Developers:Dielectric function

From OctopusWiki

This is based on G. F. Bertsch et al. Phys. Rev. B 62, 7998 (2000).

For the moment we will only consider the terms of the Lagrangean with $A^\alpha\!$ .

1 Standard formulation
- 1.1 Equation of motion for
- 1.2 The coupling with the TDKS equation
2 Exponential formulation
- 2.1 Equation of motion for
- 2.2 Change of variables
3 Notation

Standard formulation

$L=\int d^3r\sum_k\frac12\left(i\partial^\alpha\phi_k^*(r)-\frac1c A^\alpha\phi_k^*(r)\right)\left(-i\partial^\alpha\phi_k(r)-\frac1c A^\alpha\phi_k(r)\right)$

$+\int d^3r\,d^3r'\,\sum_k e^{i/cA^\alpha(r^\alpha-{r'}^\alpha)}\phi_k^*(r)V(r,r')\phi_k(r')$ $-\frac{\Omega}{8\pi c^2}\dot{A}^\alpha \dot{A}^\alpha$ $-i\int d^3r\,\sum_k\phi_k^*(r)\partial_t\phi_k(r)$

[edit]

Equation of motion for $A^\alpha\!$

$\frac{d}{dt}\frac{\partial L}{\partial \dot{A}^\beta}=-\frac{\Omega}{4\pi c^2}\ddot{A}^\beta$

$\frac{\partial L}{\partial A^\beta}=\int d^3r\sum_k\frac12\left[ \left(-\frac1c\delta^{\alpha\beta}\phi_k^*(r)\right)\left(-i\partial^\alpha\phi_k(r)-\frac1c{}A^\alpha\phi_k(r)\right) + \left(i\partial^\alpha\phi_k^*(r)-\frac1c A^\alpha\phi_k^*(r)\right)\left(-\frac1c\delta^{\alpha\beta}\phi_k(r)\right) \right]$ $+\int d^3r\,d^3r'\,\sum_k \frac{i}c\left(r^\beta-{r'}^\beta\right)e^{i/cA^\alpha(r^\alpha-{r'}^\alpha)}\phi_k^*(r)V(r,r')\phi_k(r')$

$\frac{\partial L}{\partial A^\beta}=\int d^3r\sum_k\frac12\left[ \frac{i}c\phi_k^*(r)\partial^\beta\phi_k(r)+\frac1{c^2}\phi_k^*(r)\phi_k(r)A^\beta -\frac{i}c\phi_k(r)\partial^\beta\phi_k^*(r)+\frac1{c^2}A^\beta\phi_k^*(r)\phi_k(r) \right]$ $+\frac{i}{c}\int d^3r\,d^3r'\,\sum_ke^{i/cA^\alpha(r^\alpha-{r'}^\alpha)} \phi_k^*(r)\left[r^\beta{}V(r,r')-V(r,r'){r'}^\beta\right]\phi_k(r')$

$\frac{\partial L}{\partial A^\beta}=\frac{i}c\int d^3r\sum_k\phi_k^*(r)\partial^\beta\phi_k(r) +\frac1{c^2}A^\beta\int d^3r\sum_k\phi_k^*(r)\phi_k(r)$ $+\frac{i}{c}\int d^3r\,d^3r'\,\sum_ke^{i/cA^\alpha{r^\alpha}} \phi_k^*(r)\left[r^\beta{}V(r,r')-V(r,r'){r'}^\beta\right]e^{-i/cA^\alpha{r'}^\alpha}\phi_k(r')$

$\ddot{A}^\beta=-i\frac{4\pi{c}}\Omega\int d^3r\sum_k\phi_k^*(r)\partial^\beta\phi_k(r) -\frac{4\pi{N}}{\Omega}A^\beta-i\frac{4\pi{c}}{\Omega}\int d^3r\,d^3r'\,\sum_ke^{i/cA^\alpha{r^\alpha}} \phi_k^*(r)\left[\hat{r}^\beta{},\hat{V}\right]e^{-i/cA^\alpha{r'}^\alpha}\phi_k(r')$

Finally:

$\ddot{A}^\beta=\frac{4\pi{c}}{i\Omega}\int d^3r\sum_k\phi_k^*(r)\left\{\partial^\beta\phi_k(r) +e^{i/cA^\alpha{r^\alpha}}\int d^3r'\,\left[\hat{r}^\beta{},\hat{V}\right]e^{-i/cA^\alpha{r'}^\alpha}\phi_k(r')\right\} -\frac{4\pi{N}}{\Omega}A^\beta$

[edit]

The coupling with the TDKS equation

$\frac{\delta L}{\delta \phi^*_m(r')} =\int d^3r\sum_k\frac12\left(-i\delta(r-r')\delta_{km}\partial^\alpha-\frac1c A^\alpha\delta(r-r')\delta_{km}\right)\left(-i\partial^\alpha\phi_k(r)-\frac1c A^\alpha\phi_k(r)\right)$ $+\int d^3r\,d^3r''\,\sum_k e^{i/cA^\alpha(r^\alpha-{r''}^\alpha)}\delta(r-r')\delta_{km}V(r,r'')\phi_k(r'')$

