Discrete Global Latent Symmetry

The traditional example corresponds to the case of a scalar field with the Lagrangian

$${\cal L} = \frac{1}{2} \partial_{\mu}\phi \partial^{\mu}\phi - V(\phi)$$ (234)

with the discrete symmetry induced by the transformation of reflection

$$\phi \rightarrow -\phi$$ (235)

under theassumption that the potential $V$ is an even function of $\phi$. The energy density energy is then

$${\cal H} = \frac{1}{2} (\partial_{0}\phi)^{2} + \frac{1}{2} (\vec{\nabla}\phi)^{2} + V(\phi)$$ (236)

so that the state of minimum energy (vacuum expectation value) is

$$<0\mid \phi \mid 0> = constant$$ (237)

with the $constant$ determined by the minimum of $V$ . The most simple case corresponds to the potential

$$V(\phi) = \frac{\lambda}{4} (\phi^{2} \pm \frac{m^{2}}{\lambda})^{2}$$ (238)

with $\lambda$ positive.

Case i): (+ sign)

Then the minimum of $V$ is in $\phi = 0$ so that the vacuum is

$$<0\mid \phi\mid 0> = 0$$ (239)

whose symmetry under (237) is evident. In this case it is valid to perturbate around the vacuum (241). The symmetry is realized $\grave{a}\; la$ Wigner-Weyl.

Case ii): ($-$ sign)

The potential $V$ has now two minima, absolutely equivalent, placed in

$$\phi_{min} = \pm \sqrt{\frac{m^{2}}{\lambda}}$$ (240)

In other words, $\phi = 0$ is an unstable point and there is no sense in perturbing around this value.

Which will then be the vacuum expectation value? The election of one of the two values (242) implies to take the symmetry to the latent state (or to spontaneously break the symmetry). It is observed that the minima are interchange under the operation of symmetry (237) but they are not separately symmetric.

In order to continue the study of the model one of the asymmetric vacuums is chosen in order to define the origin of perturbations. Let us take as an example

$$\phi_{min} = + \sqrt{\frac{m^{2}}{\lambda}}$$ (241)

Then a displaced field can be defined

$$\phi'(x) =\phi(x) - \phi_{min} = \phi(x) - \sqrt{\frac{m^{2}}{\lambda}}$$ (242)

so that the Lagrangian is written

$${\cal L}(\phi') = \frac{1}{2} \partial^{\mu}\phi' \partial... ...} + \sqrt{\frac{m^{2}}{\lambda}} \phi'^{3} + m^{2} \phi'^{2}$$ (243)

This expression for the Lagrangian reaffirms that the symmetry is latent as there do not appear in the theory more parameters (only the two initial ones). Besides it can be observed that the mass of the field is $m$ as it can be just now correctly defined. Finally, the other vacuum $(-\sqrt{\frac{m^{2}}{\lambda}})$ does not appear perturbatively: it is in some sense infinitely away because the previous minimum field has to change in every space-time point in order to reach it.