1 CBMS: B ham, AL June 17, 2014 On the Random XY Spin Chain Robert Sims University of Arizona
2 The Isotropic XY-Spin Chain Fix a real-valued sequence {ν j } j 1 and for each integer n 1, set acting on n 1 n H n = (σj x σj+1 x + σ y j σy j+1 ) + ν j σj z, H n = n C 2 Here, the Pauli matrices are ( ) ( σ x 0 1 =, σ y = 1 0 0 i i 0 ), σ z = ( 1 0 0 1 ) and for α {x, y, z} and 1 j n we embed these into B(H n ): σ α j = 1l 1l σ α 1l 1l with σ α in the j-th factor.
3 A Locality Review Let us denote the Heisenberg dynamics associated to H n by τt n (A) = e ithn Ae ithn for all A B(H n ) For simplicity, take A k B(H n ) to be all those observables supported at a single site 1 k n. For bounded sequences {ν j }, it is clear that Lieb-Robinson bounds apply and for any µ > 0, [τt n (A), B] C A B e µ( k k v t ) for any A A k, B A k, and t R with some v > 0 depending on µ and the sequence {ν j }. Can one prove a stronger statement if the {ν j } are random?
4 A Strong Form of Dynamical Localization Assume {ν j } is an i.i.d. random sequence with compactly supported bounded density ρ. Theorem (Hamza-S-Stolz 11) There are positive numbers C and η for which, given any integer n 1 the bound ( ) E sup [τt n (A), B] C A B e η k k t R holds for any A A k and B A k with 1 k < k n. One can think of this as a Lieb-Robinson bound with zero velocity. After reviewing some basic properties of this model, the main goal of this talk is to prove this result.
5 Diagonalizing the Hamiltonian As Gunter discussed, diagonalizing this Hamiltonian goes back to LSM 61: First, one introduces raising and lowering operators: ( ) a = 1 2 (σx + iσ y 0 1 ) = 0 0 ( ) a = 1 2 (σx iσ y 0 0 ) = 1 0 It will also be useful to observe that: ( ) ( ) a 1 0 a = and aa 0 0 = 0 0 0 1 i.e., these four operators form a basis for B(C 2 ) = C 2 2.
6 Jordan-Wigner Transform Then, one introduces the non-local Jordan-Wigner transform: c 1 = a 1 and c j = σ z 1... σ z j 1a j for each j 2. These operators are particularly useful because they satisfy the canonical anti-commutation relations (CAR): {c j, ck } = δ jk1l and {c j, c k } = {cj, ck } = 0. We often collect them as vectors: c = c 1 c 2. c n or c = (c1, c2,, cn)
7 Re-writing the Hamiltonian A short calculation shows that n 1 n H n = (σj x σj+1 x + σ y j σy j+1 ) + ν j σj z n 1 = 2 (aj a j+1 + aj+1a j ) + n 1 = 2 (cj c j+1 + cj+1c j ) + = 2c M n c E1l n ν j (2aj a j 1l) n ν j (2cj c j 1l) where ν 1 1. 1 ν.. 2 M n =...... 1 1 ν n and E = n ν j
8 The Anderson model: In the case that the {ν j } are random, M n corresponds to the well-studied Anderson model corresponding to a single quantum particle in a random environment. With our assumptions on the {ν j }, one can prove Theorem (Dynamical Localization) There exist positive numbers C and η such that for all integers n 1 and any k, k {1,..., n}, ( ) E sup (e itmn ) kk t R C e η k k. For the purposes of this talk, we will assume this is well-known; see, for example, Kunz-Souillard or Aizenman-Molchanov.
9 Summary of findings so far: We have seen that this many-body XY Hamiltonian can be expressed in terms of a single-particle Hamiltonian, i.e., H n = 2c M n c E1l In the case that the {ν j } are nice random variables, we also know that the single-particle dynamics, i.e. e itmn, is dynamically localized. We now show that this implies the desired result for the many-body dynamics, i.e. τt n ( ).
10 Bogoliubov Transformations As we saw, M n is real symmetric. In this case, there is a real orthogonal U n such that U t nm n U n = Λ = diag(λ k ) Take b 1 b 2. = Ut n c 1 c 2. i.e. b = Ut nc b n c n Observe that these b-operators also satisfy the CAR and moreover n H n = 2c M n c E1l = 2 λ k bk b k E1l k=1 Thus, this last transform expresses H n as a system of free Fermions.
11 Calculating the Many-Body Dynamics As Gunter also discussed, a simple calculation shows that τ n t (b k ) = e 2itλ k b k and τ n t (b k ) = e2itλ k b k or τ n t (b) = e 2itΛ b A further calculation, using that c = U n b, shows that τt n (c) = e 2itMn c or τt n (c j ) = ) (e 2itMn k jk c k and so the many-body dynamics (of the c-operators) can be expressed explicitly in terms of the single-particle dynamics. We can now begin the proof of the main result.
