The Erwin Schrodinger International Pasteurgasse 6/7. Institute for Mathematical Physics A-1090 Wien, Austria

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1 ESI The Erwin Schrodinger International Pasteurgasse 6/7 Institute for Mathematical Physics A-1090 Wien, Austria On the Flux{phase Problem Andras Sut}o Vienna, Preprint ESI 297 (1996) January 30, 1996 Supported by Federal Ministry of Science and Research, Austria Available via

2 On the ux-phase problem Andras Sut}o Research Institute for Solid State Physics H-1525 Budapest 114, Hungary Abstract We consider the Hubbard model H =?t P P x;y2 exp(i' xy)c y x c y + U P (n x"? 1=2)(n x#? 1=2) on rectangular domains of hypercubic lattices in and above two dimensions, with periodic boundary condition. We show that the grand-canonical partition function as a function of the phases ' xy attains its maximum when ' xy +' yu +' uv +' vx = (mod 2) on every square (xyuv). In two dimensions this corresponds to an external magnetic eld producing an odd number of ux quanta through each square. Some extensions are also discussed. 1

3 1 Introduction The ux-phase problem goes back to a 1989 paper by Hasegawa, Lederer, Rice and Wiegmann [1] and is related to the eect of a magnetic eld on the orbital motion of electrons. It can be formulated easily in lattice models, like the Hubbard model. Consider noninteracting electrons on a nite part of the square lattice Z 2 in a magnetic eld perpendicular to the plane. Neglecting the coupling of the eld to the spins, the system is described by the Hamiltonian H 0 =? X X x;y2 t xy c y xc y (1.1) Here c y x, c x are the creation and annihilation operators for a fermion at site x with spin ="; #. The hopping amplitude t xy = 0 if jx? yj 6= 1 and with t xy = jt xy j e i'xy (1.2) ' xy = Z y x A ds (1.3) if jx? yj = 1, where A is the magnetic vector potential in suitable units. The individual phases are subject to the choice of the gauge; the physically relevant quantity is xyuv = ' xy + ' yu + ' uv + ' vx (1.4) where xyuv is the number of ux quanta on the square (xyuv). The ux-phase problem consists in nding which makes the ground state energy of H 0 minimal. Let N e and jj denote the number of electrons and lattice sites, respectively. Hasegawa et al. found that the minimizing ux depended on the lling factor n = N e =jj. Their computations strongly suggested that it was uniformly min = n on every plaquette, where = 1; 1 ; 2 for square, triangular and honeycomb lattices, respectively. In particular, at half-lling, n = 1 on the square lattice hopping amplitudes 2 with Y t xy =jt xy j =?1 (1.5) s around each unit square s would yield the minimum ground state energy. [Note that our half-lling corresponds to n = 1 of Hasegawa et al., who work with spinless 2 electrons.] The ux-phase problem more generally is a search for the phases ' xy which minimize the free energy or maximize the partition function Z = Tr e?(h 0+V ) (1.6) in a system of interacting fermions on some graph. Rigorous studies, including the present one, are restricted to the case of half- lling. The rst mathematical treatment is due to Lieb [2] and Lieb and Loss [3]. They proved min = 1 for the ground state and the free energy of noninteracting 2

