Simplest model in field theory which leads to a pseudo-Goldstone boson

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.

• How about a scalar in a Mexican hat with a tiny $\phi^3$ term? – knzhou Apr 1 at 15:26
• 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 – SRS Apr 1 at 15:29
• @CosmasZachos Please see the edited question! – SRS Apr 1 at 15:35
• @SRS Yes, a complex scalar. The $U(1)$ symmetry is softly broken, as $\phi^3$ is a relevant operator. – knzhou Apr 1 at 15:42
• 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). – octonion Apr 1 at 22:13

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, $$\bbox[yellow,5px]{\epsilon= m_q \Lambda^3/\lambda f_\pi^4}$$.)

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.

• Not bad, but there are simpler ways to explain the symmetry breaking. +1 – Gareth Meredith Apr 2 at 2:31
• Should Is there a way to interpret the origin of the symmetry-breaking linear term in $\sigma$? @CosmasZachos – SRS Apr 3 at 7:48
• 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. – Cosmas Zachos Apr 3 at 11:56