$NN\pi$

Let us begin by considering the interaction strengths between nucleons and pions. We would like to know, on the basis of isospin symmetry alone (neglecting isospin breaking) the relative size of the interaction between, say, $pp\pi^0$ and $np\pi^-$, when all other quantum numbers are the same.

Using states we would go about this by asking about the overlap of states of isospin, we start by combining the $N\pi$ states into states to total isospin,

$$\vert p \pi^0\rangle~=~\vert p\rangle \vert\pi^0\rangle =~\vert{1\over 2},{1\over 2}\rangle \vert 1,0\rangle$$

$$\qquad =~\langle {1\over 2},{1\over 2},1,0\vert{3\over 2},{1\over... ...ver 2},1,0\vert{1\over 2},{1\over 2}\rangle \ \vert{1\over 2},{1\over 2}\rangle$$

$$\vert p \pi^-\rangle~=~\vert p\rangle \vert\pi^-\rangle =~\vert{1\over 2},{1\over 2}\rangle \vert 1,-1\rangle$$

$$\qquad =~\langle {1\over 2},{1\over 2},1,-1\vert{3\over 2},-{1\ov... ...\vert{1\over 2},-{1\over 2}\rangle \ \vert{1\over 2},-{1\over 2}\rangle \ \ \ ,$$ (23)

and therefore, the isospin factors of the ``matrix elements'' $\langle p\vert p \pi^0\rangle$ and $\langle n\vert p \pi^-\rangle$ are

 

$$\langle p\vert p \pi^0\rangle~=~ \langle {1\over 2},{1\over 2},1,0\vert{1\over 2},{1\over 2}\rangle~=~-{1\over\sqrt{3}}$$

$$\langle n\vert p \pi^-\rangle~=~ \langle {1\over 2},{1\over 2},1,-1\vert{1\over 2},-{1\over 2}\rangle~=~-\sqrt{2\over 3} \ \ \ ,$$ (24)

from which we see a relative factor of $\sqrt{2}$. We have used the fact that the strong interaction is a singlet under isospin transformations, and therefore does not modify the isospin structure of the states.

We can arrive at the same result by considering the tensors describing the $N$ and $\pi$ fields. In the limit we are considering the world is invariant under arbitrary isospin rotations, and therefore, the hamiltonian describing the world is a singlet in iso-space, i.e. no spare isospin indices are left floating around. The most general form for the effective interaction hamiltonian between nucleons and one pion is, considering only isospin structure and using eqs. (21) and (12),

$${\cal H}~=~\alpha\ N^{a,\dagger} \Pi_a^b N_b$$

$$=~ \alpha\ \left[ {1\over\sqrt{2}} p^\dagger p\pi^0\ +\ p^\dagger... ...i^+\ +\ n^\dagger p\pi^0\ -\ {1\over\sqrt{2}} n^\dagger n\pi^0\ \right] \ \ \ ,$$ (25)

where $\alpha$ is a reduced matrix element, independent of the isospin projections as dictated by the Wigner-Eckart theorem. Therefore, the matrix element of this effective hamiltonian between the desired states is

$$\langle p\vert{\cal H}\vert p\pi^0\rangle~=~{1\over\sqrt{2}}\alpha$$

$$\langle n\vert{\cal H}\vert p\pi^-\rangle~=~\alpha \ \ \ ,$$ (26)

which gives relative factor of $\sqrt{2}$and reproduces the result of the first method in eq. (24).

The method of tensors is much more versatile than using tabulated CG's, as it easily extends to other group structures like SU(3) etc. It is also clearly connected to the Wigner-Eckart theorem.