4 fermions on some essentially linear arrays of squares. Later Lieb [4] found a proof which worked for interacting fermions on cubic and some other bipartite lattices. Prior to us, Macris and Nachtergaele [5] simplied Lieb's proof by applying ideas of reexion positivity [6-8]. The proof presented below is independent of the formers but also uses reexion positivity and is, therefore, subject to symmetry constraints. None the less, our result, just as that of Lieb or Macris and Nachtergaele, yields (trivially) the ux minimizing the free energy at half-lling in the thermodynamic limit. 2 Results Consider the Hubbard model on a nite part of the d dimensional cubic lattice Z d : X X X H =? t xy c y xc y + U x"? 1=2)(n x#? 1=2) (2.1) x;y2 x2(n Here t xy = t yx is a nearest neighbour hopping amplitude as in (1.2), n x = c y xc x. The sign of U plays no role in the considerations below. We shall work in the grandcanonical ensemble; then the (implicit) choice of the chemical potential in Eq. (2.1), = U=2, corresponds to half-lling. Indeed, let odd = fx = (x 1 ; : : : ; x d ) 2 : X x i is oddg (2.2) The transformation interchanging c y x and c x for all x 2, followed by c x!?c x, c y x!?cy x for x 2 odd is unitary and carries all n x into 1? n x and H into itself (if all t xy 's are real). Therefore hn x i = h1? n x i or hn x i = 1 2 (2.3) for all x 2 and ="; #. Above hai = Tr Ae?H = Tr e?h (2.4) where Tr is the trace in the fermionic Fock space over. For complex t xy 's the above unitary transformation yields the complex conjugate H of H in any matrix representation in which c y x and c x are real, see e.g. Eq. (3.30) below. Because hn x i is real, it is the same for H and H, and hence Eq. (2.3) remains valid. In this equation one uses that t xy is bipartite on, i.e., t xy 6= 0 only for x 2 odd and y 2 even = n odd or vice versa. The particle-hole transformation is unitary also in the subspace with xed particle number N e = jj, therefore (2.3) holds at half-lling in the canonical ensemble as well. Although Eq. (2.3) has been obtained trivially, the constancy of hn x i for translationally noninvariant Hamiltonians is a kind of surprise (cf. Lieb, Loss and McCann [9]). For the proof of the ux-phase conjecture in nite domains we need rectangular parts of Z d (d 2) with an even number of sites along each edge: = fx = (x 1 ; : : : ; x d ) 2 Z d :?L i + 1 x i L i ; i = 1; : : : ; dg (2.5) 3

5 with all L i 2. Denote e i, i = 1; : : : ; d, the unit vectors of Z d. Write where r xy > 0 if jx? yj t xy = r xy xy (2.6) dx 1 jx i? y i j = 1 (2.7) and is zero otherwise, and xy is a complex unit, j xy j = 1. On the domain (2.5) we consider the energy operator (2.1) for xed r = (r xy ) as a function of = ( xy ) and denote it by H[]. Furthermore, H per [] denotes the same operator with periodic boundary condition, that we obtain by extending r and to pairs x; y 2 such that x? y = (2L i? 1)e i ; i 2 f1; : : : ; dg (2.8) These are nearest neighbours on per, the wrapped onto the d-torus. Z [] and Z per[] will stand for the corresponding grand-canonical partition functions, e.g. The theorem below is valid for a particular choice of r: Z per[] = Tr e?hper[ ] (2.9) r xy = r i if x? y = e i (2.10) (and also if x? y = (2L i? 1)e i, in the periodic case). We may and will suppose that xy is always dened on per. Theorem 1 Let r i > 0 be xed for i = 1; : : : ; d. (i) Z per[] attains its maximum at some such that P s [] := xy yu uv vx =?1 (2.11) around each unit square s = (xyuv). The maximizing can be chosen as real, i.e., xy = 1. (i) Suppose 0 is given on all nearest neighbour pairs of Z d and satises Eqs. (2.11) for all squares. Then Z [ 0 ] max Z per[] (2.12) and for all. If L i = 2 n i Z [ 0 ] 2 Z (?L ie i )[(+L i e i ) [ 0 ]; i = 1; : : : ; d (2.13) [ 0 ] := lim L 1 ;:::;L d!1 jj?1 ln Z [ 0 ] lim sup jj?1 ln Z [] (2.14) L 1 ;:::;L d!1 then [ 0 ] = sup n 1 ;:::;n d jj?1 ln Z [ 0 ] (2.15) 4

