# Spin projection

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 $$m_z$$?

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:
$$|j,m_x=\frac{1}{2}\rangle=\frac{1}{\sqrt{2}}\left(|j,m_z=\frac{1}{2}\rangle+|j,m_z=-\frac{1}{2}\rangle\right).$$
• 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)? Nov 4, 2020 at 8:20
• @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. Nov 4, 2020 at 8:27