Two identical particles 
A system made up of two spin-$1/2$ identical particles is prepared
  such that:
  
  
*
  
*a measurement of $\mathbf{L}_{1}^2$ and $\mathbf{L}_{2}^2$ gives $2\hbar^2$ with certainty;
  
*a measurement of $L_{1z}$ and $L_{2z}$ gives $+\hbar$ for one particle and $-\hbar$ for the other one;
  
*a measurement of $S_{1z}$ and $S_{2z}$ gives $+\hbar/2$ for one particle and $-\hbar/2$ for the other one.
  
  
  Write the most general quantum state,
  $\left|\alpha\right\rangle$, compatible with these measurements.

Since there are two fermions, I know the state must be antisymmetric.
From 1., I know $\ell_1=\ell_2=1$, so the total orbital angular momentum can be $\ell=0,1,2$.
From 2. and 3., $m_{\ell_1}=+1$ and $m_{\ell_2}=-1$, $m_{s_1}=+1/2$ and $m_{s_2}=-1/2$; so, $m_\ell=m_s=0$.
How can I write the state of this system?
 A: You know that you have $\ell_1=\ell_2=1$, and the $m_{\ell_1}=+1$ while $m_{\ell_2}=-1$.  Hence you must have (up to a factor of $1/\sqrt{2}$):
$$
\vert\psi_\pm\rangle = \vert 1 1\rangle_1\vert 1,-1\rangle_2 \pm \vert 1,-1\rangle_1\vert 1 1\rangle_2
$$
Now for the spin part, you can conclude likewise that
$$
\vert\chi_\pm\rangle = \vert 1/2,1/2\rangle_1\vert 1/2,-1/2\rangle_2\pm 
\vert 1/2,-1/2\rangle_1\vert 1/2,1/2\rangle_2\, .
$$
Note that, under the interchange of particle labels:
$$
P_{12}\vert\psi_\pm\rangle =\pm \vert\psi_\pm\rangle\, ,
\qquad 
P_{12}\vert\chi_\pm \rangle= \pm \vert\chi_\pm\rangle\, .
$$
To have overall antisymmetry you should therefore have
$$
\vert\phi\rangle=\alpha \vert\psi_+\rangle\vert\chi_-\rangle +
\beta \vert \psi_-\rangle \vert\chi_+\rangle
$$
with $\alpha$ and $\beta$ chosen so your states are properly normalized.
This way:
\begin{align}
P_{12}\vert\phi\rangle &= \alpha \left[P_{12}\vert\psi_+\rangle\right]\left[P_{12}\vert\chi_-\rangle\right]
+ \beta \left[P_{12}\vert\psi_-\rangle\right]\left[P_{12}\vert\chi_+\rangle\right]\, ,\\
&=- \alpha\vert\psi_+\rangle\vert\chi_-\rangle -\beta \vert\psi_-\rangle \vert\chi_+\rangle=-\vert\phi\rangle\, . 
\end{align}
Basically, if you pick the symmetric combination for the spatial part, it must be multiplied by the antisymmetric combinations for the spin part so the product is overall antisymmetric, and vice versa if you pick the antisymmetric combination for the spatial part you have to multiply it by the symmetric one for the spin part.
Note that the situation gets considerably more complicated with $3$ particles as the permutation group $S_3$ of three objects have a 2-dimensional representation of mixed symmetry which do not necessarily transform back to a $\pm 1$ multiple of themselves under permutation.  The combination of such types of functions is a little more delicate than in the 2-particle case.
