Goldstone mode stochastization in a quantum Hall ferromagnet S. Dickmann (Cologne, December 14, 2015) Sergey Dickmann Chernogolovka, Russia Zero and nonzero spin excitons are lowest energy excitations in a spin polarized integer quantum Hall system Goldstone mode excitation -- experiment Goldstone mode -- theory OUTLINE Goldstone mode stochastisation theory and experiment
Magneto-exciton (Gor kov, Dzyaloshinskii, 1967) electron-hole interaction electron hole drift Momentum: The Landau-level degeneration is lifted due to electron-hole interaction. Natural quantum number appears due to the translation invariance of the system. Due to translation invariance a special operator appears playing the same role as momentum operator in the absence of magnetic field:. The magneto-exciton energy:
Simplest excitation in a completely polarized QHF Spin exciton (spin wave): First-order approximation results: (Yu. A Bychkov et al, 1981; C. Kallin, B.I. Halperin 1984) q Spin exciton and other exciton-type states are suitable to study by using the excitonic basis set:, a and b are binar indexes: the LL index spin index: Important: Exciton states are eigen states of the Gor kov- Dzyaloshinski P-momentum operator of the magnetoexciton: ^P Q y abq j0i = qqy abq j0i ¾ =" = #
Zero spin exciton if : Nonzero spin exciton is eigen state for any QH system corresponding to the change of spin numbers and is eigen state for the QH ferromagnet to the leading order approximation in. Change of the spin numbers is Important: in spite of these are different states:
Physical meaning of this discrepancy Let be linear dimension of the system. is spacing for numbers means that whereas means that
Ground state: N Á z } { j0; 0i=j """ ::: " i (N Á is the degeneration number of the completely occupied Landau level). The eigenstate jn; 0i=(Q y 0 )N j0; 0i, Spin Goldstone mode where Q y 0 =S = q N Á ( S_=S x - is y ) is the Goldstone condensate: S =N Á =2 and S z =N Á =2 N cos µ =S x =S Another case of deviation is ensemble of nonzero spin excitons: e ie Nt jn; 0iE N = ² Z N quantum precession
Relaxation of spin excitations Spin stochastization or spin relaxation? Total number of spin excitons determines the component: S z = N Á =2 N S z spin Spin stochastization of the GM is a fast process without change of the spin component, i.e. at fixed number of spin excitons and at fixed Zeeman energy. N Stochastization Spin exciton relaxation is change of the S z component --- occurs much slower because related to annihilation processes of spin excitons and release of the energy --- studied experimetally and theoretically (Zhuravlev,SD, Kulik, Kukushkin, PRB 2014). The time > 100 ns.
The GM decay if spin relaxation and stochastization occur simultaneously. If a single mechanism for relaxation and stochastization -> SO coupling + smooth random potential (SD PRL 2004) S
where An elementary stochastisation process: jn; 0i jn; qi = (Q y 0 )N 1 Q y qj0; 0i, Q y q = N 1=2 Á P p e iqxp b y p+ q y a p qy 2 2, a=" and b=#. Both jn; 0i and jn; qi are eigenstates. orthogonal due to the translation invariance. The jn; 0i! jn; qi transition occurs without a change in the S z =N Á =2 N component. At q! 0, the energies of both states (E N ) are the same. However, the states jn; 0i and jn; qi remain di erent even at q! 0 and have di erent: S = N Á =2 and S = N Á =2 1, respectively.
