THE CASES OF ν = 5/2 AND ν = 12/5. Reminder re QHE:

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1 LECTURE 6 THE FRACTIONAL QUANTUM HALL EFFECT : THE CASES OF ν = 5/2 AND ν = 12/5 Reminder re QHE: Occurs in (effectively) 2D electron system ( 2DES ) (e.g. inversion layer in GaAs - GaAlAs heterostructure) in strong perpendicular magnetic field, under conditions of high purity and low ( 250mK) temperature. If df. l m (/eb) 1/2 ( magnetic length ) then area per flux quantum h/e is 2πl 2 m, so, no. of flux quanta = A/2πl 2 m (A area of sample). If total no. of electrons* is N e, define ν N e /N Φ ( filling factor ) QHE occurs at and around (a) integral values of ν (integral QHE) and (b) fractional values p/q with fairly small ( 13) values of q (fractional QHE). At ν th step, Hall conductance xy quantized to νe2 /h and longitudinal conductance xx = 0 xy 3 /(e 2/ h) 2 1 H xx Nb : (1) Fig, shows IQHE only (2) expts usually plot R xy vs B( 1 ν ) so general pattern is same but details different nh/eb * strictly, no./spin : valley (but see below)

2 Theoretical considerations : free spinless e s in magnetic field Energy levels are those of SHO with freq. ω c eb/m : E n = (n + 1/2)ω e, n = Landau level (LL) index. Massively degenerate (N Φ states per LL). Specific form of states depends on choice of gauge for A(r) magnetic v.p. e.g. with radial gauge (A r = 0, A θ = 1 rb) eigenstates of 2 n th level are approx.* of form where Ψ nl (r, θ) = exp ilθ f n (r r l ) n th SHO eigenfunction r l = (2ll 2 m )1/2 i.e. l th orbit encloses l quanta of flux (hence has zero kinematic ang. mom.). Note that states in higher LL s (n 1) overlap more that those in LLL (lowest LL, n = 0) Effects of Coulomb interaction r θ dielectric const. A crude estimate of the Coulomb interaction E coul is e 2 /4π 0 r 0 r 0 av. n. n. distance. How does this compare to the cyclotron energy (LL splitting) E cyc ω c (eb/m )? Using 1/r 0 = (νb/2π) 1/2 lm 1 and a 0 4π 0 2 /me 2, we find E coul /E cyc = ν Bohr radius 0.5Å l m 1/2 1 m 2π m (GaAs, m /m 0.07, 13) a0 = ν 1/2 l m /a Since l m 250Å for B = 1T and scales as B 1/2 this means that for µ 1 E coul /E cyc (4T/B) 1/2. Hence not entirely obvious that LL mixing is negligible for B in typical range ( 5 20T ); but numerical computation indicates that it is indeed rather small.

3 EFFECTS OF SPIN (B plane) The Zeeman energy E Z gµ B B and cyclotron energy E cyc = eb/m 2 both scale linearly with B and for an e in free space (g = 2, m = m) are comparable, but because of the small values of g and m layers E Z E cyc, eg. E Z 1 60 E cyc (GaAs) Hence if we neglect Coulomb interactions we have: n = 1, n = 1, ν < 1 : LLL partially occupied 1 < ν < 2 : LLL filled and n = 0, LLL partially occupied n = 0, E cyc } 2 nd L L E z} L L L 2 < ν < 3 : LLL and filled, n = 1 partially occupied etc. If LLL is completely filled, normally assume it is inert, so can treat n = 1 LL as if it were LLL. (e.g.) ν = 7/3 is equivalent to ν = 1/3. : Coulomb interactions may change this picture qualitatively, e.g. because of probability of exchange (spindependent) effects. Note: It is possible to enhance E Z relative to E cyc by making field inclined to r : the orbital motion sees only B, while the spin sees B. So problem is equivalent to the original one with B B, g g/ cos(θ) B θ B ~ ( : may be other effects of turning B )

4 SYSTEMATICS OF FQHE FQHE is found to occur at and near ν = p/q, where p and q are mutually prime intergers. By now, 50 different values of (p.q). Generally, FQHE with large values of q tend to be more unstable against disorder and temperature. eg. plateaux Possible approaches to identification of phases : narrower, ρ xx 0 (a) analytic, trial wf (eg Laughlin) (b) numerical, few-electron (typically N 18) (c) via CFT conformal field theory (d) experiment : alas, cannot usually measure much other than electrical props! ideally, would at least like to know total spin of sample, but... The simplest FQHE states (Laughlin states) : reminders The Laughlin states have p = 1, q = odd integer, i.e. ν = 1/(2m + 1), m = integer (e.g. ν = 1 3, 1 5,...) These are well accounted for by the Laughlin w.f. Ψ N = Π N i<j (z i z j ) q exp i z i 2 /4l 2 m q = 1 ν = 2m + 1 (z x + iy) Elementary excitations are quasiholes generated by multiplying GSWF by Π N i=1 (z i η 0 ) (hole at η 0 ). They have charge e = νe and are abelian anyons : Ψ(1, 2) = exp iπνψ(2, 1) Fairly convincing evidence for fractional charge (ν = 1/3), some evidence for fractional statistics.

5 MORE COMPLICATED (NON-LAUGHLIN) FQHE STATES A. States such as ν = 2/3, 4/5 can be viewed as the particle-hole conjugates of ν = 1/3, 1/5 (i.e. start from filled LLL and define a Laughlin wf in terms of hole operators). B. States such as ν = 4/3, 7/3 : assume 1 or more LL s completely filled, and apply Laughlin treatment to first incompletely filled LL (e.g. ν = 7/3 state has LLL filled for both spins, ν = 1/3 in n = 1 LL). ( : analytic form of Laughlin wf must be different) C. But what about say ν = 2 5 or 11 7? In literature 2 main approaches : 1. Hierarchy scheme* Suppose we start with the Laughlin state for given integer m s and change the flux so that the condition N e = N φ /m s is no longer fulfilled. The system responds by generating an appropriate number of particles or holes : define a quantity α s+1 +1 for e s, -1 for holes. It is convenient to keep a separate notation q s for the quasiparticle charge, even though for a Laughlin state it is just 1/m s. Consider for definiteness the hole case (α s+1 = 1) and now treat the hole coordinates Z i as themselves QM variables : evidently the wave formation must be symmetric between the different Z i. (though antisymmetric in the original e coordinates z i ) *F.D.M. Haldane, PRL 51, 605 (1983) B.I. Halperin, PRL 52, 1583 (1984)

6 On the other hand, we know that an adiabatic transport of a hole i around a second one j gives rise to the phase factor exp iπ/m s. We can eliminate this by applying a gauge transformation Ψ(Z) exp iq s e A(r)dr/Ψ(Z), but this then introduces an extra term into the Hamiltonian (p p q s ea). Now we consider a Laughlin-type w.f. for the quasiparticles : Ψ{Z i } = Π N s i<j (Z k Z l ) 2p s+1 exp q s i Z2 i /4l2 m integer where the factor q s enters the exponent because the effective magnetic length for the qp s is (/e B) 1/2 = (/ q s eb) 1/2 = q s 1/2 l m. By requiring consistency in the no. of quasiholes generated and the choice of the magnetic v. p. A(r ) we get the conditions (also for quasiparticle-like excitations) m s+1 = 2p s+1 α s+1 /m s ν s+1 = ν s + α s+1 q s q s m s+1 q s+1 = α s+1 q s /m s+1 (definition) If we start with the IQHE (a special case of the Laughlin state, with m 0 = q 0 = α 1 = 1, ν 0 = 0, we generate at the first iteration the Laughlin states (m 1 = odd integer ( m), q 1 = ν 1 = 1/m). At the second iteration we get, and so iteratively ν = 2p/(2mp ± 1), q = ±1/(2mp ± 1) ν = m+ 1 α 1 2p 1 + α 2 2p (α s = ±1) Note : (a) if we write ν = n 1 /n 2, then e /e q = 1/n 2 (b) denominator is always odd (c) qps still abelian

7 Alternative scheme : composite fermions. Fundamental observation : Uniform e liquid with each e carrying with it is a flux αϕ 0 is equivalent in a mean field sense to an external magnetic field which produces average flux of ϕ 0 per electron. So, start with IQHE, ν = p = integer. Now imagine attaching a flux α to each e. Then because of the AB effect, exchange of 2 e s produces a phase factor ( 1) 1+α. But if α = ±2m, m integral, this has no effect. I.e. we can write N eff ϕ = N ϕ ± 2mN e Now think of the situation as the integral QHE at N e = pn eff ϕ This corresponds to a true filling factor ν N e /N ϕ of ν = p/(2mp ± 1) Jain s trial w.f. for these states is (p > 0) Ψ 2m p = Π i<j (z i z j ) 2m Ψ p w.f. for p filled LL s. Note that for p = 1 this is just the Laughlin w.f. As in the hierarchal approach, can iterate. Note that again : (1) Since qp s feel an effective flux Nϕ eff eff. charge is e/(2mp ± 1) = N ϕ /(2mp±1), (2) only odd denominators (3) Still Abelian Either approach can explain almost all observed plateaux... * J.K. Jain, PRL 63, 199 (1989).

