# Schwartz's and Zee's proof of Goldstone theorem

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.

References:

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$$^1$$ of Goldstone's theorem at the physics level of rigor following Ref. 1:

1. We are given a self-adjoint 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 a self-adjoint observable $$\hat{A}$$ such that \begin{align} {\rm Im}a_V(t)~\stackrel{(7)}{=}~&\frac{1}{2i}\langle \Omega | [\hat{Q}_V(t),\hat{A}]|\Omega \rangle\cr \quad \longrightarrow& \quad a~\neq~0\quad\text{for}\quad V~\to ~\mathbb{R}^3. \end{align}\tag{4}

4. We may assume w.l.o.g. that $$\langle \Omega |\hat{A}|\Omega \rangle~=~0 \tag{5}$$ in eq. (4) by performing the redefinition $$\hat{A}~ \longrightarrow~\hat{A}^{\prime}~:=~\hat{A}-|\Omega \rangle \langle \Omega |\hat{A}|\Omega \rangle \langle \Omega | \tag{6}$$ (and afterwards remove prime from the notation).

5. 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{7} \cr ~\stackrel{(3)}{=}~&\int_V\! d^3{\bf x}~\langle \Omega | \hat{J}^0(x) \hat{A} |\Omega \rangle\tag{8} \cr ~=~&\int_V\! d^3{\bf x}~\sum_n\langle \Omega | \hat{J}^0(x)|n \rangle\langle n |\hat{A}|\Omega \rangle \tag{9} \cr \stackrel{(1)+(5)}{=}&\int_V\! d^3{\bf x}~\sum_{n\neq\Omega} e^{i( {\bf P}_n\cdot {\bf x}-E_nt)}c_n, \cr &\quad c_n~:=~\langle\Omega | \hat{J}^0(0)|n \rangle\langle n |\hat{A}|\Omega \rangle, \tag{10}\cr\cr ~ \longrightarrow& \sum_n (2\pi)^3 \delta^3({\bf P}_n) e^{-iE_n t}c_n \tag{11}\cr ~\stackrel{(13)}{=}~&\sum_E e^{-iE t} f(E)\tag{12}\cr &\quad\text{for}\quad V~\to ~\mathbb{R}^3, \end{align} where $$f(E)~:=~\sum_{n\neq\Omega}^{E_n=E} (2\pi)^3 \delta^3({\bf P}_n) c_n,\tag{13}$$ and where $$|n \rangle$$ are a complete set of states with definite 4-momentum $$(E_n,{\bf P}_n)$$.

6. On one hand, \begin{align} d_t a_V(t) ~\stackrel{(8)}{=}~&\int_V\! d^3{\bf x}~\langle \Omega | d_0\hat{J}^0(x) \hat{A} |\Omega \rangle \cr ~\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{14}\end{align} so that \begin{align} d_t {\rm Im}a_V(t)&\cr ~\stackrel{(14)}{=}~&-\frac{1}{2i}\int_{\partial V}\! d^2{\bf x}~\langle \Omega | [{\bf n} \cdot \hat{\bf J}(x) ,\hat{A}] |\Omega \rangle \cr \quad \longrightarrow& \quad 0 \quad\text{for}\quad V~\to ~\mathbb{R}^3,\tag{15}\end{align} 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, \begin{align} d_t a_V(t)~~\stackrel{(12)}{\longrightarrow}~~& -i \sum_E Ee^{-iE t} f(E)\cr &\quad\text{for}\quad V~\to ~\mathbb{R}^3,\end{align}\tag{16} so that \begin{align}d_t {\rm Im}a_V(t)&\cr ~~\stackrel{(16)}{\longrightarrow}~~& -\sum_E E\left\{\cos(Et) {\rm Re} f(E) +\sin(Et) {\rm Im} f(E)\right\}\cr &\quad\text{for}\quad V~\to ~\mathbb{R}^3.\end{align}\tag{17}

By comparing eqs. (15) & (17) we conclude that $$f(E)~~\stackrel{(15)+(17)}{\propto}~~ \delta_{E,0}.\tag{18}$$

