RPA in infinite systems Translational invariance leads to conservation of the total momentum, in other words excited states with different total momentum don t mix So polarization propagator diagonal in total momentum Consider creation of ph pair in an infinite system, = a p,m b p,m Total momentum N 0 = a p,m ( ) 2 +m a p, m Q = p + p q N 0 Hamiltonian commutes with total momentum so cannot change it Rewrite polarization propagator using with relative momenta of final and initial state Q = p + p p = 2 (p p ) p = 2 (p p )
Polarization propagator Keep also discrete quantum numbers for fermions (like spin) for the noninteracting polarization propagator (0) (, ;, ; E) with (0) (p m, (p m ) ; p m, (p m ) ; E) m,m m,m Q,Q p,p (0) (p; Q,E) (0) (p; Q,E) = ( p + Q/2 p F ) (p F p Q/2 ) E [ (p + Q/2) (p Q/2)] + i (p F p + Q/2 ) ( p Q/2 p F ) E +[ (p Q/2) (p + Q/2)] i For exact polarization propagator we only have (, ;, ; E) (p m, (p m ) ; p m, (p m ) ; E) Q,Q (p,m,m ; p,m,m ; Q,E) and similarly for the RPA approximation, so we can transcribe
Transcription of RPA RP A (, ;, ; E) = (0) (, ;, ; E) + (0) (, ; E) for infinite system into RP A (p,m,m ; p,m,m ; Q,E)= m,m m,m p,p + (0) (p; Q,E) V ph Q, p; m m V ph Q, p ; m m p m m RP A (p,m,m ; p,m,m ; Q,E) RP A (, ;, ; E) (0) (p; Q,E) Uncoupled ph states can be written as Q, p; m,m = a p+q/2,m b Q/2 p,m N 0 With only spin coupled states read Q, p; S, M S = ( m 2 m 2 SM S ) Q, p; m,m m,m
Coupling for propagator For interaction Coupled RPA (p,s,m S ; p,s,m S; Q,E)=m,m,m,m ( 2 m 2 m SM S ) ( m 2 m 2 S MS) (p,m,m ; p,m,m ; Q,E) Q, p; S, M S V ph Q, p ; S,MS = ( m 2 m 2 SM S )( m 2 m 2 S MS) m,m,m,m Q, p; m,m V ph Q, p ; m,m similar to angular momentum coupling in a finite system Evaluate for a local central interaction without spin dependence Remember V ph V Direct term: momentum transfer Q = p + p Exchange term: momentum transfer p p
Consider ph Transformation V ph V ( ) /2+m ( ) /2+m p + Q/2 m p Q/2 m V p Q/2 m p + Q/2 m Momentum conservation already OK Relative momentum final state p r = 2 [p + Q/2 (p Q/2)] = 2 (p p )+Q/2 Relative momentum initial state p r = 2 [p Q/2 (p + Q/2)] = 2 (p p ) Q/2 Momentum transfer p r p r = 2 (p p )+Q/2 [ 2 (p p ) Q/2] = Q Momentum transfer exchange (p r p r ) E = 2 (p p )+Q/2 [ 2 (p p)+q/2] = p p
Direct term Recoupling Q, p; S, M S V ph Q, p ; S,M S D = ( V m 2 m 2 SM S )( m 2 m 2 S M S ) m,m,m,m S p,m p for Coulomb Similar to Pandya transformation in a finite system (Ch.3.7) Using 6j symbol ( 2 m 2 m S p M p )( 2 m 2 m S p M p ) ( ) /2+m ( ) /2+m V (Q) V (Q) =4 e 2 2 /Q 2 Q, p; S, M S V ph Q, p ; S,M S D = V S,S M S,M S ( ) Sp+ (2S p + ) S p S, S p combinations (0,0), (,0), (0,), and (,) yield 2 2 S 2 2 S p V (Q) 2, 2, 2, and 6
Direct and exchange term Summing over total particle spin then yields Q, p; S, M S V ph Q, p ; S,M S D = V S,S M S,M S S,0 2 V (Q) only acting for total ph spin 0! Exchange can be calculated similarly with the result Q, p V SM S ph Q, p = V [ S,0 2 V (Q) V (p p )] suppressing spin projection (diagonal and independent) Normally total spin is conserved (special case tensor interaction) RPA in coupled spin format RP A SM S (p, p ; Q,E)= p,p + (0) (p; Q,E) p (0) (p; Q,E) Q, p V SM S ph Q, p RP A SM S (p, p ; Q,E)
Including exchange: RP A SM S (p, p ; Q,E)= p,p Keep only direct term Exchange can sometimes be neglected as for Coulomb Define in that case and + (0) (p; Q,E) RP A SM S (Q,E) (0) (Q,E) p (0) (p; Q,E) Q, p V SM S ph V p p Q, p V p (0) (p; Q,E) RP A SM S (p, p ; Q,E) to arrive at RP A SM S (Q,E)= (0) (Q,E)+ (0) (Q,E) V SM S ph (Q) Requires calculation of so-called Lindhard function RP A SM S (p, p ; Q,E) RP A SM S (Q,E)
Calculate (0) (Q,E) = Lindhard function dp ( p + Q/2 p F ) (p F p Q/2 ) (2 ) 3 E [ (p + Q/2) (p Q/2)] + i (p F p + Q/2 ) ( p Q/2 p F ) E +[ (p Q/2) (p + Q/2)] i Can be done for real and imaginary part separately Here, imaginary part (then real part from dispersion relation) Consider only energies Use E ± i = P E i (E) Only first term contributes E>0 Im (0) dp (Q,E)= (2 ) 3 ( p + Q/2 p F ) (p F p Q/2 ) (E [ (p + Q/2) (p Q/2)])
