Spin projection

Question

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 $|alpharangle $ if it is on the state $|betarangle$ is $P=|langle alpha|betarangle|^2$, but how can I project the quantum number $m_x$ on $m_z$?

ProfM · Accepted Answer

When you calculate the overlap $langlealpha|betarangle$ 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:
$$
|j,m_x=frac{1}{2}rangle=frac{1}{sqrt{2}}left(|j,m_z=frac{1}{2}rangle+|j,m_z=-frac{1}{2}rangleright).
$$
You can now calculate your overlap.

Spin projection

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