Continuous Global Latent Symmetry

The generalization of the previous model to the case of a continuous symmetry is represented by the so called Goldstone model defined by the Lagrangian

$${\cal L} = - \partial_{\mu}\phi^{\dagger}(x) \partial^{\mu}\phi(x) - \lambda (\phi^{\dagger}(x) \phi(x) - v)^{2}$$ (244)

where $\phi(x)$ is a complex scalar field. This Lagrangian is invariant under phase transformations

$$\phi(x) \rightarrow \phi'(x) = e^{i \alpha} \phi(x)$$ (245)

with $\alpha$ constant. That is to say we are in front of a global symmetry $U(1)$ which gives place to the conserved current

$$J^{\mu}(x) = i [(\partial^{\mu}\phi^{\dagger}) \phi - (\partial^{\mu} \phi) \phi^{\dagger}]$$ (246)

The potential that appears in (246)

$$V(\phi^{\dagger} \phi) = \lambda (\phi^{\dagger} \phi - v)^{2}$$ (247)

depends on the sign of the constant $v$.

So if $(v<0)$, the potential presents only one minimum at $\phi = 0$: the symmetry is realized $\grave{a}\; la$ Wigner-Weyl.

If on the contrary, $(v > 0)$ , we are in the presence of an infinite number of minima defined by

$$\phi^{\dagger} \phi = v$$ (248)

which in terms of the fields $\phi_{1}$ and $\phi_{2}$, real and imaginary part of $\phi$, respectively, it is written

$$\phi_{1}^{2} + \phi_{2}^{2} = v$$ (249)

and whose representation is the well-known ``culo de botella" or ``mexican hat".

Now, the symmetry (U(1) global) is done $\grave{a}\; la$ Nambu- Goldstone. As a consequence of the already mentioned Goldstone theorem the theory defined by the Lagrangian (246), when the symmetry is latent, will contain a massless state: the Goldstone boson.

The uniform movement (constant hunger) of the donkey in the circle of carrots correponds to the non massive excitation called Goldstone boson.

Coming back to the model, when $v>0$ in the potential (249), it has no sense the development around $\phi = 0$ as there we are in a local maximum. The perturbative development must be defined from a field

$$\phi = \sqrt{v} e^{i \theta}$$ (250)

with some $\theta$, arbitrary, due to the non unicity of the vacuum.

Certainly, the latent symmetry $U(1)$ allows to eliminate $\theta$ because

$$U^{-1}(\alpha) \phi(x) U(\alpha) = e^{i \alpha} \phi(x)$$ (251)

choosing $\alpha = - \theta$ and considering (252)

$$<0\mid U^{-1}(-\theta) \phi(x) U(-\theta) \mid 0> = \sqrt{v}$$ (252)

as $U(-\theta)\mid 0>$ is equivalent to $\mid 0>$ due to the latent character of the symmetry.

In a similar way to the case of the discrete symmetry, after choosing a perturbative vacuum the field $\chi(x)$ is defined with

$$<0\mid \chi(x) \mid 0> = 0$$ (253)

through

$$\phi(x) = \sqrt{v} + \chi(x)$$ (254)

The potential in terms of $\chi(x)$ results

$$V = \lambda v (\chi + \chi^{\dagger})^{2} + 2 \lambda \s... ... (\chi + \chi^{\dagger}) +\lambda (\chi \chi^{\dagger})^{2}$$ (255)

Introducing now the fields $\chi_{1}$ and $\chi_{2}$ , real and imaginary parts of $\chi$ respectively, it is easy to see that $\chi_{1}$ is a massive field with

$$m_{\chi_{1}} = 4 \lambda v$$ (256)

while

$$m_{\chi_{2}} = 0$$ (257)

allowing us to conclude that the field $\chi_{2}(x)$ corresponds precisely to the Goldstone boson. The results (258) and (259) show that when the symmetry (U(1) in the example) is latent or spontaneously broken it will not be present in the spectrum of the theory.

The previous results become more intuitive if the field $\chi_{1}$ is assimilated with radial excitations around a point in the minimum circle, meanwhile $\chi_{2}$ corresponds to the movement, with zero frequency, around this circle. In fact, if the field variables of angular type (polar coordinates) are introduced

$$\chi_{1} = \rho \cos \theta$$ (258)

$$\chi_{2} = \rho \sin \theta$$ (259)

The phase transformation $U(1)$ (247) corresponds then to

$$\rho \rightarrow \rho$$ (260)

$$\theta \rightarrow \theta + \alpha$$ (261)

and the Lagrangian of the model is written

$${\cal L} = \frac{1}{2} (\partial_{\mu}\rho)^{2} + \frac{\rho^{2}}{2} (\partial_{\mu}\theta)^{2} - V(\rho)$$ (262)

so that the phase invariance is shown in the independence of $V$ with the variable $\theta$. If the vacuum is chosen in

$$<0\mid \rho\mid 0> = \sqrt{v}$$ (263)

$$<0\mid \theta\mid 0> = 0$$ (264)

and the displaced fields are introduced

$$\rho' = \rho - \sqrt{v}$$ (265)

$$\theta' = \theta$$ (266)

the Lagrangian results

$${\cal L} = \frac{1}{2} (\partial_{\mu}\rho')^{2} + \frac{1}... ...^{2} (\partial_{\mu}\theta')^{2} - {\cal V}(\rho' + \sqrt{v})$$ (267)

As the field $\theta$ appears only through its derivative it is concluded that its quantums will be massless and as it was already mentioned, it corresponds to rotations around the minimal circle. The previous discussion shows also that the appearance of the Goldstone boson is directly linked to the spontaneous breaking of the symmetry and does not depend of the particular form of the potential $V$ (or ${\cal V}$).