2
$\begingroup$

There is a particle with spin of $j=\frac{1}{2}$, and it is on a state $|j,m_z=\frac{1}{2}\rangle$. What is the probability the state particle will be $|j,m_x=\frac{1}{2}\rangle$?

Well, of course the probability of being in the state $|\alpha\rangle $ if it is on the state $|\beta\rangle$ is $P=|\langle \alpha|\beta\rangle|^2$, but how can I project the quantum number $m_x$ on $m_z$?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

When you calculate the overlap $\langle\alpha|\beta\rangle$ you have to make sure that both states are written in the same basis. This means you either have to write the $|j,m_x=\frac{1}{2}\rangle$ in the $m_z$ basis or vice versa. Using standard basis transformation relations, and also trying to use your notation, we get the $m_x$ eigenstate in the $m_z$ basis as follows:

$$ |j,m_x=\frac{1}{2}\rangle=\frac{1}{\sqrt{2}}\left(|j,m_z=\frac{1}{2}\rangle+|j,m_z=-\frac{1}{2}\rangle\right). $$

You can now calculate your overlap.

$\endgroup$
2
  • $\begingroup$ Thank you! I've to use the standard basis transformation relations here because of the $\sigma_z$ Pauli matrix eigenvectors are (1 0) and (0 1)? $\endgroup$ Commented Nov 4, 2020 at 8:20
  • $\begingroup$ @KonstantinKhrizman, what I mean is that to change from one basis to another you need to use the "overlap matrix", which is true in general. In your example, you want to go from the basis of eigenstates of the $S_x$ operator to those of the $S_z$ operator. These operators are proportional to the corresponding Pauli matrices. $\endgroup$
    – ProfM
    Commented Nov 4, 2020 at 8:27

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.