= 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> matrix
$$
H(\theta)=R(\theta)\begin{pmatrix}1&0\\0&-1\end{pmatrix}R(-\theta)
=\begin{pmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{pmatrix}.
$$
Each fixed <quarter-wave plate> has $Q=\operatorname{diag}(1,i)$ in the supplied convention. Propagation through the three plates therefore gives
$$
\boxed{J(\theta)=QH(\theta)Q
=\begin{pmatrix}\cos2\theta&i\sin2\theta\\i\sin2\theta&\cos2\theta\end{pmatrix}
=e^{i2\theta\sigma_x}.}
$$
The order matters: this is a product of three propagating plates, not a similarity transformation $QHQ^{-1}$. For input Jones vector $\mathbf e_+=(1,1)^T/\sqrt2$, 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 $\theta=\Omega t$. Then
$$
\mathbf E_{\rm out}(t)\propto\mathbf e_+ e^{i2\Omega t}e^{-i\omega t}
=\mathbf e_+e^{-i(\omega-2\Omega)t}.
$$
Thus the <rotating half-wave plate frequency shift> has
$$
\boxed{\omega_{\rm out}-\omega=-2\Omega,\qquad |\Delta\omega|=2|\Omega|.}
$$
The 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 $+\hbar$ per <photon>. The rotating half-wave plate flips it to spin $-\hbar$, transferring $2\hbar$ to the rotating plate. Maintaining its rotation takes mechanical <energy> $2\hbar\Omega$ 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.
Back to article page
