How to find $p_x$ and $p_y$ components on the Bloch sphere?

Question

Consider an arbitrary state: $$|\psi\rangle = a|0\rangle+b|1\rangle,$$

where $a=\cos\left(\frac{\theta}{2}\right), b=\sin\left(\frac{\theta}{2}\right)e^{i\phi}$ (neglecting global phase), $\phi$ is the polar angle, $\theta$ is the azimutal angle on the Bloch sphere.

The Bloch sphere vectors can be found as: $$r_z=\cos\left(\theta\right)\,,$$ $$r_x=\sin\left(\theta\right)\cos\left(\phi\right)\,,$$ $$r_y=\sin\left(\theta\right)\sin\left(\phi\right)\,.$$

If I want to project this state $|\psi\rangle$ into the state $|1\rangle\langle1|$, I can get $p_z$ component of the Bloch vector:

$$p_z = |\langle1|\psi\rangle|^2 = |\langle1|\left(a|0\rangle+b|1\rangle\right)|^2 = |b|^2 = \sin^2\left(\frac{\theta}{2}\right) = \frac{1-\cos(\theta)}{2} = \frac{1-r_z}{2}$$

I know, that to get $p_x$ and $p_y$ components, I need to rotate the Bloch vector before the measurement (projection into the state $|1\rangle\langle1|$) around $x$ or $y$ axis (depending on the projection) by $\frac{\pi}{2}$, but I have no idea how to derive these components mathematically as for $p_z$

Answer 1

To make the notation a bit more precise, what you are calling $p_z$ I will call $p^{|1\rangle}_z$. Then, similarly to what you computed, we have that $$ \begin{split} p^{|0\rangle}_z &= |\langle 0|\psi\rangle|^2\\ &= \frac{1 + r_z}{2}. \end{split} $$ Note that $p^{|0\rangle}_z + p^{|1\rangle}_z = 1$, as is required for probabilities.

The general idea behind this computation is that we have the Pauli matrix $\sigma_z$ with eigenvectors $|0\rangle$ and $|1\rangle$. Thus, if a measurement of $|\psi\rangle$ is performed with respect to $\sigma_z$, then the two possible post-measurement states are $|0\rangle$ and $|1\rangle$ with respective probabilities $p^{|0\rangle}_z$ and $p^{|1\rangle}_z$. This procedure can be followed for any Hermitian matrix $H$, but you are asking specifically about $\sigma_x$ and $\sigma_y$.

I will not work out all the details, but take $\sigma_x$ for example. The eigenstates of $\sigma_x$ are \begin{split} |+\rangle &= \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \\ |-\rangle &= \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle). \end{split} From this you can compute $p^{|+\rangle}_x = |\langle +|\psi\rangle|^2$ and $p^{|-\rangle}_x = |\langle -|\psi\rangle|^2$.