6 Remarks 1. Z [] and Z [] are bounded continuous functions of each per xy on the unit circle of C, so they attain their maximum. 2. The equations (2.11) can be solved in any dimension d 2 with all xy 2 f1g. This is not a priori obvious above three dimensions when the number of squares ( 1 d(d?1) per site) is larger than the number of edges (d per site). In two dimensions, 2 a solution is, for example, x;x+e1 = 1 all x (?1 if x1 is even x;x+e2 = 1 if x 1 is odd (2.16) Given a which makes (2.11) hold,? = f? xy g also satises (2.11). With this observation, starting from (2) given by Eq. (2.16), we can obtain a solution (d) of Eqs. (2.11) in Z d inductively: If x = (x 0 ; x d ) with x 0 = (x 1 ; : : : ; x d?1 ) then (d) xy := 8 >< >: if x d = y d is even? (d?1) x 0 y if x 0 d = y d is odd 1 if jx d? y d j = 1 (d?1) x 0 y 0 (2.17) This construction of (d) works for x; y 2 Z d as well as on per, where is the domain (2.5). 3. If P s [] = P s [ 0 ] for all squares in then H[] and H[ 0 ] are unitarily equivalent, so Z [] = Z [ 0 ]. This also shows that Z [] depends on only through the products P s. To see this, for any x 2 dene a complex unit " x by setting " x = " 0 0 0x 1 0 x 1 x 2 0 x n?1 x xx n?1 x 1 0 (2.18) where (0 = x 0 ; x 1 ; : : : ; x n = x) is any path from 0 to x (that is, jx i+1? x i j = 1 for all i). Let, e.g., " 0 = 1. The denition (2.18) is meaningful because the right member of it is independent of the path: If (0; y 1 ; : : : ; y n?1 ; x) is another path, the two paths form a circle C = (0; x 1 ; : : : ; x n?1 ; x; y n?1 ; : : : ; y 1 ) (2.19) and the claim follows, if we prove Y C 0 uv = Y C Indeed, by rearranging this equation using we obtain uv (2.20) 1= uv = uv = vu; (2.21) 0 0x 1 0 x n?1 x xx n?1 x 1 0 = 0 0y 1 0 y n?1 x xy n?1 y 1 0 (2.22) 5

7 The circle C may contain `collapsed' subcircles formed by pairs of oppositely oriented coinciding edges. Due to Eq. (2.21), collapsed circles contribute a factor 1 to both sides of Eq. (2.20). In R d there are two-dimensional surfaces, composed of unit squares with vertices in, whose border is C. If S is such a surface, Eq. (2.20) follows from the hypothesis via Y C uv = Y s2s P s [] (2.23) Here it is understood that the border of each square is oriented in the same sense as C. >From Eq. (2.18) it follows that " y = " x 0 xy yx and hence 0 xy = " x xy " y (2.24) if jx? yj = 1: 0 is the `gauge-transformed' of. Equation (2.24) corresponds to adding a gradient, r, to the vector potential [cf. Eq. (1.3)] where (x) is dened through " x = e i(x) (2.25) Now H[ 0 ] is obtained from H[] by replacing c x by " x c x (and c y x by " xc y x). This is a transformation with the unitary operator exp[i(x)n x ] which does not aect the interaction term, therefore H[ 0 ] = U?1 H[]U (2.26) where U = exp [i X x2 (x)(n x" + n x# )] (2.27) The same proof fails to apply in the periodic case, because a circle of a nonzero winding number does not border two-dimensional surfaces. So if P s [] = P s [ 0 ] for all squares of per, one can dene " = (" x ) for which Eqs. (2.24) hold true if jx? yj = 1 but not necessarily if x? y = (2L i? 1)e i. Of course, if and " are given rst and 0 is dened through Eqs. (2.24), H per [ 0 ] is unitarily equivalent to H per [] and the equality of the partition functions follows. Thus, there are at least 2 jj real maximizing 's related to each other through Eqs. (2.24), where " x 2 f1g. 4. Equation (2.14) tells us that in the thermodynamic limit any with P s [] =?1 for all squares maximizes the pressure and, by the equivalence of the canonical and grand-canonical ensembles, minimizes the free energy. Let L = fx 2 : x 1 0g ; R = fx 2 : x 1 1g (2.28) and M be the reexion through the hyperplane x 1 = 1 2 : Mx = (1? x 1 ; x 2 ; : : : ; x d ) = x + (1? 2x 1 )e 1 (2.29) 6