Experiment: the time-resolved Kerr technique is used The transverse component perpendicular to magnetic field hs x (t) + is y (t)i is measured Quantum mechanical averaging A. Larionov, L. Kulik, SD, I Kukushkin. PRB 2015
How is the Goldstone mode is excited experimetally (elementary transition process) Laser pulse Valence band
The present experimental situation If a single electron, then instead of "= ³ 1 0 If one photon is absorbed in the state initially we have a combination of vectors we have % = Ã! cos 2 sin 2 j0; 0i=j """ ::: " i, then j% """:::"i, j"%"":::"i, j""%":::"i, and j""":::"%i, On the base of the indistinguishability principle one finds the initial state jii= ^L (0)j0; 0i, where ^L (0)=cos 2 ^I sin 2 Qy 0 The appearance of a zero-exciton operator is stipulated by the strict `verticality' of the transition process Lk photonk 1, where L is a linear characteristic of 2D density spatial fluctuations
Why only one `tilted electron? The initial state is not an eigenstate. It does not correspond to definite but still corresponds to definite S =N Á =2. S z N N Á, Under the experimental conditions, dephasing process is a single exciton process. the elementary A A sp =N A A sp =N A sp Consider a domain of area smaller than : ( is the area of the laser spot). No more than a single photon is absorbed within the A domain,
Kerr oscilations Our task is to study the temporal evolution of the initial state. jii= cos 2j0; 0i sin 2j1; 0i In the absence of any violation of the translation invariance, the Schroedinger equation results in state at moment t, where jti= ^L (t)j0; 0i The calculation of expectation ^L (t)=cos 2 ^I sin 2 e i² Zt Q y 0 htj ^S x +i ^S y jti= 1 2 sin qn Á e i² Zt explains the Kerr signal oscillations with frequency ² Z =¹h but does not explain the Kerr signal decay.
The stochastization process (slow compared to the precession) This is a conversion of component e i² Zt j1; 0i of state jti to component at. When calculating the we come to a zero result: e i² Zt j1; qi q!0 ht; qjs x +is y jq; ti jt; qi=cos 2 j0; 0i e ieqt sin 2j1; qi hq; tj ^S x +i^s y jt; qi 0. quantum average, at any state E q =² Z + q 2 =2M x is the spin exciton energy at small dimensionless q.) Thus the time of the Kerr signal decay is equal to the transition time of zero exciton conversion into nonzero one with the same energy: j1; 0i j1; qi q!0.
What kind of interaction is responsible for the stochastization? The perturbation responsible for the conversion must be: (i) a spin non-conserving coupling changing the the S z quantum numbers; and (ii) violating the translation invariance. j1; 0i j1; qi q!0. S, but not changing The most likely candidate is a term corresponding to the spatial fluctuations of the g-factor in 2D electron gas, i.e., the Zeeman energy is actually ² Z +g 1 (r)¹ B B, and the perturbation Hamiltinian is ^V g = 1 2 ¹ BB X i à g1 (r i ) 0 0 g 1 (r i )! i
In terms of the `excitonic representation (secondary quantization) the Hamiltonian is^v g = 1 2 N X Á g 1q ¹ B B A q B q where A y q = 1 N Á X p g 1 (r) = X q e iq xp a y p+ q y 2 q g 1q e iqr a p qy 2, and B q = A q (a! b). The `key point is calculation of the matrix element M q = h0jq q j ^V g jq y 0 j0i, then usual procedure 1 = 2¼ ¹h X q jm q j 2 ±(² Z E q )
Let us assume that the g-disorder is Gaussian and governed by correlator K(r)= R g 1 (r 0 )g 1 (r 0 +r)dr 0 =A parameterized by fluctuation amplitude length, so g K(r)= 2 g e r2 = 2 g. g and correlation The result is 1 =¼M x(¹ B B g g ) 2 =2¹hl 2 B :
Some `hidden details Really argument of the function in X is E q ² Z = q2 2M x qj~r'jl B siná where q = (q; Á) : ~r' X is smooth random electrostatic field q Á q jm q j 2 ±(² Z E q ) ~r' 0 < Á < ¼!! jm q j 2 ±(E q ² Z ) q = N Z Á 1 Z ¼ 2¼ qdq dájm qj 2 ± ³ q 2 =2M x qj~r'jl B siná 0 0 NÁ Z 1 Z ¼ 2¼ dq dájm qj 2 2M x ± ³ q 2M x j~r'jl B siná 0 0 ~r' but factor ½ appears due to 0<Á<¼: The result does not depend on