linear response theory and tddft

Claudio Attaccalitehttp://abineel.grenoble.cnrs.fhttp://abineel.grenoble.cnrs.fr/r/

Linear response theory and TDDFT

CECAM Yambo School 2013 (Lausanne)CECAM Yambo School 2013 (Lausanne)

Motivations: Absorption Spectroscopy

-+

-Many Body Effects!!!

h ν

Motivations(II):Absorption Spectroscopy

Absorption linearly related to the Imaginary part of the MACROSCOPIC dielectric constant (frequency dependent)

Outline

How can we calculate the response of the system? IP, local field effects and Time Dependent DFT

Some applications and recent steps forward

Conclusions

Response of the system to a perturbation →Linear Response Regime

Spectroscopy

From Maxwell equation to the response function

D(r , t )=ϵ0 E (r , t)+P (r , t) ∇⋅E (r , t)=4 πρtot (r , t)∇⋅D(r , t )=4πρext (r , t)

From Gauss's law:Materials equations:

Electric Displacemen

t

Electric Field

Polarization

P(r , t )=∫χ(t−t ' , r , r ')E (t ' r ')dt ' dr '+∫ dt1dt 2χ

2(...)E (t1)E (t 2

)+O (E3)

In general:

For a small perturbation we consider only the first term, the linear response regimeP(r , t )=∫χ(t−t ' ,r , r ')E (t ' r ')dt ' dr '

In Fourier space:

P(ω)=ϵ0 χ(ω)E (ω)=ϵ0(ϵ(ω)−1)E (ω) D(ω)=ϵ0ϵ(ω)E (ω)

Response Functions

ϵ(ω)=D(ω)

ϵ0 E (ω)=

δV ext (ω)

δV tot (ω)

Moving from Maxwell equation to linear response theory we define

ϵ−1

(ω)=δV tot (ω)

δV ext (ω)

V tot ( r⃗ t)=V ext ( r⃗ t )+∫ dt '∫d r⃗ ' v ( r⃗− r⃗ ')ρind ( r⃗

' t ')

V tot ( r⃗ , t)=V ext ( r⃗ , t)+V ind ( r⃗ , t ')

where

The induced charge density results in a total potential via the Poisson equation.

ϵ−1

(ω)=1−vδρind

δV ext

ϵ(ω)=1+vδρind

δV tot

Our goal is to calculate the derivatives of the induced density respect to the external potential

The Kubo formula 1/2

H=H 0+H ext (t)=H 0+∫ d rρ(r) V ext (r , t)

We star from the time-dependent Schroedinger equation:

i∂ψ

∂ t=[H 0+H ext (t)] ψ(t)

...and search for a solution as product of the solution for Ho plus an another function (interaction representation)...

ψ̃(t)=ei H 0 t ψ(t )

i∂ ψ̃(t)

∂ t=eiH 0 t H ext(t)e

−iH 0 t ψ̃(t )= H̃ ext (t) ψ̃(t )

...and we can write a formal solution as: ψ̃(t)=e−i∫t 0

t̃H ext(t)dt

ψ̃(t 0)

Kubo Formula (1957)

r t ,r ' t '=ind r , t ext r ' , t '

=−i ⟨[ r , t r ' t ' ]⟩

The Kubo formula 2/2

ψ̃(t)=e−i∫t 0

t̃H ext(t)dt

ψ̃(t 0)=[1+1i∫t 0

tdt ' H ext (t ')+O (H ext

2 )] ψ̃(t 0)

For a weak perturbation we can expand:

And now we can calculate the induced density:

ρ(t )=⟨ ψ̃(t )∣ρ̃(t)∣ψ̃(t )⟩≈⟨ρ⟩0−i∫t 0

t⟨[ρ(t) , H ext (t ')]⟩+O(H ext

2 )

ρind (t)=−i∫t0

t

∫ dr ⟨[ρ(r , t ) ,ρ(r ' t ')]⟩ϕext (r ' , t ')

...and finally......and finally...

The linear response to a perturbation is independent on the perturbation and depends only on the properties of

the sample

How to calculated the dielectric constant

i∂ρ̂k (t )

∂ t=[H k+V eff , ρ̂k ] ρ̂k (t )=∑i

f (ϵk , i)∣ψi , k ⟩ ⟨ψi , k∣

The Von Neumann equation (see Wiki http://en.wikipedia.org/wiki/Density_matrix)

r t ,r ' t '=

indr , t

ext r ' , t ' =−i ⟨[ r , t r ' t ' ]⟩We want to calculate:

We expand X in an independent particle basis set

χ( r⃗ t , r⃗ ' t ')= ∑i , j , l ,m k

χi , j , l ,m, kϕi , k (r )ϕ j ,k∗ (r )ϕl ,k (r ' )ϕm ,k

∗ (r ')

χi , j , l ,m , k=∂ρ̂i , j , k

∂V l ,m ,k

Quantum Theory of the Dielectric Constant in Real Solids

Adler Phys. Rev. 126, 413–420 (1962)

What is Veff ?