8 To any associate L and R (all the three dened on nearest neighbour pairs of per ) by setting and L xy = 8 >< >: R xy = 8 > < >: xy if x; y 2 L? Mx;My if x; y 2 R 1 if x 2 L, y 2 R or vice versa xy if x; y 2 R? Mx;My if x; y 2 L 1 if x 2 L, y 2 R or vice versa (2.30) (2.31) So both L and R assign 1 to all edges (uv) with u 1 = 0 and v 1 = 1 or u 1 =?L 1 +1 and v 1 = L 1. >From the denition it follows that and Lemma 1 and ( L ) L = ( L ) R = L (2.32) ( R ) L = ( R ) R = R (2.33) Z per[] 2 Z per[ L ] Z per[ R ] (2.34) Z L [ ] Z R [ ] Z [ ] Z per [ ] (2.35) where = L; R. The inequalities hold if in the denition of L;R the number 1 is replaced by any other constant of modulus 1. Because of the periodic boundary condition and the choice (2.10) of the moduli jt xy j, the lemma implies altogether P i L i inequalities of the type (2.34). The role of the hyperplane x 1 = 1 2 can be played by any other hyperplane x i = k? 1 2 where i = 1; :::; d and k = 1; :::; L i, and L = fx 2 :?L i + k x i < kg ; R = n L (2.36) shifted, if necessary, by?2l i e i. 3 Proofs Mx = x + (k? 2x i )e i (2.37) Without explicit mentioning of reexion positivity, we shall employ that H[ L ] and H[ R ] with a satisfying Eqs. (2.11) possess this symmetry. Our proof is inspired by that of Lemma 4.1 and Theorem 4.2 in Dyson, Lieb and Simon's paper [7]. An application of reexion positivity to a system of fermions (\quantum monopole gas") appeared in Frohlich et al.'s article [10]. Although that model proves not to be reexion positive and therefore the argument for the phase transition fails to work, some part of it (in particular, Proposition 5.4 about the matrix representation) is useful for us. 7

9 3.1 Proof of the Theorem We shall make repeated use of the Lemma. For any maximizing, the inequality (2.34) holds with the equality sign, so both L and R are maximizing. Suppose now that a maximizing does not satisfy Eq. (2.11) for all squares of per. Choose a hyperplane cutting a square s 0 for which P s 0[] 6=?1. Because P s [ L;R ] =?1 for each square cut by this hyperplane, either L or R, e.g. L contains strictly more squares with a product?1 than. Replacing by L and repeating the procedure, after a nite number of steps we get a maximizing such that P s [] =?1 for all squares in per. If xy is nonreal, choose the hyperplane which separates x and y. Then in at least one of L and R, for example in L, the number of nonreal hopping amplitudes is strictly less than in. Replace by L and repeat the procedure. After a nite number of steps one gets a maximizing real. This proves the rst part of the theorem. Consider now a 0 such that P s [ 0 ] =?1 for all squares in Z d. Then P s [( 0 ) L;R ] =?1 for all squares, so ( 0 ) L and ( 0 ) R are gauge-transformed of 0. Combined with translation invariance, this implies Z [ 0 ] = Z [( 0 ) ] = Z L ie i [( 0 ) ] ( = L; R) (3.1) Using this equation, the inequality (2.12) follows from the second inequality of (2.35) (which does not depend on the extension of 0 to per ), and (2.13) comes from the rst of the inequalities (2.35). The limit dening the grand potential [ 0 ] exists because 0 can be chosen as periodic. For any of the form (2.5) let be given on per, satisfy (2.11) and maximize Z []. Then per Z [ ] = Z [ 0 ] (3.2) and therefore lim jj?1 ln Zper[ ]? jj?1 ln Z [ 0 ] = 0 (3.3)!Z d This proves the inequality (2.14). Equation (2.15) follows from the inequality (2.13) by a standard argument on sub- (or super-) additive sequences. With this, the theorem is proven, provided that the lemma is known. 3.2 Proof of the Lemma Let us start with rewriting the Hamiltonian in the following form. A[] = X B[] = X?H per [] = A[] + B[] + KX X X t xy c y xc y? U x;y2 L X (t i c y id i + t i d y ic i ) (3.4) (n x"? 1=2)(n x#? 1=2) (3.5) x2 X L t xy c y xc y? U (n x"? 1=2)(n x#? 1=2) (3.6) x;y2 R x2 R 8

