Quantum Computing 1.3 -- Polarizers

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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