7. Finally \begin{align}0~\neq~ a~ \stackrel{(4)}{\longleftarrow} ~ {\rm Im} a_V(t) ~ \stackrel{(12)}{\longrightarrow}&~ {\rm Im}\sum_E e^{-iE t} f(E)\cr ~\stackrel{(18)}{=}~&{\rm Im}f(E\!=\!0)\cr &\text{for}\quad V~\to ~\mathbb{R}^3.\end{align} \tag{19} In order to have SSB, we must have $$f(E\!=\!0)\neq 0$$, i.e. there exists a massless mode $$|n \rangle\neq|\Omega \rangle$$ with $$(E_n,{\bf P}_n)=(0,{\bf 0})$$ that couples $$c_n\neq 0$$ between the current $$\hat{J}^0$$ and the observable $$\hat{A}$$. $$\Box$$

References:

1. C. Itzykson & J.B. Zuber, QFT, 1985, Section 11-2-2, p. 520.

2. S. Weinberg, Quantum Theory of Fields, Vol. 2, 1995; Section 19.2.

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$$^1$$ Cartoon version of the proof of Goldstone's theorem (ignoring the Fabri–Picasso theorem):

• $$\quad |{\bf 0}\rangle ~:=~\hat{Q}|\Omega\rangle~\neq ~0.$$ $$\quad\hat{H}|\Omega\rangle~=~ 0.$$ $$\quad [\hat{H},\hat{Q}]~=~ 0.$$

• $$\quad \hat{H}|{\bf 0}\rangle~=~\hat{H}\hat{Q}|\Omega\rangle ~=~\hat{Q}\hat{H}|\Omega\rangle~=~ 0.$$

• $$\quad \hat{Q}~:=~\int \! d^3{\bf x}~\hat{J}^0(x).$$ $$\quad |{\bf k}\rangle ~:=~\int \! d^3{\bf x} ~e^{-i{\bf k}\cdot{\bf x}}\hat{J}^0(x)|\Omega\rangle.$$ $$\quad |{\bf 0}\rangle~=~|{\bf k}\!=\!{\bf 0}\rangle.$$

• $$\quad \hat{H}|{\bf k}\rangle ~=~ \sqrt{{\bf k}^2+m^2}|{\bf k}\rangle.$$ $$\quad \Rightarrow \quad m~=~0.$$ $$\Box$$

• Idea: 1. Goldstone theorem via 1PI effective action, cf. Ref. 2. 2. Non-relativistic Goldstone theorem. Still have spacetime translation symmetry but not Lorentz symmetry. We could have $\quad \langle J^0_a \rangle ~\neq~ 0$ and $\quad \langle [Q_a,J^0_b ]\rangle ~\neq~ 0.$ Do we really need the assumption $d_{\mu}\hat{J}^{\mu}(x)=0\quad (2)?$ Apparently it is used in eq. (14). Nov 4, 2018 at 16:05
• Notes to self: 1. Lagrangian path integral version $\quad \delta \phi~=~f.$ $\quad \langle f \rangle ~\neq~ 0.$ $\quad \Rightarrow \quad$ Goldstone mode (basically $\phi \sim \hat{A}$). 2. Interestingly, definition (7) seems to be correct normalized wrt. volume $V$, cf. eq. (12). 3. Third assumption: no long-range forces involved, because that ensures that commutators fall off sufficiently rapidly at infinity that the surface integral can be dropped. 4. Eq. (4) smells like Schur lemma. Nov 5, 2018 at 12:42
• 1. I complement the way you used the equation label in the equals stack. That is sharp notation! 2. Can you say why you chose an arrow instead of $=$ for (11)? Dec 11, 2020 at 22:32
• Hi @hodop smith: 1. Thanks. 2. It is a limit $V\to \mathbb{R}^3$. Dec 11, 2020 at 22:42
• How do we know that the massless mode $|n\rangle$ is a new state and not the vacuum (which also has zero energy)? Jun 17, 2021 at 2:30