10 In Eq.(3.4) the sum goes over K = 4(2L 2 ) (2L d ) terms, each corresponding to a spin and an edge (xy) of per cut by the hyperplane x 1 = 1 2 : either x 1 = 0 and y 1 = 1 or x 1 =?L 1 +1 and y 1 = L 1. Now t i, c i and d i are shorthands for t xy, c x and c y, respectively, where x 2 L and y 2 R. The notation is unambiguous because x 2 L with x 1 = 0 or?l i + 1 has a unique nearest neighbour in R. Remark that t i = r 1 i where r 1 was introduced in Eq. (2.10). The next step is to apply the Lie (Trotter) product formula [11] to obtain where Let Y Q N [] = e A=N e B=N K e?hper[ ] = lim N!1 Q N[] (3.7) exp h N (t i c y id i + t i d y ic i ) i N (3.8) a = cosh[r 1 =N]? 1 ; b = sinh[r 1 =N] (3.9) Expanding the exponentials in the product, using? i = c y ic i ; i = d y i d i (3.10) [ i c y id i + i dy i c i ] 2k =? i (1? i ) + (1?? i ) i (3.11) for any integer k > 0 and resumming, we nd exp h N (t i cy id i + t i dy ic i ) i = 1 + a[? i (1? i ) + (1?? i ) i ] + b[ i c y id i + i dy ic i ] (3.12) Therefore KY exp h N (t i c y id i + t i d y ic i ) i = X a jj+jj b jj+jj ( )? (1??) (1? ) (c y d) (d y c) (3.13) The summation goes over all four-vectors (; ; ; ) whose elements are disjoint - maybe empty - subsequences of (1; :::; K); jj; jj; jj and jj are their respective lengths, so jj+jj+jj+jj K. In Eq. (3.13) the powers with exponents ; ; ; have the following meaning. If then = (k 1 ; : : : ; k j ) ; 1 k 1 < < k j K (3.14) = k1 kj ; (c y d) = c y k 1 d k1 c y k j d kj (3.15) and similar holds for the other powers. Our aim is to bring all operators with support in L to the left of all operators with support in R. Doing this, we have to count the number of interchanged pairs of creation and annihilation operators. When moving all c y i to the left of all d i in (c y d), one performs 1+ +(jj?1) pair changes. To move all c i to the left of all d y i in (d y c) one has to permute 1+ +jj 9

