Introduction to quantum spin systems

Transcription

1 Exercise.. Show that {S i x} satisfy the usual spin commutation relations: Contents Introduction to quantum spin systems Daniel Ueltschi June, 07 Setting ressure and magnetisation Correlation functions 7 4 High temperature expansions A Matrix inequalities 5 Setting Let b Z d denote the domain. Recall that = {, } is the set of classical spin configurations of the Ising model. We consider here the complex Hilbert space H =span, (.) that is, H consists of all linear combinations of elements of ;dimh =. Equivalently, H = `( ). We use Dirac s notation: If!, the corresponding element in H is written!i. The Hilbert space H is the state space of spin systems. The counterparts to the local functions { i } i are the spin operators {Sx} i i=,, x. Their action on the basis elements {!i}! proceeds as follows: S x!i =!(x) i, S x!i = i! x! (x) i, S x!i =! x!i. We state important properties in the form of exercises, which we encourage the readers to do. (.) [S x,s y]=i x,y S x + cyclic permutations of (,,), (S x) +(S x) +(S x) = 4 l. Exercise.. Consider = {x}, so that H ' C with basis { i, +i}. Show that the spin operators are given by (one half of) the auli matrices, namely 0 0 i 0 S =, S = 0, S = i 0. 0 Exercise.. Spin operators on C and rotations in R. For ~a R, let S ~a = ~a ~S = a S + a S + a S, where S i is the auli matrix of the previous exercise. Show that (i) [S ~a,s ~b ]=is ~a ~b. (ii) Let R ~a ~ b denote the vector ~ b rotated around ~a by the angle k~ak. Then e is~a S ~b e is~a = S R ~a ~b. The energy of the system is given by a self-adoint operator on H. It involves interactions between nearest-neighbours and interactions with an external magnetic field. We list here five important hamiltonians; recall that E denote the set of edges (unordered pairs of nearest-neighbours) of. Ising: H Ising,,h = SxS y h Sx. {x,y}e x Quantum Ising: H qu.ising,,h = Quantum Y: H Y,,h = Heis. ferromagnet: H Heis.F,,h = SxS y h Sx. {x,y}e x SxS y + SxS y h Sx. {x,y}e x SxS y + SxS y + SxS y h Sx. {x,y}e x Heisenberg antiferromagnet: H Heis.AF,,h = H Heis.F,,h.

2 The partition function for the hamiltonian H,,h is Z,,h =Tr H e H,,h. (.) We only consider free boundary conditions (or perhaps also periodic boundary conditions); unlike in the classical situation, one cannot play with boundary conditions here. In the case of the Ising hamiltonian, all operators are diagonal, and Z Ising HIsing,,h =Tre,,h = e 4! {x,y}!x!y+ h x!x, (.4) and we recognise the partition function of the usual Ising model with inverse temperature 4 and magnetic field h. ressure and magnetisation The pressure is defined as the logarithm of the partition function divided by the volume, as in the classical case. It is also convex in and h; theproofismore di cult and it involves the Hölder and Golden-Thompson inequalities; to prove the latter, we need the Trotter product formula. The pressure is (,h)= log Z,,h. (.) roof. We use the Golden-Thompson inequality (roposition A.6) and then the Hölder inequality (roposition A.). sa+( s)b f(sa +( s)b) =logtre apple log Tr e sa e ( s)b h apple log Tr e sa s s Tr e ( s)b s = sf(a)+( s)f(b). We now turn to the thermodynamic limit (i.e. infinite-volume limit) of the pressure. Recall that * Z d denote the limit in the sense of van Hove. Theorem.. There exists a function (,h), that is convex in (,h) and even in h, such that (,h)= lim n! n (,h) along all sequences of domains such that n * Z d. r s i (.) Theorem.. The pressure (,h) is ointly convex in (,h) and even in h. There is no spin flip symmetry in the quantum case, but we can use spin rotations. Let U = Q x ei S x be the unitary that rotates the spin operators around the first axis by angle ; it sends (Sx,S x,s x)to(s x, Sx, Sx). All five hamiltonians listed above satisfy U H,,h U = H,, h. Then Z,,h =Tre H,,h =TrU e H,,h U =Tre H,, h = Z,, h. (.) Then (,h)= (, h). Regarding convexity, it is more elegant to prove the following, much more general statement. Theorem. (Convexity of the abstract pressure). The function f(a) =logtre A is a convex function on the space of hermitian matrices. Figure.: The large box of size n is decomposed in k d boxes of size m; there are no more than drn d remaining sites in the darker area. artial proof. We consider the hamiltonian H, Heis.F,h, but the modifications for the other models are straightforward. We only consider the sequence ( n )ofboxesof size n. We use a subadditive argument. Notice that the inequality Tr e A+B Tr e A holds for all self-adoint operators A, B with B 0. (This follows e.g. from the minimax principle, or from Klein s inequality, roposition A.7.) We rewrite the Hamiltonian so as to have only positive definite terms. Namely, let m n h x,y = S ~ x ~S y + l. (.4) 4 4

