Explicit plane-wave reduction to transverse-traceless gauge ($H_{+}=(H_{11}-H_{22})/2,H_{\times }=H_{12}$)

For a plane gravitational wave in linearized gravity with $k_{\mu }=(-\omega ,0,0,\omega )$, Lorenz gauge in linearized gravity gives $H_{3\nu }=-H_{0\nu }$. Under $\delta H_{\mu \nu }=i(k_{\mu }{X}_{\nu }+k_{\nu }{X}_{\mu }-{\eta }_{\mu \nu }{k}^{\rho }{X}_{\rho })$, choose $X_i=H_{0i}/(i\omega )$ for $i=1,2$, $X_3-X_0=i{H}_{00}/\omega$, and $X_0+X_3=(H_{11}+H_{22})/(2i\omega )$. This residual gauge symmetry of linearized gravity removes all time and longitudinal entries and the transverse trace. The remaining amplitudes $H_{+},H_{\times }$ are the two gravitational wave polarizations. The formula uses the plus-sign convention for $\delta h=2{\partial }_{(\mu }{\xi }_{\nu )}$; reversing the definition of $\xi$ reverses the gauge change.