more Lindhard Consider momentum dependence of ph energy difference (p + Q/2) (p Q/2) = p Q pq cos = m m using only kinetic energies Total ph momentum along z-axis then p cos = p z So energy conservation yields E = p zq m Step functions in numerator represent two spheres in the relative ph momentum variable displaced from the origin by Two cases: spheres overlap don t overlap Q<2p F Q>2p F ±Q/2
a) overlap b) no overlap Energy condition shows there is a minimum and maximum energy for given total momentum For b) min at p z = Q/2 p F max at p z = Q/2+p F Spheres Illustration of the constraints imposed by the step functions. The condition p+q/2 >p F corresponds to the area outside the lower sphere in both figures, while the condition p Q/2 <p F only allows contributions from inside the top sphere. Part a) illustrates the case Q<2p F, with overlapping spheres, while b) is appropriate for Q>2p F, when there is no overlap. The gray area indicates the allowed region for the integration over p. The dashed lines in part a) and b) identify possible energy values for which a nonzero imaginary part is obtained. This condition expresses energy conservation, which shows that the only contributions to the imaginary part of the propagator correspond to the part of the dashed line inside the gray area of the top sphere. Limiting values of the integration variable p are indicated by the arrows with the corresponding labels.
Development for nonoverlapping spheres Energy condition for b) therefore for nonvanishing imaginary part given by Q 2 Qp F Q2 <E< 2m m 2m + Qp F m Integrations straightforward Azimuth angle yields Other angle: use delta function E pq m cos = m pq Remaining integration Im (0) 2 m p+ (Q,E)= 3 (2 ) 3 Q p with p = me Q and p + = p 2 F Q 2 /4+mE /2 2 me pq dp p = 3 cos m 8 Q (see spheres) p 2 F me Q Q 2 2
Energies that allow Overlapping spheres The corresponding energy domain is Remaining interval =0 0 <E< Qp F m generate the same result as before Q 2 2m Q 2 2m Qp F m Lower limit in momentum integral changes to Final result Im (0) (Q,E)= m 3 4 Q me Q2 <E< 2m + Qp F m p = p 2 F Q 2 /4 me /2 For real part see FW or use dispersion relation in the form (0) (Q,E)= with limits E = E E + de Im (0) (Q,E ) E E + i Q 2 2m 0 Q<2p F Qp F Q>2p F m + and Inserting imaginary part confirms validity of dispersion relation E de Im (0) (Q,E ) E + E E i E + = Q2 2m + Qp F m
Result Re Special limits Real part Lindhard function (0) (Q,E)= 3 mp F + p F 2Q p F 2Q 4 2 me Qp F me Qp F + Fixed momentum, limit energy to zero yields no imaginary part Q 2p F Q 2p F 2 2 but real part is given by Re (0) (Q, 0) = mp F 3 4 2 + p 2 F Q Q 2p F Fixed energy, momentum to zero also no imaginary part Real part in that limit Re (0) mp F 2 Q 2 p 2 F (Q,E) 3 4 2 3 m 2 E 2 ln 2Qp F +2mE Q 2 2Qp F 2mE + Q 2 ln 2Qp F +2mE + Q 2 2Qp F 2mE Q 2 ln 2Qp F Q 2 2Qp F + Q 2
Different shapes for the imaginary part as a function of energy depending on the magnitude of the momentum Energy in units of the Fermi energy Im part divided by factor mp F /(8 3 ) Inverted parabola when spheres don t overlap (widening with Q) otherwise straight line Plots Illustrates probability for noninteracting system to absorb energy and momentum
How about this? Quasielastic electron scattering (read from right to left) width proportional to Fermi momentum --> ~260 MeV
Real and imaginary parts im -> full line re E>0 term -> dashed re E<0 -> dotted re total -> long-dashed Factor ±mp F /(8 included (- for im) Scales are not all identical 3 ) More plots
Plasmons in the electron gas Solution of RPA equation generates important physics for condensed matter systems Coulomb interaction (note units) Q, p; S, M S V C ph Q, p ; S,M S = V S,S M S,M S { S,0 2V (Q) V (p p )} Since Q is conserved, direct term dominates Neglect exchange OK for small So solving yields here = S,S MS,M S 4 e 2 2 RP A SM S (Q,E)= (0) (Q,E)+ (0) (Q,E) V SM S ph (Q) RP A S=0 (Q, E) = Q (0) (Q, E) 2V (Q) (0) (Q, E) V S,0 2 Q 2 p p 2 RP A SM S (Q,E)
