Question: I have an example that I don't understand, especially 2 expressions.

Given a spin in the state $|z + \rangle$, i.e., pointing up along the z-axis what are the
probabilities of measuring $\pm \hbar/2$ along $\hat{n}$?

I have these 2 expressions

$$|+\rangle\_n = cos(\theta/2)e^{-i\phi/2}|+\rangle +sin(\theta/2)e^{i\phi/2}|-\rangle$$
$$|-\rangle\_n = sin(\theta/2)e^{-i\phi/2}|+\rangle -cos(\theta/2)e^{i\phi/2}|-\rangle$$

I'm not sure to fully understand what that means.
Is $|+\rangle\_n$ only used for a particle in the spin $|z + \rangle$ and $|+\rangle\_n$ is a linear combination of $\hbar/2$ and $- \hbar/2$.
Hence, the probability to have $\hbar/2$ is $|cos(\theta/2)e^{-i\phi/2}(1,0)|^2$, knowing that $|+\rangle = (1,0)$

I'm not that convince about my explanation. This is not based on anything.

Answer: 
$\def \b {\mathbf}$

[![enter image description here](https://i.stack.imgur.com/oYYtO.png)](https://i.stack.imgur.com/oYYtO.png)

[![enter image description here](https://i.stack.imgur.com/agfvO.png)](https://i.stack.imgur.com/agfvO.png)

In both cases the center of rotation is the same ,but you are looking at the object towards the vector $~\mathbf u~$, from differet perspective .

The equations

**I)**

$$\b u=\b r\_o(\psi)-\b r(\omega\,t)$$

**II)**

$$\b u=\b r\_o(\omega\,t)+\b r(\varphi)$$
I think the notion of "actual centre" depends on the frame in which you are considering your rotation motion.

For an observer outside of the rotating frame, the merry-go-round actually has a centre about which it rotates. But when we shift frames to the person on the mere-go-round, the centre in the rotating frame would be the point where all the fictitious forces which are experienced by a person in the rotating frame vanish (i.e the actual centre of rotation).

Thus the centre of the rotation is the same irrespective of who is viewing the motion, the difference is only how you describe that point.

I hope this helps.