$NN$ Systems

The isospin analysis of the two nucleons sector is similar to the above. Each nucleon is a doublet of SU(2), denoted by ${\bf 2}$, and the isospin structure of the two nucleon system is ${\bf 2}\otimes {\bf 2}~=~{\bf 3}\oplus {\bf 1}$. Keeping track of only the isospin indices, we see that the ${\bf 3}$ ($I=1$, denoted by a symmetric two-index tensor) and the ${\bf 1}$ ($I=0$, denoted by a zero-index tensor) are

$$T_{ij}\sim N_iN_j\ +\ N_j N_i$$

$$T\sim \varepsilon^{kl} N_k N_l \ \ \ ,$$ (37)

and we have not shown the spin structure associated with each object. In order to understand better how things are working, we must include the spin structure. Each element of the nucleon field has an index that denotes the spin, e.g. $p^{a}$, where $p^a~=~p\uparrow$ and $p^2~=~p\downarrow$. Therefore, we really need to denote the nucleon field by $N^a_i$, where the up index is the spin index, and the down index is the isospin index. The anticommutation of the nucleon field (it is a fermion) means that for nucleons in an even relative orbital angular momentum state, the only possible configurations are $I~=~1, S~=~0$ and $I~=~0, S~=~1$, denoted by the operators

$$T_{ij}\sim \left[ N_i^a N_j^b\ +\ N_j^a N_i^b\right] \varepsilon_{ab}$$

$$T^{ab}\sim \left[ N_i^a N_j^b\ +\ N_i^b N_j^a\right] \varepsilon^{ij} \ \ \ .$$ (38)

Returning to the isospin structure of the wavefunction of two nucleons, it is easy to see that the $I=1$ part of two-nucleon wavefunction are

$$\vert(NN), I=1, I_z=1\rangle \ =\ \vert pp\rangle$$

$$\vert(NN), I=1, I_z=0\rangle \ =\ {1\over\sqrt{2}}\left[ \vert pn\rangle\ +\ \vert np\rangle \right]$$

$$\vert(NN), I=1, I_z=-1\rangle \ =\ \vert nn\rangle \ \ \ ,$$ (39)

while the $I=0$ part of a two-nucleon wavefunction is

$$\vert(NN), I=0, I_z=0\rangle \ =\ {1\over\sqrt{2}}\left[ \vert pn\rangle\ -\ \vert np\rangle \right] \ \ \ .$$ (40)

The analogy with two-component spin wavefunctions that you will have encountered in nonrelativistic quantum mechanics is exact, for obvious reasons.

As nucleons are fermions, with $J^\pi~=~{1\over 2}^+$, nucleon wavefunctions must be totally antisymmetric under pair-wise interchange. For two nucleons in an even orbital angular momentum state, $L~=~0,2,...$, the spin-isospin part of the wavefunction must be antisymmetric, as the space part is symmetric, and hence $I~=~1, S~=~0$ or $I~=~0, S~=~1$, as mentioned above. The s-wave $I~=~1, S~=~0$ channel is called the ${}^1\kern-.14em S_0$ channel, while the s-wave $I~=~0, S~=~1$ channel is called the ${}^3\kern-.14em S_1$ channel. Apriori the physics in these two channels are unrelated, and in fact we will see that they are different, but yet have similarities that are very suggestive of deeper physics. In particular, there is a bound state in the ${}^3\kern-.14em S_1(-{}^3\kern-.14em D_1)$ channel, while there is not one in the ${}^1\kern-.14em S_0$ channel. However, they share a common feature of having very large scattering amplitudes near zero energy.