Universal Sound Attenuation in Amorphous Solids at Low-Temperatures

Transcription

1 Universal Sound Attenuation in Amorphous Solids at Low-Temperatures Dervis C. Vural Anthony J. Leggett Physics Department, University of Illinois at Urbana Champaign May 17, 2011

2 Outline Introduction Review of the Tunneling Two-Level Model (TTLS) Problems of the Tunneling Two-Level Model (TTLS) Universalities A More Generic Model Formulation of the Problem Solution to the Two Block Problem Solution to the N-Block Problem Summary and Conclusion Inputs and Outputs of the Theory

3 Review of the Tunneling Two-Level Model (TTLS) The Tunneling Two-Level Model (TTLS) Objects ( with two metastable ) ( configurations ) ɛ 0 γ 0 H = + e 0 ɛ 0 γ

4 Review of the Tunneling Two-Level Model (TTLS) The Tunneling Two-Level Model (TTLS) Objects ( with two metastable ) ( configurations ) ɛ 0 γ 0 H = + e 0 ɛ 0 γ 0 Ωe ξ, ξ = d 2M 0 V

5 Review of the Tunneling Two-Level Model (TTLS) The Tunneling Two-Level Model (TTLS) Objects ( with two metastable ) ( configurations ) ɛ 0 γ 0 H = + e 0 ɛ 0 γ 0 Ωe ξ, ξ = d 2M 0 V Distribution of TTLS parameters: P(E, 0 ) = P 0

6 Review of the Tunneling Two-Level Model (TTLS) Predictions of the TTLS theory Can explain below 1K:

7 Review of the Tunneling Two-Level Model (TTLS) Predictions of the TTLS theory Can explain below 1K: c v T, K T 2

8 Review of the Tunneling Two-Level Model (TTLS) Predictions of the TTLS theory Can explain below 1K: c v T, K T 2 T and ω dependence of sound attenuation and dispersion

9 Review of the Tunneling Two-Level Model (TTLS) Predictions of the TTLS theory Can explain below 1K: c v T, K T 2 T and ω dependence of sound attenuation and dispersion Non-linear acoustic properties (saturation, echoes,... )

10 Problems of the Tunneling Two-Level Model (TTLS) Problems of the Tunneling Two-Level Model (TTLS) Cannot explain behavior T > 1 K.

11 Problems of the Tunneling Two-Level Model (TTLS) Problems of the Tunneling Two-Level Model (TTLS) Cannot explain behavior T > 1 K. What is tunneling? Why 2-states? Why the same P(E, 0 )?

12 Problems of the Tunneling Two-Level Model (TTLS) Problems of the Tunneling Two-Level Model (TTLS) Cannot explain behavior T > 1 K. What is tunneling? Why 2-states? Why the same P(E, 0 )? Interactions only added when necessary. Do they preserve the TTLS structure?

13 Problems of the Tunneling Two-Level Model (TTLS) Problems of the Tunneling Two-Level Model (TTLS) Cannot explain behavior T > 1 K. What is tunneling? Why 2-states? Why the same P(E, 0 )? Interactions only added when necessary. Do they preserve the TTLS structure? Cannot explain universality

14 Universalities Universalities c v, K, Echoes, hole burning, saturation...

15 Universalities Universalities c v, K, Echoes, hole burning, saturation... Scaled thermal conductivity l/λ 150 for T c < 10K l/λ 1 for T > T c

16 Universalities Universalities c v, K, Echoes, hole burning, saturation... Scaled thermal conductivity l/λ 150 for T c < 10K l/λ 1 for T > T c c l /c t, χ l /χ t

17 Formulation of the Problem Phonons and Non-phonon Stuff Ĥ = Ĥphonons + Ĥstuff

18 Formulation of the Problem Phonons and Non-phonon Stuff Ĥ = Ĥphonons + Ĥstuff The stuff couples linearly to strain e ij (=thermal phonons or experimenter s probe) Ĥ stuff = Ĥ 0 + e ij ˆT ij.

