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: 
> 
> If I'm sitting on the edge of a rotating merry-go-round, then from my perspective the world is rotating around me.
> 
> 
> 

No, it isn't. Think about this carefully (or find a merry-go-round and do a practical experiment). From your perspective the merry-go-round is stationary and the rest of the world is rotating *about the centre of the merry-go-round*.

Think about the point in the outside world that is nearest to you. When the merry-go-round has rotated through $180^o$ this point is now on the opposite side of you *and the opposite side of the merry-go-round*. If the world appeared to be rotating about you then this point would stay the same distance away from you as the merry-go-round turned, which it clearly does not.

Therefore, whether you think of yourself as stationary and the world as rotating or vice versa, there is only one fixed centre of rotation, which is always the centre of the merry-go-round.
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.