3 Then Z n,,h =e 4 E n Tr exp h x,y + h {x,y}e n x n S x. (.5) Let m, n, k, r be integers such that n = km + r and 0 apple r<m. The box n is the disoint union of k d boxes of size m, and of some remaining sites (fewer than drn d ); see Figure. for an illustration. We get an inequality for the partition function in n by dismissing all h x,y where {x, y} are not inside a single box of size m. The boxes m become independent, and Z n,,h e 4 E n htr H( m) exp =[Z m,,h] kd e 4 E n e kd 4 E m. h x,y + h {x,y}e m S x () x m We have neglected the contribution of e hs x for x outside the boxes m, which is possible because their traces are greater than. It is not hard to check that i k d (.6) E n apple k d E m + k d dm d + d rn d. (.7) We then obtain a subbaditive relation for the free energy, up to error terms that will soon disappear: Then, since km n n (,h) (km) d! asn!, n d m (,h) k d dm d 4n d lim inf n! n (,h) m (,h) d r 4n. (.8) d 4m. (.9) Taking the lim sup over m in the right side, we see that it is smaller or equal to the lim inf, and so the limit necessarily exists. Corollary.4 (Thermodynamic limit with periodic boundary conditions). Let ( per n ) be the sequence of cubes in Z d of size n with periodic boundary conditions and nearest-neighbor edges. Then ( per (,h)) n n converges pointwise to the same function (,h) as in Theorem., uniformly on compact sets. The finite-volume magnetisation m (,h)isdefinedas m (,h)= D h (,h)= x S x E. (.),,h Theorem.5. First, assume that (,h) is di erentiable in h at (,h); then (a) The limit m(,h)= lim *Z d m (,h) exists. (b) We have m(,h)= h (,h). Second, without assuming that (,h) is di erentiable in h at (,h), we have D E (c) (,h)= lim lim inf S h + h 0!h + *Z d x.,,h 0 art (c) also holds with lim sup instead of lim inf. We have proved parts (a) and (b) of this theorem in the case of the Ising model. The proof actually extends to the quantum case without modifications. roof. We use the fact that lim sup i lim inf i inf a i sup x apple inf sup lim sup a i, i a i, lim inf i and the following expressions for left- and right-derivatives of convex functions: df f(h) f(h s) (h) =sup, dh s>0 s df f(h + s) f(h) (h) =inf. dh + s>0 s (.) (.) This follows from per (,h) n n (,h) apple d, which is not too hard to prove, 4n and Theorem.. A finite-volume Gibbs state is a positive, normalised, linear map on the space of operators on H of the form hai,,h = Z,,h Tr H A e H,,h. (.0) 5 6