19 Formulation of the Problem Phonons and Non-phonon Stuff Ĥ = Ĥphonons + Ĥstuff The stuff couples linearly to strain e ij (=thermal phonons or experimenter s probe) Ĥ stuff = Ĥ 0 + e ij ˆT ij. Stuff exchanges virtual phonons ˆV = Λ ijkl ˆT ij 1 ˆT kl 2.

20 Formulation of the Problem Phonons and Non-phonon Stuff Ĥ = Ĥphonons + Ĥstuff The stuff couples linearly to strain e ij (=thermal phonons or experimenter s probe) Ĥ stuff = Ĥ 0 + e ij ˆT ij. Stuff exchanges virtual phonons ˆV = Λ ijkl ˆT ij 1 ˆT kl 2. Full Non-phonon Hamiltonian: Ĥ many stuff = N s Ĥ (s) 0 + N s<s 3 ijkl (s) Λ ijkl ˆT ij ˆT (s ) ij.

21 Formulation of the Problem Objective: Phonons go infinitely far in a crystal = Contribution to Q 1 comes from stuff.

22 Formulation of the Problem Objective: Phonons go infinitely far in a crystal = Contribution to Q 1 comes from stuff. For N coupled blocks, solve the ultrasonic absorption Q 1 ijkl (ω) n ˆT ij kl 0n ˆT n0δ(e n E 0 ω) T ij mn n T ij m, ˆT ˆT 1 + ˆT 2 + ˆT 3... Many-body non-phonon stress. E n E n1 + E n2 + E n3... Many-body levels.

23 Solution to the Two Block Problem Argument - 1 Consider the average Q 1 1 U UN m 0 dωq 1 m (ω E m )

24 Solution to the Two Block Problem Argument - 1 Consider the average Q 1 1 U UN m 0 dωq 1 m (ω E m ) Consider TrV 2 : Basis Invariant

25 Solution to the Two Block Problem Argument - 1 Consider the average Q 1 1 U UN m 0 dωq 1 m (ω E m ) Consider TrV 2 : Basis Invariant Evaluate it in the uninteracting eigenbasis ijkl i j k l Λ ijkl Λ i j k l T m 1 n 1 m 2 n 2 ij m 1 n 1 Tm kl 1 n 1 T i j m 2 n 2 T k l m 2 n 2 = K Q 0 2 U2 0 Ns 2

26 Solution to the Two Block Problem Argument - 1 Consider the average Q 1 1 U UN m 0 dωq 1 m (ω E m ) Consider TrV 2 : Basis Invariant Evaluate it in the uninteracting eigenbasis ijkl i j k l Λ ijkl Λ i j k l T m 1 n 1 m 2 n 2 and in the interacting eigenbasis ij m 1 n 1 Tm kl 1 n 1 T i j m 2 n 2 T k l m 2 n 2 = K Q 0 2 U2 0 Ns 2 K Q 2 U 2 N 2 s

27 Solution to the Two Block Problem Argument - 1 Consider the average Q 1 1 U UN m 0 dωq 1 m (ω E m ) Consider TrV 2 : Basis Invariant Evaluate it in the uninteracting eigenbasis ijkl i j k l Λ ijkl Λ i j k l T m 1 n 1 m 2 n 2 and in the interacting eigenbasis ij m 1 n 1 Tm kl 1 n 1 T i j m 2 n 2 T k l m 2 n 2 = K Q 0 2 U2 0 Ns 2 K Q 2 U 2 N 2 s Q 1 / Q 1 0 = U 0 /U.

28 Solution to the Two Block Problem Argument - 2 Ĥ = Ĥ0 + ˆV. No correlations between ˆT s TrĤ2 TrĤ2 0 = Tr ˆV 2 LHS: U 0 ω2 g(ω)dω U 0 0 ω 2 g 0 (ω)dω

29 Solution to the Two Block Problem Argument - 2 Ĥ = Ĥ0 + ˆV. No correlations between ˆT s TrĤ2 TrĤ2 0 = Tr ˆV 2 LHS: U 0 ω2 g(ω)dω U 0 0 ω 2 g 0 (ω)dω Expand {g(ω), g 0 (ω)} = { k c kω k, k c 0kω k }: Quantum Many-Body System = Dominating Powers are Large.

