Spin excitation anisotropy in the optimally isovalent-doped superconductor BaFe2(As0.7P0.3)2

We use neutron polarization analysis to study spin excitation anisotropy in the optimally isovalent-doped superconductor ${\mathrm{BaFe}}₂$(${\mathrm{As}}_0.7{\mathrm{P}}_0.3$)${}₂$ (${T}_c=30$ K). Different from optimally hole- and electron-doped ${\mathrm{BaFe}}₂{\mathrm{As}}₂$, where there is a clear spin excitation anisotropy in the paramagnetic tetragonal state well above ${T}_c$, we find no spin excitation anisotropy for energies above 2 meV in the normal state of ${\mathrm{BaFe}}₂$(${\mathrm{As}}_0.7{\mathrm{P}}_0.3$)${}₂$. Upon entering the superconducting state, significant spin excitation anisotropy develops at the antiferromagnetic (AF) zone center ${\mathbf{Q}}_{\mathrm{AF}}=(1,0,L=\mathrm{odd})$, while the magnetic spectrum is isotropic at the zone boundary $\mathbf{Q}=(1,0,L=\mathrm{even})$. By comparing the temperature, wave vector, and polarization dependence of the spin excitation anisotropy in ${\mathrm{BaFe}}₂$(${\mathrm{As}}_0.7{\mathrm{P}}_0.3$)${}₂$ and hole-doped ${\mathrm{Ba}}_0.67{\mathrm{K}}_0.33{\mathrm{Fe}}₂{\mathrm{As}}₂$ (${T}_c=38$ K), we conclude that such anisotropy arises from spin-orbit coupling and is associated with the nearby AF order and superconductivity.