above. The all-spins-up state would be a bad choice since it has a very small overlap with
the cyclicly symmetric ground state, in other words, it is very far away from being the
configuration preferred by the probability distribution e-s
So proi. ,-tii in* out the unwanted states is not a viable solution. Instead we try
to consider cyclicly symmetric observables, for example the sum of the spins at a
time-slice. Although such an amplitude would still get contributions from the intermediate
non-cyclicly symmetric states, these can be suppressed. Consider for example cyclicly
symmetric operators 01 and 02 at times ji and j2, in the presence of a state 1p). The
quantity in question is
Q(i2)- = iTJ 21) (4-19)
Q(jlj2) TrTjl+j2
As before we write |Q) = a, In) but now we assume that even and odd n label the
cyclicly symmetric and ..i-,iin,. I ic sectors respectively. The ta-s are now ordered in
decreasing size within each sector separately. We write O = (mnOi In) for all n, and
i 1,2 and we consider the case where El < E2 (see figure (4-7)). We have
QJ1,J2) Tanakr nOnkt0t^
TT1J2 (AtJ+2 + B't t2 + Btt]2 + C +tj+j2 + DI 2 + DT )
A + B1ea2 + B2eaAIjI + CeaAI(j1+j2) + DieaA2J2 + D2aAZ1j + ... (420)
(we use capital Latin letters for constants with respect to ji and j2). But we have
O',2m+i = 0 because of cyclic symmetry and therefore B1 = B2 = 0. We can see
that the unphysical gap A1 only contributes to Q as an exponent with ji + j2. In the limit
where the sheet is large N = ji + j2 compared to interspacing of operators j = \j* j1
we see that artifacts of A1 are suppressed compared to A2. Of course, this relies on A1
not being too small. We will see how a degeneracy between E0 and the unphysical El can