Local Latent Symmetry: Higgs Mechanism

When changing a latent global symmetry into a (gauge) local one, the Goldstone boson (or bosons disappear) from the theory and the corresponding gauge fields acquire mass: this is the Higgs mechanism. From a more formal point of view, it happens that the degrees of freedom of the Goldstone bosons are transferred to the longitudinal polarizations of the gauge bosons which are essential after acquiring mass.

Let us see the example U(1)

The Lagrangian with global symmetry is as before

$${\cal L} = - \partial_{\mu}\phi^{\dagger} \partial^{\mu}\phi - \lambda (\phi^{\dagger} \phi - v)^{2}$$ (268)

with $v>0$ in order to trigger the latent symmetry, that is to say $\grave{a}\; la$ Nambu-Goldstone. If the symmetry $U(1)$ is local, there appears the necessity of including the massless gauge field so that the derivative becomes a covariant one, that is

$$\partial_{\mu} \rightarrow D_{\mu} = \partial_{\mu} - i g a_{\mu}(x)$$ (269)

so that the symmetric Lagrangian is

$${\cal L} = - (D_{\mu}\phi)^{\dagger} (D^{\mu}\phi) - \lambd... ...^{\dagger} \phi - v)^{2} - \frac{1}{4} F_{\mu\nu} F^{\mu\nu}$$ (270)

when the field transformations are

$$\phi(x) \rightarrow e^{i \alpha(x)} \phi(x)$$ (271)

$$a_{\mu}(x) \rightarrow a_{\mu}(x) + \frac{1}{g} \partial_{\mu}\alpha(x)$$ (272)

The presence of the covariant derivative implies the interactions

$${\cal L}_{INT}^{(1)} = g a_{\mu}(x) J^{\mu}(x)$$ (273)

$${\cal L}_{INT}^{(2)} = - g^{2} a_{\mu}(x) a^{\mu}(x) \phi^{\dagger}(x) \phi(x)$$ (274)

Due to the election $v>0$, the symmetry is latent, that is to say

$$<0\mid \phi \mid 0> =\sqrt{v} \neq 0$$ (275)

and we must introduce the shifted field $\chi(x)$ by

$$\phi(x) = \sqrt{v} + \chi(x)$$ (276)

so that

$$<0\mid \chi(x) \mid 0> = 0$$ (277)

When doing the traslation (278), the interaction Lagrangian (276) provides

$${\cal L}_{M} = - g^{2} v a^{\mu} a_{\mu}$$ (278)

$$= - \frac{1}{2} m_{a}^{2} a^{\mu} a_{\mu}$$ (279)

that is, the gauge field acquires mass. This is the Higgs mechanism in action.

It is interesting to verify that during the realization of the mechanism there is an explicit conservation of degrees of freedom. That is to say, we are going to prove that, as in the Lagrangian (272), there are four degrees of freedom (two from the complex scalar field and two from the massless gauge field) after the gauge field acquires mass (three degrees of freedom). This is because only one real scalar field remains in the theory (one degree of freedom).

For this purpose it is useful to make an exponential parametrization of the field $\phi(x)$:

$$\phi(x) = [\rho(x) + \sqrt{v}] e^{i \frac{\sigma(x)}{\sqrt{v}}}$$ (280)

where the fields $\rho(x)$ and $\sigma(x)$ are real. We will see then that the phase field $\sigma(x)$ is the field of the Goldstone boson which uncouples after the Higgs mechanism is completed.

If we write the quartic $V(\phi^{\dagger} \phi)$ in terms of the parametrization (282) it results

$$V= \lambda (\rho^{2} + 2 \rho \sqrt{v})^{2}$$ (281)

which clearly shows that

$$m_{\rho}^{2} = 8 \lambda v$$ (282)

But what happens with the field $\sigma(x)$. Let us write the covariant derivative of the field $\phi(x)$

$$D_{\mu}\phi = e^{i \frac{\sigma}{\sqrt{v}}} [\partial_{\mu... ...) g (a_{\mu} + \frac{1}{g \sqrt{v}} \partial_{\mu}\sigma)]$$ (283)

It is clear that in the kinetic term of the Lagrangian the exponential factor $e^{i \frac{\sigma}{\sqrt{v}}}$ dissappears. On the other hand, the term

$$a_{\mu} + \frac{1}{g \sqrt{v}} \partial_{\mu}\sigma$$ (284)

which appears in (285) is just the transformed gauge field of $a_{\mu}(x)$ with

$$\alpha(x) = \frac{\sigma(x)}{\sqrt{v}}$$ (285)

Then, fixing appropriately the gauge everything can be written in terms of the gauge field

$${\cal A}_{\mu}(x) = a_{\mu}(x) + \frac{1}{g \sqrt{v}} \partial_{\mu}\sigma(x)$$ (286)

and the Goldstone field $\sigma(x)$ disappears from the theory (is gauged away) so that the Lagrangian can be written (in this particular gauge)as

$${\cal L} = - \frac{1}{2} \partial_{\mu}\rho \partial^{\mu}\rho - \frac{1}{... ..._{\mu\nu} F^{\mu\nu} - \frac{1}{2} m_{a}^{2} {\cal A}_{\mu} {\cal A}^{\mu}$$

$$- g^{2} (\rho^{2} + 2 \sqrt{v} \rho) {\cal A}_{\mu} {\cal A}^{\mu} - \lambda (\rho^{4} + 2 \sqrt{v} \rho^{3})$$ (287)

where there are clearly the degrees of freedom (four) that we must have.

In summary, as we previously said, when a local gauge symmetry is spontaneously broken, when it is latent, everything happens as if the gauge field acquires mass from the Goldstone boson which ``should" appear and uncouples. This exchange of degrees of freedom, implied by the Higgs mechanism, does not mean a violation of the Goldstone theorem because it cannot be strictly applied when there are long range forces as the ones due to a gauge field.

Certainly the theory described by the Lagrangian (289) is not explicitly renormalizable as the propagator of ${\cal A}_{\mu}$, in that gauge is

$$[g_{\mu\nu} + \frac{k_{\mu} k_{\nu}}{m_{a}^{2}}] \frac{1}{k^{2} + m_{a}^{2}}$$ (288)

Nevertheless, the theory is gauge equivalent to the described by the Lagrangian (272) (from the one it was obtained fixing a gauge) which is renormalizable. Therefore the Lagrangian (229) can be used in practical calculations.

The important conclusion is that a spontaneously broken Yang- Mills theory is renormalizable although it contains massive vectorial bosons.