Broken Symmetry: Examining Lagrangian with $\Phi$

[tirrel]

As an exemple of borken symmetry the teacher showed what happens with this lagrangian:

$$\Lambda=\frac{1}{2}\partial_\mu \Phi^{+}\partial^{\mu}\Phi - \left( \frac{m^2}{2} \partial_\mu \Phi^{+}\partial^{\mu}\Phi +\frac{g^2}{4}(\partial_\mu \Phi^{+}\partial^{\mu}\Phi )^2 ) \right)$$
where
$$\Phi=(\phi_1,\phi_2)$$
the components are two real fields and the plus sign means making the transposed conjugate of the vector (actually $$\Phi$$ should be a column vector)... of course everything depends on the space time point $$x$$

ok... the approach was, as far as I've understood it: let's expand $$\Lambda(\phi_1,\phi_2)$$ in one of the minima diagonalizing also the quadratic term and recognize frome the quadratic term what can be the physical meaning of the field, that is: two particle, and among them a goldstone boson... the new fields were so done:

$$\sigma=\phi_1-\frac{m}{g}$$

$$\eta=\phi_2$$

the new lagrangian (can't make the calculation here, but it's not important) has no more an easy symmetry related to the new fields and the void is not invariant through the action of an operator implementing the symmetry... this is (as far as I've understood) the broken symmetry)

ok, going on... one realizes that one could have chosen other minima to expand the hamiltonian... this is my first question: choosing other minima has a physical meaning? I guess so, and the result is describing the same particles in a more difficult way: I expect the lagrangian for a single field to contain terms of self interaction that don't permit to identify the quadratic part and the relative vectors in Hilbert space as observed particles... I don't bother of the interaction terms becuase I expect them to count only when the particles are interacting...

do you agree with the red sentence? It's an idea of mine so I wanted to have an exchange of opinions...

Going on, my teacher says, let's try this:

$$\phi_1=\rho cos(\theta)$$

$$\phi_2=\rho sen(\theta)$$

writing the hamiltonian this way he finds that the goldstone boson has no self interacting term (I call this way terms who contain only one field but are not quadratic)... and since the other particle has a mass he says (this is a remembering, avtually I didn't take notes on this point) that at low energy the effective lagrangian is:

$$\Lambda=\frac{1}{2}\partial_\mu \alpha\partial^{\mu}\alpha$$

having relabled $$\theta=\alpha\frac{g}{m}$$

but now I think... why not consider also the other lagrangian the one with $$\phi_1,\phi_2$$ for low energies? if the intuitive reason above is correct, I expect the mass-particle not to count and conserve only terms with the goldstone boson, and find:

$$\Lambda=\frac{1}{2}\partial_\mu \eta\partial^{\mu}\eta-\frac{g^2}{4}\eta^4+\frac{m^4}{4g^2}$$

so the second question is: at low energy we have these lagrangians, of course the second more complicated then the second one... but I expect them to describe exactly the same particles (at low energy)! and so having the same low energy "eigenstates"... and so solving the second should be in some sense equivalent to solving the first one...

ok... i said a lot of things... am I totally out of way with my understanding?... is there some correct idea?

thanks to anybody who's going to help me...

 

[eljose79]

Hello,

It seems like you have a good understanding of the concept of broken symmetry and its implications in quantum field theory. To answer your first question, choosing other minima does indeed have physical meaning. This is because the choice of minima determines the vacuum state of the system, which in turn affects the behavior of the particles in the system. Different choices of minima can lead to different vacuum states and therefore different physical predictions.

Regarding your second question, it is possible to consider the lagrangian with \phi_1,\phi_2 at low energies, but it may not be the most convenient or useful approach. This is because the goldstone boson is the massless particle and therefore dominates the low energy behavior of the system. The other particle, with mass, will have a much smaller contribution at low energies and can be neglected in certain calculations. However, as you mentioned, the two lagrangians should describe the same particles at low energies, so solving one should be equivalent to solving the other.

Overall, it seems like you have a good understanding of the concept of broken symmetry and its implications in quantum field theory. Keep exploring and asking questions to deepen your understanding. Good luck with your studies!

 

[Entropia]

I would say that your understanding of the broken symmetry in this example is largely correct. The Lagrangian provided does indeed exhibit a broken symmetry, as seen by the fact that it is not invariant under a transformation of the fields. This broken symmetry is manifested by the presence of a Goldstone boson, which arises from the spontaneous breaking of the symmetry.

To answer your first question, choosing different minima to expand the Hamiltonian does have a physical meaning. This is because the choice of minima corresponds to different vacuum states of the system, and different vacuum states can have different physical properties. So, expanding the Hamiltonian around different minima can provide different insights into the system.

Your idea about the interaction terms only being relevant when the particles are interacting is also correct. In this case, the interaction terms appear when the fields are expanded around their non-zero vacuum expectation values. These terms are only relevant when the particles are interacting with each other, and at low energies, they can be neglected.

In regards to your second question, it is possible to consider the other Lagrangian with the fields \phi_1 and \phi_2 for low energies. However, the resulting effective Lagrangian will be different from the one obtained by expanding around the minimum of the potential. This is because the non-zero vacuum expectation values of the fields will introduce new terms in the effective Lagrangian. So, while both Lagrangians may describe the same particles at low energies, they will have different forms due to the different vacuum states.

In summary, your understanding of the broken symmetry and its consequences in this example is on the right track. It is important to keep in mind that different choices of minima can lead to different physical insights and that the effective Lagrangian will depend on the chosen vacuum state. I hope this helps to clarify your understanding of the concept of broken symmetry.