Independent Particle

Independent Particle Veff = V

ext

∂

∂V l ,m,keff i

∂ρi , j ,k

∂ t= ∂

∂V l ,m, keff [H k+V eff , ρ̂k ]i , j , k

Using: {H i , j ,k = δi , j ϵi(k)

ρ̂i , j , k = δi , j f (ϵi ,k)+∂ ρ̂k

∂V eff⋅Veff+....

And Fourier transform respect to t-t', we get:

χi , j , l ,m, k (ω)=f (ϵi ,k)−f (ϵ j ,k)

ℏω−ϵ j ,k+ϵi ,k+iηδ j ,lδi ,m

i∂ρ̂k (t )

∂ t=[H k+V eff , ρ̂k ]

χi , j , l ,m , k=∂ρ̂i , j , k

∂V l ,m ,k

Optical Absorption : IP

Non Interacting System

δρNI=χ0δV tot χ

0=∑

ij

ϕi(r)ϕ j*(r)ϕi

*(r ' )ϕj(r ')

ω−(ϵi−ϵ j)+ iη

Hartree, Hartree-Fock, dft...

=ℑχ0=∑ij

∣⟨ j∣D∣i⟩∣2δ(ω−(ϵ j−ϵi))

ϵ''(ω)=

8π2

ω2 ∑i , j

∣⟨ϕi∣e⋅v̂∣ϕ j ⟩∣2δ(ϵi−ϵ j−ℏω)

Absorption by independent Kohn-Sham particles

Particles are interacting!

Time-dependent Hartree (local fields)

Time-dependent Hartree(local fields effects)

Veff = V

ext + VH

V tot r t =V ext r t ∫dt '∫ d r ' v r−r 'ind r' t '

The induced charge density results in a total potential via

the Poisson equation.

r , r ' , t−t ' =r , t

V ext r ' , t ' =

r , t V tot r ' ' , t ' '

V tot r ' ' , t ' '

V ext r ' , t '

χ( r⃗ t , r⃗ ' t ')=χ0( r⃗ t , r⃗ ' t ')+∫ dt1dt2∫ d r⃗1d r⃗2χ0 ( r⃗ t , r⃗1 t1)v (r⃗1−r⃗2)χ(r⃗2 t2 , r⃗' t ')

ind

V indV tot

0r ,r '=

ind r , t

V tot r' t '

Screening of the external perturbation

Time-dependent Hartree (local fileds)

PRB 72 153310(2005)

Macroscopic Perturbation....

ϵ−1(ω)=1+v

δρind

δV ext

ϵ(ω)=1−vδρind

δV tot

Which is the right equation?

...microscopic observables

Not correct!!

Macroscopic averages 1/3

In a periodic medium every function V(r) can be represented by the Fourier

series:V (r)=∫dq∑G

V (q+G)ei(q+G)r

orV (r)=∫dqV (q , r)eiqr=∫ dq∑G

V (q+G)ei(q+G )r

Where: V (q ,r )=∑GV (q+G)eiGr

The G components describe the oscillation in the cell while the q components the oscillation larger then L

Macroscopic averages 2/4

Macroscopic averages 3/4

Macroscopic averages 4/4

The external fields is macroscopic, only components G=0

Macroscopic averages and local fields

If you want the macroscopic response use the first equation and then invert the dielectric

constant

ϵ−1

(ω)=1+vδρind

δV ext

ϵ(ω)=1−vδρind

δV tot

Local fields are not enough....

What is missing?

Two particle excitations, what is missing?Two particle excitations, what is missing?electron-hole interaction, exchange, higher order effects......

The DFT and TDDFT way

DFT versus TDDFT

DFT versus TDDFT

V ext=0 V extV HV xc

q ,=0q ,

0q ,vf xcq , q ,

TDDFT is an exact theory for neutral

excitations!

Time Dependent DFT

V eff (r , t )=V H (r , t)+ V xc(r , t)+ V ext (r , t)

Interacting System

Non Interacting System

Petersilka et al. Int. J. Quantum Chem. 80, 584 (1996)

I=NI= I

V ext

0=NI

V eff

... by using ...

=01

V H

V ext

V xc

V ext

vf xc

i∂ρ̂k (t )

∂ t=[H KS , ρ̂k ]=[H k

0+V eff , ρ̂k ]

Time Dependent DFT

Choice of the xc-functional

...with a good xc-functional you can get the right spectra!!!

Summary

● How to calculate linear response in solidsmolecules

● The local fields effects: time-dependent Hartree

● Correlation problem: TD-Hartree is not enough!

● Correlation effects can be included by mean of TDDFT

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References!!!

Electronic excitations: density-functional versus many-body Green's-function approachesRMP, vol 74, pg 601, (2002 ) G. Onida, L. Reining, and A. Rubio

On the web:● http://yambo-code.org/lectures.php● http://freescience.info/manybody.php● http://freescience.info/tddft.php● http://freescience.info/spectroscopy.php

Books: