Asymmetric Fermi surfaces for magnetic Schrödinger operators

Transcription

1 Research Collection Reort Asymmetric Fermi surfaces for magnetic Schrödinger oerators Author(s): Feldman, Joel S. Trubowitz, Eugene Knörrer, Horst Publication Date: 998 Permanent Link: htts://doi.org/0.399/ethz-a Rights / License: In Coyright - Non-Commercial Use Permitted This age was generated automatically uon download from the ETH Zurich Research Collection. For more information lease consult the Terms of use. ETH Library

2 Asymmetric Fermi Surfaces for Magnetic Schrodinger Oerators Joel Feldman Deartment of Mathematics University of British Columbia Vancouver, B.C. CANADA V6T Z Horst Knorrer, Eugene Trubowitz Mathematik ETH-Zentrum CH-809 Zurich SWITZERLAND Abstract. We consider Schrodinger oerators with eriodic magnetic eld having zero ux through a fundamental cell of the eriod lattice. We show that, for a generic small magnetic eld and a generic small Fermi energy, the corresonding Fermi surface is convex and not invariant under inversion in any oint. Research suorted in art by the Natural Sciences and Engineering Research Council of Canada, the Schweizerischer Nationalfonds zur Forderung der wissenschaftlichen Forschung and the Forschungsinstitut fur Mathematik, ETH Zurich

3 and IR d F (A V )= k k F (A V ) I. Introduction Let be a lattice in IR d d and let r>d. Dene A = A =(A A d ) L r IR(IR d =) d R For (A V ) AV set V = V L r= IR (IRd =) R V (x) dx =0 IR d = H k (A V ) = ir + A(x) k + V (x) A(x) dx =0 IR d = When d = 3, the oerator H k (A V ) describes an electron in IR d with quasimomentum k moving under the inuence of the magnetic eld with eriodic vector otential A(x) = (A (x) A d (x)) and electric eld with eriodic otential V (x). In general let e (k A V ) e (k A V ) be the eigenvalues of the oerator H k (A V ) on L (IR =). The restriction of e n (k A V )to the rst Brillouin zone B of is called the n-th band function of A. Observe that e (k 0 0) = jkj The Fermi surface of (A V ) with energy is dened as Because H k (A V )=H k (A V ) for all n. In articular, when A =0, F (A V ) = k B en (k A) = for some n e n (k A V )=e n (k A V ) e n (k 0 V)=e n (k 0 V) for all n, so that F (0 V)=F (0 V) for all and V. For all (A V ) AV, IR The main result of this aer is

4 Theorem There is a neighbourhood A 0 V 0 of the origin in AV and 0 > 0 such that (i) for all ( a V ) ( 0 ) A 0 V 0, F (A V ) is either a strictly convex (d ){ dimensional real analytic submanifold of B, or consists of one oint, or is emty. (ii) there isanoen dense subset S of ( 0 )A 0 V 0 such that for all ( A V ) S and all IR d F (A V ) \ F (A V ) has dimension at most d. Furthermore S\ ( 0 ) A 0 f0g is oen and dense in ( 0 ) A 0 f0g. The Theorem shows that, for generic small eriodic magnetic elds of mean zero and generic small Fermi energies, the Fermi surface is strictly convex and does not have inversion symmetry about any oint. In articular, when d =, the intersection of the Fermi surface and its inversion in any oint is generically a nite set of oints. The same statements hold if one considers both electric and magnetic elds. This inversion asymmetry suresses the Cooer channel in weakly{couled short{range many Fermion systems. See [FKLT]. There are real materials containing a eriodic array of magnetized ions with zero ux through a fundamental cell. See the review [O, -48]. In xii we show that the Fermi surface is always a real analytic subvariety of IR d, and that it deends holomorhically on (A V ) AV. The roof of the main Theorem, resented in xiii, is based on a third order erturbation calculation around (A V )=(0 0).

5 II. Analyticity of the Fermi surfaces Let A C = A =(A A d ) L r (IR d =) d R V C = V L r= (IR d =) R V (x) dx =0 IR d = be the comlexications of A res. V. A(x) dx =0 IR d = Theorem. There exists an analytic function F on C d C A C V C such that, for k A V real, Sec H k (A V ) () F (k A V ) = 0 Corollary Fixanoen ball D in the rst Brillouin zone B that contains 0 such that D B. There is a neighbourhood U of the origin in AV and there is 0 > 0 such that i) For all (A V ) U and all k D e (k A V ) <e (k A V ) ii) The ma D U! IR (k A V ) 7! e (k A V ) is real analytic. iii) For each xed (A V ) U the Hessian of the ma k 7! e (k A V ) is ositive denite. Furthermore inf kd e (k A V ) < 0. iv) For each (A V ) U and each < 0 the Fermi surface F (A V ) is either emty, or consists of one oint only, or is a real analytic smooth strictly convex (d ) {dimensional real analytic manifold that is comletely contained in D. Proof of the Corollary: The sectrum of H k (0 0) is jk + bj b ]. Hence, for k D, e (k 0 0) = jkj and e (k 0 0) > e (k 0 0). The Corollary follows from reeated alication of the Imlicit Function Theorem and continuity. 3

6 Proof of the Theorem: Write ir + A(x) k + V (x) =l +u(k )+w(k A V ) with u(k ) =ik r+ k l At the end of this Section we rove w(k A V )=ira + ia rk A + A + V Lemma. Let kbk r = [tr (B B) r= ] =r where tr denotes the trace dened on trace class oerators on L (IR d =). There is a constant const r d a) b) l w(k A V ) such that r l const r d ( + jkj)kakl r + kak L + kv k r L r= l u(k ) r l const r d +jkj + jj c) Let 0 rd r. There is a constant const r d k A V such that (u(k )+w(k A V ) const r d k A V k(l ) ()= kk(l ) = for all L (IR d =). k + k(l ) = kk(l ) ()= k Because L s (IR d =) L s0 (IR d =) for all s s 0, we may assume, without loss of generality, that r d +. Then the Lemma imlies that F (k A V ) = det d+ l + l u(k ) l + l w(k A V ) l is a well{dened analytic function on C d C A C V C. regularized determinant which, for matrices, is dened by det d+ (l+b) = ex dp i= This regularized determinant is dened for B with kbk d+ () i i tr B i det(l+b) is analytic since one can take limits of nite rank aroximations of B. 4 Here, det d+ (l +B) is the nite. See [S], Theorem 9.. It

7 Let D be the domain of l. By the Lemma, ir + A(x) k + V (x) =l +u(k )+w(k A V ) gives a well{dened quadratic form on DD. Furthermore, for D l +u(k )+w(k A V ) k l k const k(l ) ()= kk l k For any >0 there is a constant c such that k(l ) ()= kk l k + c kk Choosing aroriately, l +u(k )+w(k A V ) k l k const c kkk l k and the form is semibounded. It is closed since const c kk k l k const kk l +u(k )+w(k A V ) 4 const k l k Hence there is a unique associated self adjoint semibounded oerator H k (A V ). The resolvent (l ) is comact. Hence by the resolvent identity and art a) of the Lemma the resolvent of H k (A V ) is also comact. Hence the sectrum of H k (A V ) is discrete. Then Sec H k (A V ) if and only if there exists D Hk (A V ) D such that Hk (A V ) l l = 0 This is the case if and only if l Hk (A V ) l Theorem 9. (e), this is the case if and only if F (k A V )=0. has a nontrivial kernel. By [S], Proof of the Lemma: f `r( # ) and g L r (IR d =), a) We reeatedly aly the result that, for any r and any kf(ir)g(x)k (IR d =) =r kfk`r( # )kgk L r (IR d =) () 5

8 This is roven just as in [S] Theorem 4., the corresonding bound for oerators on L (IR d ). One rst roves easily that the Hilbert-Schmidt norm of f(ir)g(x) is bounded by (IR d =) = kfk`( # )kgk L (IR d =) and that the oerator norm of f(ir)g(x) is bounded by kfk` ( # )kgk L (IRd =). One then interolates using [S], Theorem.9. As the oerator norms l r, l k jkj and kbb 0 k r kbk r kb 0 k, we have l w l r +jkj l a r + l A r + l V l r Write l = f(ir) with f(b) = if b=0 +b if b 6= 0 This f `r( # ) for all r>d, so the desired result follows from () with g = A A d V. b) The sectrum of l u(k ) l is kb+k +b b # For any r > d, the `r( # ){norm of kb+k +b, which is also the k k r norm of l u(k ) l, is bounded by const r +jkj + jj. c) Denote D = (l ). The condition on imlies that r( ) d + rd > d so that (+b ) r()= Consequently, is still summable. So, as in art a), k D Akconst r kak L r k D V D kconst rkv k L r= ku D kconst r( + jkj + jj) h(ira) i const r kak L rkd kkd k h(air) i const r kak L rkdkkd k h(k A) i const r jkjkak L rkd kk k h(a A) i const r kak L rkd kkd k hv i const r kv k L r=kd kkd k hu i const r ( + jkj + jj)kkkd k 6