Plasmons Probability density for absorption of energy and momentum where there is imaginary part so similar domain as noninteracting propagator Divergence of Coulomb at small momentum leads to an untrapped solution as in the schematic model Condition for collective state Note: only real part of noninteracting propagator involved so a real pole in propagator Think of Lehmann representation written as RP A S=0 (Q, E) = E A p (Q) E p + i 2V (Q) Re B p (Q) E + E p (0) (Q, E p )=0 Insert in RPA equation and consider energies near plasmon pole Result eigenvalue equation for plasmon energy in the form of i + continuum
Figure contains ph continuum boundaries and location of plasmon for r s =2 Insert small momentum limit of real part of polarization propagator for fixed plasmon energy: E p = 4 3 e 2 2 p 3 F m 3 /2 = 4 e2 2 Coulomb divergence has been cancelled Appearance of collective state above continuum unique for Coulomb interaction Location of plasmon m /2
Results for r s = Plasmon at zero momentum RPA response E p F 0.94 r s Merges into continuum at Q c /p F 0.56 Noninteracting dashed Interacting solid Plasmon at small momentum collects all the EWSR (f-sum rule)
f-sum rule Operator that transfers momentum Second quantization Calculate EWSR S(ˆ(Q)) = n N(Q) = N i= e iq r i/ Q ˆ(Q) = (E N n E N 0 ) N n ˆ(Q) to the system pm s a pm s a p Qms N 0 2 as double commutator S(ˆ(Q)) = N 0 ˆ (Q), Ĥ, ˆ(Q) 2 N 0 Use e.g. parity to show S(ˆ(Q)) = S(ˆ( Q)) noting ˆ (Q) =ˆ( Q) No velocity-dependent interaction implies only contribution of kinetic energy in double commutator Use first quantization S(ˆ(Q)) = N Q2 2m
Relation with polarization propagator Rewrite excitation operator ˆ(Q) = 2 ( m 2 p m 2 h 0 0) a pm p b Q pm h p m p m h so S=0 ph states are excited Use Lehmann representation to obtain S(ˆ (Q)) = 2 de E Im (p, p ; S = 0; Q,E) p p 0 Define dynamic structure function for S=0 S S=0 (Q,E)= V Im S=0 (Q,E) then f-sum rule S(ˆ(Q)) = keeping in mind SM S (Q,E) 0 de E S S=0 (Q,E)=N Q2 2m V p p SM S (p, p ; Q,E)
Dielectric constant connection See Pines (963) for connection of dielectric constant (connecting displacement field and electric field) and f-sum rule Im (Q,E) = 4 2 e 2 2 Q 2 S(Q,E) and therefore f-sum rule can be written as de E Im (Q,E) = 2 E2 p 0 using classical plasmon energy Determination of transition strength to plasmon requires numerators at plasmon pole RP A A p (Q) B p (Q) S=0 (Q, E) = + continuum E E p + i E + E p i In standard fashion one obtains with E p (Q) plasmon energy A p (Q) = Re (0) (Q, E p (Q)) 2V (Q) Re (0) (Q, E) E E=E p (Q)
Contribution of plasmon to f-sum rule In RPA plasmon exhausts the f-sum rule for Q 0 Still true at Q/p F =0. At 0.5 22% in the continuum At 0.6 it has merged Difference between RPA and noninteracting response small at and decreasing
Inelastic electron scattering on metals Global features consistent with RPA plasmons Li has r s =3.25 energy 4.74 ev E p with Fermi Using 0.94 r s generates F 8 ev in reasonable agreement with experiment Narrow peak and increasing energy are also seen Transition to the continuum expected at 0.9 Å and the width correspondingly increases Width at small momentum...
Phys. Rev. B40, 08 (989) Other alkali system
Screening of the effective e-e interaction Sum all divergent contributions that modify Coulomb interaction W (0) (Q, E) =V (Q)+V (Q)2 (0) (Q, E)V (Q) Use limit of real part of noninteracting polarization propagator for small energy W (0) 4 e 2 2 (Q, 0) = Q 2 + p 2 TF g(q/p F ) with and +V (Q)2 (0) (Q, E)V (Q)2 (0) (Q, E)V (Q)+... = V (Q)+V(Q)2 (0) (Q, E)W (0) V (Q) (Q, E) = 2V (Q) (0) (Q, E) p 2 TF = 4r s(9 /4) /3 p 2 F 2x g(x) = 2 4 x2 ln 2 x + 2 x Leads to finite-range, Yukawa-type interaction but is in general retarded