12 Recall the Main Result: Theorem Assume {ν j } is an i.i.d. random sequence with compactly supported bounded density ρ. There are positive numbers C and η for which, given any integer n 1 the bound ( ) E sup [τt n (A), B] C A B e η k k t R holds for any A A k and B A k with 1 k < k n.
13 Proof of Main Result: Fix 1 k < k n as indicated and take B A k. Consider first the non-local A = c k, i.e.: [τ n t (c k ), B] = = n ) (e 2itMn n ) (e 2itMn j=k kj [c j, B] kj [c j, B] Using the single-particle dynamical localization result, we find that ( ) E sup [τt n (c k ), B] t 2C B n j=k e η (j k) 2C B 1 e η e η (k k) This is an estimate of the type desired; excepting that it is for the non-local observable c k. By taking adjoints, it is clear that a similar result holds for c k.
14 Proof of Main Result (cont.): Now take A = a k. Recall a k = σ z 1 σz k 1 c k. Observe that [τ n t (a k ), B] = τ n t (σ z 1) τ n t (σ z k 1 )[τ n t (c k ), B] + +[τ n t (σ z 1) τ n t (σ z k 1 ), B]τ n t (c k ) where we have used the automorphism property of τ n t ( ), i.e. and the Leibnitz rule: It is clear then that τ n t (AB) = τ n t (A)τ n t (B) [AB, C] = A[B, C] + [A, C]B [τt n (a k ), B] [τt n (c k ), B] + [τt n (σ1) z τt n (σk 1 z ), B]
15 Proof of Main Result (cont.): Now, for any j, the quantity appearing above satisfies [τ n t (σ z 1) τ n t (σ z j ), B] [τ n t (σ z j ), B] + [τ n t (σ z 1) τ n t (σ z j 1), B] again by Leibnitz. Moreover, σ z j = 2a j a j 1l = 2c j c j 1l and so [τt n (σj z ), B] = 2[τt n (cj ), B]τt n (c j ) + 2τt n (cj )[τt n (c j ), B] and, in fact: [τt n (σj z ), B] 2 [τt n (cj ), B] + 2 [τt n (c j ), B] Consequently, k [τt n ( (a k ), B] 2 [τ n t (cj ), B] + [τt n (c j ), B] )
16 Proof of Main Result (cont.): Using our previous result, it is clear that ( ) E sup [τt n (a k ), B] t which is the result for A = a k. 8C B 1 e η k e η (k j) 8C B (1 e η ) 2 e η (k k) By taking adjoints, a similar result holds for A = ak. Using Leibnitz again, it is clear that a similar result holds for both A = ak a k and A = a k ak. Since these operators form a basis for A k, this completes the proof.
17 A Generalization The isotropic XY model is not the only spin chain that reduces to a system of free Fermions. Consider e.g. the anisotropic XY Spin Hamiltonian n 1 n H n = µ j [(1 + γ j )σj x σj+1 x + (1 γ j )σ y j σy j+1 ] + ν j σj z with real parameters given by: interaction strengths {µ j }, anisotropy {γ j }, and field strengths {ν j }. Introducing the same raising and lowering operators and then the Jordan-Wigner transform, this many-body operator can also be written in terms of an effective single-particle Hamiltonian.
18 Diagonalizing the Hamiltonian As is discussed in the notes, H n = C M n C where C = (c 1, c 2,, c n, c 1, c 2,, c n) t is a column vector and C = (c1, c2,, cn, c 1, c 2,, c n ) In this case, the single particle Hamiltonian is a block-matrix ( ) M n = A n B n B n A n
19 Diagonalizing the Hamiltonian(cont.) with ν 1 µ 1. µ 1 ν.. 2 A n =...... µn 1 µ n 1 ν n and 0 γ 1 µ 1. γ 1 µ..... 1 B n =............... γn 1 µ n 1 γ n 1 µ n 1 0
20 The More General Result (at least in words) In our paper, we prove an analogous result: If the single particle Hamiltonian M n is dynamically localized, then the many body Hamiltonian satisfies dynamical localization as well, in the sense that we establish a zero-velocity Lieb-Robinson bound (in average). What we do not quantify (and what remains an interesting open question) is: Under what conditions is this more general, random single particle system dynamically localized?
21 For the Experts: By re-ordering the basis vectors, the block-matrix above is easily seen to be unitarily equivalent to ν 1 σ z µ 1 S(γ 1 ) µ 1 S(γ 1 ) t ν 2 σ z... M n =............... µn 1 S(γ n 1 ) µ n 1 S(γ n 1 ) t ν n σ z where ( ) 1 γ S(γ) = γ 1 The question now becomes: If some of these coefficients are random, is such a one-dimensional, single-particle model dynamically localized?