Schwartz's and Zee's proof of Goldstone theorem

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 $$begin{align} {rm Im}a_V(t)~stackrel{(7)}{=}~&frac{1}{2i}langle Omega | [hat{Q}_V(t),hat{A}]|Omega ranglecr quad longrightarrow& quad a~neq~0quadtext{for}quad V~to ~mathbb{R}^3. end{align}tag{4}$$

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

6. On one hand, $$begin{align} d_t a_V(t) ~stackrel{(8)}{=}~&int_V! d^3{bf x}~langle Omega | d_0hat{J}^0(x) hat{A} |Omega rangle cr ~stackrel{(2)}{=}~&-int_V! d^3{bf x}~langle Omega | {bf nabla} cdot hat{bf J}(x) hat{A} |Omega rangle cr ~=~&-int_{partial V}! d^2{bf x}~langle Omega | {bf n} cdot hat{bf J}(x) hat{A} |Omega rangle,tag{14}end{align}$$ so that $$begin{align} d_t {rm Im}a_V(t)&cr ~stackrel{(14)}{=}~&-frac{1}{2i}int_{partial V}! d^2{bf x}~langle Omega | [{bf n} cdot hat{bf J}(x) ,hat{A}] |Omega rangle cr quad longrightarrow& quad 0 quadtext{for}quad V~to ~mathbb{R}^3,tag{15}end{align}$$ 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, $$begin{align} d_t a_V(t)~~stackrel{(12)}{longrightarrow}~~& -i sum_E Ee^{-iE t} f(E)cr &quadtext{for}quad V~to ~mathbb{R}^3,end{align}tag{16} $$ so that $$ begin{align}d_t {rm Im}a_V(t)&cr ~~stackrel{(16)}{longrightarrow}~~& -sum_E Eleft{cos(Et) {rm Re} f(E) +sin(Et) {rm Im} f(E)right}cr &quadtext{for}quad V~to ~mathbb{R}^3.end{align}tag{17} $$

   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 $|n rangleneq|Omega rangle$ with $(E_n,{bf P}_n)=(0,{bf 0})$ that couples $c_nneq 0$ 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):

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

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

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

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