Unformatted text preview: The
q,
q
ground state for free fermions is just the simple spherical Fermi sea, ﬁlled up to exactly to the
Fermi energy. The excited states are of the form
k,
kq
n� � = hole at � e− at � + ��
k (1.43) These single particlehole excitations are the only�
types of excitations possible in this case,
†
since the external ﬁeld U couples to the density ρq = � c� c� . The matrix elements we need
ˆ�
k k −q k
�
are simple to calculate as well, since all that is required is a ﬁlled initial state below the Fermi
k
kq
sea (with wave vector � and an open state above the Fermi sea with wave vector � + � to jump
into. Thus
�n� ρ† φ0 � = (1 − f� +q )f�
�
k ˆq
k� k (1.44) where f� is 1 if the state with momentum � is occupied in the ground state, and 0 if it is empty.
k
k
Substituting this in, we get
�
�
�
(1 − f� −q )f�
(1 − f� +q )f�
k� k
k� k
−
Π0 (� ω ) =
q,
(1.45)
ω − (�� +q − �� ) + iη ω + (�� +q − �� ) + iη
k
k�
k
k�
�
k Letting
� − � = ��
kqk
� = �� + �
kk
q
we can switch the dummy summation variables on the second term and combine both terms into
one:
Π0 (� ω ) =
q, � (1 − f� +q )f� − (1 − f� )f� +q
k� k
k k�
�
k = �
�
k ω − (�� +q − �� ) + iη
k
k�
f� − f� +q
k
k�
ω − (�� +q − �� ) + iη
k
k� (1.46) (1.47)
(1.48) This is exactly the Lindhardt formula that we derived in 8.511. 1.5 The Correlation Function S (�, t)
r Let’s switch gears now and talk about another object that we will see is related to the response
function. We deﬁne the correlation function
S (�, t) = �φ0  ρH (�, t)ˆH (0, 0) φ0 �
r
ˆrρ (1.49) S (�, t) described ﬂuctuations of the electron density across the sample in space and time. Due
r
�
to the translational invariance of the sample, we arbitrarily set one of arguments to (r� , t� ) = (0, 0) The Correlation Function S (�, t)
r 8 and observe the density correlation with another point (�, t). What we want to show next is that
r
there is a relationship between dissipation and ﬂuctuations.
Fourier transforming S (�, t) in space yields
r
S (� t) = �φ0  ρH (�, t)ρH (−� 0) φ0 �
q,
ˆqˆ
q,
�
†
=
�nρq φ0 �2 e−i(En −E0 )t
ˆ� (1.50)
(1.51) n where the second line follows by inserting a complete set of states between the density operators
ˆ
and acting the e±iHt operators on the eigenstates to the left and the right. Notice that this is
very similar with what we did earlier on our way to deriving the form of the response function.
Now we take the Fourier transform in time:
�
�
qr
S (� ω ) =
q,
d�
r
dt eiωt e−i�·� S (� t)
q,
(1.52)
�
= 2π
�nρ† φ0 �2 δ (ω − (En − E0 ))
ˆq
(1.53)
�
n This expression for S (�, ω ) is identical to the ﬁrst (absorptive) term in the expression for the
q
q
imaginary part of the response function χ�� (�, ω ). This can be restated as th...
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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

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