Total angular momentum operator How do the eigenfunctions of the total angular momentum operator analytically look like? 
I mean the operator is given by $J = L+S$ so the eigenfunctions have to be tensor-product states, right? Can we explicitely say what they are?
I should add that I am particularly interested in $L$ to be orbital angular momentum operator and $S$ the spin-operator for electrons.
 A: 
How do the eigenfunctions of the total angular momentum operator
analytically look like?
I mean the operator is given by $J = L+S$ so the eigenfunctions have
to be tensor-product states, right? Can we explicitely say what they
are?

The eigenfunctions of $J$ are going to be made up of linear combinations of tensor-product states of the eigenfunctions of $L$ and the eigenfunctions of $S$. In general, the linear combinations will involved more than just one tensor-product of the eigenfunctions of $L$ and $S$, however the "stretch-state" is the exception and is the starting point for constructing the others.
For example, if $L$ and $S$ are both spin 1/2 (yes, I know that $L$ usually stands for "orbital" angular momentum, but in this example $L$ and $S$ are both spin 1/2) then total $J$ can be either spin 1=|1/2+1/2| or spin 0=|1/2-1/2|. One of the eigenfunctions of $J=1$ is given by a tensor-product state
$$
|1,1\rangle=|1/2,1/2\rangle\otimes|1/2,1/2\rangle\;,
$$
which is the "stretch-state". The other states can be obtained by applying the lowering operator
$$
J_- = L_{-}\otimes 1 +1\otimes S_{-}\;,
$$
to the stretch-state and normalizing. E.g., we find (I'm putting in the square root of two on the LHS by hand to indicate that the RHS is not generated as a normalized state)
$$
\sqrt{2}|1,0\rangle=|1/2,-1/2\rangle\otimes|1/2,1/2\rangle+|1/2,1/2\rangle\otimes|1/2,-1/2\rangle
$$
and
$$
|1,-1\rangle=|1/2,-1/2\rangle\otimes|1/2,-1/2\rangle\;.
$$
And the $|0,0\rangle$ state is generated by creating a state with $J_z=0$ that is orthogonal to the $|1,0\rangle$ state. It is
$$
\sqrt{2}|0,0\rangle=|1/2,-1/2\rangle\otimes|1/2,1/2\rangle-|1/2,1/2\rangle\otimes|1/2,-1/2\rangle\;.
$$
So, you see that two $(|1,1\rangle$ and $|1,-1\rangle)$ of the eigenfunctions of J in this case are simple tensor-products of the eigenfunctions of L and S, and the other two $(|1,0\rangle$ and $|0,0\rangle)$ are linear combinations of more than one tensor-product of the eigenfunctions of L and S.
A: The trick is to expand one basis (say the uncoupled one with elements $\{\vert LM_L\rangle \vert SM_s\rangle:= \vert L M_L;SM_S\rangle \}$) in terms of another (say the coupled one with elements $\{\vert JM_J\rangle\}$.)
The assumption is that the $\{\vert JM_J\rangle\}$ form a complete set in the sense that the identity 
$$
\hat 1=\sum_{JM_J}\vert JM_J\rangle \langle J M_J\vert\, .
$$
Hence:
\begin{align}
\vert LM_L;  S M_S\rangle=
\sum_{J(M_J)}\vert JM_J\rangle \langle J M_J\vert LM_L;SM_S\rangle\, .  \tag{1}
\end{align}
The overlap coefficients $\langle J M_J\vert L M_L;SM_S\rangle$ are known as  Clebsch-Gordan coefficients, sometimes also written as $C^{JM_J}_{LM_L;SM_S}$ or variations on that theme.  The coefficients are easiest to calculate from recursion relations but the recursion has been solved and the coefficients have been reduced to summation form ; the simplest cases are often tabulated.
The possible values of $J$ in the sum of Eq.(1) are in the range $L+S, L+S-1, L+S-2, \ldots, \vert L-S\vert$, often written more compactly as $L+S\le J\le \vert L-S\vert$.
In addition, since the total projection $\hat J_z=\hat L_z+\hat S_z$, the eigenvalue $M_J=M_L+M_S$, further restricting the summation in (1).  This restricted sum is indicated with the parenthesis around $(M_J)$.
Because they are transition coefficients from one orthonormal basis to another, the CG coefficients satisfy a number of orthonormality conditions, such as
$$
\sum_{J } \vert \langle JM_J\vert LM_L;S M_S\rangle \vert^2=1\, .
$$
There are additional such formulae.  Starting from $\langle JM_J\vert J’M’_{J}\rangle=\delta_{JJ’}\delta_{M_J M’_J}$ and inserting
$$
\hat 1=\sum_{M_LM_S}\vert LM_L;SM_S\rangle \langle LM_L;SM_S\vert
$$
one gets
$$
\sum_{M_LM_S}\langle JM_J\vert L M_L;SM_S\rangle \langle LM_L;SM_s\vert J’M’_J\rangle=\delta_{JJ’}\delta_{M_JM’_J}
$$
etc.
