Including $\Delta$'s and $\Delta \rightarrow N\pi$

As we know the nucleon is not a fundamental object, it will come as no surprise to learn that it has excited states. One lowest lying excitation that couples incredibly strongly to $N\pi$is the $\Delta$, which is shorthand for $\Delta^{++}$, $\Delta^+$, $\Delta^0$and $\Delta^{-}$, an isospin quartet of particles. They have masses $\sim~1230~{\rm MeV}$ and widths $\Gamma~\sim~110~{\rm MeV}$. When we neglect electromagnetic and weak interactions these hadrons are almost identical, and can be assigned to an irreducible representation of SU(2).

To figure out how the $\Delta$'s are assigned to the totally symmetric tensor $T_{abc}$, with $a,b,c~=~1,2$ (it is easy to show that this has four independent entries, further if it had a component that was antisymmetry under interchange of two indices, then it could be contracted with $\epsilon^{ab}$), we look at the free hamiltonian for non-relativistic particles, which for nucleons is

$${\cal H}_{\rm free}~=~N^{a, \dagger} \left[ -{\nabla^2\over 2 M_N}\right] N_a \ \ \ ,$$ (27)

and for the $T_{abc}$ field is

 

$${\cal H}_{\rm free}~=~T^{abc, \dagger} \left[ -{\nabla^2\over 2 M_\Delta}\right] T_{abc} \ \ \ .$$ (28)

Explicitly multiplying out eq. (28), and using the symmetry of $T_{abc}$ under interchange of $abc$we find

 

$${\cal H}_{\rm free}~=~T^{111, \dagger} \left[ -{\nabla^2\over 2 M... ...111} \ +\ 3\ T^{112, \dagger} \left[ -{\nabla^2\over 2 M_\Delta}\right] T_{112}$$

$$\ +\ 3\ T^{122, \dagger} \left[ -{\nabla^2\over 2 M_\Delta}\right] T_{122} \ +\ T^{222, \dagger} \left[ -{\nabla^2\over 2 M_\Delta}\right] T_{222}$$

$$= \Delta^{++ \dagger}\left[ -{\nabla^2\over 2 M_\Delta}\right]\De... ...{++} \ +\ \Delta^{+ \dagger}\left[ -{\nabla^2\over 2 M_\Delta}\right]\Delta^{+}$$

$$\ +\ \Delta^{0 \dagger}\left[ -{\nabla^2\over 2 M_\Delta}\right]\... ...\ \Delta^{- \dagger}\left[ -{\nabla^2\over 2 M_\Delta}\right]\Delta^{-} \ \ \ ,$$ (29)

which tells us that

 

$$T^{111}\ = \Delta^{++} \ ,\ T^{112}\ =\ {1\over\sqrt{3}} \Delta^{... ... ,\ T^{122}\ =\ {1\over\sqrt{3}} \Delta^{0} \ ,\ T^{222}\ =\ \Delta^{-} \ \ \ .$$ (30)

The relation between strong decay rates $\Delta \rightarrow N\pi$ can be found be the two methods we have used previously. Let us determine the ratio of decay rates for $\Delta^{++}\rightarrow p\pi^+$and $\Delta^{+}\rightarrow n\pi^+$. Firstly, using states

$$\langle p\pi^+\vert\Delta^{++}\rangle~=~ \left[\langle {1\over 2},{1\over 2}\vert\langle 1,1\vert\right]\vert{3\over 2}, {3\over 2}\rangle$$

$$\langle {1\over 2},{1\over 2},1,1\vert{3\over 2},{3\over 2}\rangl... ...\ =\ \langle {1\over 2},{1\over 2},1,1\vert{3\over 2},{3\over 2}\rangle \ =\ +1$$

$$\langle n\pi^+\vert\Delta^{+}\rangle~=~ \left[\langle {1\over 2},-{1\over 2}\vert\langle 1,1\vert\right]\vert{3\over 2}, {1\over 2}\rangle$$

$$\ =\ \left[ \langle {1\over 2},-{1\over 2},1,1\vert{3\over 2},{1\... ...langle {1\over 2}, {1\over 2}\vert \right] \ \vert{3\over 2}, {1\over 2}\rangle$$

$$\ =\ \langle {1\over 2},-{1\over 2},1,1\vert{3\over 2},{1\over 2}\rangle \ =\ +{1\over\sqrt{3}} \ \ \ ,$$ (31)

which leads to $\Gamma(\Delta^{++}\rightarrow p\pi^+)~=~3\Gamma( \Delta^{+}\rightarrow n\pi^+)$.

Solving the same problem using tensors we have and effective hamiltonian of the form

$${\cal H}~=~\beta \epsilon^{cd} N^{a \dagger} \Pi_c^b T_{abd}$$

$$=~\beta\ \left[ \ p^\dagger\left( \sqrt{2\over 3}\Delta^+\pi^0 \ +\ {1\over\sqrt{3}} \Delta^0\pi^+ \ -\ \Delta^{++}\pi^-\right) \right.$$

$$\left.\ +\ n^\dagger \left( \sqrt{2\over 3}\Delta^0\pi^0 \ +\ {1\... ...sqrt{3}}\Delta^-\pi^+ \ -\ {1\over\sqrt{3}}\Delta^+\pi^- \right)\ \right] \ \ ,$$ (32)

where $\epsilon$ is the totaly antisymmetric tensor, $\epsilon^{12}~=~-\epsilon^{21}~=~+1$. It follows that

$$\langle p\pi^+\vert{\cal H} \vert\Delta^{++}\rangle ~=~-\beta \ \... ...le n\pi^+\vert{\cal H} \vert\Delta^{+}\rangle ~=~-{1\over\sqrt{3}}\beta \ \ \ ,$$ (33)

and therefore $\Gamma(\Delta^{++}\rightarrow p\pi^+)~=~3\Gamma( \Delta^{+}\rightarrow n\pi^+)$.