Spin Chains with Dynamical Lattice Supersymmetry

Abstract

Spin chains with exact supersymmetry on finite one-dimensional lattices are considered. The supercharges are nilpotent operators on the lattice of dynamical nature: they change the number of sites. A local criterion for the nilpotency on periodic lattices is formulated. Any of its solutions leads to a supersymmetric spin chain. It is shown that a class of special solutions at arbitrary spin gives the lattice equivalents of the $\mathcal{N}=(2,2)$ superconformal minimal models. The case of spin one is investigated in detail: in particular, it is shown that the Fateev-Zamolodchikov chain and its off-critical extension possess a lattice supersymmetry for all its coupling constants. Its supersymmetry singlets are thoroughly analysed, and a relation between their components and the weighted enumeration of alternating sign matrices is conjectured.

Appendices

Appendix A: Trigonometric Supercharges

In this appendix we explain a constructive way to arrive at (18). Our strategy is the following. We specialise first the matrix elements h _(r,s),(m,n) to the diagonal case (r,s)=(m,n), and construct a particular solution which is invariant under spin reversal. Second, in the main text, we show that this solution satisfies indeed the spin-reversal symmetry even for off-diagonal matrix elements. We start thus by considering the matrix element b _m,n =h _(m,n),(m,n), given by

$$ b_{m,n} = a_{m+n+1,m}^2 + \frac{1}{2} \Biggl(\, \sum_{k=0}^{m-1}a_{m,k}^2 +\sum_{k=0}^{n-1}a_{n,k}^2 \Biggr). $$

First, set m=n=0. We impose b ₀₀=b _ℓℓ , and thus have

$$ a_{1,0}^2 = \sum_{k=0}^{\ell-1}a_{\ell,k}^2. $$

Next, consider the equation b _0,ℓ−n =b _ℓ,n . Using the previous result we find that

$$ a_{\ell-n+1,0}^2 + \frac{1}{2}\sum _{k=0}^{\ell-n-1}a_{\ell-n,k}^2 = \frac{1}{2}a_{1,0}^2 +\frac{1}{2}\sum _{k=0}^{n-1}a_{n,k}^2. $$

Replacing now n→ℓ−n and subtracting both equations, we find a useful symmetry property:

$$ a_{m,0}^2+a_{\ell+2-m,0}^2 = a_{1,0}^2, \quad m = 2,\dots\ell. $$

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Now impose b _m,1=b _{ℓ−m,ℓ−1}. For m=0,…,ℓ−2 it follows that

$$ a_{m+2,1}^2=\frac{1}{2} \Biggl(\,\sum _{k=0}^{\ell-m-1}a_{\ell -m,k}^2+\sum _{k=0}^{\ell-2}a_{\ell-1,k}^2- \sum_{k=0}^{m-1}a_{m,k}^2 \Biggr)-\frac{1}{2}a_{1,0}^2. $$

In particular, the case m=0 leads to $a_{2,1}^{2}= a_{2,0}^{2} = \sum_{k=0}^{\ell-2}a_{\ell-1,k}^{2}/2$, and thus to

$$ a_{m+2,1}^2=\frac{1}{2} \Biggl(\,\sum _{k=0}^{\ell-m-1}a_{\ell -m,k}^2-\sum _{k=0}^{m-1}a_{m,k}^2 \Biggr)+a_{2,0}^2-\frac{1}{2}a_{1,0}^2. $$

Again, we notice that the expression in brackets is antisymmetric under m→ℓ−m. This allows to derive the equation

$$ a_{m+2,1}^2+ a_{\ell+2-m,1}^2 = 2a_{2,0}^2-a_{1,0}^2. $$

The left-hand side can be reduced with the help of the recursion relation (16), and thus everything can be rewritten in terms of $a_{m,0}^{2}$. After a little algebra, we find the non-linear recursion relation

$$ a_{m+2,0}^2a_{m+1,0}^2 + \bigl(a_{1,0}^2-a_{m+1,0}^2\bigr) \bigl(a_{1,0}^2-a_{m,0}^2\bigr) = a_{1,0}^2\bigl(2a_{2,0}^2-a_{1,0}^2 \bigr). $$

We will now show that it admits a particular solution related to a linear recursion relation. To this end, set $a_{m,0}^{2} = z_{m}/z_{m-1}$ and normalise z ₀=1. We find that

$$ z_{m-1}(z_{m+2}-z_1z_{m+1}+z_m) + z_{m}\bigl(z_{m+1}-z_1z_m + \bigl(2\bigl(z_1^2-z_2\bigr)-1 \bigr)z_{m-1}\bigr)=0. $$

This relation holds certainly if we choose for z _m as solution of the equation

$$ z_{m+2}-z_1z_{m+1}+z_m = 0, \quad \mbox{with} \ z_0 = 1,\ z_0=0. $$

This is the recursion relation for the Chebyshev polynomials of the second kind. We prefer to use q-integers and thus write

$$ z_m = [m+1], \qquad[n] = \frac{q^n-q^{-n}}{q-q^{-1}},\quad\mbox {with}\ q=e^{i\eta}. $$

The phase η is not arbitrary as we have to respect the symmetry relation (37). After some algebra, one finds that this relation implies sin(ℓ+2)η=0. The unique choice for η to keep all the $a_{m,0}^{2}$ positive is η=π/(ℓ+2). Hence, q takes the root-of-unity value

$$ q = e^{\mathrm{i}\pi/(\ell+2)}. $$

Thus, we have [m+ℓ+2]=−[m]=[−m]. Hence, we find

$$ a_{m,0}^2= \frac{[m+1]}{[m]}. $$

The general coefficients a _m,k are obtained from the recursion relation (16). This completes the derivation of the constants (18).

