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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.

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1 Answer 1

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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 $\hat{A}$. $\Box$

See also this related Phys.SE post.

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):

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

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

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

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

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  • $\begingroup$ 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). $\endgroup$
    – Qmechanic
    Nov 4, 2018 at 16:05
  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Nov 5, 2018 at 12:42
  • $\begingroup$ 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)? $\endgroup$ Dec 11, 2020 at 22:32
  • $\begingroup$ Hi @hodop smith: 1. Thanks. 2. It is a limit $V\to \mathbb{R}^3$. $\endgroup$
    – Qmechanic
    Dec 11, 2020 at 22:42
  • $\begingroup$ How do we know that the massless mode $|n\rangle$ is a new state and not the vacuum (which also has zero energy)? $\endgroup$
    – ersbygre1
    Jun 17, 2021 at 2:30

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