8 The ν = 5/2 STATE First seen in 1987 : to date the only even-denom. FQHE state reliably established* (some ev. for ν = 19/8). Quite robust : xy /(e2 /h) = 5/2 to high accuracy, excluding e.g. odd-denominator values ν = 32/13 or 33/13*, and xx vanishes within exptl. accuracy. The gap 500 mk. WHAT IS IT? First question : is it totally spin-polarized (in relevant LL)? Early experiments showed that tilting B away from r destroyed it suggests spin singlet. But later exptl. work, and numerics, suggests this may be tilted field changes orbital behavior and hence effective Coulomb interaction. So general belief is that it is totally spinpolarized (i.e. LLL, both filled, n = 1, half-filled, no filling of n = 1, ). (but it would be nice to have unambiguous exptl. evidence of this!). Thus, it is the n = 1 analog of ν = 1/2. However, the actual ν = 1/2 state does not correspond to a FQHE plateau. In fact the CF approach predicts that for this ν N eff φ = N φ 2N e = 0 and hence the CF s behave as a Fermi liquid : this seems to be consistent with expt. If LLL, both filled, this argt. should apply equally to ν = 5/2 (since (N e /N φ ) eff = 1/2). So what has gone wrong? One obvious possibility : Cooper pairing of composite fermions! since spins, must pair in odd-l state, e.g. p-state. *except for ν = 7/2 which is the corr. state with n = 1, filled. *Highest denominator seen to date 19 Moore & Read, Nuc. Phys. B 360, 362 (1991): Greiter et. al. 66, 3205 (1991)

9 THE PFAFFIAN ANSATZ Consider the Laughlin ansatz formally corresponding to ν = 1/2: Ψ (L) N = Π i<j(z i z j ) 2 exp i z i 2 /4l 2 m (z i=electron coor.) This cannot be correct as it is symmetric under i j. So must multiply it by an antisymmetric function. On the other hand, do not want to spoil the exponent 2 in numerator, as this controls the relation between the LL states and the filling. Inspired guess (Moore & Read, Greiter et. al.):(n = even) Ψ N = Ψ (L) N P f 1 z i z j P f(f(ij)) f(12)f(34)... f(13)f(24) ( Pfaffian) antisymmetric under ij This state is the exact GS of a certain (not very realistic) 3 - body Hamiltonian, and appears (from numerical work) to be not a bad approximation to the GS of some relatively realistic Hamiltonians. With this GS, a single quasihole is postulated to be created, just as in the Laughlin state, by the operation Ψ qh = Π N i=1 (z i η 0 ) Ψ N It is routinely stated in the literature that the charge of a quasihole is e/4, but this does not seem easy to demonstrate directly: the argts are usually based on the BCS analogy (quasihole h/2e vortex, extra factor of 2 from usual Laughlin-like considerations) or from CFT. conformal field theory 2 qps are more interesting.

10 Ansatz for 2-quasihole states: Ψ 2qh = Ψ (L) N P f (zi η 1 )(z j η 2 )+(z i η 2 )(z j η 1 ) z i z j Note that this state cannot in general be expressed in the form f(z i, z j ) Ψ GS. Consider now a 4-qh state (i.e 2 qh pairs).* Start with N = 2, then a possible w.f. is Ψ 4qh = Ψ (L) (z 1 z 2 ) 1 {(z 1 η 1 )(z 1 η 2 )(z 2 η 3 )(z 2 η 4 ) + (z 1 z 2 )} (12)(34) But at first sight two other possibilities, namely (13)(24) and (14)(23). However, now we note the identity (12)(34) (13)(24) = (z 1 z 2 ) 2 (η 1 η 4 )(η 2 η 3 ) from which it follows that (12)(34)(η 1 η 2 )(η 3 η 4 ) + (13)(24)(η 1 η 3 )(η 2 η 4 ) + (14)(23)(η 1 η 4 )(η 2 η 3 ) = 0 i.e. only 2 linearly independent functions. This result also holds for the generalization for N > 2: Ψ 4qh = Ψ (L) N P f (13)(24) z 1 z 2 (13)(24) z 3 z 4... and it can be generalized to the case of 2n quasiholes : for 2n quasiholes, 2 n 1 linearly independent states NW exhibit an explicit form of a possible choice of basis states: * Nayak & Wilczek, Nuc. Phys. B 479, 529 (1996).

