What can be a simple (if not simplest) continuum field theory model that gives rise to a pseudo Goldstone boson (doesn't matter if it is a toy model)? For example, I would be very happy if one can show that it it is possible to arrive at the notion of pseudo Goldstone bosons in an approximate global $U(1)$ invariant theory of a complex scalar field.

Please note I know nontrivial physical examples such as QCD axion. But I am looking for an even simpler model to introduce the idea to someone who has basic understanding of $\rm U(1)$ global symmetry breaking.

  • $\begingroup$ How about a scalar in a Mexican hat with a tiny $\phi^3$ term? $\endgroup$
    – knzhou
    Commented Apr 1, 2019 at 15:26
  • $\begingroup$ By including $\phi^3$ term, you mean a real scalar field in which the discrete $Z_2$ symmetry is broken? But that is a discrete symmetry: no Goldstone. If you had $|\phi|^2\phi$ in mind then it is indeed not $U(1)$ invariant. But pseudo Goldstone require the symmetry to be softly broken. Right? Do you mean one should set the coupling of $|\phi|^2\phi$ term to be small? @knzhou $\endgroup$
    – SRS
    Commented Apr 1, 2019 at 15:29
  • $\begingroup$ @CosmasZachos Please see the edited question! $\endgroup$
    – SRS
    Commented Apr 1, 2019 at 15:35
  • $\begingroup$ @SRS Yes, a complex scalar. The $U(1)$ symmetry is softly broken, as $\phi^3$ is a relevant operator. $\endgroup$
    – knzhou
    Commented Apr 1, 2019 at 15:42
  • 2
    $\begingroup$ Here's probably the simplest: A free massless scalar field has a shift symmetry, and the conserved current is given by its conjugate momentum density. The scalar field itself can be thought of as a goldstone boson of this symmetry. If you give the scalar field a mass, it is like a pseudo-goldstone boson. It sounds trivial but you can write formulas that look a lot like those for partially conserved axial current (in this case it is the conjugate momentum current). $\endgroup$
    – octonion
    Commented Apr 1, 2019 at 22:13

1 Answer 1


Pseudo Goldstone bosons are exceptionally light scalars in systems with both spontaneous and explicit symmetry breaking, the latter serving as a small perturbation to the former.

I'll vulgarize the standard SU(2) σ-model of pions, the original pseudogoldstons, serving as the prototype of all that followed, the serious student should find, in, e.g., Itzykson & Zuber, Section 11-4.

To this end, I'll just keep one σ and one π, rotating into each other in a plain O(2), the real field presentation of U(1), and skip kinetic and Yukawa terms to focus on the celebrated Goldstone sombrero potential (1961). In suitable over-all normalizations, $$ V_0= \frac{\lambda}{4} (\sigma^2 + \pi^2 -v^2)^2, $$
which is manifestly invariant under field plane rotations, $$ \delta \pi = \sigma, \qquad \delta \sigma = -\pi~. $$ The standard SSB choice dictates that, at the minimum, $$ \langle \sigma\rangle=v ~, \qquad \langle \pi\rangle=0~, ~\Longrightarrow \langle \delta \pi\rangle=v, $$ identifying the π as a goldston. Further defining $\sigma'\equiv \sigma -v$, so that $\langle \sigma'\rangle=0$, it is evident the potential expands to interaction terms, no π mass term and a σ' mass term with $m_{\sigma'}^2= \lambda v^2$, so the σ' is a dull conventional field. This is plain SSB.

Suppose now the gods of current quark masses break the symmetry explicitly by a small amount proportional to the dimensionless parameter $\epsilon$, a small perturbation whose $O(\epsilon^2)$ expressions we'll systematically discard. That is, they tilt the sombrero slightly in the σ direction, favoring the original vacuum, $$ V_\epsilon= \frac{\lambda}{4} (\sigma^2 + \pi^2 -v^2)^2 -\bbox[yellow,5px]{\epsilon \lambda v^3 \sigma }~. $$ It is evident the above rotation transformation does not leave the perturbation invariant, so the original O(2) symmetry is explicitly broken to $O(\epsilon)$ in the potential, on top of the huge SSB.

To lowest order in $\epsilon$, the minimum of the potential is at $\langle \pi\rangle = O(\epsilon)$, with mostly ignorable effects, and $$ \langle \sigma\rangle (\langle \sigma ^2\rangle -v^2) -\epsilon v^3=0 ~~~\Longrightarrow ~~~ \langle \sigma\rangle = v(1+\bbox[yellow,5px]{\epsilon /2}). $$ Again, defining $\sigma ''\equiv \sigma -v(1+\epsilon /2)$ and not bothering with shifting π, since such a shift would not result in order $O(\epsilon)$ consequences to it (check!), we have, up to a constant, $$ V_\epsilon = \frac{\lambda}{4} \left (\left (\sigma'' + v (1+\epsilon/2) \right )^2 + \pi^2 -v^2\right )^2 -\epsilon \lambda v^3 \sigma'' \\ \approx \frac{\lambda}{4} (\sigma'' ^2 + \sigma '' v (2+\epsilon) +\epsilon v^2 + \pi^2)^2 -\epsilon \lambda v^3 \sigma '', $$ which now displays a mass for the approximate goldston, $$ m_\pi^2= \bbox[yellow,5px]{\epsilon\lambda v^2 /2 }, $$ while shifting the mass of the σ" to $m^2_{\sigma''}=\lambda v^2 (1+3\epsilon /2)$.

  • The cognoscenti and lattice consumers might discern the flotsam of Dashen's formula here, $\bbox[yellow,5px]{\epsilon= m_q \Lambda^3/\lambda f_\pi^4}$ : the square of the pseudogoldston mass goes as the quark mass, not its square!

Note there is no linear term in σ" left over (no tadpoles).

Tilting the sombrero has given the Goldstone mode a minuscule σ" component, $\langle \delta \pi \rangle=v(1+\epsilon/2)$ and now $\langle \delta \sigma'' \rangle=-\langle \pi\rangle =O(\epsilon) $.

The symmetry current of O(2), $J_\mu= \sigma \partial_\mu \pi - \pi \partial_\mu \sigma$ is now evidently only Partially Conserved, on-shell, $$ \partial \cdot J=\bbox[yellow,5px]{ \epsilon \lambda v^2 ~\pi }~. $$

This is essentially the linchpin formula invented by Feynman and understood by Gell-Mann as PCAC, eqn (5) in the game-changing Gell-Mann, M., & Lévy, M. (1960), "The axial vector current in beta decay", Nuov Cim (1955-1965), 16(4), 705-726.

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    $\begingroup$ Not bad, but there are simpler ways to explain the symmetry breaking. +1 $\endgroup$ Commented Apr 2, 2019 at 2:31
  • $\begingroup$ Should Is there a way to interpret the origin of the symmetry-breaking linear term in $\sigma$? @CosmasZachos $\endgroup$
    – SRS
    Commented Apr 3, 2019 at 7:48
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    $\begingroup$ Yes, in chiral hadron dynamics, the linear term summarizes the quark mass terms explicitly breaking chiral symmetry, with $\epsilon\propto m_q$, as above, so the square of pion masses is proportional to the quark masses. The uniform tilting of the sombrero makes a marble roll to the unique true vacuum by itself--it is not metastable. The science of achieving this is Dashen's celebrated "vacuum alignment formula", of use in technicolor modeling. $\endgroup$ Commented Apr 3, 2019 at 11:56

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