magnetic moment of proton I just tried to calculate the magnetic moment of a proton. I took the 
proton g-Factor of $g=5.585694$ 
nuclear magneton of $\mu_k = 5.050783 * 10^{−27}$ J/T 
proton spin of $I=1/2$ 
At first I calculated the norm of the proton spin $|\vec{I}|=\hbar \sqrt{I*(I+1)}=\hbar \frac{1}{2}\sqrt{3}$ 
And then I put everything together in $\mu=g\mu_k\frac{|\vec{I}|}{\hbar}$ and obtain 2,44134228 × 10^-26 instead of 1.410606 × 10^-26 ...
Interesting enough, I obtain the correct value if I devide by $\sqrt{3}$. But I see no reason to do this... 
It would be great if you could help me.
Thanks in advance
ftiaronsem
 A: No spin measurement of proton can give a value more or less than $\hbar/2$. But what do we mean when we say that spin of proton is $\hbar/2$ ? Spin is a 'vector' quantity (at least this is what it is classically). So one should also specify its direction. The thing is that in this case direction doesn't matter much. If you think of proton as some sphere and choose any arbitrary axis passing through its center and carry our an experiment to measure spin along that axis you will see that it is always $\hbar/2$ (or negative of it). You will never find spin of any proton to be $\hbar\sqrt{3}/2$ no matter along which axis you measure it. For the same reason you will never find a proton with magnetic moment $2.44134228 × 10^{-26}$ $JT^{-1}$ along any axis. 
Edit : vector addition in QM
Suppose you choose three perpendicular directions $i,j,k$; and suppose in three successive experiments to measure spin of proton you find its value to be : 
$\hbar/2$ along $i$ axis in experiment 1.
$\hbar/2$ along $j$ axis in experiment 2.
$\hbar/2$ along $k$ axis in experiment 3.
Now you may tend to conclude (using vector addition) that "total spin" of proton should be $\hbar/2(i+j+k)$ or equivalently $\hbar\sqrt{3}/2$ along unit direction $(i+j+k)/\sqrt{3}$. But if you carry out a fourth experiment to measure spin along this direction somewhat magically you will again find that spin is $\hbar/2$ (or - $\hbar/2$). Hence ordinary rules of vector addition do not apply in this case. 
A: It is only a question of definition. There is the operator of interaction of particle with an externally-produced magnetic field:
$\hat{H}_{int}=-\hat{\boldsymbol{\mu}}\cdot\mathbf{H}$,
where $\mathbf{H}$ is a magnetic field and $\hat{\boldsymbol{\mu}}$ is an operator: 
$\hat{\boldsymbol{\mu}}=\displaystyle \frac{g e}{2 m} \hat{\mathbf{s}}$
By the «value» of the magnetic moment of particle, people usually imply the maximum of the following diagonal matrix element: 
$\mu=\langle\psi\vert \hat{\boldsymbol{\mu}}_{z} \vert\psi\rangle,$
which is of course $g \mu_N/2$ 
