Solve Quantum Mechanics: Probability of Spin-1/2 Particle

[ProjectionSpin]

Quantum Mechanics Help [Probability of a Spin-1/2 Particle]

Part 1) \hat{S} is a spin-1/2 operator, \vec n is a unit vector, \vert{\psi_\pm}\rangle are normalized eigenvectors of n\cdot\hat{S} with eigenvectors \pm\frac12. Write \vert{\psi_\pm}\rangle in terms of \vert{+z}\rangle and \vert{-z}\rangle.

Part 2) \vec n_1 and \vec n_2 are unit vectors. A measurement found the projection of spin-1/2 on the direction \vec n_1 to be 1/2. Use the results of the previous part to show that a subsequent measurement of the projection of spin on the direction \vec n_2 will give 1/2 with probability

P=\frac12(1+\vec n_1\cdot\vec n_2)







The attempt at a solution
First I declare what my normal vector \vec n will be, \vec n = (sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta)

I then solve for \sum_i\vec n\cdot\hat{S}_i and started looking for the eigenstates, which lead to

\mu=\pm 1

Setting this up lead me to
\langle -z\vert\psi_+\rangle=-e^{i\phi}\frac{\cos\theta-1}{\sin\theta}\langle+z\vert\psi_+\rangle
which, after normalizing, I then find
\vert\psi_+\rangle=\cos\frac\theta2\vert+z\rangle+e^{-i\phi}\sin\frac\theta2\vert-z\rangle

Likewise, when I use \mu=-1 I find

\vert\psi_-\rangle=\sin\frac\theta2\vert+z\rangle-e^{i\phi}\cos\frac\theta2\vert-z\rangle

**This officially marks the end of part 1**

For part 2, I assumed that "the projection of spin-1/2 on the direction n⃗ 1 to be 1/2" was equivalent to
\langle \vec n\vert\vec n_1\rangle=\frac12

But supposably this is not the case, which leaves me to just being confused on where to go.

 

[DrClaude]

For part 2, I assumed that "the projection of spin-1/2 on the direction n⃗ 1 to be 1/2" was equivalent to
\langle \vec n\vert\vec n_1\rangle=\frac12

There is no $\vec{n}$ in part b. The problem mentions the measurement of the projection of the spin in the direction of $\vec{n}_1$. How do you translate that into mathematical terms?