# Where do the Goldstone Boson degrees of freedom come from?

The punchline of Goldstone's theorem is well known. When a continuous symmetry breaks

necessarily, new massless (or light, if the symmetry is not exact) scalar particles appear in the spectrum of possible excitations. There is one scalar particle—called a Nambu–Goldstone boson—for each generator of the symmetry that is broken, i.e., that does not preserve the ground state. The Nambu–Goldstone mode is a long-wavelength fluctuation of the corresponding order parameter.

Where do the degrees of freedom of the Goldstone bosons come from?

• "Come from"? You mean you want a transition history of a combination of massive degrees of freedom before the SSB transition and a Morse explication of how they morph into a goldston after it? Simple toy examples won't suffice? – Cosmas Zachos Nov 16 '17 at 14:47
• @CosmasZachos a broad outline would suffice. Are the Goldstone bosons massive d.o.f. before the phase transition? E.g. in a simple field theory, what are the Goldstones before the symmetry breaking? – jak Nov 16 '17 at 15:09
• In your U(1) model of your wikilink, φ=ρ exp(i θ) , before and after SSB. After SSB, θ is promoted to goldston, and ρ is fixed/banished to infinitely massive, so only v, its v.e.v. survives. In the sigma model, with σ and three πs, the σ remains massive, and the three πs are promoted to goldstons. – Cosmas Zachos Nov 16 '17 at 15:20
• The number of d.o.f. is infinite before and after the continuous symmetry be broken. Actually, is the $N\rightarrow \infty$ that is responsible for the superselection rule behind the symmetry breaking mechanism. The goldstone boson is just an excitation around this superselected groundstate. Turns out that the excitations that goes towards another superselection sector are massless and are called Goldstone boson. – Nogueira Nov 16 '17 at 20:17
• – Cosmas Zachos Nov 19 '17 at 22:02

## 1 Answer

Well, Goldstone's 1961 sombrero potential model amply illustrates the basics. Let me vulgarize them.

In O(2) language, cavalierly about normalizations, he thinks of a complex scalar field theory, whose real and imaginary components resolve to a two real scalar d.o.f. system, with prototype potential $$\frac{\lambda}{4} (A^2 + B^2 +\epsilon v^2)^2 .$$ The O(2) isorotation symmetry sends A to $A\cos \theta- B \sin \theta$; and B to the orthogonal combination. The conserved current is $J_\mu = A\partial _\mu B - B\partial_\mu A$, i.e. $\partial \cdot J=0$, and $i\theta [Q, A]=\delta A ~,$ and likewise for B, so then $$\delta A =- \theta B, ~~~~\delta B=\theta A ~.$$

Slide the parameter ε from 1 (generic for non vanishing positive) ; to 0; to -1 (generic for non vanishing negative), and monitor the qualitative fate of the two fields along these three cases.

For ε=1, A and B are twins. They have the same mass, $\sqrt{\lambda} v$, as the minimum of the potential is at $\langle A\rangle =\langle B\rangle = 0$, and so $\langle \delta A\rangle= \langle \delta B\rangle= 0$, that is, $Q|0\rangle=0$, so the vacuum is invariant under isorotation, $e^{i\theta Q}|0\rangle=|0\rangle$. The potential looks like so: Decreasing ε to 0, the mass of A and B decreases to 0, but they remain twins, and their rotation into each other is still linear, and the vacuum is still symmetric—all the above relations (except for the vanished mass) are the same as above.

As soon as ε goes negative, something catastrophically, qualitatively different happens: SSB. Take ε to be -1 for simplicity. Now the quartic potential morphs into the iconic Goldstone sombrero, and the minimum is this entire flat circle on the A-B plane. The symmetry slides you around that degenerate bottom (orbit) without resistance. A choice is thus forced for the ground state: suppose you pick, arbitrarily, $\langle B\rangle=0$ and $\langle A\rangle =v$. Since you are interested in excitations around this vacuum, change to variables of convenience, $A\equiv v+h$, so h is the excitation around this vacuum with $\langle h\rangle=0$.

The quartic potential now expands to $$\frac{\lambda}{4} \left ( B^4+ h^4 +2h^2B^2 +4vh(B^2+h^2) +4v^2 h^2 \right).$$ There is no mass term for B, but a hefty mass for h (formerly known as A ; you could take this mass to infinity, if inclined, by sending λ there), similar to the mass it used to have before the SSB.

• The crucial part is $\langle B \rangle= \langle h \rangle= \langle \delta A \rangle=0$ , but $\langle \delta B \rangle= \theta v$ , non-vanishing : the hallmark of the Goldstone boson, since $\delta B= \theta v+ \theta h$ ; call v the order parameter. This shift nonlinearity in the goldston's transform precludes the existence of a mass term for it, as that term would not be invariant under the symmetry, still all-powerful, but a bit hidden (whom are we fooling? This is dubbed the Nambu-Goldstone realization).

The current $J_\mu = v\partial_\mu B+ h\partial _\mu B - B\partial_\mu h$ , is, of course, still conserved, but, check that now $Q|0\rangle\neq 0$ : the symmetry shifts the vacuum around the bottom of the sombrero, exciting sloshing Bs out of it—it taps the bowl of jello with a spoon. $|B(p=0)\rangle$ is degenerate with $|0\rangle$, as $\langle B(p)| J_\mu (x)|0\rangle\sim e^{ip\cdot x} v p_\mu$ .

Check that $[Q,H]=0$ , so $H (Q|0\rangle)= Q(H|0\rangle)= QE_0|0\rangle= E_0(Q|0\rangle)$ .

By contrast, oscillations of the massive h (the σ, or "Higgs") correspond to rolling up and down the walls of the valley of the sombrero, transversely to the valley axis.

• The takeaway: As ε slides from 1 to 0 to -1, the mass of the "higgs", A/h, decreases to 0 and then back to above its former value; by contrast, the mass of B decreases to 0, and stays there: but, suddenly, for ε<0, it morphs into a goldston.