4 Since is convex, we have h (,h)=sup (,h) (,h s) lim inf s>0 *Z d s apple lim inf *Z d h (,h) apple lim sup *Z h (,h) d =limsupinf *Z d s>0 = h + (,h). (,h+ s) (,h) s But is di erentiable, so that inequalities are identities and we get h (,h)=liminf *Z d h (,h)=limsup *Z d (.4) (,h). (.5) h Since (,h)=m h (,h), this shows that the infinite volume limit of the latter exists, and is equal to the derivative of. This proves (a) and (b). For (c), let h n! h + such that is di erentiable in h at (,h n ). We have ust proved that (,h h + n )=lim m (,h n ). Since right-derivatives of convex functions are right-continuous, and since m is nondecreasing in h, we get the result. Exercise.. Let " be the proector onto the subspace of H spanned by configurations! such that x! x / ( ", "). Show that if (,h) is di erentiable in h at (, 0), then h " i,,0 apple e c for some c>0 that is uniform in. (Hint: Show that a Chernov inequality holds.) vanish, one talks about long-range order and this indeed implies the existence of several Gibbs states. We focus in this section on the Heisenberg ferromagnet. We expect the Gibbs state to be unique when h 6= 0,orwhen is small. The most interesting case is then when h =0,inwhichcase apple,,0 (x, y) =hs xs yi,,0. (.) In order to relate correlations to magnetisation, let M = x S x denote the operator for the rd component of the total spin. roposition.. For all 0, we have D (a) h (, 0) M E lim inf. + *Z d,,0 (b) D M E apple,,0 D M E apple,,0 D M E.,,0 roof. For (a), we start with Theorem.5 (c) to get (, 0) = lim lim inf h + h!0 + *Z d hm i,,h. (.) Let {' } be an orthonormal basis of eigenvectors of H,,0 and M with eigenvalues e and m, respectively. Since H,,h = H,,0 hm and [H,,0,M ]=0,the eigenvalues of H,,h are e hm. Because of symmetry (rotation around st direction of spin by angle ), the set {m } is symmetric around 0 (with multiplicity). Let h>0. We have :m hm i,,h = m >0 e e (e hm e hm ) :m e e >0 (e hm +e hm )+. (.4) e :m =0 e Let h i 0,,h be the Gibbs state with hamiltonian H,,0 h M.Wehave Correlation functions Correlation functions give useful information about the state of the system; they usually characterise phases. The most natural truncated correlation function deals with rd components of spins and is given by apple,,h (x, y) =hs xs yi,,h hs xi,,h h S yi,,h. (.) It is a general fact that, when the Gibbs state is unique, truncated correlations decrease to zero as kx yk becomes large. When truncated correlations do not 7 Then h M i 0 :m,,h = m >0 e e+hm :m e e+hm >0 +. (.5) e :m =0 e hm i,,h h M i 0,,h :m m >0 e e hm :m e. (.6) e+hm >0 The last term goes to 0 in the limit * Z d. Indeed, the sum of terms with m apple p contribute less than / p, and the sum of terms with m > p contribute 8

5 less than e hp. By convexity of the pressure of the model with hamiltonian H,,0 h M (Theorem.), we get h M i 0,,h = h log Tr e H,,0+h M h M i,,0. (.7) For the first inequality in (b), we can use M = M l and the following Cauchy-Schwarz inequality: ha Bi appleha AihB Bi, (.8) where h i = Tr e H denotes a Gibbs state, and A, B are any operators that commute with H. For the second inequality in (b), observe that M apple l implies that M apple M, and use the fact that the Gibbs state is a positive linear functional. Exercise.. Let h DY 0. Show that x S x E,,h 0 for all. We now prove a variant of the Mermin-Wagner theorem; in broad terms, it states that continuous symmetries cannot be broken in two dimensions. More precisely, we prove that spin correlations have (at least) power-law decay in the quantum Heisenberg model. This result goes back to McBryan and Spencer (977) in the case of the classical Heisenberg model, and to Koma and Tasaki (99) in the quantum case. We only consider the Heisenberg model, but the method and the result apply to general models with U() symmetry. The decay of correlations is measured by the following expression: apple (x) = sup ( y)r x=0 0 {y,z}e cosh( y z ). (.9) The solution of this variational problem is essentially a discrete harmonic function. We can estimate it explicitly in the case of D-like graphs with nearest-neighbor couplings. Let denote a graph, i.e. a finite set of vertices and a set of edges, and let d(x, y) denotethegraphdistance,i.e.thelengthoftheshortestpaththatconnects x and y. Lemma.. Assume that there exists a constant K such that, for any ` N, # {x, y} : d(0,x)=` and d(0,y)=` + apple K`. Then there exists C = C(,K), which does not depend on x, such that roof. With c to be chosen later, let ( c log d(0,x)+ if d(0,y) apple d(0,x), d(0,y)+ y = 0 otherwise. Then d(0,x) (x) c log(d(0,x)+) K `=0 (.0) cosh(c log `+ ) `. (.) `+ From Taylor expansions of the logarithm and of the hyperbolic cosine, there exist C, C 0 such that The optimal choice is c =( K). (x) c log(d(0,x)+) Kc Theorem.. For i =,,, we have d(0,x) `= ` c Kc log(d(0,x)+) C. hs i 0S i xi,,0 apple e (x). C 0 (.) In the case of D-like graphs, we can use Lemma. and we obtain algebraic decay with a power greater than ( K). roof of Theorem.. We use the method of complex rotations. Let One can check that for any a C, wehave S ± y = S y ± is y. (.) e as y S ± y e as y =e ±a S ± y. (.4) The Heisenberg hamiltonian can be rewritten as H,,0 = S y + S z + S y S z + + SyS z (.5) {y,z}e Given numbers y, let (x) log d(0,x)+ C. K A = Y y e ys y. (.6) Then Tr S + 0 S x e H,,0 =TrAS + 0 S x A e AH,,0A. (.7) 9 0

