How to determine the isospin of a pair of nucleons? While I was reading from the book of Brussaard on shell-model applications in nuclear spectroscopy, he mentioned that isospin of a neutron-neutron pair couple to $T=1$. On the other hand, if it was neutron-proton pair, the isospin couples to either $T=0$ or $T=1$. Now, my question is I can't see how this coupling works, i.e., how to couple isospin in this case? 
 A: The isospin formalism was put by Heisenberg after the discovery of the neutron by J. Chadwick. Heisenberg proposed that proton and neutron are different states of charge of the same particle, the nucleon, assuming the two states carry the same mass. Introducing the variable $t_z$ to label these states, where at low energies, he assigned $t_z = 1/2$ for neutron and $t_z=-1/2$ for proton. We point out that the mathematics of isospin is analogous to the intrinsic spin, and state it as the following.
The proton and neutron states are
$|p\rangle=\left(\begin{array}{l}{0} \\ {1}\end{array}\right), \quad|n\rangle=\left(\begin{array}{l}{1} \\ {0}\end{array}\right)$, 
respectively.
Defining the isospin $\vec{t}$ by its three components operators
$t_{x}=\frac{1}{2}\left(\begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right), \quad t_{y}=\frac{1}{2}\left(\begin{array}{cc}{0} & {-i} \\ {i} & {0}\end{array}\right), \quad t_{z}=\frac{1}{2}\left(\begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right)$.
And we associate dimensionless $t(t+1)$ as eigenvalues to $\vec{t}^2$.  
The eigenvalues of $t_z$ are obtained once acted on the proton and neutron states, 
$
t_{z}|p\rangle=\frac{1}{2}\left(\begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right)\left(\begin{array}{c}{1} \\ {0}\end{array}\right)=-\frac{1}{2}|p\rangle, \quad t_{z}|n\rangle=\frac{1}{2}\left(\begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right)\left(\begin{array}{l}{1} \\ {0}\end{array}\right)=\frac{1}{2}|n\rangle
$.   
In nuclear physics, we often deal with nucleus characterized by $Z$ protons and $N$ neutrons, i.e., many-body system. Therefore, the total isospin obeys the relations,
$
\vec{T}_{\text {total }}=\sum_{i=1}^{A} \vec{t}_{i}, \quad T_{z}=\sum_{i=1}^{A} t_{z_{i}}
$
where $A$ is the total number of nucleons within the nucleus.
Similarly, the operator $\vec{T}^2$ admits $T(T+1)$ as eigenvalues. 
Now, we distinguish two different representations, the first is the m-scheme space spanned by the states $|t_1,t_2,t_{1z},t_{2z}\rangle$ and represented by a set of individual operators, $\vec{t_1}^2,\vec{t_2}^2,t_{1z},t_{2z}$. In this case, each $t_z$ takes values within the range, 
$-t_1 \leq t_{1z} \leq t_1$
$-t_2 \leq t_{2z}\leq t_2$
The second is, is the Coupling-scheme by which we associate its own space spanned by the states $|{T,T_z}\rangle$ and the total isospin operators, $T$ and $T_z$. The same applies as above,
$-T \leq T_{z} \leq T$
However, we joint the two spaces by the same rules of addition of angular momenta,
$T_z= t_{1z}+t_{2z}$
$|t_1-t_2| \leq T \leq t_1+t_2$
Application
-(neutron-neutron) state
As an application, we consider a nucleus with only two neutrons and zero protons with $t_1=t_2=1/2$ and $t_{1z}=t_{2z}=1/2$. According to what stated before, it follows that, 
$0 \leq T \leq 1$ $\rightarrow T=0,1$ and $T_z=1$, $\Rightarrow T=1$.
The allowed value for the total isospin is one because the z-th component is generated by $T=1$ only!
- (neutron-proton) state
The nucleus is considered with only one neutron and one proton, with $t_1=t_2=1/2$ and t_1z=1/2,t_2z=-1/2$.
$0  \leq T \leq 1$ $\rightarrow T=0,1$ and $T_z=0$.
Here both total isospin values are allowed since the z-th component is generated by both values, i.e., by the singlet $T=0$, and the triplet $T=1$.
References
Lam, Y.L.. (2011). Isospin Symmetry Breaking in sd Shell Nuclei. 
