Deuteron wave function The deuteron wave function is given by $$|\psi _d\rangle  = a|^3S_1\rangle+b|^3D_1\rangle$$
where all states are normalized. How do we find $b^2$ s.t. the wave function reproduces the magnetic moment of the deuteron:
$$\mu = 0.857 \mu _N$$ ?
 A: One can write the magnetic momentum operator in the following form:
$$\mu=g_p\bf{s_p} +g_n\bf{s_n}+\frac{l}{2}$$ where $\bf l$ is a angular momentum, $g_p$ and $g_n$ are gyromagnetic ratio for proton and neutron respectively, $\bf{s}_p$ and $\bf{s}_n$ are sin operators.The coefficient $\frac{1}{2}$ in front of $\bf l$ is appear because of the contribution in magnetic momentum is given by $[r_p\times p_p]=\frac{1}{2}[r\times p]=\frac{1}{2}\bf l$. 
Let's rewrite this formula via full deuteron momentum $J$.
$$\mu=\frac{g_p+g_n}{2}\bf{S}+\frac{g_p-g_n}{2}(s_p-s_n)+\frac{1}{2}\bf l$$
$$<\mu>=\frac{g_p+g_n}{2}\bf{<S>}-\frac{g_p+g_n-1}{2}\bf{<l>}$$
$$\mu=(\mu_p+\mu_n)\bf{J}-(\mu_p+\mu_n-\frac{1}{2})\frac{<Jl>}{J(J+1)}$$
$$\mu=(\mu_p+\mu_n)-(\mu_p+\mu_n-\frac{1}{2})\frac{3}{2}b^2$$
Where $\mu_p=g_p/2$.After substituting numbers we will set b^2=0.04
But it is not exactly number because of we use magnetic momentum operator is used for free nucleons.There is relativistic corrections for magnetic momentum operator.
