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 $p_z$

  • $\begingroup$ What you are calling $p_z$ is the probability of measuring $|\psi\rangle$ in the $|1\rangle$ eigenstate of the Pauli matrix $\sigma_z$. Are you asking how to compute the probabilities of measuring $|\psi\rangle$ in the eigenstates of $\sigma_x$ and $\sigma_y$? $\endgroup$ Dec 19, 2023 at 10:10
  • $\begingroup$ @NickMertes $p_z$ is the expectation value; yes, exactly, actually I just want to know, how to calculate $p_x$ and $p_y$ expectation values as for $p_z$ $\endgroup$
    – Curious
    Dec 19, 2023 at 10:28

1 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 $p^{|-\rangle}_x = |\langle -|\psi\rangle|^2$.

  • $\begingroup$ thank you for your answer!) but I think it's my fault I didn't explain my question more correctly: I'd like to find $p_x$ and $p_y$ from the measurement of the state $|1\rangle$ $\endgroup$
    – Curious
    Dec 19, 2023 at 12:38
  • $\begingroup$ and what about $p_y$? I tried to do the same and get almost the same result as for $p_x$ $\endgroup$
    – Curious
    Dec 19, 2023 at 19:14
  • 1
    $\begingroup$ If we call $|0\rangle$ and $|1\rangle$ the standard basis, then it is indeed possible to obtain the $x$ and $y$ probabilities by making only standard basis measurements. This procedure is described in the IBM Basics of quantum information course in the section Implementing projective measurements using standard basis measurements. $\endgroup$ Dec 20, 2023 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.