The Nucleon Doublet

In the absence of electromagnetic interaction the proton and the neutron are practically identical and can be considered as states of a more fundamental entity, the nucleon. For this one the quantum state can be written

$$\mid N> = a \mid p> + b \mid n>$$ (192)

where $\mid p>$ is the proton state and $\mid n>$ the neutron state, respectively. The constants $\mid a \mid ^{2}$ and $\mid b \mid ^{2}$ measure the probability of finding a nucleon as proton or as neutron respectively. The two different charge states $\mid p>$ and $\mid n>$ are eigenstates of

$$\frac{1}{2} \tau_{3} = \frac{1}{2} \left( \begin{array}{cr} 1 & 0 \\ 0 & -1 \end{array} \right)$$ (193)

with eigenvalues $+1/2$ and $-1/2$ respectively.

The following operators are also introduced

$$\tau_{+} = \frac{1}{2} (\tau_{1} + i \tau_{2})$$ (194)

$$\tau_{-} = \frac{1}{2} (\tau_{1} - i \tau_{2})$$ (195)

which change from the neutron state to the proton state and vice versa.

The complete state of the nucleon will be

$$\mid Nucleon> = \mid m,\ell,s,\cdots> \otimes \mid N>$$ (196)

where $\mid m,\ell,s,\cdots>$ is the state defined by the corresponding dynamical attributes and $\mid N>$ the nucleon state of the isospace defined in (194) and determined by two quantum numbers: $I^{2}$ and $I_{3}$.

The isospin symmetry - present when there is no electromagnetism - implies that there exists a unitary group $SU(2)$, of transformations in the Hilbert space of the state vectors so that

$$\mid N>\;and\;\mid N'> = U \mid N>$$ (197)

or in the general case

$$\mid \Phi>\;and\;\mid \Phi'> = U \mid \Phi>$$ (198)

define the same phenomenon when only the strong interactions are present.

If now the case is a physical observable represented by an operator $0$ and $0'$ is the observable after an isospin transformation, one must have

$$<\Phi\mid O \mid \Phi> = <\Phi'\mid O'\mid \Phi'>$$ (199)

that using the definition of $\mid \Phi'>$ leads to

$$O' = U O U^{\dagger}$$ (200)

which is the typical form of transformation of an operator.

If we deal now with the field theory of a nucleon, its state will be presented by the field

$$\psi = \left( \begin{array}{c} \psi_{p} \psi_{n} \end{array} \right)$$ (201)

which is a double spinor - 8 components - composition of the proton spinor with the neutron spinor. As the field has the character of operator, it is transformed under isospin as

$$\left( \begin{array}{c} \psi'_{p} \psi'_{n} \end{array}... ...}{l} \psi_{p} \psi_{n} \end{array} \right) U^{\dagger}$$ (202)

but as the fields of the proton and of the neutron are components of an irreducible tensorial operator of order $j=1/2$, the transformation (204) has the form

$$\left( \begin{array}{c} \psi'_{p} \psi'_{n} \end{array} ... ...t( \begin{array}{l} \psi_{p} \psi_{n} \end{array} \right)$$ (203)

Note: $2j+1$ operators $O_{m}^{(j)}$ which act in the vector space of the irreducible representation of dimension $2j+1$ form the components of an irreducible tensorial operator of order $j$ if they are transformed as

$$U O_{m'}^{(j)} U^{\dagger} = O_{m}^{(j)} U_{mm'}^{(j)}$$ (204)

where $U_{mm'}^{(j)}$ is the representation of the transformation group.

The unitarity of $U$ implies that the anticommutation relations of the fermionic fields of the nucleon are conserved in a transformation of isospin.

Using now

$$U = e^{i I_{j} \alpha_{j}} = e^{i \vec{I}\cdot\vec{\alpha}}$$ (205)

which for the 1/2 representation is

$$U = e^{i \vec{\alpha}\cdot\frac{\vec{\tau}}{2}}$$ (206)

the isotopic infinitesimal transformation of the nucleon finally results

$$\psi \rightarrow \psi' = \psi + i \vec{\alpha}\cdot \frac{\vec{\tau}}{2} \psi$$ (207)

Clearly, any further generalization to include $SU(N)$ can be done inmediatly.