Collective Modes. Feb 20, 2008

Transcription

1 Collectve Modes Feb 20, 2008

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3 Learnng Goals Idea of probng collectve modes phonons, plasmons,... Equatons of moton (collectve coordnate) approach Vral Theorem Sum Rules Lnear Response Very specfc example

4 Setup N atoms n harmonc trap (ansotropc) H = [ p 2 2m + α ] 1 2 mω2 αrα 2 Measure n an experment: + U 0 2 Change omega s and watch response δ(r r j ) <j Q α = j r 2 jα (breathng modes, quadrupole modes) Goal today: calculate frequences

5 Equaton of moton approach Q α = 1 [Q α, H] = 1 m Q α = 1 [ Q α, H] = 4 m r α p α + p α r α (T α V α + 12 U ) T α = p 2 α 2m V α = 1 2 mω2 αq α U = 1 2 U 0 δ(r r j ) (AKA collectve coordnates)

6 Equaton of moton approach Q α = 1 [Q α, H] = 1 m Q α = 1 [ Q α, H] = 4 m r α p α + p α r α (T α V α + 12 U ) = vral T α = p 2 α 2m V α = 1 2 mω2 αq α U = 1 2 U 0 δ(r r j ) (AKA collectve coordnates)

7 Q α = 1 [ Q α, H] = 4 m Vral Theorem (T α V α + 12 U ) In equlbrum Q s tme ndependant T α V α U = 0 Also useful to note H = α T α + α V α + U

8 Closng the equatons Q α = 1 [ Q α, H] = 4 m Case 1: U=0 H α = T α + V α [ Q α = 4ωα 2 Q α H ] α mωα 2 (T α V α + 12 U ) s constant of moton Shft Q: Qα = Q α H α mω 2 α Q α = 4ω 2 α Q α Oscllates at twce trap frequency

9 Closng the equatons Q α = 1 [ Q α, H] = 4 m (T α V α + 12 U ) Case 2: T=0 (good approxmaton for a BEC) U = H β V β d 2 dt 2 Q x Q y Q z = 2 m H 3ω 2 x ω x ω y ω x ω z ω x ω y 3ω 2 y ω y ω z ω x ω z ω y ω z 3ω 2 z Q x Q y Q z constant (shft away) egenvalues gve oscllaton frequences

10 Closng the equatons Q α = 1 [ Q α, H] = 4 m (T α V α + 12 U ) Q: f all omegas are equal, what are oscllaton frequences? Case 2: T=0 (good approxmaton for a BEC) U = H β V β d 2 dt 2 Q x Q y Q z = 2 m H 3ω 2 x ω x ω y ω x ω z ω x ω y 3ω 2 y ω y ω z ω x ω z ω y ω z 3ω 2 z Q x Q y Q z constant (shft away) egenvalues gve oscllaton frequences

11 Closng the equatons Genercally the equatons do not close so easly -- ntroduce more formal tools for dealng wth t Changng trap constants for short tme: H pert = λ α (t)q α Response Q β (t) = T e R t dτ(h 0 +H pert ) Q β e R t dτ(h 0 +H pert ) = Q β 0 + dt χ R αβ(t t )λ α (t ) + χ R αβ = θ(t) [Q β (t), Q α (0)] 0

12 Interacton pcture (only f requested) U = T e R t dτ(h 0 +H ) Ū = e H 0t U t Ū = e H 0t H U = H (t)ū so Ū(t) 1 t dτh (τ)

13 χ R αβ = θ(t) 2 t χ αβ = δ (t) Equatons of moton [Q β (t), Q α (0)] 0 0 [Q β (0), Q α (0)] 0 + δ(t) [ Q β (0), Q α (0)] + θ(t) [ Q β (t), Q α (0)] Substtute n EOM for Q When EOM for Q close -- so do EOM for ch [ Q β (0), Q α (0)] = δ αβ 4 Q α m Q α = 1 [Q α, H] = 1 m r α p α + p α r α

14 Expected Structure: Damped Harmonc oscllator χ(t) Im[χ(ω)] t ω Probe and response commute at t=0 Sum rules: Make Ansatz for ch(t) Ft parameters from t=0 ch and dervatves

15 χ(t) Sum Rules Im[χ(ω)] t ω f-sum rule: M 1 = (always set by knematcs) dω 2π (ω)χ(ω) = [Q (0), Q(0)] = 4 m Q α δ αβ M 3 = dω 2π (ω)3 χ(ω) = [Q (0), Q (0)] = 8 m 2 [ δ αβ (2 T α + 2 V α ) U ]

16 χ(t) Sum Rules Im[χ(ω)] t ω Estmate frequency Other tools: ω 2 egenvalues(m 1 1 M 3) dω χ(ω) M 1 = lm eωt t 0 2π ω = χ(ω = 0) (or use Kramers-Krong) Compressblty sum rule = lm t 0 t Statc Response χ(t )dt

17 2-component Ferm Gas !! /! z 1.555! + /! ! (k f a s )!1 1.7! (k f a s )!1

18 Measurement So What Agreement Calculaton Asde from learnng that our sum rules work -- what do we learn? A: Modes depend on equaton of state -- learn about t

19 Smpler Case Dpole Mode (Kohn Mode) X = x P = p x Shft Trap center: χ = θ(t) Equatons of moton: H pert = λ(t)x X(t)X(0) X(0)X(t) t X = P/m t P = mω 2 xx Soluton X(t) = cos(ωt)x(0) + (m/ω) sn(ωt)p (0)

20 χ = θ(t) Response functon X(t)X(0) X(0)X(t) χ = m ω θ(t) sn(ωt) Useful to ntroduce structure factors χ > = X(t)X(0) (we wll see these next when we dscuss Scatterng) = X 2 cos(ωt) + (m/ω) XP sn(ωt) = m (coth(βω/2) cos(ωt) + sn(ωt)) 2ω χ < = X(0)X(t) = m 2ω (coth(βω/2) cos(ωt) sn(ωt)) know about T

21 Fourer Transform: Detaled Balance χ > (t) = Tre βh e Ht Xe Ht X = Tre βh Xe βh e Ht Xe Ht e βh = χ < (t + β) χ > (ν) = χ < ( ν) = e βν χ > ( ν) Energy added to probe (vanshes at T=0) χ > Energy taken from probe ν