11 pairs. Furthermore, jj jj pairs are to be permuted when moving jj operators c i to the left of jj operators d i. Thus, the left-right separation can be done by s = s(; ) = 1 2 (jj?1)jj+ 1 2 jj(jj+1)+jj jj = 1 2 (jj+jj)2 + 1 (jj?jj) (3.16) 2 pair permutations, so that KY exp h N (t i c y id i + t i d y ic i ) i = X (?1) s a jj+jj b jj+jj ( )? (1??) (c y ) c (1? ) d (d y ) (3.17) Multiply this with e A=N e B=N, take the Nth power, expand it and, by permutations, rearrange each term in such a way that all operators with support in L will be regrouped on the left side. Because of the vanishing of the commutators [A; B]; [A; d i ]; [B; c i ]; [ i ; c j ]; [? i ; d j ], the sign changes are easy to follow. The result is X X Y N Q N [] = : : : (?1) S a 1 b 2 i ( ) i GD (3.18) with and ( 1 ; 1 ; 1 ; 1 ) ( N ; N ; N ; N ) G = e A=N? 1 (1??) 1 (c y ) 1 c 1 e A=N? N (1??) N (c y ) N c N (3.19) D = e B=N (1? ) 1 1 d 1 (dy ) 1 eb=n (1? ) N N d N (d y ) N (3.20) For every i, i ; i ; i ; i are pairwise disjoint subsequences of (1; : : : ; K), and S = where NX 1 = s( i ; i ) + NX j=2 NX (j i j + j i j) ; 2 = (j j j+j j j) = NX j?1 X NX (j i j + j i j) (3.21) (j i j+j i j) = 1 2 ( + )2 + 1 (? ) (3.22) 2 j i j ; = NX j i j (3.23) The operators G and D have support in L and R, respectively. Therefore Tr GD = 0 if 6= (3.24) Indeed, if we expand exp A=N and exp B=N, Tr GD is obtained as a sum of traces of monomials of creation and annihilation operators. Each monomial conserves particle 10

12 number but, if 6=, no monomial conserves particle number locally, i.e., for every (x), which is necessary for getting a nonvanishing trace. If = then S is even, therefore we obtain X Y N Tr Q N [] = a 1 b 2 i ( ) i Tr GD (3.25) where the summation is the same as in Eq. (3.18), but only the terms with NX j i j = NX are nonvanishing. Hence, if i is constant, we get j i j (3.26) Tr Q N [] = X a 1 b 2 Tr GD ( i = const) (3.27) In particular, this holds for, = L; R. It will be seen that in this case each term of the sum is nonnegative. The rest of the work is easier to do in an explicit matrix representation [10]. Order the pairs (x) for x 2, ="; # in the following way. (i) Put the jj pairs with x 2 L at the rst jj places. (ii) If x 2 L and (x) is at the n th place, put (Mx; ) at the (jj + n) th place. Write the Fock space over as a 2jj-fold tensor product of C 2 with itself, where each factor corresponds to a one-particle subspace at a site x for a spin, the order being determined above. Use the canonical basis 1 2jj, where i = 1 0! or 0 1! (3.28) Let the vacuum state be!! vac = (3.29) If (x) is at the n th place, the 2 2jj dimensional matrix representation of c y x will be R(c y x) = (? 3 ) (? 3 ) (3.30) where 3 = 1 0! 0?1 + = 0 1! 0 0? = 0 0! = 1 0! 0 1 (3.31) and + is at the n th place. R(c x ) is obtained by replacing + with?. Let us return to Eq. (3.25). If + is even, each of the rst jj factors in R(D) is (? 3 ) raised to an even power, so R(G) = R 0 (G) I R(D) = I R 0 (D) R(GD) = R 0 (G) R 0 (D) (3.32) 11

