Yang-Mills Fields

The free Lagrangian for ``nucleons" (considered as point particles) reads:

$${\cal L} = \bar{\psi} (i \gamma^{\mu}\partial_{\mu} - m) \psi$$ (208)

with

$$\psi = \left( \begin{array}{l} \psi_{p} \psi_{n} \end{array} \right)$$ (209)

is invariant under a transformation of isospin (208) with $\vec{\alpha}$ arbitrary and constant. Therefore the isotopic current

$$J_{k}^{\mu} = \bar{\psi} \gamma^{\mu} \tau_{k} \psi$$ (210)

is conserved and it is concluded that there is no term in the Lagrangian (210) allowing to physically distinguish the proton and neutron states. So we can think (with Yang and Mills [1954]) that: ``... The difference between a proton and a neutron is then a purely arbitrary process. Nevertheless, as it is generally conceived, this arbitrarity is subject to the following limitation: once it is chosen in a space-time point how to label a proton, or a neutron, there is no more freedom to choose in other points of the space-time..."

``...It seems that this is not consistent with the concept of the localized field which supports the usual physical theories. In the present work we want to explore the possibility of requiring all interactions to be invariant under isotopic spin rotations which are independent in each point of the space-time".

What happens then if different isotopic transformations are implemented in different points of the space-time? That is to say, what is the consequence of a local isotopic phase transformation with

$$\alpha_{k} = \alpha_{k}(x)$$ (211)

Certainly, the mass term of ${\cal L}$, does not represent any problem, as it was the case in $U(1)$ because it is invariant even when the freedom (213) is choosen. Nevertheless, for the term with derivative one has

$$\partial_{\mu}\psi \rightarrow \partial_{\mu}\psi + i \partial_{\mu}\vec{\alpha}(x) \frac{\ve... ...au}}{2} \psi + i \vec{\alpha}(x)\cdot\frac{\vec{\tau}}{2} \partial_{\mu}\psi$$ (212)

$$\simeq e^{i \vec{\alpha}(x)\cdot\frac{\vec{\tau}}{2}} (\partial_{\mu} + i \partial_{\mu}\vec{\alpha}(x)\cdot\frac{\vec{\tau}}{2}) \psi$$ (213)

with $\vec{\alpha}$ infinitesimal. This result indicates the necessity of introducing a compensating field or gauge field (in fact, three fields) which in this case is called Yang-Mills field

$$\vec{b}_{\mu} \equiv (b_{\mu k} ; k=1,2,3)$$ (214)

It is an isovector field which is added to the derivative in ${\cal L}$ so that the covariant derivative is defined

$$D_{\mu} = \partial_{\mu} - i g \vec{b}_{\mu}\cdot\frac{\vec{\tau}}{2}$$ (215)

Then, it results

$${\cal L} = i \bar{\psi} \gamma^{\mu} (\partial_{\mu} - i\... ...}_{\mu} \cdot\frac{\vec{\tau}}{2}) \psi - m \bar{\psi} \psi$$ (216)

Again the requirement of local invariance determines the interaction. Certainly, the gauge invariance of second class of the Lagrangian imposes a precise transformation law for the gauge field $\vec{b}_{\mu}$. If ${\cal L}$ is supposed invariant, we can write

$${\cal L} = \bar{\psi}' [i \gamma^{\mu} (\partial_{\mu} - i g \vec{b}'_{\mu} \cdot\frac{\vec{\tau}}{2}) - m] \psi'$$ (217)

$$= \bar{\psi} e^{-i \vec{\alpha}\cdot\frac{\vec{\tau}}{2}} [i \g... ...ot\frac{\vec{\tau}}{2}) -m] e^{i \vec{\alpha}\cdot\frac{\vec{\tau}}{2}} \psi$$ (218)

$$= \bar{\psi} [i \gamma^{\mu} (\partial_{\mu} - i g\vec{b}_{\mu}\cdot\frac{\vec{\tau}}{2}) - m] \psi$$ (219)

and therefore it is valid the identification

$$\partial_{\mu} - i g \vec{b}_{\mu}\cdot\frac{\vec{\tau}}{2... ...vec{\tau}}{2}) e^{-i \vec{\alpha}\cdot\frac{\vec{\tau}}{2}}$$ (220)

considering now infinitesimal transformations results

$$-i g \vec{b}'_{\mu}\cdot\frac{\vec{\tau}}{2} = i \partial... ...rac{\vec{\tau}}{2} , \vec{b}_{\mu}\cdot\frac{\vec{\tau}}{2}]$$ (221)

But remembering (186) the transformation law of the Yang-Mills field emerges

$$\delta b_{\mu k} = b'_{\mu k} - b_{\mu k} = -\frac{1}{g... ...{\mu}\alpha_{k} - 2 \alpha_{i} b_{\mu j} \epsilon_{ijk}$$ (222)

The clear difference with the electromagnetic ($U(1)$) case is in the last term of this expression that comes from the fact that $SU(2)$ is a non abelian group.

In conclusion, the Lagrangian (218) is invariant under local isospin phase transformations - transformations (209) with (213) - if the gauge field is simultaneously transformed as (224).

Once the Yang-Mills field is introduced in the game, it is clear that the complete description of the theory implies establishing the free Lagrangian of the gauge field. For the electromagnetic case it is known that

$${\cal L}_{0}^{(em)} = - \frac{1}{4} F_{\mu\nu} F^{\mu\nu}$$ (223)

with

$$F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$$ (224)

is local phase invariant $U(1)$.

For the case of isotopic invariance ${\cal L}$ will be, as usual, a function of the field $b_{\mu k}$ and of its derivatives. The different fact is that in order to satisfy local invariance, there must appear in the more involved combination

$$f_{\mu\nu j} = \partial_{\mu} b_{\nu j} - \partial_{\nu} b_{\mu j} + 2 g \epsilon_{jk\ell} b_{\mu k} b_{\nu \ell}$$ (225)

to define the field intensity tensor. As a consequence, the simplest free lagrangian for the Yang-Mills field is

$${\cal L}_{0}^{(YM)} = -\frac{1}{4} \vec{f_{\mu\nu}}\cdot\vec{f^{\mu\nu}}$$ (226)

Summarizing, the local $SU(2)$ invariant Lagrangian for the field of the ``nucleon" is

$${\cal L}^{(YM)} = -\frac{1}{4} \vec{f_{\mu\nu}}\cdot\vec{f^... ...i \bar{\psi} \gamma^{\mu} D_{\mu}\psi - m \bar{\psi} \psi$$ (227)

which shows that due to the structure of $\vec{f_{\mu\nu}}$ given in (227), even in the absence of the matter field $\psi$, the pure Yang-Mills part of the Lagrangian contains interactions. That is to say, the fields $b_{\mu k}$ suffer selfinteractions. This is due to the fact that the quanta of the Yang-Mills fields, carry isospin, unlike the electromagnetic case where the the photon is neutral. This is the essential distinction between a local phase invariance linked to a non abelian group as $SU(2)$ and an abelian one as the electromagnetic $U(1)$.