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?