9 III. Proof of the Main Theorem For simlicity we write e(k A V )=e (k A V ). By art (iii) of the Corollary in xii, for each (A V ) U the band function e( A V) has a unique extremum k min (A V ). This extremum is a nondegenerate minimum. It follows from the imlicit function theorem that k min (A V ) deends analytically on (A V ). The same is true for the corresonding critical value min (A V )=e k min (A V ) A V. Observe that min (A V ) < 0 Corollary in xii. We set by art (iii) of the P = ( A V ) IR U min (A V ) << 0 Then for each ( A V ) P the Fermi surface F (A V ) is a smooth, real analytic, strictly convex (d ){dimensional manifold which is not emty. For k F (A V ) denote by n(k) the outward unit normal vector to F (A V )atk. If ( A V ) P then for each on the unit shere S d there is a unique oint k ( A V ) F (A V ) such that n(k ( A V )) =. Again it follows from the imlicit function theorem that S d P! D ( A V ) 7! k ( A V ) is a real analytic ma. To rove the Theorem stated in the Introduction we have toshow that for ( A V ) in an oen dense subset of P and all IR d the intersection F (A V ) \ F (A V ) has dimension at most d. Since for all ( A V ) P the manifold F (A V ) is real analytic, smooth and strictly convex, one has for all ( A V ) P and all IR d dim F (A V ) \ F (A V ) d or F (A V ) = F (A V ) If F (A V ) = F (A V ) then inversion in the oint = mas the oint of F (A V ) with normal vector to the oint of F (A V ) with normal vector. In other words, F (A V )= F (A V ) =) k ( A V )+k ( A V )= for all S d Therefore the set of all ( A V ) P for which there is a oint IR d dim F (A V ) \ F (A V ) > d is contained in such that S 0 = ( A V ) P r k ( A V )+k ( A V ) =0for all S d 7

10 Observe that S 0 is the intersection of the analytic hyersurfaces ( A V ) P r k ( A V )+k ( A V ) =0 S d Thus to show that the comlement of S 0 is oen and dense it suces to exhibit one trile ( A V ) P that does not lie in S 0. We will do this by choosing V = 0, choosing a articular two dimensional vector otential A and showing that for small t and aroriate the trile ( t A 0) does not lie in S 0. This then also shows that the comlement of S 0 \ ( A V ) P V =0 is oen and dense in ( A V ) P V =0. In the following calculation we will only consider the oints ( t A 0) of P with a articular two dimensional vector otential A. Therefore we restrict ourselves to the case d = and delete the V {variable in the notation. First we comute the rst three t- derivatives of e(k t A) at the origin for arbitrary (two-dimensional) A. We use the notation _ f(k) = d dt f(k t) t=0. Lemma. Fix A A. Put (k t) = e(k t A) Then there is a constant C such that for all k D _(k) = 0 (k) = C... (k) = Re 6 b # f0g b c # f0g b c # f0g b +k b (k + b) ^A(b) b +k b [ ^A(c) ^A(c b)] [(k + b) ^A(b)] [(k + c) ^A(c)] c +k c Here ^A(b) =(^A (b) ^A (b)) are the Fourier coecients of A, i.e A(x) = [(k + b + c) ^A(c b)] [(k + b) ^A(b)] b +k b b # ^A(b)e ibx and # = b IR <b > Z for all 8

11 is the dual lattice to. Furthermore for each (0 0 ) and every S d dt k ( t A) t=0 = 0 d dt k ( t A) t=0 = ( ) d3 dt 3 k ( t A) t=0 =... ( ) Proof: Let k (t) be the eigenfunction of eigenvalue (k t) for the oerator H k (t A) normalized by the conditions that k(0) = where is the ume of IR =, and < k (0) k (t) > = for all t. Then, for small t and k D, k (t) is an analytic function of t and k. For convenience we suress the argument k in the following comutation. From the identity H = one gets by dierentiation _H + H _ = _ + _ () H + H + H = + + ()... H +3 H _ +3 _H + H... = _ +3_ +... (3) Forming the inner roduct of these equations with (0) gives _ = < _ H > (4) = < H > + < _H _ > (5) = < H > + 3 < H _ > + 3 < _H > (6) 9

12 since < _ >=< >=<... >= 0 and < H _ >=< H >=< H... >= 0. However Therefore for t =0 d dt H k(t A) = A ir + t A k + ir + t A k = A ir + t A k + i + _H k = _H k = k A @x (8) Taking inner roduct with (0) = we get < _H > = D k A @x E = 0 and hence _ = 0 From () we now deduce _ = (H ) _H (9) where, by denition, (H ) vanishes on (0) and is the inverse of (H ) when restricted to the orthogonal comlement of (0). Substituting into (5) = < H > < _H (H ) _H > Now by (7) H = (A + A ) so that < H > is a constant C = < (A + A ) > indeendent of k. Using (8) we therefore get = C D Hk k A @x k A + @x Since and k A @x = b # (k + b) ^A(b)e ibx (H k (0 0) (0)) e ibx = b +k b e ibx 0

