Polarizing beam-splitter observable at an arbitrary angle
In my previous post here, I discussed polarizing beam-splitters aligned with the x-axis. Now, I extend this analysis to consider a polarizing beam-splitter oriented at an arbitrary angle \theta from the x-axis. The quantum observable for this measurement, denoted as \mathbf A_\theta, is defined by its eigenstates | \mathbf 1_\theta \rangle and | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle, corresponding to polarization directions at angles \theta and \theta + \frac{\pi}{2} from the x-axis, with eigenvalues +1 and -1 respectively:
\mathbf A_\theta = | \mathbf 1_\theta \rangle \langle \mathbf 1_\theta | - | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle \langle \mathbf 1_{\theta + \frac{\pi}{2}} |
To express \mathbf A_\theta in the (| \mathbf 1_x \rangle, | \mathbf 1_y \rangle) basis, I first express the eigenstates | \mathbf 1_\theta \rangle and | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle in terms of | \mathbf 1_x \rangle and | \mathbf 1_y \rangle:
\begin{aligned} & | \mathbf 1_\theta \rangle = \cos\left(\theta\right)| \mathbf 1_x \rangle + \sin\left(\theta\right) | \mathbf 1_y \rangle \\ & | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle = -\sin\left(\theta\right)| \mathbf 1_x \rangle + \cos\left(\theta\right) | \mathbf 1_y \rangle \end{aligned}
By expanding the outer products and simplifying using trigonometric identities, I arrive at the matrix form of \mathbf A_\theta:
\mathbf A_\theta = \begin{bmatrix} \cos\left(2\theta\right) & \sin\left(2\theta\right) \\ \sin\left(2\theta\right) & -\cos\left(2\theta\right) \end{bmatrix}
To verify the consistency of this result, I compute the eigenvalues of \mathbf A_\theta. The characteristic equation is given by \left|\mathbf A_\theta - \lambda \mathbf I\right| = 0, which leads to:
\lambda^2 - 1 = 0
The eigenvalues are indeed \lambda = \pm 1, as expected for a polarization measurement.
Next, I determine the eigenvectors corresponding to these eigenvalues. For \lambda = 1, solving (\mathbf A_\theta - \mathbf I) \mathbf v = \mathbf 0 yields the eigenvector proportional to:
\begin{bmatrix} \cos\left(\theta\right) \\ \sin\left(\theta\right) \end{bmatrix}
This corresponds to the eigenstate | \mathbf 1_\theta \rangle. For \lambda = -1, solving (\mathbf A_\theta + \mathbf I) \mathbf v = \mathbf 0 gives the eigenvector proportional to:
\begin{bmatrix} -\sin\left(\theta\right) \\ \cos\left(\theta\right) \end{bmatrix}
This corresponds to the eigenstate | \mathbf 1_{\theta + \frac{\pi}{2}} \rangle. These results confirm that the derived matrix \mathbf A_\theta correctly represents the polarizing beam-splitter observable at an arbitrary angle \theta, with the expected eigenvalues and eigenstates.
For more insights into this topic, you can find the details here.