30 Solution to the Two Block Problem Argument - 2 Ĥ = Ĥ0 + ˆV. No correlations between ˆT s TrĤ2 TrĤ2 0 = Tr ˆV 2 LHS: U 0 ω2 g(ω)dω U 0 0 ω 2 g 0 (ω)dω Expand {g(ω), g 0 (ω)} = { k c kω k, k c 0kω k }: Quantum Many-Body System = Dominating Powers are Large. N 2 s (U 2 U 2 0 ) = K Q 2 U 2 N 2 s, regardless of c k, c 0k. Now we know U/U 0.

31 Solution to the Two Block Problem Combine Argument-1 and Argument-2 Q 1 can be related to Q 1 0 : Q 1 = [ Q K] 1/2 CK: Found from the tensor components in the coupling constant ˆV = Λ ijkl T ij T kl.

32 Solution to the Two Block Problem Combine Argument-1 and Argument-2 Q 1 can be related to Q 1 0 : Q 1 = [ Q K] 1/2 CK: Found from the tensor components in the coupling constant ˆV = Λ ijkl T ij T kl. 2-Blocks complete. Do the same for 8 blocks.

33 Solution to the Two Block Problem Combine Argument-1 and Argument-2 Q 1 can be related to Q 1 0 : Q 1 = [ Q K] 1/2 CK: Found from the tensor components in the coupling constant ˆV = Λ ijkl T ij T kl. 2-Blocks complete. Do the same for 8 blocks. We know how to iterate blocks to superblocks: The rest is RG.

34 Solution to the N-Block Problem Combine two Argument-1 and Argument-2 RG flow gives logarithmically vanishing Q 1 Q z z Volume dependence Q 1 1/ log 1/2 v/v 0 For v 1m 3, Q

35 Solution to the N-Block Problem Combine two Argument-1 and Argument-2 RG flow gives logarithmically vanishing Q 1 Q z z Volume dependence Q 1 1/ log 1/2 v/v 0 For v 1m 3, Q Universal Q or universal sample size?

36 Solution to the N-Block Problem Microscopic Q 0 dependence Dependence on microscopic parameters: Q Q 0 1 For realistic volumes Q 1 cannot exceed regardless of Q 1 0.

37 Solution to the N-Block Problem Microscopic Q 0 dependence Dependence on microscopic parameters: Q Q 0 1 For realistic volumes Q 1 cannot exceed regardless of Q 1 0. We need Q 1 in the ω MHz-GHz range. What is the ω dependence?

38 Solution to the N-Block Problem Frequency Dependence of Attenuation

39 Solution to the N-Block Problem Frequency Dependence of Attenuation Q 1 (ω) n T 2 n0 δ(e n E 0 ω). Assume that Q 1 changes mostly due to change in E n.

40 Solution to the N-Block Problem Frequency Dependence of Attenuation Q 1 (ω) n T n0 2 δ(e n E 0 ω). Assume that Q 1 changes mostly due to change in E n. Input Q 0 (ω) const. Second order perturbation theory output: Q 1 1 (ω) A ln U 0 /ω Alarm: Interaction too strong!

41 Solution to the N-Block Problem Frequency Dependence of Attenuation Q 1 (ω) n T n0 2 δ(e n E 0 ω). Assume that Q 1 changes mostly due to change in E n. Input Q 0 (ω) const. Second order perturbation theory output: Q 1 1 (ω) A ln U 0 /ω Alarm: Interaction too strong! Heuristic arguments Q 1 (ω) = 1 B + A log U 0 /ω B A Use the value of Q 1 = to find A 350.

