# Goldstone theorem for classical and quantum potential

Consider a quantum theory $$\mathcal{L}(\phi^a) = \mathcal{L_{kin}}(\phi^a)-V(\phi^a),\tag{11.10}$$ depending on any type of fields $$\phi^a$$. The VEV of this theory are constant fields $$\phi_0^a$$ such that $$V(\phi^a)$$ is minimized and if some continuous symmetry of $$\mathcal{L}$$ is broken in the VEV we know that due to Goldstone theorem $$$$\left(\frac{\partial^2}{\partial \phi^a \partial \phi^b}V\right)_{\phi_0}$$\tag{11.11}$$ must have some zero eigenvalues. This is proved in chapter 11.1 of Peskin & Schroeder. What I don't understand is the claim that this is true only for classical theory or only at tree-level.

The quantum version of Goldstone theorem takes into account the effective potential $$V_{eff}$$ and the claim now is that for a continuous broken symmetry it's
$$$$\left(\frac{\partial^2}{\partial \phi^a \partial \phi^b}V_{eff}\right)_{\phi_0}$$\tag{p.388}$$ that has at least one zero eigenvalue. Where does the reasoning with the potential $$V$$ fail (i.e. why do we need to introduce the effective action $$\Gamma$$?) It seems to me that the proof of Goldstone theorem should work for both classical and quantum potential.

• The linear algebra is the same for both the classical and quantum cases. Your text illustrates the point for the simple classical case, and you are meant to apply the argument to the generic quantum case, summarized by the effective potential for it, $V_{eff}$, which might well have different minimum/symmetry properties than the classical one, so different Goldstonery... Commented Jun 13 at 16:33

OP asks a good question about the interplay between the classical and the quantum theory in Goldstone's theorem. In this answer we review Goldstone's theorem following Refs. 1 & 2.

1. The typical starting point is an (infinitesimal) zero-mode quasi-symmetry transformation $$\delta\phi^{\alpha}(x)~=~\epsilon c^{\alpha}{}_{\beta}\overline{\phi^{\beta}}, \qquad \delta\widetilde{\phi}^{\alpha}(k)~=~(2\pi)^d\delta^d(k)\epsilon c^{\alpha}{}_{\beta}\overline{\phi^{\beta}},\tag{1}$$ of the classical action $$S_V[\phi]~:=~\int_V\! d^dx~{\cal L}$$ (where we assume that the Lagrangian density $${\cal L}(\phi(x),\partial\phi(x))$$ has no explicit $$x$$-dependence)$$^1$$: \begin{align}0~\sim~\delta S_V[\phi]~\sim~& \int_V\! d^dx~\frac{\delta S[\phi]}{\delta\phi^{\alpha}(x)}\delta\phi^{\alpha}(x)\cr ~\stackrel{V=\mathbb{R}^d}{=}&~\int\! \frac{d^dk}{(2\pi)^d}~\frac{\delta S[\phi]}{\delta\widetilde{\phi}^{\alpha}(k)}\delta\widetilde{\phi}^{\alpha}(k).\end{align}\tag{2} Here we have introduced the spacetime-averaged quantity $$\overline{f(\phi)}~:=~\frac{1}{|V|}\int_V\! d^dx~f(\phi(x))\tag{3}$$ of a spacetime region $$V\subseteq\mathbb{R}^d$$.

2. Now the main point is that the 1PI quantum effective action$$^2$$ $$\Gamma_V[\phi_{\rm cl}]$$ inherits$$^3$$ the zero-mode quasi-symmetry (2) of the classical action \begin{align}0~\stackrel{(2)}{\sim}~\delta \Gamma_V[\phi_{\rm cl}]~\sim~& \int_V\! d^dx~\frac{\delta \Gamma_V[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^{\alpha}(x)}\delta\phi_{\rm cl}^{\alpha}(x)\cr ~\stackrel{V=\mathbb{R}^d}{=}&~\int\! \frac{d^dk}{(2\pi)^d}~\frac{\delta \Gamma_V[\phi_{\rm cl}]}{\delta\widetilde{\phi}_{\rm cl}^{\alpha}(k)}\delta\widetilde{\phi}_{\rm cl}^{\alpha}(k).\end{align}\tag{4}

