \rangle$$

$$= - i \omega \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \omega t} \langle [ P(t), j ] \rangle$$

$$= -i \omega \rho_{Pj} (\omega ),$$ (4)

where

$$\rho_{Pj} (\omega ) = \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \omega t} \langle [ P(t), j ] \rangle .$$ (5)

Then,

$$\int_{- \infty}^{\infty} d \omega \sigma (\omega ) = - \frac{\pi i}{V} \int_{- \infty} ^{\infty} d \omega \rho_{Pj} (\omega )$$

$$= - \frac{\pi i}{V} \langle [P(0), j ] \rangle .$$ (6)

The current operator is given by $j = \sum_{i = 1}^N (e p_i)/m$, where $p_i$ is the momentum of the i-th particle. The polarization operator is given by $P = \sum_{i = 1}^N e r_i$, where $r_i$ is the position of the i-th particle. Since,

$$[P, j]= \frac{e^2}{m} \sum_{ij} [ r_i , p_j] = \frac{N e^2 i}{m},$$

(in units where $\hbar = 1$), we finally get the f-sum rule as

$$\int_{- \infty}^{\infty} \frac{d \omega } {\pi} \sigma (\omega ) = \frac{n e^2}{m},$$ (7)

where $n = N/V$ is the density of particles.