42 Solution to the N-Block Problem Frequency Dependence of Attenuation Q 1 (ω) n T n0 2 δ(e n E 0 ω). Assume that Q 1 changes mostly due to change in E n. Input Q 0 (ω) const. Second order perturbation theory output: Q 1 1 (ω) A ln U 0 /ω Alarm: Interaction too strong! Heuristic arguments Q 1 (ω) = 1 B + A log U 0 /ω B A Use the value of Q 1 = to find A 350. Then Q 1 (ω = 1MHz)

43 Solution to the N-Block Problem Revisit K Q 1 depends sensitively on K(c l /c t, χ l /χ t ). c l /c t and χ l /χ t is measurable.

44 Solution to the N-Block Problem Revisit K Q 1 depends sensitively on K(c l /c t, χ l /χ t ). c l /c t and χ l /χ t is measurable. Theoretical justification: Assume u x = e xx r x, u x = e xy r y. Assume central potential i<j φ(r ij).

45 Solution to the N-Block Problem Revisit K Q 1 depends sensitively on K(c l /c t, χ l /χ t ). c l /c t and χ l /χ t is measurable. Theoretical justification: Assume u x = e xx r x, u x = e xy r y. Assume central potential i<j φ(r ij). = [ N a<b χ 0t,l = 2 r ab φ(r ab ) eij φ(r ab ) r ab rab 2 N φ(r ab ) eij 2 a<b ( rab e ij ) 2 ] e ij =0 e ij =0 (1) (2)

46 Solution to the N-Block Problem Revisit K χ 0t = 1 φ (r) r 2 dr r 2 L 3 r 2 x ry 2 dω.

47 Solution to the N-Block Problem Revisit K χ 0t = 1 φ (r) r 2 dr r 2 L 3 r 2 x ry 2 dω. χ 0l = 1 φ (r) r 2 dr r 4 L 3 r 2 x dω.

48 Solution to the N-Block Problem Revisit K χ 0t = 1 φ (r) r 2 dr r 2 L 3 r 2 x ry 2 dω. χ 0l = 1 φ (r) r 2 dr r 4 L 3 r 2 x dω. ct c l = χ0t χ 0l = 1 3

49 Inputs and Outputs of the Theory Input parameters No free parameters. No assumptions on universal something else.

50 Inputs and Outputs of the Theory Input parameters No free parameters. No assumptions on universal something else. Experimentally measured c l /c t and χ l /χ t to find K.

51 Inputs and Outputs of the Theory Input parameters No free parameters. No assumptions on universal something else. Experimentally measured c l /c t and χ l /χ t to find K. The microscopic upper cut-off energy U 0 10K (for which V becomes oscillatory). Q(ω) depends on U 0 only logarithmically.

52 Inputs and Outputs of the Theory Input parameters No free parameters. No assumptions on universal something else. Experimentally measured c l /c t and χ l /χ t to find K. The microscopic upper cut-off energy U 0 10K (for which V becomes oscillatory). Q(ω) depends on U 0 only logarithmically. No assumptions regarding microscopic nature: Arbitrary stuff with arbitrary number of levels.

53 Inputs and Outputs of the Theory Predictions Universality of Q 1

54 Inputs and Outputs of the Theory Predictions Universality of Q 1 T-dependence of Q 1 (ω, T ), K(T ), and c(t )/c fit better than TTLS. l 1 Λ Β c 10 4 c Β K T K Figure: Q vs β = 1/T (left) and c/c vs T (right). Dashed line: TLS (without ad-hoc fit functions). Solid line: Present Theory. Dots: Golding&Graebner (1976)

55 Inputs and Outputs of the Theory Predictions K(T ), fits better than TTLS. K T W cm 1 K T K Figure: K(T ) vs T. Dashed line: TTLS prediction. Solid line: Present Theory. Dots: Zeller and Pohl (1972)

56 Inputs and Outputs of the Theory Acknowledgements Thanks NSF-DMR and NSF-DMR Thanks to Doug Osheroff, Pragya Shukla, Karin Dahmen, Alexander Burin, Zvi Ovadyahu and Philip Stamp for helpful discussions over the years. Thanks to University of British Columbia, University of Waterloo, and Harvard University for generously providing accommodation during long visits.