One Pion Exchange Potential properties for a two-nucleon system I'm going through my Nuclear Physics book, and has come across a section called "Properties of OPEP for the two-nucleon system".
It start out by considering the n-p system in a singlet spin state $(S=0)$. It says that from the generalized Pauli principle, the n-p system must be in a triplet isotopic spin state. Thus, $\overrightarrow{\sigma}_1 \cdot \overrightarrow{\sigma}_2 = -3$, while $\overrightarrow{\tau}_1 \cdot \overrightarrow{\tau}_2 = 1$ - which I assume is just the Pauli spin and isospin matrices.
Then this is used to determine the OPEP via a big equation, where stuff vanishes and such.
My question is: How is these determined ? Why is it a triplet isospin state, where do the -3 and 1 come from ?
I've looked through the whole chapter, but I can't seem to figure it out :/
 A: I suppose that what is meant is the following:
We can consider the neutron and proton as 2 states of the same particle, the nucleon N (regarding the strong interaction, not electromagnetism). Since neutrons and protons are fermions, the wave function of 2 identical particles (here 2 nucleons) must be anti-symetric because of Pauli principle. If the 2 nucleons are in S-state for the spin, this state is antisymetric by interchange of the 2 spins ($|0,0> = \frac{1}{\sqrt{2}}(\uparrow \downarrow- \downarrow\uparrow)$). That means that the interchange of isospin number must be symmetric so that the total wave function:
$\psi (NN) = \psi_{spin} \otimes \psi_{isospin}$
is antisymetric. Now the maths for isospin is exactly the same as the maths for spin. When 2 spins 1/2 are combined, you can get a total spin 0 which is antisymetric (as mentioned above) or a total spin 1 (which is symmetric). Similarly when 2 isopsins are combined, you can get a total isospin 0 (which is anti-symmetric) or a total isospin 1 (which is symmetric). In order to satisfy the Pauli pinciple, only the total isospin 1 is allowed. Such state is a triplet, because when the total isospin $I=1$, you can have 3 isospin projections $I_3 = 1,0,-1$.
A: The only thing I have to add is to answer your question about where the -3 and 1 come from. First, the spin operator is related to the Pauli matrices by
$$s_1 = \frac{1}{2}\sigma_1$$
and so we have the relation
$$\sigma_1 \cdot \sigma_2 = 4 s_1\cdot s_2$$
Next, we can evaluate $s_1\cdot s_2$ using
$$ S^2 = (s_1+s_2)^2 = s_1^2 + s_2^2 + 2 s_1\cdot s_2$$
$$ s_1 \cdot s_2 = \frac{1}{2} (S^2 - s_1^2 - s_2^2)$$
and so we have
$$ \sigma_1\cdot \sigma_2 = 2(S^2 - \frac{3}{2} ) = 2S(S+1) - 3$$
which yields -3 for $S=0$ and 1 for $S=1$.