Appendix B: Special Polynomials

Bazhanov and Mangazeev introduced in [11, 75] a series of polynomials s _n (z), n∈ℤ, which appear as components in the ground-state vectors of the spin-1/2 XYZ Hamiltonian (13a), (13b), and sum rules (see also the related work of Razumov and Stroganov [84]). A sum rule for the square norm was recently proved by Zinn-Justin [98].

These polynomials appear to be not only off-critical generalisations of the combinatorial numbers found in the ground states of various variants of the spin-1/2 XXZ chain at Δ=−1/2, but display also an interesting relation to classical integrability, namely to certain tau-function hierarchies associated to the Painlevé VI equation. They are solutions to the non-linear difference-differential equation

with s ₀(z)=s ₁(z)≡1. This equation can be understood as a result of special Bäcklund transformations for Painlevé VI [74, 78]. It is far from being evident that its solutions are polynomials for the given initial conditions. Various properties of the solutions to this equation were conjectured in [75]. For our purposes, the factorisation properties are relevant. In fact, the authors claim that for any k∈ℤ one has

$$ s_{2k+1}\bigl(w^2\bigr) = s_{2k+1}(0)p_k(w)p_k(-w), \quad p_k(0)=1, $$

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where the p _k (w) are polynomials of degree degp _k (w)=k(k+1) with positive integer coefficients. We have the transformation symmetry

$$ p_k(w)= \biggl(\frac{1+3w}{2} \biggr)^{k(k+1)}p_k \biggl(\frac {1-w}{1+3w} \biggr). $$

The p _k (w)’s are the polynomials appearing in the zero-energy states of the twisted elliptic Fateev-Zamolodchikov chain at x=±1/2, y=±ζ/2.

Appendix C: Counting Momentum States

In this appendix, we use the classical theory of necklace enumeration in order to count certain momentum states for a spin-ℓ/2 chain with N sites [51].

We consider thus a chain/necklace with N beads or equivalently a periodic lattice with N sites. Each site of this graph, labelled by some integer j=1,…,N, can be coloured with a colour m _j =0,1,…,ℓ. Thus, a colouring/configuration is given by an N-tuple μ=(m ₁,…,m _N−1,m _N ). We denote by X the set of all possible configurations. There is a natural action of the cyclic group, generated by the cyclic shift T, on X through T⋅(m ₁,…,m _N−1,m _N )=(m _N ,m ₁,…,m _N−1). We will study the fixed points of this group action. This is motivated by the fact that the elements of X are labels of the Hilbert space of the spin chain.

Momentum States

To each μ=(m ₁,…,m _N−1,m _N )∈X we associate a basis vector |μ〉=|m ₁,…,m _N−1,m _N 〉 in V ^⊗N where V≃ℂ^ℓ+1 as in Sect. 2. The translation operator on V ^⊗N acts according to T|μ〉=|T⋅μ〉 (we use the same notation for it acting on the vector space and the set X). Out of any basis state |μ〉 we build a momentum state

$$ |\psi_\mu\rangle= \sum_{j=0}^{N-1}e^{-\mathrm{i}j k} T^j|\mu \rangle ,\qquad T|\psi_\mu\rangle= e^{\mathrm{i}k}|\psi_\mu\rangle, $$

where k=2πm/N, m=0,1,…,N−1 is the momentum. |ψ _μ 〉 may be zero what implies that the configuration μ is incompatible with the given momentum k. In fact, for each configuration μ∈X there is a minimal positive integer r such that T ^r⋅μ=μ. We call r the symmetry factor of μ. For the momentum state to be non-zero we need

$$ r \in p\mathbb{N} \quad\mbox{with} \ p = \frac{N}{\gcd(N,m)}. $$

Motivated by the main text, our aim is to count all momentum states with k=0 for N odd, and k=π for N even, with given total particle number

$$ M = \sum_{j=1}^N m_j. $$

Chains of Odd Length

We start with N odd. A momentum state may be generated by two different configurations μ,μ′ if they can be mapped onto each other through a suitable translation. This induces an equivalence relation on the set of configurations, and thus our task consists of counting the number ν _MN of equivalence classes (or equivalently the number of orbits in the set of configurations generated by the action of T).