11 Ψ (0) 4qh = (η 13η 24 ) 1/4 1+ (1 x) Ψ (1/2) 4qh = (η 13η 24 ) 1/4 1 (1 x) 1/2 Ψ(13)(24) + 1 xψ (14)(23) 1/2 Ψ(13)(24) 1 xψ (14)(23) (?) x η 12η 34 η 13 η 24 η 12 η 1 η 2, etc. 1 Let s take x 1 ( i.e. ) so Ψ (0) 4qh = 2 1/2 (η 13 η 24 ) 1/4 Ψ (13)(24) + Ψ (14)(23) Ψ (1/2) 4qh = 2 1/2 (η 13 η 24 ) 1/4 Ψ (13)(24) Ψ (14)(23) Then it is clear that interchange of 1 and 3 (or 2 and 4) affects only the prefactor, so gives a phase factor exp iπ/4. However, interchange of (e.g.) 2 and 3 gives a nontrivial rotation* in the space of Ψ (0) 0 and Ψ (1/2). Thus, the states and Ψ(1/2) 4qh can in principle be used as the basis for a qubit. More generally, Ψ (0) 4qh 2n anyons n qubits *confirmed by numerical calculations: Tserkovnyak & Simon, PRL 90, (2003).

12 IS THE ν = 5/2 FQHE STATE REALLY THE PFAFFIAN STATE? Problem : Several alternative identifications of the ν = 5/2 state (331, partially polarized, anti-pfaffian... ). Some are abelian, some not : all however predict e = e/4 [does this follow from general topological considerations? ]. Numerical studies tend to favor the Pfaffian, but... 2 very recent experiments: A. Dolev et. al., Nature (2008) Shot-noise expt., similar to earlier ones on ν = 1/3 FQHE state. Interpretations needs some nontrivial assumptions about the states neighboring the edge channels through which cond. n takes place. Conclusion: data consistent with e = e/4, inconsistent with e = e/2 unfortunately, doesn t discriminate between Pfaffian and other identifications. Radu et. al., Science (2008) Tunnelling expt., measures T- dependence of tunnelling current across QPC quantum point contact. Fits to theory of Wen for general FQHE state, which involves 2 characteristic number of states, e and g: for Pfaffian, e = e/4, g = 1/2 (also for 2 other nonabelian candidates : abelian candidates have e = e/4 but g = 3/8 or 1/8). Conclusion : best fit to date is e /e = 0.17, g = 0.35 which is actually closer to the abelian (331) state (g = 0.375) than to the Pfaffian.

13 If the ν = 5/2 FQH state is indeed Pfaffian. (a) how would we tell? (b) how could we use it for TQC? Proposal voltage/ current lead N B 1 2 Q voltage/ current lead M A P ν = 5/2 Longitudinal conductance xx is entirely determined by the back-scattering across MN and QP contacts, and we look for their interference. In general, xx t MN + e iϕ t P Q 2, where ϕ is the phase picked up on the circuit NMPQ. (1) Put qubit (2 anyons, so charge e/2) on 1. If in state 0. ( no fermionic zero mode ) ϕ = π/2, if 1 then ϕ = π/2. So if arg t MN /t P Q = 0, can detect which. (2) transfer single anyon (charge e/4) from 1 to 2. State of qubit is unchanged. (3) Now transfer single anyon from A to B. If ν = 5/2 state is Pfaffian, then will act as ˆσ x operator on qubit, so get ϕ ϕ π : if ν = 5/2 abelian, unchanged. (4) Re-measure xx. If different from 1st. measurement, Pfaffian (or at least nonabelian) : if the same, abelian. [for a 2-qubit gate (eg. CNOT) need at least 6 anyons] * Das Sarma et. al., PRL (2005)

14 THE ν = 12/5 STATE This state has so far seen in only one experiment*: it is quite fragile (short plateau, R xx 0). It could perfectly well be the n = 1 LL analog of the 2/5 state, which fits in the CF picture (p = 2, m = 1), and would of course be Abelian. Why should it be of special interest? In 1999 Read & Rezayi speculated that the ν = 1/3 Laughlin state and the Pfaffian ν = 1/2 state are actually the beginning of a series of parafermion states with ν = k/(k + 2) The ansatz for the wave function is where Ψ k:n = ps N Π 0<r<s<N/k χ(z p(kr+1)...z p(k(r+1)) : z p(ks+1)...z p(k(s+1)) χ(z 1...z k : z k+1...z 2k ) (z 1 z k+1 )(z 1 z k+2 )(z 2 z k+2 ) (z 2 z k+3 )...(z k z 2k )(z k z k+1 ) The state ψ k:l can be shown to be the exact groundstate of the (highly unrealistic!) Hamiltonian H = i<j<l<... δ(z i z j )δ(z j z l )δ(z l z m )... (k + 1) terms The quasiholes generated by this state have charge e = e/(k + 2) and are nonabelian for k 2; for k = 3 they are Fibonacci anyons, which permit universal TQC. Of course, the no. 12/5 = k/(k + 2). However, it is possible that the ν = 12/5 state is the n = 1, particle-hole conjugate of ν = 3/5. In this context it is intriguing that the ν = 13/5 state has never been seen... How would we tell? Interference methods? * Xia et. al., PRL (2003)