Let's say we are considering the case of Abelian symmetry, with the Lagrange density given by:

$$\mathcal{L}=\frac{1}{2}(\partial_\lambda \sigma)^2 + \frac{1}{2} (\partial_\lambda \pi)^2-V(\sigma^2 +\pi^2),$$

where $$V(\sigma^2+\pi^2)=-\frac{\mu^2}{2}(\sigma^2 +\pi^2)+\frac{\lambda}{4}(\sigma^2 +\pi^2 )^2. $$

The Lagrangian has $U(1)\cong O(2)$ symmetry. By minimizing the potential we get the minimum of the potential given by (for $\mu^2>0)$:

$$\sigma^2+\pi^2 =v^2,\qquad v=\sqrt{\frac{\mu^2}{\lambda}}.$$

In my literature it then says:

The minima consist of points on a circle with radius $v$ in the ($\sigma$,$\pi$) plane. They are related to each other through $O(2)$ rotations. Hence they are all equivalent and there are an infinite number of degenerate vacua. Any point on this circle may be chosen as the true vacuum. We can pick, for example, $$\langle 0|\sigma|0\rangle =v,\qquad\langle 0|\pi|0\rangle =0.$$

Then small oscillations around the true minimum are considered and a shifted field is defined: $\sigma'=\sigma-v$, which then leads to massless particles, those are the result of vacuum symmetry-non-invariance.

My questions:

1. Why do we get only one Goldstone particle? Why can't we take some other two points on the mentioned circle, for example,

$$ \langle 0|\sigma|0\rangle =\alpha, \qquad\langle 0|\pi|0\rangle =\beta$$

so that $\alpha^2+\beta^2=v^2$. Then we could define: $\sigma'=\sigma-\alpha$ and $\pi'=\pi-\beta$ and get two different types of Goldstone, massless particles? Or more in general, we know that: #broken symmetry operators (charges) = #Goldstone bosons, why is that so?

2. How are oscillations around the minimum connected to particle creation?

Remark: I know that the result of massless particles can be obtained more formally using the Goldstone theorem which leads to the same result, i.e. we only get one massless particle.

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    $\begingroup$ Hint: take a linear combination of your two coordinates with a vanishing vev and note the vev of the orthogonal one. Observe their respective masses and non/linearity of transformations. What hits you? $\endgroup$ Commented Sep 21, 2019 at 22:18
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    $\begingroup$ 2. Basic second quantization: inspect your resulting lagrangian in variables you recognize! $\endgroup$ Commented Sep 21, 2019 at 22:38
  • $\begingroup$ @CosmasZachos About the first part... If I take a linear combination $v_1$ of the "0 vev" fields ($\sigma',\pi'$) it would have the same vev expectation value, as would an orthogonal combination $v_2$. This means that the fields $v_1$ and $v_2$ are an equivalent choice to $\sigma'=\sigma-\alpha$ and $\pi'=\pi-\beta$. If I plug the $v_1$ and $v_2$ fields I should get (again) one massless particle? And that no matter which combination of the original fields I choose, I will allways get the same result? $\endgroup$
    – Luka8281
    Commented Sep 22, 2019 at 10:46
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    $\begingroup$ You got it. A rotational change of field coordinates cannot alter the physics. It should have been part of your symmetry breaking intro. $\endgroup$ Commented Sep 22, 2019 at 20:34
  • $\begingroup$ @CosmasZachos Thank you! $\endgroup$
    – Luka8281
    Commented Sep 23, 2019 at 7:59

1 Answer 1



The Goldstone theorem is the short answer.

The number of Goldstone bosons is equal to the number of broken generators of the continuous symmetry.
Your initial symmetry is $U(1)$, which has one generator. Having broken that, since your post-SSB Lagrangian depends on real fields only, you can expect at most one Goldstone mode.

The Goldstone mode is the gapless excitation, i.e. the one that may also cost no energy at all. This is why they only come out of a continuous symmetry, and not a discrete one - for instance, in a $\mathbb{Z}_2$ symmetric system you'd need to spend a minimum energy $E$ to flip the particles from one orientation to the other, etc.

This Goldstone mode corresponds to an oscillation along the ring of minima in your potential (green, below, reference):

enter image description here

The other one is gapped, i.e. it costs energy to excite that oscillation. If you approximate the cross-section of the potential as a parabola, you'd need a minimum of $\hbar \omega$ to excite the system.

We'll see later that the gapped (Higgs) mode is the one that is massive, whereas the Goldstone mode is the massless one.


As suggested in the comments, and done in many resources online, the gist is as follows.

Your new minimum, post-SSB, is $(0,v)$ (say). You then write your original field $\phi = (\pi, \sigma)$ as $(\pi, v+\sigma')$, i.e. you expand the field about the new minimum.

Plug this back into the Lagrangian density $\mathcal{L}$ and you get something like:

$$ \mathcal{L} = \frac{1}{2} (\partial_\mu \pi)^2 + \frac{1}{2}(\partial_\mu \sigma')^2 - \mu^2\sigma'^2. $$

You see that $\sigma'$ has both kinetic and mass terms, and corresponds to the Higgs/massive mode (with the correct sign on the mass term, as opposed to the pre-SSB $\mathcal{L}$). On the other hand, $\pi$ is the massless Goldstone mode.

You can finally quantise your theory, write $\pi$ in terms of $a$ and $a^\dagger$.

  • $\begingroup$ "The number of Goldstone bosons is equal to the number of broken generators of the continuous symmetry" can you please elaborate on that. Why is that so in general? $\endgroup$
    – Luka8281
    Commented Sep 22, 2019 at 10:48
  • $\begingroup$ It’s the Goldstone theorem, there are proofs online. Each generator produces a continuous (not gapped) symmetry. When you break it, the system looses that symmetry but gains a gap less excitation instead. Degrees of freedom stay the same. $\endgroup$ Commented Sep 22, 2019 at 15:48

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