Represent a pure state in terms of 2 antipodal points on the Bloch sphere

Question

I recently had an assignment where the question is based on the assumption that we can write any pure state qubit $|\phi \rangle$ as: $$|\phi \rangle = \gamma |\psi\rangle + \delta |\psi^\perp \rangle$$ Where $|\psi\rangle$ and $|\psi^\perp \rangle$ are 2 antipodal points on the Bloch sphere: $$ |\psi\rangle = \cos \frac{\theta}{2} |0\rangle +e^{i\varphi}\sin \frac{\theta}{2} |1\rangle$$ $$ |\psi^\perp\rangle = \cos \frac{\theta + \pi}{2} |0\rangle +e^{i\varphi}\sin \frac{\theta + \pi}{2} |1\rangle$$

I have a lingering question about how this actually works. So far I got: $$|\phi\rangle= \gamma |\psi\rangle + \delta |\psi^{\perp}\rangle$$ $$= \gamma \left(\cos \frac{\theta}{2} |0\rangle +e^{i\varphi}\sin \frac{\theta}{2} |1\rangle \right) + \delta \left(\cos \frac{\theta + \pi}{2} |0\rangle +e^{i\varphi}\sin \frac{\theta + \pi}{2} |1\rangle \right)$$ $$ = \left(\gamma \cos \frac{\theta}{2} + \delta \cos \frac{\theta + \pi}{2}\right)|0\rangle + \left(\gamma e^{i\varphi}\sin \frac{\theta}{2} + \delta e^{i\varphi} \sin \frac{\theta + \pi}{2}\right)|1\rangle$$ $$\Rightarrow \alpha = \gamma \cos \frac{\theta}{2} + \delta \cos \frac{\theta + \pi}{2}$$ $$\Rightarrow \beta = \gamma e^{i\varphi}\sin \frac{\theta}{2} + \delta e^{i\varphi} \sin \frac{\theta + \pi}{2}$$ So $\alpha^2 + \beta^2 = 1$. I'm not sure if I can solve this equation. I wonder if it's solvable or is there a better way to go about understanding writing a pure state in $|\psi\rangle$ and $|\psi^\perp \rangle$ basis. I know that they are orthonormal so intuitively it should work.

Answer 1

Two antipodal states in the Bloch sphere (note that $0 \leq \theta \leq \pi$):

\begin{equation} |\psi \rangle = \cos \frac{\theta}{2} |0 \rangle + e^{i\varphi}\sin \frac{\theta}{2} |1 \rangle \\ |\psi^\perp \rangle = \cos \frac{\pi - \theta}{2} |0 \rangle + e^{i\varphi + \pi}\sin \frac{\pi - \theta}{2} |1 \rangle = \sin \frac{\theta}{2} |0 \rangle - e^{i\varphi}\cos \frac{\theta}{2} |1 \rangle \end{equation}

The expression for $|\psi^\perp \rangle$ from the question is different from this $|\psi^\perp \rangle$ by a global phase. The reason why I prefer this notation is that I want to keep Bloch sphere formalism (e.g. $0 \leq (\pi - \theta) \leq \pi$ constrant that is true for $|\psi^\perp \rangle$ presented here). Note that:

$$\langle \psi | \psi^\perp \rangle = \cos \frac{\theta}{2} \sin \frac{\theta}{2} - \sin \frac{\theta}{2} \cos \frac{\theta}{2} = 0$$

By doing the same calculations we can obtain for $|\phi\rangle= \gamma |\psi\rangle + \delta |\psi^{\perp}\rangle = \alpha |0\rangle + \beta |1\rangle$:

\begin{align*} &\alpha = \gamma \cos \frac{\theta}{2} + \delta \sin \frac{\theta}{2} \\ &\beta = \gamma e^{i\varphi}\sin \frac{\theta}{2} - \delta e^{i\varphi} \cos \frac{\theta}{2} \end{align*}

Then (here I take into account that $|e^{i\varphi}| = 1$):

\begin{equation} |\alpha|^2 = |\gamma|^2 \cos^2 \frac{\theta}{2} + |\delta|^2 \sin^2 \frac{\theta}{2} + 2 Re(\gamma) Re(\delta) \cos \frac{\theta}{2}\sin \frac{\theta}{2} + 2 Im(\gamma) Im(\delta) \cos \frac{\theta}{2}\sin \frac{\theta}{2} \end{equation}

\begin{equation} |\beta|^2 = |\gamma|^2 \sin^2 \frac{\theta}{2} + |\delta|^2 \cos^2 \frac{\theta}{2} - 2Re(\gamma) Re(\delta)\cos \frac{\theta}{2} \sin \frac{\theta}{2} - 2Im(\gamma) Im(\delta)\cos \frac{\theta}{2} \sin \frac{\theta}{2} \end{equation}

Because if we have two complex numbers $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, then:

$$|z_1 \pm z_2|^2 = (x_1 \pm x_2)^2 + (y_1 \pm y_2)^2 = |z_1|^2 + |z_2|^2 \pm 2 x_1 x_2 \pm 2 y_1 y_2$$

After summing the expressions for $|\alpha|^2$ and $|\beta|^2$ we will obtain:

$$|\alpha|^2 + |\beta|^2 = \left(|\gamma|^2 + |\delta|^2 \right) \left(\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} \right) =1$$