13 we get = C = C * b # f0g b # f0g b +kb (k + b) ^A(b) e ibx b +k b From () we deduce that (k + b) ^A(b) = (H ) H + _H _ b # (k + b) ^A(b) e ibx = (H ) H + (H ) _H (H ) _H (0) + Using (6), (9), (0) and the fact that..ḣ =0,this gives Since we get... = 3 < H (H ) _H > 3 < _H(H ) H > +6< _H(H ) _H (H ) _H > = 6 Re < (H ) _H H >+6 < _H (H ) _H (H ) _H >... = Re (H ) _H = H = b # f0g b c # b +kb (k + b) ^A(b) e ibx ^A(b c) ^A(c)e ibx D He _ ibx e icxe = (k + b + c) ^A(c b) 6 b c # f0g b c # f0g b +k b [(k + b) ^A(b)] [ ^A(c) ^A(c b)] [(k + b) ^A(b)] b +k b This roves the statement about the derivatives of. [(k + b + c) ^A(c b)] [(k + c) ^A(c)] c +k c We now rove the statement about the derivatives of t 7! k ( t A) for xed (0 0 ) and S. To simlify notation ut ( t) = k ( t A)

14 Dierentiating the identity ( t) t = we get r k ( ( t) t = 0 () Since _ =0 and r k ( 0) 0 =, setting t =0gives _() = 0 (a) If? denotes the vector ( ) erendicular to =( ), then by the denition of k we have? r k (( t) t)=0. Dierentiating this identity we get? Hessian() _()+r k _ = 0 Since for t =0we have Hessian() = l and _ =0,we get? _() = 0 (b) Putting (a) and (b) together gives _ = 0 () Dierentiating () again and setting t = 0 gives _() Hessian k () _()+r k _(( 0)) _()+ ()+(( 0)) = 0 Using () we get () = (( 0)) = ( ) Dierentiating () twice, setting t = 0 and using _ = 0 3r k _(( 0)) ()+... ()+... (( 0)) = 0 Since _ =0... () =... (( 0)) =... ( )

15 We now continue the roof of the main Theorem, and consider again a xed vector otential A and a xed (0 0 ). If F (t A) = ( t) F (t A) for all small t then k ( t A)+k ( t A) =( t) for all S and all small t. Multilying this equality with, dierentiating three times with resect to t and setting t =0gives d 3 dt 3 k ( t A) t=0 + d3 dt 3 k ( t A) t=0 for all S. By the revious Lemma, this is equivalent to =... () (k) (k) = k () for all k with jkj = or, since... (k) =... (k), to (k) = k () for all k with jkj = (3) As said above, it suces to secify one vector otential A and one (0 0 ) such that (3) fails. We now give a concrete examle of a vector otential A for which (3) does not hold any >0. Fix a nonzero vector d #. Without loss of generality wemay assume that d =( 0). Let ^A(d) = ^A(d) = ^A(d) = ^A(d) =d? and ^A(b) =0 for b 6= d d where d? =(0 ). Then, by the revious Lemma,... (k) = Re 6 b c # f0g b c # f0g = 4jdj k d? b +k b [ ^A(c) ^A(c b)] [(k + b) ^A(b)] [(k + c) ^A(c)] c +k c b cfd dg cbfd dg 48(k d? ) 3 b cfd dg cbfd dg b +k b c +k c 3 [(k + b + c) ^A(c b)] [(k + b) ^A(b)] b +k b b +k b

16 Inserting the allowed values for b and c... (k) = 4k 48k 3 +k + k + 4(+k + ) 4(k ) (+k )(+k + ) (+k )(k + ) (k )(k ) If (3) were to hold, that is if the above quantity were of the form k... () for all k with jkj = then one would have... () = () d?, because the right hand side vanishes for k =0. Therefore () = 4 k 4k (+k )(k ) k (k )(k ) + +k + 4(+k ) + 4(k ) k (+k )(+k ) If (3) were to hold, the right hand side of (4) would have to be constant on the circle (k k ) IR k + k =. Since the right hand side of (4) is a meromorhic function of f(k k ), it would then be constant on the comlex quadric Q = (k k ) C k + k = On the other hand, f(k k ) has a ole with residue 4( k ) along the comlex line L = (k k ) C k =. Consequently f Q is innite on the oints of Q \L dierent from ( i ). This shows that f Q >0, this is not the case. (4) cannot be constant unless Q \ L f( i )g. For 4

17 References [FKLT] J. Feldman, H. Knorrer, D. Lehmann, E. Trubowitz, Fermi Liquids in Two-Sace Dimensions, Constructive Physics V. Rivasseau ed. Sringer Lecture Notes in Physics 446, (995). [O] H. R. Ott, Ten Years of Suerconductivity: 980{990, Kluwer, Dordrecht, 993. [S] B. Simon, Trace ideals and their alications, Cambridge University Press,