I am a little confused about symmetry breaking - in particular, what I see to be two different interpretations of it.

First, what I have seen taken to be the definition of a broken symmetry - we start with a Lagrangian with symmetry group $G$, but when we chose a vacuum state, this vacuum state has a different symmetry group $H$. That is, if $Q$ generates a broken symmetry (in $G$, not in $H$), then $Q|\phi_{vac}\rangle\neq0$.

So far, so good. However, I have also seen (lectures on The Standard Model by C. E. Thomas) the following example of a broken symmetry. Suppose we start with a Lagrangian, where $\phi=(\phi_1, ..., \phi_N)$ is a real scalar field, with global $O(N)$ symmetry:

$$L = \frac{1}{2}(\partial\phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4}\phi^4.$$

$m^2<0$, so $\phi$ is minimised for $v=-\frac{m^2}{\lambda}\geq0$.

Choosing the vacuum to be $\phi_0 = (\pi_1, ..., \pi_{N-1}, v + \sigma)$, the Lagrangian can be rewritten

$$L = \frac{1}{2}(\partial\pi)^2+\frac{1}{2}(\partial\sigma)^2 - \frac{1}{2}(\sqrt{2\lambda v})^2\sigma^2 - \lambda v (\sigma^2+\pi^2)\sigma -\frac{\lambda}{4}(\sigma^2+\pi^2)^2.$$

Under varying $\pi$, this Lagrangian has a $O(N-1)$ symmetry (although there must be a $O(N)$ symmetry hidden in there somewhere). At the bottom of the page here, it says that in this example, then, we have $G=O(N)$, and $H=O(N-1)$.

So how are these two concepts related? I don't see an obvious reason why the symmetry group of the second Lagrangian is the same as the symmetry group of the vacuum?

  • 2
    $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, page, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Sep 10, 2020 at 9:39
  • $\begingroup$ Monitor the symmetries in G but not H, which are realized nonlinearly in this coordinate system you chose. They still keep your action invariant. Do this. $\endgroup$ Commented Sep 10, 2020 at 10:53
  • $\begingroup$ @CosmasZachos so how can you see in the new coordinate system that $H=O(N-1)$? The text seems to imply this is an obvious fact. Is there any way to see this from the second Lagrangian, or is it just that $(0, ..., 0, v)$ has a $O(N-1)$ symmetry group? $\endgroup$
    – awsomeguy
    Commented Sep 10, 2020 at 12:44
  • $\begingroup$ Clearer now, or not ? $\endgroup$ Commented Nov 4, 2021 at 10:15

2 Answers 2


In the second example, you have expressed the Lagrangian in terms of some "shifted coordinates" with origin at the minimum of the potential. You can think of it like expanding the Lagrangian around the (classical) ground state.

So the Lagrangian in terms of $(\vec{\pi},\sigma)$ coordinates is manifestly $O(N-1)$-symmetric corresponding to rotations among the $\vec{\pi}$ components. This same rotation preserves the minimum of the potential, and hence, amounts to a rotation among the degenerate ground states. This is your $O(N-1)$ symmetry of the ground state from the first example.

  • $\begingroup$ @ValterMoretti How are the fundamentals of perturbative QFT related to the question being asked? I must be misunderstanding the question (or lacking understanding of some deep mathematical truth). Is not awesomeguy just asking about how the symmetry of the ground state is related to the symmetry corresponding to rotating amongst the space of Goldstone modes? I don't see a need to invoke unitary inequivalence to understand simple physics. $\endgroup$
    – user167506
    Commented Sep 11, 2020 at 12:34
  • $\begingroup$ I decided to remove my initial comment since I have had the impression that it is a source of confusion. $\endgroup$ Commented Sep 11, 2020 at 14:22

So how are these two concepts related? I don't see an obvious reason why the symmetry group of the second Lagrangian is the same as the symmetry group of the vacuum?

A change of variables is the most obvious way a Lagrangian stays the same although looks different in form; SSB means the Lagrangian's potential terms dictate a certain type of vacuum requiring a slightly different realization of the same symmetry; see below. The Lagrangian is the same object, written in two different ways, and so it has the same symmetries, O(N) in your example, of course.

Since these two forms have the same symmetries, by Noether's theorem they have the same conserved currents, N(N-1)/2 in number, in your case. I gather you want to know why the second set of variables is superior to the first, and what it is you may see better by using it: you see the realization of this symmetry better.

In short, since fields normally describe fluctuations/excitations around the vacuum, you may want to have fields with vanishing vacuum expectation values (recall your QFT course?). You rewrote $$\phi=(\phi_1, ..., \phi_N) = (\pi_1, ..., \pi_{N-1}, v + \sigma).$$

You already ascertained that it is impossible to have all N $\phi$s have vanishing vacuum values. You may use the O(N) symmetry to choose the v.e.v., v, reside in the last component; and the change of variables makes it easy for you to recognize that, in the shifted variables, all N-1 πs and the σ have vanishing v.e.v.s, so they are good excitation generators acting on the vacuum state, unlike the inferior $\phi$s. (Nothing wrong with keeping them, if you can keep these tallies in your head without undue bemusement.)

Now consider the N(N-1)/2 currents of the theory, presumably the very first exercise you did in that chapter, otherwise one wonders why you took that course.

The N(N-1)/2 symmetry infinitesimal rotations with N(N-1)/2 independent angles $\theta_{ij}=-\theta_{ji}$, $$ \delta \phi_i= \theta_{ij} \phi_j $$
correspond to the manifestly on-shell conserved (unnormalized) currents, $$ \phi_i\partial_\mu \phi_j - \phi_j\partial_\mu \phi_i . $$ But note then the integrals of their zero components, the charges, will not all annihilate the vacuum.

By contrast, the shifted language is manifestly superior. Defining for $i,j=1,2,...,N-1$, $$\theta_{ij}\equiv \vartheta_{ij}, ~~~~~~ \theta_{iN}\equiv \epsilon_i, $$ the symmetry transformations of the Lagrangian devolve to $$ \delta \pi_i= \vartheta_{ij} \pi_j ~~;~~~\delta \pi_i= \epsilon_i (v+\sigma), ~~\delta \sigma =-\epsilon_i \pi_i, $$
and their conserved currents to $$ \pi_i\partial_\mu \pi_j - \pi_j\partial_\mu \pi_i ~; \\ (v+\sigma)\partial_\mu \pi_i - \pi_ i\partial_\mu \sigma. $$

So the O(N-1) linear transformations with $\vartheta_{ij}$ (rotations, in H) before the semicolons correspond to the customary bilinear currents, and hence charges which annihilate the vacuum, and their action is dull, unbroken rotations of the π fields, (N-1)(N-2)/2 of them.

But the fun starts after the semicolons. These N-1 transformations (in G/H) are nonlinear: they are shifts by v, plus higher order stuff. They kind of rotate the σ into the πs and vice versa, but, crucially, they also shift the πs by a constant: this means the corresponding charge does not annihilate the vacuum, but, instead, pumps πs into and out of it! (This sort of current realization was discovered by Feynman "with some assistance by Gell-Mann" before 1960.) Some peculiar nontrivial vacuum this must be, then...

The term for this is SSB, or, more descriptively, nonlinear realization, or Nambu-Goldstone mode realization of the symmetry.

As you saw, its hallmark is $\langle \delta\pi_i\rangle =\epsilon_i v\neq 0$, even though $\langle \pi_i\rangle = \langle \sigma \rangle=0$. This would be harder, but not impossible, to see in the original $\phi$ notation, but why use them? You wish to focus on the coset space.


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