Qubit Exercises

Part (i)

Find the scalar product ⟨Ψ−(θ)∣Ψ+(θ)⟩\langle \Psi_{-}(\theta) | \Psi_{+}(\theta) \rangle⟨Ψ−​(θ)∣Ψ+​(θ)⟩ and discuss.

Given: ∣Ψ+(θ)⟩=cos⁡(θ)∣0⟩+sin⁡(θ)∣1⟩|\Psi_{+}(\theta)\rangle = \cos(\theta) |0\rangle + \sin(\theta) |1\rangle∣Ψ+​(θ)⟩=cos(θ)∣0⟩+sin(θ)∣1⟩ ∣Ψ−(θ)⟩=cos⁡(θ)∣0⟩−sin⁡(θ)∣1⟩|\Psi_{-}(\theta)\rangle = \cos(\theta) |0\rangle – \sin(\theta) |1\rangle∣Ψ−​(θ)⟩=cos(θ)∣0⟩−sin(θ)∣1⟩

To find the scalar product:

⟨Ψ−(θ)∣Ψ+(θ)⟩=(cos⁡(θ)⟨0∣−sin⁡(θ)⟨1∣)(cos⁡(θ)∣0⟩+sin⁡(θ)∣1⟩)\langle \Psi_{-}(\theta) | \Psi_{+}(\theta) \rangle = (\cos(\theta) \langle 0| – \sin(\theta) \langle 1|)(\cos(\theta) |0\rangle + \sin(\theta) |1\rangle)⟨Ψ−​(θ)∣Ψ+​(θ)⟩=(cos(θ)⟨0∣−sin(θ)⟨1∣)(cos(θ)∣0⟩+sin(θ)∣1⟩)

Expanding the product, we get:

⟨Ψ−(θ)∣Ψ+(θ)⟩=cos⁡2(θ)⟨0∣0⟩+cos⁡(θ)sin⁡(θ)⟨0∣1⟩−sin⁡(θ)cos⁡(θ)⟨1∣0⟩−sin⁡2(θ)⟨1∣1⟩\langle \Psi_{-}(\theta) | \Psi_{+}(\theta) \rangle = \cos^2(\theta) \langle 0|0\rangle + \cos(\theta)\sin(\theta) \langle 0|1\rangle – \sin(\theta)\cos(\theta) \langle 1|0\rangle – \sin^2(\theta) \langle 1|1\rangle⟨Ψ−​(θ)∣Ψ+​(θ)⟩=cos2(θ)⟨0∣0⟩+cos(θ)sin(θ)⟨0∣1⟩−sin(θ)cos(θ)⟨1∣0⟩−sin2(θ)⟨1∣1⟩

Using the orthonormality of the basis states ($⟨0|0⟩=1$, $⟨1|1⟩=1$, and $⟨0|1⟩=⟨1|0⟩=0$), we have:

⟨Ψ−(θ)∣Ψ+(θ)⟩=cos⁡2(θ)⋅1+0−0−sin⁡2(θ)⋅1=cos⁡2(θ)−sin⁡2(θ)\langle \Psi_{-}(\theta) | \Psi_{+}(\theta) \rangle = \cos^2(\theta) \cdot 1 + 0 – 0 – \sin^2(\theta) \cdot 1 = \cos^2(\theta) – \sin^2(\theta)⟨Ψ−​(θ)∣Ψ+​(θ)⟩=cos2(θ)⋅1+0−0−sin2(θ)⋅1=cos2(θ)−sin2(θ)

Therefore:

⟨Ψ−(θ)∣Ψ+(θ)⟩=cos⁡(2θ)\langle \Psi_{-}(\theta) | \Psi_{+}(\theta) \rangle = \cos(2\theta)⟨Ψ−​(θ)∣Ψ+​(θ)⟩=cos(2θ)

Discussion: The scalar product ⟨Ψ−(θ)∣Ψ+(θ)⟩=cos⁡(2θ)\langle \Psi_{-}(\theta) | \Psi_{+}(\theta) \rangle = \cos(2\theta)⟨Ψ−​(θ)∣Ψ+​(θ)⟩=cos(2θ) tells us about the overlap between the states ∣Ψ−(θ)⟩|\Psi_{-}(\theta)\rangle∣Ψ−​(θ)⟩ and ∣Ψ+(θ)⟩|\Psi_{+}(\theta)\rangle∣Ψ+​(θ)⟩. When $\theta =0$, $\cos (2\theta )=1$, so the states are identical. When θ=π4\theta = \frac{\pi}{4}θ=4π​, $\cos (2\theta )=0$, indicating that the states are orthogonal.