13 whence Tr GD = Tr R 0 (G) Tr R 0 (D) Tr 0 G Tr 0 D (3.33) In these equations R 0 and Tr 0 denote, resp., the 2 jj dimensional matrix representation and its trace, I is the 2 jj dimensional unit matrix. Again, we see that Tr 0 G = Tr 0 D = 0 (3.34) if 6= (i.e., if G and D do not conserve the particle number; in fact, if they do not conserve the number of particles with a given spin). Our nal formula for Tr Q N [] is X Y N Tr Q N [] = a 1 b 2 i ( ) i Tr 0 G[] Tr 0 D[] (3.35) Clearly, G depends on xy (through A) only if x; y 2 L, and D depends on xy (through B) only if x; y 2 R. Therefore G[] = G[ L ] and D[] = D[ R ] (3.36) Let U L be the unitary operator which exchanges the two basis vectors (3.28) for every (x) such that x 2 L. Carried over to the operators, U L exchanges c x and c y x for all x 2 L ; therefore (half-lling!) U L A[]U?1 L = A[? ] (3.37) Due to the choice (2.10), the denition of L and the above ordering of the pairs (x), we get R 0 (U L G[ ]U L?1 ) = R 0 (D[ L ]) (3.38) Thus, Similarly, (Tr 0 G[]) = Tr 0 G[ ] = Tr 0 U L G[ ]U?1 L = Tr0 D[ L ] (3.39) (Tr 0 D[]) = Tr 0 D[ ] = Tr 0 G[ R ] (3.40) For the rst equalities in the above two equations the reality of R(c x ) was essential. Writing, for example, Eq. (3.39) for =, ( Tr 0 G[ ]) = Tr 0 D[ ] ( = L; R) (3.41) where Eqs. (2.32) and (2.33) were used. It follows, then, Tr Q N [ ] = X a 1 b 2 jt r 0 G[ ]j 2 = X a 1 b 2 j Tr 0 D[ ]j 2 ( = L; R) (3.42) Now the Schwarz inequality applied to Eq. (3.35) yields (Tr Q N []) 2 X a 1 b 2 j Tr 0 G[]j 2 X a 1 b 2 j Tr 0 D[]j 2 = Tr Q N [ L ] Tr Q N [ R ] (3.43) 12

14 Taking the limit N! 1, we obtain the inequality (2.34). We may number the K terms of the sum in Eq. (3.4) so that for i = 1; :::; K=2, c i = c x where x 1 = 0. Consequently, i > K=2 corresponds to c i = c x with x 1 =?L Then Tr 0 e A[ ] Tr 0 e B[ ] < X 0 a 1 b 2 Tr0 G[ ] Tr 0 D[ ] < X a 1 b 2 Tr 0 G[ ] Tr 0 D[ ] = Tr Q N [ ] (3.44) where the primed sum is restricted to subsequences of (1; :::; K=2). The left member of this inequality equals the left member of (2.35) while the middle and right members tend to the middle and right members of (2.35), respectively, when N goes to innity. This nishes the proof of the lemma. 4 Extensions 1. In the Hamiltonian the hopping amplitude may depend on the spin, i.e., one may replace t xy by t xy; = r xy; xy; (4.1) A maximizing will satisfy P s; := Y s xy; =?1 (4.2) for each unit square and. The proof needs no modication. Varying r xy;# between r xy;" and zero, one can interpolate between the Hubbard and the Falicov-Kimball models. 2. One can add any `reexion negativ' interaction to the Hamiltonian, cf. Ref. [8]. Here are two simple examples of nearest neighbour interactions: (i) X X V 1 =? K xy; n x (1? n y ) (4.3) x;y2 where K xy; = k i; 0 if x? y = e i (4.4) (also if x? y = (2L i? 1)e i, in the periodic case) and vanishing otherwise. If all k i; > 0, the ground states of V 1 are the two Ising antiferromagnetic states and the two singlet states with alternating empty and doubly occupied sites. In the Lie formula ( ) written down for exp?fh[] + V 1 g there appear additional factors of the type exp ( N K i[? i (1? i )+(1?? i ) i ]) = 1+(e K i=n?1)[? i (1? i )+(1?? i ) i ] (4.5) where K i 0 is one of the k 1; 's. Since such terms occured already earlier [see Eq. (3.12)], the proof applies without modication. 13

