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.

  • $\begingroup$ I think $\varphi$ in the $|\psi^\perp\rangle$ definition should be with "$-$" sign, am I right? The main question/problem is about how to prove $\alpha^2 + \beta^2 = 1$? Or is it about finding $\gamma$ and $\delta$ for given $\alpha$ and $ \beta$? $\endgroup$ – Davit Khachatryan Apr 6 at 20:48
  • $\begingroup$ @DavitKhachatryan hm why should it be $-\varphi$? I thought it's only negative when you take the complex conjugate i.e. $\langle \psi^\perp|$. And yes, the question is to prove that. I could have been clearer about it. $\endgroup$ – Cat Mai Apr 6 at 20:58
  • $\begingroup$ @DavitKhachatryan I guess once I prove $\alpha^2 + \beta^2 = 1$ then it's should be easy to find $\gamma$ and $\delta$. $\endgroup$ – Cat Mai Apr 6 at 21:00
  • $\begingroup$ But antipodal points don't have the same $\varphi$ in the Bloch sphere (here I am thinking just geometrically). $\endgroup$ – Davit Khachatryan Apr 6 at 21:07
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    $\begingroup$ Wouldn't it be $\langle \psi^\perp| = \cos \frac{\theta + \pi}{2} \langle 0| -e^{i\varphi}\sin \frac{\theta + \pi}{2} \langle 1|$? Then when you take $\langle\psi^\perp | \psi\rangle = 0$ $\endgroup$ – Cat Mai Apr 6 at 21:11

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:

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

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  • $\begingroup$ I don't really follow the calculation. Why is there 2 terms of $Im$? I think the first $Re$ term is from $|\alpha^2|$ and the other one is from $|\beta^2|$. But I don't get where the 2 $Im$ terms come from. Also, for 2 complex number, did you do the calculation directly instead of expanding $e^{i\varphi} = \cos\varphi + i\sin\varphi$? $\endgroup$ – Cat Mai Apr 7 at 14:35
  • $\begingroup$ @CatMai $2 Re(\gamma)Re(\delta)... + 2Im(\gamma)Im(\delta)...$ is coming from $|\alpha|^2$. $ - 2 Re(\gamma)Re(\delta)... - 2Im(\gamma)Im(\delta)...$ is coming from $|\beta|^2$. $\endgroup$ – Davit Khachatryan Apr 7 at 14:49
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    $\begingroup$ For example, for $\alpha$: $z_1 = \gamma \cos(\theta/2)$, $z_2 = \delta \sin(\theta/2)$ , $x_1 = Re(\gamma cos(\theta/2))$, $x_2 = Re(\delta \sin(\theta/2)$ , $y_1 = Im(\gamma cos(\theta/2))$ and $y_2 = Im(\delta \sin(\theta/2)$. $\endgroup$ – Davit Khachatryan Apr 7 at 14:49
  • $\begingroup$ For $|\beta|^2 = |e^{i\varphi}|^2|\gamma \sin(\theta/2)- \delta \cos(\theta/2)|^2 $ and $|e^{i\varphi}|^2 = 1$. So we shouldn't worry about $\varphi$. $\endgroup$ – Davit Khachatryan Apr 7 at 14:53
  • $\begingroup$ @CatMai I edited my answer in order to show the calculations of $|\alpha|^2$ and $|\beta|^2$ separately. $\endgroup$ – Davit Khachatryan Apr 7 at 15:16

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