An angular momentum projection method is applied to the rotational states of even-mass tin isotopes. It is found that the alignment of two neutrons in hlll2 orbit explains the recently observed discontinuity in the moment of inertia of the rotational band in 112Sn.Rotational bands built on deformed excited states have been observed in tin isotopes, though they have been limited to rather low spins (I~10). A recent experiment has shown that there exist higher spin levels in the bands in lighter tin isotopes. I) Since the higher spin states have clearly larger moment of inertia than the previously known lower spin states, their intrinsic structure must be different from that of the lower spin members. Similar discontinuities in the moment of inertia are widely observed in the yrast bands of rare-earth nuclei, where detailed studies have been done both experimentally and theoretically. Few theoretical calculations, however, have been reported so far on the band crossings in tin isotopes. Since the ground states of tin isotopes are spherical, we do not have, contrary to the case of rare-earth nuclei, much information on the mean fields for the deformed bands. Therefore, theoretical calculations for tin isotopes may not be so straightforward as for the rare-earth nuclei. In other words, we can test theoretical models more seriously by applying them to the band crossings in tin isotopes.In the present paper we investigate these rotational bands of tin isotopes using an angular momentum projection method which is suited for the description of band crossing phenomena. 2 ),3) The method has been successfully applied to the backbending phenomena (crossing of zero-quasiparticle (OQP) and 2QP bands) in doublyeven mass rare-earth nuclei 4 ) as well as to the Coriolis attenuation phenomena (crossing of IQP bands) in the odd-mass rare-earth nuclei. 3 ),5) Detailed description of the calculation method is seen in these references.The Hamiltonian employed in the present calculations is the same as the one which we used in the previous works,3)-5) (1) where fi is the spherical single-particle Hamiltonian with the Nilsson [. sand [2 parameters, KlI=0.066, ,u1l=0.35. 6 ) For the pairing force strengths we take the values determined in Ref. 6), Gn-=/·22.5/A MeV, GlI=/·(22.5 -18.0·(N-Z)/A)/AMeV, with a reduction factor /=0.87. The reduction factor is introduced because our single particle spaces (3-major shells both for protons and neutrons) are larger than theirs. Since we assume the same deformation for protons and neutrons, we take the isoscalar Q. Q force whose strength is X=233A -5/3b-4 MeV, at University of Ulster at Coleraine on March 11, 2015 http://ptp.oxfordjournals.org/ Downloaded from