15 where (ii) Nearest neighbour antiferromagnetic Ising interaction V 2 = 1 2 X x;y2 J xy (n x"? n x# )(n y"? n y# ) (4.6) J xy = J i 0 if x? y = e i (4.7) (also if x? y = (2L i? 1)e i, in the periodic case) and vanishing otherwise. Now the additional factors entering the Lie formula for exp?fh[] + V 2 g are e (Jxy=N )(n x#?n x" )(n y"?n y# ) = 1 + [cosh(j xy =N)? 1]jn x"? n x# j jn y"? n y# j + sinh(j xy =N)(n x#? n x" )(n y"? n y# ) (4.8) where x 2 L and y = Mx. When applying U L, the factor n x#? n x" changes sign, therefore the left-right symmetry is restored and each term of the sum composing Tr Q N [ L;R ] remains positive. 3. Instead of rectangular domains of Z d wrapped onto the d-torus we may consider other graphs. Let G be a graph and C a circle of even length jcj = n, i.e., C = (x 1 ; :::; x n ) is a sequence of vertices without repetition such that (x 1 x 2 ); :::; (x n x 1 ) are edges of G. Suppose that G can be embedded in R D for some D 2 in such a way that C is cut by a hyperplane which does not contain any vertex and divides G into two symmetric halves. Such a circle will be called symmetric. Dene the Hubbard model H G [] on G with jt xy j = jt x 0 y 0j if (xy) and (x0 y 0 ) are related by symmetry. In Section 3 we proved, in fact, the more general Theorem 2 There is a which maximizes the partition function and on any symmetric circle C, Z G [] = Tr exp?h G [] (4.9) Y C uv = (?1) jcj=2?1 (4.10) Furthermore, if reexion hyperplanes go through every edge of a circle C then can be chosen real on C. The simplest graph to which this assertion applies is a closed chain of even length. For the unit cube of Z d in any dimension d 2 this theorem shows that any giving P s [] =?1 for all squares maximizes the partition function. It also applies to parts of the honeycomb lattice wrapped onto the 2-torus, for which it yields that an even number of ux quanta per hexagon will maximize the grand potential. 14

16 5 Limitations 1. We could not prove that P s [] =?1 for all squares is necessary in order that maximizes the partition function. This is true for the approximants Tr Q N [] of any order because the inequalities of the Lemma hold for them in the strict sense. We may loose the strict inequalities when taking the limit N! In nite volume the canonical and grand-canonical ensembles are not fully equivalent. We found which maximizes the grand potential or pressure but have not shown that the same exactly minimizes the free energy. The proof does not extend to the canonical partition function because we do not have the analog of the factorization (3.33) in the canonical ensemble. 3. Not all bipartite lattices contain symmetric circles. For example, if we `decorate' every edge of the square lattice by a single vertex, we obtain a bipartite lattice, all symmetry axes of which pass through vertices. The symmetry restriction for the proof has to be considered as a failure of the method. Our theorem does not cover certain nonsymmetric cases in which Lieb and Loss [2], [3] could prove the conjecture. Therefore, the ux-phase problem even at half-lling cannot be considered as completely solved. Acknowledgements. This work was started in preparation to the Heisenberg- Hubbard Workshop held recently at the Erwin Schrodinger Institute in Vienna. I thank J. Avron, V. Bach and R. Seiler for the invitation and the ESI for nancial support and hospitality during my stay in Vienna. This work was supported by the Hungarian Scientic Research Fund (OTKA) through Grant No. T

17 References [1] Y. Hasegawa, P. Lederer, T.M. Rice, P.B. Wiegmann, Phys. Rev. Lett. 63, 907 (1989) [2] E.H. Lieb, Helv. Phys. Acta 65, 247 (1992) [3] E.H. Lieb and M. Loss, Duke Math. J. 71, 337 (1993) [4] E.H. Lieb, Phys. Rev. Lett. 73, 2158 (1994) [5] N. Macris and B. Nachtergaele, to be published [6] J. Frohlich, B. Simon and T. Spencer, Commun. Math. Phys. 50, 79 (1976) [7] F.J. Dyson, E.H. Lieb and B. Simon, J. Stat. Phys. 18, 335 (1978) [8] J. Frohlich, R. Israel, E.H. Lieb and B. Simon, Commun. Math. Phys. 62, 1 (1978) [9] E.H. Lieb, M. Loss and R.J. McCann, J. Math. Phys. 34, 891 (1993) [10] J. Frohlich, R. Israel, E.H. Lieb and B. Simon, J. Stat. Phys. 22, 297 (1980) [11] M. Reed and B. Simon, Methods of modern mathematical physics I: Functional analysis. Academic Press