A two-state quantum system is called a **qubit**. A qubit can be realized by a photon which has two different polarization states \(\ket{0^{\circ}}\) and \(\ket{90^{\circ}}\) which correspond to polarization along the vertical and horizontal axes, respectively. The ability to precisely and reliably manipulate qubits is key to the advent of large-scale quantum computing

A linearly polarized photon aligned with angle \(\theta\) can be split into its \(\ket{0^{\circ}}\) and \(\ket{90^{\circ}}\) components with amplitudes corresponding to the projection of the wave on the measurement axis. These states are **orthogonal**, so a pure \(\ket{0^{\circ}}\) photon has no \(\ket{90^{\circ}}\) component, and *vice versa:*
\[ \ket{\theta ^ \circ} = \cos(\theta) \ket{0^{\circ}} + \cos(90 ^ \circ -\theta) \ket{90^{\circ}}, \quad \theta \in [0^\circ,360^\circ].\]

One way to manipulate a photon qubit is with polarization filters. Each filter measures and absorbs all photons in the \(\ket{\alpha+90 ^ \circ}\) state, while all others pass through. Consider a series of three linear polarization filters, each rotated by an angle of \(\alpha = 30 ^ \circ, 60 ^ \circ, \) and \(90 ^ \circ, \) respectively, with respect to the \(\ket{0^{\circ}}\) polarization direction.

What is the probability \(P \) that a \(\ket{0^{\circ}}\) photon passes all three polarizing filters and emerges as a \(\ket{90^{\circ}}\) photon?

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