ar X iv : h ep - l at / 9 80 80 12 v 1 1 2 A ug 1 99 8 1 Distinguishing J = 4 from J = 0 on a cubic lattice

1. THE PROBLEM On a square lattice (we consider here a D=2+1 lattice for simplicity) the exact rotational symmetry is confined to rotations that are integer multiples of π/2. If we attempt to build operators of spin J from basic components that are identical – that is, related by a lattice symmetry – then there is an ambiguity: exp(iJθ) ≡ exp(iJ ′ θ) ∀θ = nπ 2 (1) if J ′ = J + 4N ∀N (2) Thus, for example, we cannot distinguish J=0 from J=4. It is conventional to assign the lowest value of continuum J to such an operator. However even if this choice is correct for the lightest state of a given 'J' – see the discussion in [1] – the ambiguity cannot be avoided once we start calculating several excitations of the same 'J'. Of course this is a self-inflicted problem. As a → 0, we know that we recover full rotational invariance on physical length scales. Therefore we should be able to construct operators that become as close as we like to spin J, for any J, as a → 0. This will require using linear combinations of operators that are approximate rotations of each other, where the approximation is such that it becomes exact as a → 0. In this poster we are going to explore how well the simplest embodiment of this idea works. We shall do so by attempting to calculate the J=0 and J=4 glueball masses in the 2+1 dimensional SU(2) lattice gauge theory. We start with some closed loop on the lattice that is symmetric about the x-axis. Call it φ Ax. There is a corresponding loop about the y-axis, φ Ay , which is identical in the sense that e.g. φ Ay (t)φ Ay (0) = φ Ax (t)φ Ax (0) ∀t (3) If we sum these two, φ A = φ Ax +φ Ay , then we obtain an operator φ A that we would normally call 'J=0'. (As usual we always take zero-momentum sums of such operators.) We now construct a loop that is symmetric around the diagonal which is at π/4 to the x-axis. We choose a loop that 'looks' as though it is roughly a rotation of φ Ax. There will be an identical loop that is rotated by π/2. Summing these two loops gives us the diagonal operator φ D. Clearly our trial J=0 …