In Refs. 1 & 2 the Goldstone theorem is proven with a rather short proof which I paraphrase as follows.

Proof: Let $Q$ be a generator of the symmetry. Then $[H, Q] = 0$ and we want to consider the case in which $Q | 0 \rangle \neq 0$. As a consequence of the null commutator the state $Q | 0 \rangle$ has 0 energy. We know that $Q = \int d^{D} x ~J^{0} ( \vec{x}, t )$. Then we consider the state $| s \rangle = \int d^{D} x ~e^{- i \vec{k} \vec{x}} J^{0} ( \vec{x}, t )| 0 \rangle$ which has spatial momentum $\vec{k}$. In the zero momentum limit this state goes to $Q |0 \rangle$ which we know has 0 energy. We thus conclude that $| s \rangle$ describe a massless scalar particle with momentum $\vec{k}$. $\Box$

The problem with this proof is that the operator $Q$ is not well-defined because of the Fabri-Picasso theorem. So $Q |0 \rangle$ is not even a state of the Hilbert space. Is it possible to fix this proof so that it becomes rigorous maybe through the use of some regularization of the charge?

I must say I'm not asking for alternative rigorous derivation of the theorem such as the original one or something that exploits the effective action. I'm asking to provide a rigorous proof along the line of the Zee one.


  1. M.D. Schwartz, QFT & the standard model, 2014, section 28.2, p.563-64.

  2. A. Zee, QFT in a nutshell, 2010, p. 228.


In this answer we give a proof of Goldstone's theorem at the physics level of rigor following Ref. 1:

  1. We are given a spacetime-translation-covariant 4-current $$\hat{J}^{\mu}(x)~=~e^{i(\hat{H}t-\hat{\bf P}\cdot {\bf x})} \hat{J}^{\mu}(0)e^{i(\hat{\bf P}\cdot {\bf x}-\hat{H}t)} \tag{1}$$ that satisfies the continuity equation
    $$d_{\mu}\hat{J}^{\mu}(x)~=~0. \tag{2}$$ It is furthermore assumed that the vacuum state $|\Omega \rangle$ is spacetime-translation-invariant.

  2. In order to avoid the fallacy of the Fabri–Picasso theorem, let us introduce a bounded spatial integration region $V \subseteq \mathbb{R}^3$. Define a volume-regularized charge operator $$\hat{Q}_V(t)~:=~\int_V\! d^3{\bf x}~\hat{J}^0(x), \qquad V~\subseteq ~\mathbb{R}^3. \tag{3}$$

  3. The assumption of spontaneous symmetry breaking (SSB) is implemented via the existence of an observable $\hat{A}$ such that $${\rm Im}a_V(t)~\stackrel{(5)}{=}~\frac{1}{2i}\langle \Omega | [\hat{Q}_V(t),\hat{A}]|\Omega \rangle\quad \longrightarrow \quad a~\neq~0\quad\text{for}\quad V~\to ~\mathbb{R}^3. \tag{4}$$ On the rhs. of eq. (4) we have defined $$\begin{align} a_V(t)~&:=~\langle \Omega | \hat{Q}_V(t)\hat{A}|\Omega \rangle\tag{5} \cr ~&\stackrel{(3)}{=}~\int_V\! d^3{\bf x}~\langle \Omega | \hat{J}^0(x) \hat{A} |\Omega \rangle\tag{6} \cr ~&=~\int_V\! d^3{\bf x}~\sum_n\langle \Omega | \hat{J}^0(x)|n \rangle\langle n |\hat{A}|\Omega \rangle \tag{7} \cr ~&\stackrel{(1)}{=}~\int_V\! d^3{\bf x}~\sum_n e^{i( {\bf P}_n\cdot {\bf x}-E_nt)}c_n, \qquad c_n~:=~\langle\Omega | \hat{J}^0(0)|n \rangle\langle n |\hat{A}|\Omega \rangle, \tag{8}\cr ~& \longrightarrow \sum_n (2\pi)^3 \delta^3({\bf P}_n) e^{-iE_n t}c_n \tag{9}\cr ~&\stackrel{(11)}{=}~\sum_E e^{-iE t} f(E) \quad\text{for}\quad V~\to ~\mathbb{R}^3,\tag{10} \end{align}$$ where $$ f(E)~:=~\sum_n^{E_n=E} (2\pi)^3 \delta^3({\bf P}_n) c_n,\tag{11}$$ and where the state $|n \rangle$ is assumed to have definite 4-momentum $(E_n,{\bf P}_n)$.