6 We now compute the rotated hamiltonian. AH,,0 A = e y z S y + S z +e z y S y S z + + SyS z {y,z}e = H,,0 cosh( y z ) (S y + S z + S y S z + ) {y,z}e sinh( y z )(S y + S z S y S z + ) {y,z}e H,,0 + B + C. Notice that B = B and C = C. Weobtain (.8) Tr S + 0 S x e H,,0 =e 0 x Tr S + 0 S x e H,,0 B C. (.9) We now estimate the trace in the right side using the Trotter product formula and the Hölder inequality for traces. Recall that kbk s =(Tr B s ) /s, with kbk = kbk being the usual operator norm. N Tr S 0 + S x e H,,0 B C = lim Tr N! S+ 0 S x e N H e N B e N C apple lim N! ks+ 0 S x k e N H,,0 N N e N B N e N C N. (.0) Observe now that k e N H k N N = Z, k e N B k N apple e kbk,andke N C k =. Wealso have ks 0 + S x k =S (check it!). The theorem then follows from kbk apple cosh( y z ). (.) {y,z}e 4 High temperature expansions Let us recall the method in the case of the Ising model, since it can be essentially generalised in the quantum case. We started by writing e!x!y =cosh ( +! x! y tanh ), and then expanded the Gibbs factor so as to get a sum of sets of edges with nonnegative weights. This allowed to get h x y i?,,0 apple (tanh ) E ; (4.) E E connected,x,ye Lemma 4.. The number of connected sets E E, with E = k, and that contain the origin, is less than (4d ) k. roof. Given any connected graph with k edges, and any vertex, there exists a sequence of edges (e,...,e k )suchthate contains this vertex, e i+ \ e i 6= ;, and each edge of the graph appears exactly twice. (This can be easily proved by induction, adding edges or vertices.) The number of connected sets is then less than the number of such sequences. Given an edge, there are 4d overlappingedges, hence the bound. One then gets exponential decay for small. Indeed, with d(x, y) beingthe graph distance between x, y, wehave e cd(x,y) h x y i?,,0 apple e c tanh E E E connected,x,ye apple k 0 h (4d ) e c tanh i k. (4.) The right side is bounded if (4d ) tanh <, and if c is small enough. Then h x y i?,,0 apple C e cd(x,y) with a constant C independent of, x, y. We now turn to quantum spin systems. We consider here a hamiltonian of the form H 0 = {x,y}e b x,y, where the operators b x,y satisfy h! b x,y! 0 i = h! b x,y! 0 i ( =0 ifw z 6=! 0 z for some z 6= x, y, 0 always, In words, we suppose that b x,y a ects the sites x, y only, that it has nonnegative matrix elements, and that its elements are invariant under spin flips. Without loss of generality, we can suppose that the matrix elements of b x,y are less than (we can rescale otherwise). We expand the exponential in Taylor series and get Tr S xs y e H0 n = Tr S n! xs yb e...b en n 0 e,...,e ne = n n!! 4 x! y h! b e...b en!i. n 0 e,...,e ne! In order for the sum over configuration to di er from zero, the sites x, y must be connected by a path of overlapping edges. Summing first over edges forming a (4.) (4.4) the sum is restricted over sets E are such that the graphs with vertices [ {u,v}e {u, v}, and edges E, is connected and it contains the sites x, y.