$\frac{\delta L}{\delta \phi^*_k(r)} =\frac12\left(-i\partial^\alpha-\frac1c A^\alpha\right)\left(-i\partial^\alpha\phi_k(r)-\frac1c A^\alpha\phi_k(r)\right)$ $+\int\,d^3r'\,e^{i/cA^\alpha(r^\alpha-{r'}^\alpha)}V(r,r')\phi_k(r')$

$\frac{\delta L}{\delta \phi^*_k(r)}=-\frac12\partial^\alpha\partial^\alpha\phi_k(r)+\frac1{2c^2}A^\alpha A^\alpha + \frac{i}c{A}^\alpha\partial^\alpha\psi_k(r)$ $+e^{i/cA^\alpha r^\alpha}\int\,d^3r'\,V(r,r')e^{-i/cA^\alpha{r'}^\alpha}\phi_k(r')$

[edit]

Exponential formulation

We define

$\psi_k(r)=e^{-i/cA^\alpha{r}^\alpha}\phi_k(r)$

so the Langrangean reads

$L=\int d^3r\sum_k\frac12\partial^\alpha\psi_k^*(r)\partial^\alpha\psi_k(r)$ $+\int d^3r\,d^3r'\,\sum_k\psi_k^*(r)V(r,r')\psi_k(r')$ $-\frac{\Omega}{8\pi c^2}\dot{A}^\alpha \dot{A}^\alpha$ $-i\int d^3r\,\sum_k\psi_k^*(r)\partial_t\psi_k(r) + \frac{\dot{A}^\alpha}{c}\int d^3r\,\sum_k\psi_k^*(r)r^\alpha\psi_k(r)$

[edit]

Equation of motion for $A^\alpha\!$

$\frac{\partial\psi_k(r)}{\partial A^\alpha}=-\frac{i{r}^\alpha}{c}\psi_k$

We will not consider the last two terms, as in the original Lagrangean this term doesn't depend on $A$ , so its contributions will vanish.

$\frac{d}{dt}\frac{\partial L}{\partial \dot{A}^\beta}=-\frac{\Omega}{4\pi c^2}\ddot{A}^\beta$

$\frac{\partial L}{\partial A^\beta}=\frac{i}c\int d^3r\sum_k\frac12\left[ \partial^\alpha\left(r^\beta\psi_k^*(r)\right)\partial^\alpha\psi_k(r)-\partial^\alpha\psi_k^*(r)\partial^\alpha\left(r^\beta\psi_k(r)\right) \right]$ $+\frac{i}c\int d^3r\,d^3r'\,\sum_k\left[\psi_k^*(r)r^\beta{}V(r,r')\psi_k(r')-\psi_k^*(r)V(r,r'){r'}^\beta\psi_k(r')\right]$

$\frac{\partial L}{\partial A^\beta}=\frac{i}c\int d^3r\sum_k\frac12\left[ r^\beta\partial^\alpha\psi_k^*(r)\partial^\alpha\psi_k(r) +\delta^{\alpha\beta}\psi_k^*(r)\partial^\alpha\psi_k(r) -r^\beta\partial^\alpha\psi_k^*(r)\partial^\alpha\psi_k(r) -\partial^\alpha\psi_k^*(r)\delta^{\alpha^\beta}\psi_k(r) \right]$ $+\frac{i}c\int d^3r\,d^3r'\,\sum_k\psi_k^*(r)\left[\hat{r}^\beta, \hat{V}\right]\psi_k(r')$

$\frac{\partial L}{\partial A^\beta}=\frac{i}c\int d^3r\sum_k\frac12\left[ \psi_k^*(r)\partial^\beta\psi_k(r)-\partial^\beta\psi_k^*(r)\psi_k(r) \right]$ $+\frac{i}c\int d^3r\,d^3r'\,\sum_k\psi_k^*(r)\left[\hat{r}^\beta, \hat{V}\right]\psi_k(r')$

$\frac{\partial L}{\partial A^\beta}=\frac{i}c\int d^3r\sum_k\psi_k^*(r)\partial^\beta\psi_k(r)$ $+\frac{i}c\int d^3r\,d^3r'\,\sum_k\psi_k^*(r)\left[\hat{r}^\beta, \hat{V}\right]\psi_k(r')$

Finally:

$\ddot{A}^\beta=\frac{4\pi{c}}{i\Omega}\int d^3r\sum_k\psi_k^*(r)\left\{\partial^\beta\psi_k(r) +\int d^3r'\,\left[\hat{r}^\beta{},\hat{V}\right]\psi_k(r')\right\}$

[edit]

Change of variables

What happens if we consider an actual change of variables $\phi\to\psi$ ?

$\frac{d}{dt}\frac{\partial L}{\partial \dot{A}^\beta}=-\frac{\Omega}{4\pi c^2}\ddot{A}^\beta + \frac1c\frac{d}{dt}\int d^3r\,\sum_k\psi_k^*(r)r^\beta\psi_k(r)=0$