 Solution

Let $R(\theta )$ be the real rotation matrix with columns along the plate's fast and slow axes. Expressing the input in those axes, applying the principal retardance, then rotating back gives the rotated half-wave plate matrixEach fixed quarter-wave plate has $Q=\mathrm{diag}(1,i)$ in the supplied convention. Propagation through the three plates therefore givesThe order matters: this is a product of three propagating plates, not a similarity transformation $QH{Q}^{-1}$. For input Jones vector ${\mathbf{e}}_{+}=(1,1{)}^T/\sqrt{2}$, the Pauli matrix has eigenvalue $+1$, so $J{\mathbf{e}}_{+}=e^{i2\theta }{\mathbf{e}}_{+}$. The final polarization remains at 45 degrees, but its phase depends on the rotating angle.

Take the physical field as the real part of $\mathbf{E}{e}^{-i\omega t}$ and let $\theta =\Omega t$. ThenThus the rotating half-wave plate frequency shift hasThe magnitude is the stated $2\Omega$ shift. Its sign follows the explicit time and rotation conventions above; reversing the physical rotation or using the orthogonal linear input reverses the shift. A time-phasor convention must be changed consistently with the plate matrices, rather than used alone to change a physical redshift into a blueshift.

The physical origin is a rotational Doppler effect and angular-momentum exchange. In the stated convention, the first fixed quarter-wave plate changes the 45-degree input into circular polarization of spin $+\hslash$ per photon. The rotating half-wave plate flips it to spin $-\hslash$, transferring $2\hslash$ to the rotating plate. Maintaining its rotation takes mechanical energy $2\hslash \Omega$ from each photon, agreeing with the redshift. The final fixed plate restores linear polarization; fixed plates can exchange angular momentum with their supports without the rotating element's corresponding mechanical work. The instantaneous Jones treatment assumes slow rotation compared with optical and transit-time scales and ideal retardances over the small shifted bandwidth.