Linear response and Time dependent Hartree-Fock

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where the ɛ p , ɛ h are the Hartree-Fock single particle energies. I calculate only one of the half-off-diagonal terms, p ≠ p ′ , h = h ′ . In a uniform system, it is clear that these matrix elements must vanish because the left state has a different momentum as the right state. In a inhomogeneous system, calculate 〈φ HF | a † h a pH 1 a † p ′a h |φ HF〉 = ∑ αβ = ∑ αβ = ∑ αβ 〈α| h |β〉〈φ HF | a † h a pa † αa β a † p ′a h |φ HF〉〈α| h |β〉〈φ HF | δ αp a † h a β a† p ′a h + a† h a† αa β a p a † p ′a h |φ HF〉〈α| h |β〉 δ αp δ βp ′ = 〈p| h |p ′ 〉 (43) because, to get from the second to the third line, we could use that p ≠ p ′ . The potential term is a little tedious: 〈φ HF | a † h a pV a † p ′a h |φ HF〉 = 1 2 = 1 2 = ∑ βδ ∑ αβγδ ∑ βγδ 〈αβ| v |γδ〉〈φ HF | a † h a pa † αa † β a δ a γa † p ′a h |φ HF〉〈pβ| v |γδ〉〈φ HF | a † h a† β a δ a γa † p ′a h |φ HF〉〈pβ| v |p ′ δ〉 a 〈φ HF | a † h a† β a δ a h |φ HF〉 = ∑ h ′ 〈ph ′ | v |p ′ h ′ 〉 a . (44) The last line (i.e. that we can restrict the sum over the δ states to occupied states follows because of δ were a particle state it would destroy the ground state |φ HF 〉. Thus, the result is 〈φ HF | a † h a pHa † p ′a h |φ HF〉 = 〈p| h |p ′ 〉 + ∑ h ′ 〈ph ′ | v |p ′ h ′ 〉 a (45) In the above, we recover the matrix element of the Hartree-Fock operator. Thus, if all states are generated by the Hartree-Fock equation, then these “half-off-diagonal” term vanish. Since this is a reasonable assumption (how else would one generate the particle wave functions ?) we can go on and assume either p = p ′ , h = h ′ or p ≠ p ′ , h ≠ h ′ . In the latter case, we can have only contributions from the interaction: 〈φ HF | a † h a pV a † p ′a h ′ |φ HF〉 = 1 2 ∑ αβγδ 〈αβ| v |γδ〉〈φ HF | a † h a pa † αa † β a δ a γa † p ′a h ′ |φ HF〉 8
= 1 ∑ 〈pβ| v |γδ〉〈φ HF | a † h 2 a† β a δ a γa † p ′a h |φ HF〉 ′ βγδ = ∑ 〈pβ| v |p ′ δ〉 a 〈φ HF | a † h a† β a δ a h |φ HF〉 ′ βδ = − 〈ph ′ | v |p ′ h〉 a = 〈ph ′ | v |hp ′ 〉 a . (46) This derivation deviates from the previous one only in the last line: Now we can not have β = δ because that would imply h = h ′ which contradicts our assumption. With these manipulations, we have derived our Lagrangian (32) to second order: L(t) = ∑ [ c ∗ ph 〈p| δh(t) |h〉 + c ph 〈h| δh(t) |p〉 ] + i¯h ∑ [ c ∗ ph 2 ph (t)ċ ph (t) − ċ ∗ ph(t)c ph (t) ] ph + 1 ∑ [ ] c ∗ 2 ph c ∗ p ′ h ′ 〈pp′ | v |hh ′ 〉 a + c ph c p ′ h ′ 〈hh′ | v |pp ′ 〉 a pp ′ hh ′ + ∑ c ∗ phc p ′ h ′ [(ɛ p − ɛ h )δ pp ′δ hh ′ + 〈ph ′ | v |hp ′ 〉 a ] (47) pp ′ hh ′ 3 Equations of Motion We are now ready to derive the equations of motion. The action integral (47) has been expanded to second order in the particle–hole amplitudes c ph , c ∗ ph. The linear equations of motion are now obtained from the stationarity principle δS [ c ph (t), c ∗ ph(t) ] = 0 . (48) Recall that the Lagrangian (47) appears under the time integral, we can therefore integrate the time coordinate by parts. Doing the variation with respect to c ph and c ∗ ph gives two equations: 0 = i¯hċ ph + 〈p| δh ext (t) |h〉 + ∑ p ′ h ′ [ 〈pp ′ | v |hh ′ 〉 c ∗ p ′ h ′ + [(ɛ p − ɛ h )δ pp ′δ hh ′ + 〈ph ′ | v |hp ′ 〉 a ] c p ′ h ′ ] 0 = −i¯hċ ∗ ph + 〈p| δh ext (t) |h〉 + ∑ p ′ h ′ [ 〈pp ′ | v |hh ′ 〉 c p ′ h ′ + [(ɛ p − ɛ h )δ pp ′δ hh ′ + 〈ph ′ | v |hp ′ 〉 a ] c ∗ p ′ h ′ ] (49) 9
Page 1 and 2: Linear response and Time dependent
Page 3 and 4: with ω n0 = ω n − ω 0 . We can
Page 5 and 6: or, in second quantized form H 1 =
Page 7: + 1 ∑ c ∗ 4 phc p ′ h ′ 〈
Page 11 and 12: first order in the particle-hole am
Page 13: χ (−) 0 (r, r ′ , ω) = 1 ∑

Linear response and Time dependent Hartree-Fock

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