7 connected set that contains x, y, and then over the remaining edges, we get Tr S xs y e H0 Next we use We obtain = k! k 0 = k n 0 n n! n n k k=0 e 0,...,e0 n k E \[ i e i e,...,e k E! [i e i connected,x,y[ ie i! [i e i e,...,e k E! [i e i connected,x,y[ ie i! 0 \[i e i h! 0 b e 0...b e 0 n k!i! 4 x! y h! b e...b ek!i! 4 x! y h! b e...b ek!itr H \[i e i e H0 \[ i e i,.! 4 x! y h! b e...b ek!i apple 4 [iei, [iei Tr H \[i e i e H0 \[ i e i, apple Tr H e H0,. hs xs yi 0, apple 4 k 0 k k! (4.5) (4.6). (4.7) e,...,e k E connected,x,y[ ie i By first summing over sets of edges, and then over their number of occurrences, we get k 0 k k! = e,...,e k E connected,x,y[ ie i = E E n,...,n E connected,x,ye E E connected,x,ye e E. n + + n E n...n E (n + + n E )! n+ +n E (4.8) Theorem 4.. Let h i 0, be the Gibbs state with hamiltonian H 0, = {x,y}e, where the operators b x,y satisfy the proprieties (4.), and with matrix elements less than. Then, if (4d ) (e ) <, there exist constants C, c that are uniform in b Z d and x, y, such that hs xs yi 0, apple C e cd(x,y). The method uses the fact that the hamiltonian can be written using matrices with nonnegative elements. There is a powerful method, called cluster expansions, that allows to obtain exponential decay more generally. Exercise 4.. Let T x,y denote the transposition operator on sites x, y, namely 8 ><! y if z = x; T x,y!i =! 0 i, where! z 0 =! x if z = y; >:! z if z 6= x, y. Show that S ~ x ~S y = T x,y. 4 Exercise 4.. Let H,,0 = SxS y + usxs y + SxS y, where u [0, ] is a {x,y}e fixed parameter. Find operators b x,y that satisfy (4.), and such that the Gibbs state h i 0, with hamiltonian H, 0 = {x,y}e b x,y has identical spin correlations, i.e. hsxs yi,,0 = hsxs yi 0,. Finally, we can proceed as in the case of Ising, namely e cd(x,y) hsxs yi 0, apple e c (e 4 ) E apple k 0 E E connected,x,ye h (4d ) e c (e ) i k. (4.9) Let us summarise the results obtained in this section. 4

8 A Matrix inequalities We collect here a series of useful properties of square matrices. Recall that the absolute value of a matrix is A =(A A), where the square root of a nonnegative hermitian matrix can be defined by diagonalising and taking the square root of the eigenvalues. The p-norm of a matrix is then defined as Exercise A.. kak p =(Tr A p ) /p. (i) Show that kak p is decreasing in p, and that lim p! kak p = kak. (ii) rove that kak p is a norm for all apple p apple. (iii) Use Hölder inequality to show that kak p is submultiplicative, that is, kabk p apple kak p kbk p. roposition A. (Hölder inequality for matrices). If apple p, q, r apple with p + q = r, we have kabk r applekak p kbk q. It follows from a simple induction that ny ny A apple ka k p r = whenever apple r, p,...,p n with n = p = (A.) (A.) =. There are no short proofs of Hölder s r inequality for matrices. The proof is due to Fröhlich [978] and it uses chessboard estimates. The proof of roposition A. can be found after Lemma that of A.4. Lemma A. (Chessboard estimate). For any n N and any matrices A,...,A n, we have Tr A...A n apple ny i= /n. Tr (A i A i ) n roof. Since (A, B) 7! Tr A B is an inner product, the following inequality follows from Cauchy-Schwarz: Tr A...A n apple Tr A...A n A n...a Tr A n...a n+a n+...a n. (A.) This allows to use a reflection positivity argument. It is enough to prove the inequality for matrices that satisfy Tr (A i A i ) n =;thegeneralresultfollowsfrom scaling. Let A,...,A n be matrices that maximise Tr A...A n, with maximum number of matching neighbours A i+ = A i. Suppose there exists an index such that A + 6= A. Using cyclicity, we can assume that = n. By the inequality (A.), A,...,A n,a n,...,a and A n,...,a n+,a n+,...,a n are also maximisers. At least one has strictly more matching neighbours, hence a contradiction. The maximum is then Tr (AA ) n for some matrix A, which is equal to. Chessboard estimates allow to prove what is essentially the case r = of Hölder s inequality. Corollary A.. We have Tr A...A n apple ny ka i k pi i= for all n and all rational p i s such that n i= p i =. roof. Let ` be a positive integer such that `/p i is integer for all i. Let A i = U i A i be the polar decomposition of A i, and let Then A i = ˆB i B (`/pi) i,andwehave B i = A i pi/`, ˆBi = U i A i pi/`. (A.4) Tr A...A n = Tr ˆB B...B {z }... ˆB n B n...b {z n} (`/p ) (`/p n) ny apple (Tr A i pi ) /pi = i= ny ka i k pi. The inequality follows from Lemma A. and from the identities i= (A.5) Tr (B i B i )` =Tr(ˆB i ˆB i )` =Tr A i pi. (A.6) 5 6

