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52 2 n 0 2 en e0 q 153 n this expression

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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, � † = |�n|ρq |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 |�n|ρ† |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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