  We use neutron polarization analysis to study spin excitation anisotropy in
the optimal-isovalent-doped superconductor BaFe2(As0.7P0.3)2 (Tc = 30 K).
Different from optimally hole and electron-doped BaFe2As2, where there is a
clear spin excitation anisotropy in the paramagnetic tetragonal state well
above Tc, we find no spin excitation anisotropy for energies above 2 meV in the
normal state of BaFe2(As0.7P0.3)2. Upon entering the superconducting state,
significant spin excitation anisotropy develops at the antiferromagnetic (AF)
zone center QAF = (1, 0, L = odd), while magnetic spectrum is isotropy at the
zone boundary Q = (1, 0, L = even). By comparing temperature, wave vector, and
polarization dependence of the spin excitation anisotropy in BaFe2(As0.7P0.3)2
and hole-doped Ba0.67K0.33Fe2As2 (Tc = 38 K), we conclude that such anisotropy
arises from spin-orbit coupling and is associated with the nearby AF order and
superconductivity.
