Unformatted text preview: )]
ˆr (1.30) n and noting that
ˆ ˆ ρI = eiHt ρe−iHt
ˆ
ˆ (1.31) and
ˆ e−iHt n� = e−iEn t n� (1.32) Electron Density Response to an Applied Electric Potential 6 we obtain
χ(�, �� , ω ) = −i
rr � 0 dt�� −∞ � �� �φ0 ρ(�)n��nρ(�� )φ0 � ei(En −E0 )t
ˆr
ˆr −(iω −η )t�� (1.33) n − � �� �φ0 ρ(�� )n��nρ(�)φ0 � e−i(En −E0 )t
ˆr
ˆr −(iω −η )t�� (1.34) n All of the time dependence has now been brought up into the exponentials, so it is trivial to
perform the integration over time. This yield the spectral representation of χ(�, r� , ω ):
r�
� � �φ0 ρ(�)n��nρ(�� )φ0 � �φ0 ρ(�� )n��nρ(�)φ0 � �
ˆr
ˆr
ˆr
ˆr
�
χ(�, � , ω ) =
rr
−
(1.35)
ω − (En − E0 ) + iη
ω + (En − E0 ) + iη
n
If there is translational invariance in the sample, then the response function χ(�, �� , ω ) should
rr
be simply a function of the diﬀerence � − �� . In this case, the spatial Fourier transform is simple:
rr
�
1
q r r�
χ(�, ω ) =
q
d� d�� e−i�·(�−� ) χ(� − �� , ω )
rr
rr
(1.36)
V
�
�
� �φ0 ρ(�)n��nρ(−�)φ0 � �φ0 ρ(−�)n��nρ(�)φ0 �
ˆq
ˆq
ˆq
ˆq
=
−
(1.37)
ω − (En − E0 ) + iη
ω + (En − E0 ) + iη
n
where ρ(�) ≡ ρq is given by equations (1.7) and (1.8) in ﬁrst quantized or second quantized
ˆq
ˆ�
notation, respectively.
Since the electron density ρ(�) is a real function, we have the important relation
ˆr
ρ−q = ρ†
ˆ � ˆq
� (1.38) which is a simple consequence of the nature of the Fourier transform. This implies that
�φ0 ρ(�)n��nρ(−�)φ0 � = �nρ† (�)φ0 �2
ˆq
ˆq
ˆq (1.39) Using this along with the relation
�
Im 1
x + iη �
= −πδ (x) (1.40) we arrive at the next important result:
� {�nρ† (�)φ0 �2 δ (ω − (En − E0 ))
ˆq (1.41) −�nρ(�)φ0 �2 δ (ω + (En − E0 ))}
ˆq Im{χ(�, ω )} = −π
q (1.42) n Why are we interested in the imaginary part of χ? The imaginary part of χ gives us
information about dissipation, i.e. the absorption and loss of energy as a result of the interaction
with the probe. We will often use the notation
χ�� (� ω ) = Im{χ(� ω )}
q,
q,
We can plot χ�� (�, ω ) as a function of ω for ﬁxed � (see plot). The location of the peaks
q
q
tells us about the types of excitations being produced. As we will see shortly, it actually turns
out that knowledge of χ�� (�, ω ) is all we need; the real part of χ(� ω ), denoted χ� (� ω ), can be
q,
q
q,
reconstructed from χ�� (�, ω ) alone.
q Sanity Check: Free Fermions 1.4 7 Sanity Check: Free Fermions To convince ourselves that of this formalism is really working, we will try it out on the case of
free fermions, which we studied last semester in 8.511. Now, χ(�, ω ) is simply Π0 (� ω )....
View
Full
Document
This note was uploaded on 02/07/2014 for the course PHYS 8.512 taught by Professor Patricklee during the Fall '09 term at MIT.
 Fall '09
 PatrickLee
 Charge, Polarization

Click to edit the document details