Physical interpretation of triplet state When we look at a triplet state we know its spin part $\sigma(1,2)$ must be symmetric. Then it must be proportional to:
$\sigma_{\uparrow}(1)\sigma_{\uparrow}(2)$
$\sigma_{\downarrow}(1)\sigma_{\downarrow}(2)$
$\sigma_{\uparrow}(1)\sigma_{\downarrow}(2) + \sigma_{\downarrow}(1)\sigma_{\uparrow}(2)$
I can understand what first two microstates represent (particles have same spin). What I can't tell is the physical interpretation of the third one. If the first two represent that the particles have the same spin does it then imply that the third one represents the mixture of two states both with different spin? 
 A: Why does the third state (which looks like it has spin $0$) actually have spin $1$?
To see what is going on, you should look at the state
$$
\frac{1}{\sqrt{2}}( \sigma_{\uparrow\downarrow} + \sigma_{\downarrow\uparrow} )
$$
in the basis of $\sigma_x$ states. We have
$$
\sigma_{\uparrow\downarrow} = \frac{1}{2}(\sigma_{\rightarrow}+\sigma_\leftarrow)(\sigma_\rightarrow - \sigma_{\leftarrow}),
$$
$$
\sigma_{\downarrow\uparrow} = \frac{1}{2}(\sigma_\rightarrow - \sigma_{\leftarrow})(\sigma_{\rightarrow}+\sigma_\leftarrow).
$$
So
$$
\frac{1}{\sqrt{2}}( \sigma_{\uparrow\downarrow} + \sigma_{\downarrow\uparrow} ) = \frac{1}{\sqrt{2}} (\sigma_{\rightarrow\rightarrow}-\sigma_{\leftarrow\leftarrow})
$$
So the triplet state with spin $0$ in the $z$ direction has spin $1$ if you measure in the $x$ or $y$ direction. Similarly, if you measure the state $
\frac{1}{\sqrt{2}}( \sigma_{\uparrow\uparrow} - \sigma_{\downarrow\downarrow} )$ in the $\sigma_x$ basis, both particles will have opposite spins.
On the other hand, the singlet state $\frac{1}{\sqrt{2}}( \sigma_{\uparrow\downarrow} - \sigma_{\downarrow\uparrow} )$ has spin $0$ no matter which basis you use to measure both particles.
A: There is nothing to prevent a state with $L=1$ to have $M=0$ since the possible values of $M$ range from $-L$ to $L$.  The state
$$
\sigma_{\uparrow}(1)\sigma_{\downarrow}(2) + \sigma_{\downarrow}(1)\sigma_{\uparrow}(2)
$$
is simply proportional to the $L=1, M=0$ state.  
You can verify it has $M=0$ by checking the eigenvalue of $L_z=L_z(1)+L_z(2)$ and you can verify is has $L=1$ by 
checking it is an eigenstate of $L_x^2+L_y^2+L_z^2$ (with $L_k=L_k(1)+L_k(2)$) with the appropriate eigenvalue.  
Alternatively, you can use $L_+=L_+(1)+L_+(2)$ to connect this state with the $L=1,M=1$ state, meaning it must have the same value of $L$ as $\sigma_\uparrow(1)\sigma_\uparrow(2)$.
A: All the three states correspond to the angular momentum of magnitude 1, i.e. $S^2 = \hbar^2 n(n+1)$, where $n=1$. The projection of this angular momentum on $z$-axis can take three values: $S_z = 0,\pm \hbar$.
It is easy to see which state corresponds to which projection by acting on them with the operator of total spin projection on $z$-axis, $\hat{S}_z = \frac{\hbar}{2}(\hat{\sigma}_z^{(1)} + \hat{\sigma}_z^{(2)})$:
\begin{array}
\hat{S}_z |\uparrow\uparrow\rangle = (\frac{\hbar}{2} + \frac{\hbar}{2})|\uparrow\uparrow\rangle = \hbar|\uparrow\uparrow\rangle,\\
\hat{S}_z |\downarrow\downarrow\rangle = (\frac{\hbar}{2} + \frac{\hbar}{2})|\downarrow\downarrow\rangle = \hbar|\downarrow\downarrow\rangle,\\
\hat{S}_z \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle  + |\downarrow\uparrow\rangle) = 
(\frac{\hbar}{2} - \frac{\hbar}{2} - \frac{\hbar}{2} + \frac{\hbar}{2}) \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle  + |\downarrow\uparrow\rangle) = 0.
\end{array}
The state with zero projection on the $z$-axis is very similar to the singlet state, which has zero momentum and zero projection on the $z$-axis, and the wave function $$\frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle  - |\downarrow\uparrow\rangle).$$
In my opinion, the easiest way to get the grasp of this is by diagonalizing the Hamiltonian for two interacting spins:
$$\hat{H} = \hat{\vec{S}}_1\cdot \hat{\vec{S}}_2 = 
\hat{S}_{1x}\hat{S}_{2x} + \hat{S}_{1y}\hat{S}_{2y} + \hat{S}_{1z}\hat{S}_{2z}$$
in the basis $|\downarrow\downarrow\rangle, |\uparrow\downarrow\rangle, |\downarrow\uparrow\rangle,|\uparrow\uparrow\rangle$.  
A: The physical interpretation of the third state is that the two particles have opposite spin projections along an axis. If one is up, the other is down and vice-verse. 
There are two ways in which that can happen, $|ud\rangle$ and $|du\rangle$ (I have used a place value system and u=up, d=down). Since the two particles are identical, the state must be either symmetric or antisymmetric. And for triplet, it turns out to be the symmetric combination of $|ud\rangle+|du\rangle$ up to a normalisation.
Remember though that in this state, once you measure the spin on one particle, the state will either be $|ud\rangle$ or $|du\rangle$
A: It is not correct to interpret $$\sigma_{\uparrow}(1)\sigma_{\downarrow}(2) + \sigma_{\downarrow}(1)\sigma_{\uparrow}(2)$$ as describing opposite spins. The arrows indicate alignment with a quantisation axis, say z. The expression describes parallel spins, oriented in the xy plane with equal probability in any direction, in terms of $S_z$ eigenfunctions.