  4. On one hand, $$\begin{align} d_t a_V(t) ~&\stackrel{(6)}{=}~\int_V\! d^3{\bf x}~\langle \Omega | d_0\hat{J}^0(x) \hat{A} |\Omega \rangle ~\stackrel{(2)}{=}~-\int_V\! d^3{\bf x}~\langle \Omega | {\bf \nabla} \cdot \hat{\bf J}(x) \hat{A} |\Omega \rangle \cr ~&=~-\int_{\partial V}\! d^2{\bf x}~\langle \Omega | {\bf n} \cdot \hat{\bf J}(x) \hat{A} |\Omega \rangle,\tag{12}\end{align}$$ so that $$d_t {\rm Im}a_V(t)~\stackrel{(12)}{=}~-\frac{1}{2i}\int_{\partial V}\! d^2{\bf x}~\langle \Omega | [{\bf n} \cdot \hat{\bf J}(x) ,\hat{A}] |\Omega \rangle \quad \longrightarrow \quad 0 \quad\text{for}\quad V~\to ~\mathbb{R}^3,\tag{13}$$ because we assume that the observable $\hat{A}$ has a compact spatial support, and commutes with spatially separated(=causally disconnected) operators.

    On the other hand, $$ d_t a_V(t)~~\stackrel{(10)}{\longrightarrow}~~ -i \sum_E Ee^{-iE t} f(E)\quad\text{for}\quad V~\to ~\mathbb{R}^3,\tag{14} $$ so that $$ d_t {\rm Im}a_V(t)~~\stackrel{(14)}{\longrightarrow}~~ \sum_E E\left\{ \sin(Et) {\rm Im} f(E)-\cos(Et) {\rm Re} f(E)\right\}\quad\text{for}\quad V~\to ~\mathbb{R}^3.\tag{15} $$

    By comparing eqs. (13) & (15) we conclude that $$f(E)~~\stackrel{(13)+(15)}{\propto}~~ \delta_{E,0}.\tag{16}$$

  5. Finally $$0~\neq~ a~ \stackrel{(4)}{\longleftarrow} ~ {\rm Im} a_V(t) ~ \stackrel{(10)}{\longrightarrow}~ {\rm Im}\sum_E e^{-iE t} f(E)~\stackrel{(16)}{=}~{\rm Im}f(E\!=\!0) \quad\text{for}\quad V~\to ~\mathbb{R}^3. \tag{17} $$ In order to have SSB, we must have $f(E\!=\!0)\neq 0$, i.e. there exists a massless mode $|n \rangle$ with $(E_n,{\bf P}_n)=(0,{\bf 0})$ that couples $c_n\neq 0$ between the current $\hat{J}^0$ and the observable $\hat{A}$. $\Box$

See also this related Phys.SE post.


  1. C. Itzykson & J.B. Zuber, QFT, 1985, Section 11-2-2, p. 520.
  • 1
    $\begingroup$ Cartoon version of proof: $\quad |{\bf 0}\rangle ~:=~\hat{Q}|\Omega\rangle~\neq ~0.$$\quad \hat{H}|\Omega\rangle~=~ 0.$ $\quad [\hat{H},\hat{Q}]~=~ 0.$ $\quad \hat{H}|{\bf 0}\rangle~=~\hat{H}\hat{Q}|\Omega\rangle ~=~\hat{Q}\hat{H}|\Omega\rangle~=~ 0.$ $\quad \hat{Q}~:=~\int \! d^3{\bf x}~\hat{J}^0(x).$ $\quad |{\bf k}\rangle ~:=~\int \! d^3{\bf x} ~e^{-i{\bf k}\cdot{\bf x}}\hat{J}^0(x)|\Omega\rangle.$ $\quad |{\bf 0}\rangle~=~|{\bf k}\!=\!{\bf 0}\rangle.$ $\quad \hat{H}|{\bf k}\rangle ~=~ \sqrt{{\bf k}^2+m^2}|{\bf k}\rangle.$ $\quad \Rightarrow \quad m~=~0.$ $\endgroup$ – Qmechanic Nov 5 '18 at 11:49

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