We denote by $X^{m}_{M} \subset X$ the set of all configurations with given M which are stable under the action of T ^m. According to Burnside’s lemma, we have

$$ \nu_{MN} = \frac{1}{N}\sum_{m=0}^{N-1}\big|X^m_M\big|. $$

(39)

Therefore, we need to evaluate $|X^{m}_{M}|$. The case m=0 is the simplest: we just need to count all possible configurations for given M. Let us thus consider m=1: in this case all sites need to have equal colour because a single translation (cyclic shift) has only one cycle. Hence, $X^{1}_{M}$ contains a single configuration of colour j if and only if M=jN for some integer j=0,1,…,ℓ, in other words if N|M as long as M≤ℓN. We abbreviate $a_{N}(M)=|X^{1}_{M}|$. For m>1 there various distinct cycles. It is a classical result that their number is given by gcd(N,m), and therefore they have length r _m =N/gcd(N,m). For each cycle, we have however the same problem as for m=1. Hence, we find the general result

$$ \big|X_M^{m}\big| = \sum_{M_1,\dots,M_{\gcd(N,m)}=0}^\infty \Biggl(\,\prod_{j=1}^{\gcd(N,m)}a_{r_m}(M_j) \Biggr)\delta_{M,\sum _{j=1}^{\gcd(N,m)} {M_j}}. $$

(40)

We notice a typical convolution structure. It is therefore useful to introduce the generating function $\mathcal{A}_{N}(z)$ for the numbers a _N (M) where z is the conjugate variable of M. We obtain

$$ \mathcal{A}_N(z) = \sum_{M=0}^\infty a_{N}(M)z^M = \frac{1-z^{(\ell +1) N}}{1-z^N}. $$

We use this in (40) and find the generating series

$$ \sum_{M=0}^\infty\big|X_M^m\big|z^M = \mathcal{A}_{r_m}(z)^{\gcd(N,m)}= \biggl(\frac{1-z^{(\ell+1)N/\gcd(N,m)}}{1-z^{N/\gcd(N,m)}} \biggr)^{\gcd(N,m)}. $$

Performing the sum with respect to m leads to the generating function f _N (z) given in the main text for odd N.

Chains of Even Length

Next, we treat the case of even N and momentum k=π. Thus, we want to count the number of all representatives with even symmetry factor r. It seems natural to modify (39) in such a way that only fixed points of T ^m with even m are retained. Thus, we attempt to discard all contributions with odd m and write $1/N \sum_{m=0}^{N/2-1}|X_{M}^{2m}|$. Yet, in this way we count some unwanted configurations. Indeed for m>0, if we decompose $2m = 2^{k_{m}}b_{m}$ with b _m an odd integer, and k _m ≥1 some integer, then the fixed points of T ^2m contain the fixed points of $T^{b_{m}}$. These need to be subtracted. For m=0 we need to subtract the fixed points of $T^{b_{N/2}}$. Thus we arrive at

$$ \nu_{MN} = \frac{1}{N} \Biggl(\,\sum _{m=0}^{N/2-1}\big|X_M^{2m}\big|-\sum _{m=1}^{N/2}\big|X_{M}^{b_m}\big| \Biggr) \quad\mbox{for even } N. $$

The sum can be rearranged in a convenient way. To this end observe in (40) that $|X_{M}^{m}|$ depends on m only through gcd(N,m). In order to rearrange the sum we need the following lemma:

Lemma 1

Let N be even and consider c _m =gcd(N,b _m ) for m=1,2,…,N/2. We have c _m+N/2=c _m . Furthermore, let p=b _N/2 then it follows that c _m+(p−1)/2=gcd(N,2m−1).

Proof

The proof is entirely based on two simple identities. Let m and n be positive integers. (i) if m>n then we may write gcd(m,n)=gcd(m−n,n); (ii) if n is odd then gcd(m,n)=gcd(2m,n). From the second identity it follows in particular that gcd(b _m ,n)=gcd(m,n) for odd n. Furthermore we find gcd(b _m ,n)=gcd(m,b _n )=gcd(b _m ,b _n ).

For the first part, use b _N =b _N/2 and write

For the second part, use b _N =b _N/2=p and write

This proves the claim. □

Combining the implicit dependence of $|X_{M}^{b_{m}}|$ on c _m =gcd(N,b _m ) with the periodicity and shift properties given by this lemma, we obtain

$$ \sum_{m=1}^{N/2}\big|X_{M}^{b_m}\big| = \sum_{m=1}^{N/2}\big|X_{M}^{2m-1}\big|. $$

Putting everything together we thus find

$$ \nu_{MN} = \frac{1}{N} \Biggl(\,\sum _{m=0}^{N/2-1}\big|X_M^{2m}\big|-\sum _{m=1}^{N/2}\big|X_{M}^{b_m}\big| \Biggr) = \frac{1}{N}\sum_{m=0}^{N-1}(-1)^m \big|X_M^{m}\big|\quad\mbox{for even } N. $$

Hence, we proved the extra factor (−1)^m for even N. Computing the generating function we obtain the result given in the main text.