Spontaneous symmetry breaking in a scalar field

Question

In Modern Particle Physics, Thomson explains the symmetry breaking for a scalar field in the following way:

1. Starting with the Lagrangian (density) $$L=\frac{1}{2}(\partial^\mu\phi)(\partial_\mu\phi)-\frac{1}{2}\mu^2\phi^2 - \frac{1}{4}\lambda \phi^4,$$ where the first term is the KE and the other two the potential energy (V($\phi$)).

2. The vacuum is given by the lowest level of $V(\phi).$ $\lambda$ must be positive for the potential to have finite minimum.

3. This can happen either for $\mu^2>0$ or $\mu^2<0.$ The first implies $V(\phi)$ minimum at $\phi=0$, but in the second case this happens when $$ \phi=\pm v=\pm\left| \sqrt{\frac{-\mu^2}{\lambda^{\prime}}} \right|.$$

4. He concludes by saying that "The choice of the vacuum state [$\pm v$] breaks the symmetry of the Lagrangian, a process known as spontaneous symmetry breaking."

Main question: This may seem obvious for those acquainted with the issue, but I can't see the relation between having two possible $\phi$s (again $\pm v$) and the breaking of the symmetry. What is such relation?

Side question: Why is this process called spontaneous?

Answer 1

Thre is a minimum at $\phi $ = 0, but this an unstable point, so a pertubative expansion around this point will not converge.

I hope that my interpretation of ACuriousMind's comment concurs with my statement above.

(although it does when you expand the field around a minimum)

In order for the field $\phi $ to be real, it must be true that $m^2$ < 0

The graph corresponding to $\phi $ = 0 is shown below.

Now look at the graphs for $\phi $ = $+v $ or $\phi$ = $-v $

A perturbative expansion around these points $\phi $ = $+v $ or $\phi$ = $-v $ will converge, due to the well they are trapped in.

When $\phi $ = 0, symmetry existed, but when $\phi $ = $+v $ or $\phi$ = $-v $, there is no longer just one ground state, but two, the mimina at the latter two points. Choosing one of these points will break the symmetry.

A mass term in the Langranian is one that is quadratic in the fields, which is a term of the form $\alpha^2\phi^2$ for some $\alpha $.

So the Langranian is no longer invariant under $\phi $ goes to $-\phi $, but although this symmetry is lost, we can get some knowledge about the true mass of the particle associated with the field $\phi $.

Explicit symmetry breaking differs from spontaneous symmetry breaking. In the latter case, the defining equations respect the symmetry but the ground state (vacuum) of the theory breaks it, as illustrated above. Spontaneous symmetry breaking is a mode of realization of symmetry breaking in a physical system, where the underlying laws are invariant under asymmetry transformation, but the system as a whole changes under such transformations, in contrast to explicit symmetry breaking.

It is a spontaneous process by which a system in a symmetrical state ends up in an asymmetrical state. It thus describes systems where the equations of motion or the Lagrangian obey certain symmetries, but the lowest-energy solutions do not exhibit that symmetry.

Consider a symmetrical upward dome with a trough circling the bottom. If a ball is put at the very peak of the dome, the system is symmetrical with respect to a rotation around the center axis. But the ball may spontaneously break this symmetry by rolling down the dome into the trough, a point of lowest energy. Afterward, the ball has come to a rest at some fixed point on the perimeter. The dome and the ball retain their individual symmetry, but the system does not.

For more on this, see Spontaneous Symmetry Breaking, which is the source of the above extracts.

Your original Langrangian:

$L=\frac{1}{2}(\partial^\mu\phi)(\partial_\mu\phi)-\frac{1}{2}\mu^2\phi^2 - \frac{1}{4}\lambda \phi^4$,

Rescale the field to give $\phi(x) = v + \eta (x)$

the $\eta (x)$ represents field fluctuations around $v$

The original Langrangian now has to be rewritten using this scale term, and remember that $v$ is just a number.

$L=\partial_\mu\phi (x) = \partial_\mu[v + \eta (x)] = \partial_\mu\eta (x)$

Square $\phi(x) = v + \eta (x)$ to give $\phi^2 = v+2v\eta + \eta^2$

Then this in turn leads to $\phi^4 = (v +\eta)^4 =v^4 +4v^3\eta + 6v^2\eta^2 +4v\eta^3 + \eta^4$

Now subsitute the $\phi^2$ and $\phi^4$ potential terms into the original Langrangian.

Without working through every step, this eventually leads to a new Langrangian, which I hope you can find in your book (probably with different indices), but the same form.

$L = \frac{1}{2}(\partial_\mu\eta)^2 -\lambda v^2\eta^2 -\lambda v\eta^3-\frac{1}{4}\lambda\eta^4$

Again noting that terms that are quadratic in the fields are associated with mass terms $m = \sqrt{2\lambda v^2} = \sqrt{2\lambda} v$. This can be seen by rewriting $\lambda v^2\eta^2 = \frac{1}{2}\times 2\lambda v^2\eta^2$, and comparing to $\frac{1}{2}m^2 \eta^2$.

The first term is the kinetic energy, the second is the mass term (now incorporating $\eta$), the the third term is the self interaction (three legs in a Feynman diagram) and the last term is the self interaction (four legs).

The point of all this is to show that SSB leads you to an expression for mass and also, (but not in this particular case) you may expand the Langrangian to ensure there are no other hidden quadratic terms associated with mass.