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?

  • $\begingroup$ Try applying the symmetry to one of the $\phi$s. What happens? $\endgroup$
    – ACuriousMind
    Commented Aug 21, 2016 at 0:54
  • $\begingroup$ @ACuriousMind 'Cause I understand it as the breaking of the symmetry of the Lagrangian. Since $V(\phi)$ is the same for either $\phi,$ L doesn't depend on which one we choose. $\endgroup$
    – Patrick
    Commented Aug 21, 2016 at 1:00
  • 4
    $\begingroup$ Oh, you're confused by the wording: It's not that the Lagrangian suddenly becomes asymmetric (although it does when you expand the field around a minimum), it's that those solutions are not invariant "under the symmetry of the Lagrangian". $\endgroup$
    – ACuriousMind
    Commented Aug 21, 2016 at 1:06
  • $\begingroup$ It is perhaps interesting noticing that however the set of the two solutions is still invariant under the symmetry of the action...In this sense the symmetry is restored at a weaker classical level. However if quantizing the system, I think that the symmetry is not restored. I mean the two quantum theories relying upon the two different vacua are not unitarily equivalent. $\endgroup$ Commented Aug 21, 2016 at 9:41

1 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.

enter image description here

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

enter image description here

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.

  • $\begingroup$ The dome + ball system is very good. It helps visualize the breaking of the symmetry, but in the Lagrangian where does it break? If I do $\phi \rightarrow -\phi$, then I get $V(\phi) \rightarrow V(\phi)$ due to the even powers of $\phi$. So is the Kinetic term the one where symmetry is broken? $\endgroup$
    – Patrick
    Commented Aug 21, 2016 at 2:48
  • 1
    $\begingroup$ You are looking for an expression for the mass in terms of $\lambda $ and $v $, this is what, in this case anyway, the SSB allows you to find. I am not familiar with your text, but I would expect in a few pages for $\phi (x)$ to be replaced by a small rescaling factor $\phi (x)$ = $v + \eta (x) $ and this will transform the langranian. Just read on a few pages and see does this makes sense to you. You end up with a revised langranian with a k.e term, a mass term, and two self interaction terms. Sorry, it's easier to read on in your book than write it out, it's a bit of a slog. $\endgroup$
    – user108787
    Commented Aug 21, 2016 at 3:08
  • 1
    $\begingroup$ I have done a little more work on my answer, in the hope you will recognize the new form of the Langrangian in your textbook. I have skipped the intermediate steps in the transformation. The new part starts from Your original Langrangian: onwards. Best of luck with it. $\endgroup$
    – user108787
    Commented Aug 21, 2016 at 10:07
  • $\begingroup$ I haven't totally understood, but I'll keep your answer as a reference. It has already helped. $\endgroup$
    – Patrick
    Commented Aug 21, 2016 at 15:37
  • $\begingroup$ TBH, I self study, so attempting an answer, which I am 90 percent is correct, is a way for me to learn, ( I even had to learn mathjax this morning) so it was as much selfishness on my part. The book I used was QFT Demystified, by McMahon and I would say I follow it, until a question like yours appears :) I recommend the book though. If I get any comments questioning it's validity, I will let you know. There's a hell of a lot of new concepts to learn at the same time. Drink lots of coffee. Regards $\endgroup$
    – user108787
    Commented Aug 21, 2016 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.