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ProjectionSpin

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

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

**Part 2) [itex]\vec n_1[/itex] and [itex]\vec n_2[/itex] are unit vectors. A measurement found the projection of spin-1/2 on the direction [itex]\vec n_1[/itex] 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 [itex]\vec n_2[/itex] will give 1/2 with probability**

[itex]P=\frac12(1+\vec n_1\cdot\vec n_2)[/itex]

[itex]P=\frac12(1+\vec n_1\cdot\vec n_2)[/itex]

**The attempt at a solution**

First I declare what my normal vector [itex]\vec n[/itex] will be, [itex]\vec n = (sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta)[/itex]

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

[itex]\mu=\pm 1[/itex]

Setting this up lead me to

[itex]\langle -z\vert\psi_+\rangle=-e^{i\phi}\frac{\cos\theta-1}{\sin\theta}\langle+z\vert\psi_+\rangle[/itex]

which, after normalizing, I then find

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

Likewise, when I use [itex]\mu=-1[/itex] I find

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

**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

[itex]\langle \vec n\vert\vec n_1\rangle=\frac12[/itex]

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

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