9 Lemma A.4. Let r, r 0 [, ] such that + =. Then for any square r r 0 matrix A, we have kak r = sup kck r 0 = Tr C A. roof. The right side is smaller by Corollary A.: Tr C A applekck r 0kAk r = kak r. In order to check that this inequality is saturated, let A = U A be the polar decomposition of A, and choose C = kak r r U A r. Then kck r 0 =andtrc A = kak r. roof of roposition A.. Starting with Lemma A.4 and then using Corollary A., we have kabk r = apple sup Tr C AB kck r 0 = sup kck r 0kAk p kbk q. kck r 0 = roposition A.5 (Trotter formula). Let A, B be square matrices. Then n h i n. e A+B = lim e n A e n B = lim + A e n B n! n! n roof. We prove the second formula the mild changes for the first formula are straightforward. Let K n be the matrix such that (A.7) (A.8) + n A e n B =+ n (A + B)+K n. (A.9) roposition A.6 (Golden-Thompson inequality). Let A, B be hermitian matrices. Then Tr e A+B apple Tr e A e B. roof. Hölder s inequality, roposition A., implies that Tr (AB) n applekabk n n. The latter is equal to Tr (A B ) n/ since A, B are hermitian. If n is a power of, we can iterate and we get Tr (AB) n apple Tr A n B n. (A.) We use this inequality with A 7! e n A and B 7! e n B, which gives n Tr e n A e n B apple Tr e A e B. The left side converges to Tr e A+B as n!by the Trotter formula. roposition A.7 (Klein inequality). Let f be a convex di erentiable function, and A, B be hermitian matrices with eigenvalues in the domain of f. Then Tr f(a) f(b) (A B)f 0 (B) 0. With f(s) = e s, exchanging A and B, weget (A.) Tr e A e B apple Tr (A B)e A. (A.4) roof. Let ( i )and( i )beorthonormalbasesofeigenvectorsofa and B, and let (a i )and(b i )theeigenvalues.letc i =( i, ). Then i, f(a) f(b) (A B)f 0 (B) i = f(a i ) = c i f(b ) c i (a i b )f 0 (b ) c i f(a i ) f(b ) (a i b )f 0 (b ) It is clear that kk n k = O( n ). We have h i n n+rn + A e n B n = + (A + B), (A.0) n where R n is a matrix whose norm satisfies kr n kapple n k=0 n k k+ (A + n B)kk kk n k n k = O( ). (A.) n The first term in the right side of (A.0) converges to e A+B. 0. (A.5) 7 8

10 Index chessboard estimate, 5 convexity of the pressure, correlation function, 7 Gibbs state, 5 Golden-Thompson inequality, 8 Hölder inequality for matrices, 5 hamiltonian: Ising, quantum Ising, Y, Heisenberg, Heisenberg hamiltonian, Ising hamiltonian, Klein inequality, 8 magnetisation, 6 Mermin-Wagner theorem, 9 partition function, auli matrices, pressure, quantum Ising hamiltonian, spin auli matrices, rotations, Trotter formula, 7 truncated correlations, 7 Y hamiltonian, 9