3. Functional differentiation then yields$$^4$$ $$0~\stackrel{(4)}{=}~ \int_V\! d^dx~\frac{\delta^2 \Gamma_V[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^{\beta}(x^{\prime})\delta\phi_{\rm cl}^{\alpha}(x)}c^{\alpha}{}_{\gamma}\underbrace{\overline{\phi_{\rm cl}^{\gamma}}}_{=\overline{\langle\phi^{\gamma}\rangle_J}}+\underbrace{\frac{\delta \Gamma_V[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^{\alpha}(x^{\prime})}}_{=-J_{\alpha}(x^{\prime})\approx 0}c^{\alpha}{}_{\beta}\frac{1}{|V|} .\tag{5}$$ The Fourier transform reads: $$\frac{\delta^2 \Gamma_V[\phi_{\rm cl}]}{\delta\widetilde{\phi}_{\rm cl}^{\beta}(k^{\prime})\delta\widetilde{\phi}_{\rm cl}^{\alpha}(0)}\underbrace{\overline{c^{\alpha}{}_{\gamma}\langle\phi^{\gamma}\rangle_{J=0}}}_{\neq 0}~\stackrel{(5)}{\approx}~0.\tag{6}$$

4. Let us now rename one of the (possible several) Goldstone fields as $$\pi:=c^{\alpha}{}_{\beta}\phi^{\beta}$$ with non-zero VEV $$\overline{\langle \pi\rangle_{J=0}}~\neq~ 0.\tag{7}$$

5. It follows from eqs. (6) & (7) that the Hessian $$\left.\frac{\delta^2\Gamma[\phi_{\rm cl}]}{\delta \widetilde{\pi}_{\rm cl}(k_1)\delta\widetilde{\pi}_{\rm cl}(k_2)}\right|_{k_1=0=k_2}~\stackrel{(6)+(7)}{=}~0\tag{8}$$ has a zero, or its inverse$$^5$$, the Fourier-transformed connected propagator/2-point function $$\langle\widetilde{\pi}(k_1)\widetilde{\pi}(k_2)\rangle^c~\stackrel{(8)}{\propto}~\frac{1}{k_1^2}\delta^4(k_1\!+\!k_2)\tag{9}$$ has a pole, i.e. $$\pi$$ is massless, thereby proving the Goldstone theorem. $$\Box$$

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; Section 11.1 p. 351-352 + Section 11.6 p. 388.

2. S. Weinberg, Quantum Theory of Fields, Vol. 2, 1995; Section 19.2 p. 167-168.

3. S. Weinberg, Quantum Theory of Fields, Vol. 2, 1996; Section 16.4 p. 77 + Section 17.2 p. 84.

$$^1$$ Notation: The $$\sim$$ symbol means equality modulo boundary terms. The $$\approx$$ symbol means equality modulo eqs. of motion.

$$^2$$ At tree-level/zeroth order in $$\hbar$$, the quantum effective action $$\Gamma_V[\phi_{\rm cl}]$$ is the classical action $$S_V[\phi_{\rm cl}]$$, cf. e.g. eq. (12) in my Phys.SE answer here.

Much of the construction can equivalently be reformulated in terms of the quantum effective potential density $${\cal V}_{\rm eff}(\overline{\phi_{\rm cl}})~:=~-\frac{1}{|V|}\Gamma_V[\phi_{\rm cl}\!=\!\overline{\phi_{\rm cl}}]. \tag{10}$$

$$^3$$ More generally, the generating functional $$W_{c,V}[J]$$ of connected diagrams and the 1PI quantum effective action $$\Gamma[\phi_{\rm cl}]$$ inherits affine quasi-symmetries of the classical action $$S_V[\phi]$$ and the path integral measure, cf. e.g. Ref. 3. In the present case the sources transform as $$\delta J_{\alpha}(x)~=~-\epsilon \overline{J_{\beta}} c^{\beta}{}_{\alpha}, \qquad \delta\widetilde{J}_{\alpha}(k)~=~-(2\pi)^d\delta^d(k)\epsilon \overline{J_{\beta}} c^{\beta}{}_{\alpha},\tag{11}$$ and the classical fields as $$\delta\phi_{\rm cl}^{\alpha}(x)~=~\epsilon c^{\alpha}{}_{\beta}\overline{\phi_{\rm cl}^{\beta}}, \qquad \delta\widetilde{\phi}_{\rm cl}^{\alpha}(k)~=~(2\pi)^d\delta^d(k)\epsilon c^{\alpha}{}_{\beta}\overline{\phi_{\rm cl}^{\beta}}.\tag{12}$$

$$^4$$ Recall that $$\phi_{\rm cl}^{\alpha}=\langle \phi^{\alpha}\rangle_J$$, cf. e.g. eq. (4) in my Phys.SE answer here.

$$^5$$ See e.g. eq. (8) in my Phys.SE answer here.