I am reading Matthew Schwartz's wonderful QFT book, but I have stumbled in my attempts to understand his rather brief proof of Goldstone's theorem. The preliminary information is that $$HQ|\Omega\rangle= E_0Q|\Omega\rangle,\tag{28.7}$$ where $Q$ is the conserved charge corresponding to the spontaneously broken symmetry $$Q\equiv\int d^3x\;j^0(x).\tag{28.5}$$ This is fine. However, from this, he claims that one can create states $$|\pi(\vec{\bf{p}})\rangle\equiv\frac{-2i}{F}\int d^3x e^{-i\,\vec{\bf{p}}\cdot\vec{\bf{x}}}j^0(x)|\Omega\rangle\tag{28.8}$$ that have energy $E(\vec{\bf{p}}) + E_0$. If this is true, then it is clear that the state must obey a massless dispersion relation. However, I do not understand how one determines the energy to be $E(\vec{\bf{p}}) +E_0$.

I have seen a similar proof on p. 228 in Zee's book, but there he shows in footnote 3 that $$\hat{P}^i|\pi(p)\rangle = p^i|\pi(p)\rangle,$$ which is not quite the same although it has the same implication. Any ideas?

  • $\begingroup$ Have you try to use the decomposition of $j_0$ in terms of creation/annihilation operators ? $\endgroup$
    – Adam
    Commented Apr 3, 2017 at 8:32
  • $\begingroup$ Well, I guess I could do it explicitly in terms of the operators for a simple complex scalar field, but he claims this is true in general. $\endgroup$ Commented Apr 3, 2017 at 9:41
  • $\begingroup$ Related. $\endgroup$ Commented Nov 21, 2017 at 15:40
  • $\begingroup$ More on Schwartz's book Goldston theorem: physics.stackexchange.com/q/535771/226902 $\endgroup$
    – Quillo
    Commented Jul 30, 2023 at 13:14

2 Answers 2


I think you got the sign of the exponential in Schwartz's (28.8) wrong here, which I reverse, but I am cavalier with signs and factors, as I assume you just want to see the principle (the forest, not the trees).

The key point is that the SSB current is basically always of the form $j_\mu(x)\propto F \partial_\mu \pi (x) +...$ where the omitted terms are of higher order in the fields (and so do not contribute to the particle state defined); and, of course, group indices have been omitted as well.

This is codified in the appellation "Nambu-Goldstone nonlinear realization of the symmetry". This is the only way to have the v.e.v. of the transform of this Goldstone field not vanish, while the v.e.v.s of all the fields themselves vanish--after shift redefinitions. (See this answer .) It is illustrated later on in (28.13).

The Fourier transform of the Goldstone field π is $$ \pi(\vec{\bf{p}}) = \frac{-2i}{F}\int d^3x ~e^{i\,\vec{\bf{p}}\cdot\vec{\bf{x}}}j^0(x), $$
with $$ [H,\pi(\vec{\bf{p}})] = E(\vec{\bf{p}}) ~ \pi(\vec{\bf{p}}) . $$ The SSB charge Q is essentially the space integral of the canonical momentum of π, so that $\langle \Omega |[Q,\int d^3 p ~ \pi(\vec{\bf{p}})]|\Omega\rangle \propto F\neq 0 $ , cf. his (28.9).

As a consequence, decompressing your text's one liner, $$H|\pi(\vec{\bf{p}})\rangle= ( [H,\pi(\vec{\bf{p}})] + \pi(\vec{\bf{p}}) H ) ~|\Omega\rangle = ( E(\vec{\bf{p}}) + E_0 )\pi(\vec{\bf{p}})|\Omega\rangle= ( E(\vec{\bf{p}}) + E_0 )|\pi(\vec{\bf{p}})\rangle , $$ so $E(0)\to 0$ .

Still, for the more conventional general proof of the theorem, bypassing your perceptual snag, consider Kibble's summary.

  • $\begingroup$ How would you prove that $[H,\pi(\vec p)]=E(\vec p)\pi(\vec p)$? $\endgroup$ Commented Nov 21, 2017 at 17:57
  • $\begingroup$ This is the standard 2nd quantization expression... I'm sure that text covers it in previous chapters... $\endgroup$ Commented Nov 21, 2017 at 18:01
  • $\begingroup$ It would be a standard 2nd qn. expression if $\pi(\vec p)$ were a standard creation operator. But here $\pi(\vec p)$ is defined from $j^0$, so it's far from clear whether it satisfies standard expressions or not. In principle, it should be possible to prove that $[H,\pi]=E\pi$ from the properties of $j^\mu$ alone, that is, from $\partial\cdot j=0$ and perhaps something else that I am missing right now. $\endgroup$ Commented Nov 21, 2017 at 18:06
  • $\begingroup$ That's the very point. For a N-G realization, the current is the gradient of a standard field operator times the decay constant F--this must be so. $\endgroup$ Commented Nov 21, 2017 at 19:14
  • $\begingroup$ Thanks @CosmasZachos! However, I am a bit confused in the same way as @AccidentalFourierTransform. I do not see how this commutation relation follows. $\endgroup$ Commented Nov 24, 2017 at 9:53

I guess this follows from the Heisenberg equation of motion for $\pi$: $$i\partial_t \pi = [\pi, H] $$ Calculating it, I found this works if

1) his sign convention is such that the coefficient of E in the exponent is -i so that $i\partial_t$ brings down a factor -E

2) $H|\Omega\rangle =0$

3)Perhaps I'm mistaken, but it looks as if your equation is missing a factor of $Q$ so that $$HQ|\Omega\rangle = E_0 Q|\Omega\rangle$$ -otherwise I have a hard time getting energy eigenstates here.

  • $\begingroup$ yea thanks you are right about the typo. I actually also screwed up the -2i/F factor which I changed now in the question. This explains the messed up sign with the energy. Thanks a lot for the response! However, I don't think H|omega> =0, it should equal E_0|omega>. $\endgroup$ Commented Apr 7, 2017 at 15:16
  • $\begingroup$ So $E_0$ is the vacuum energy... What's F? $\endgroup$
    – rwold
    Commented Apr 7, 2017 at 15:24
  • $\begingroup$ Yea exactly E_0 is the vacuum energy. I think F is just a constant with mass dimension 1 to give everything the right dimensions. The current has dimension 2, and I think you want you pi field to have dimension 1 because it should be a scalar. $\endgroup$ Commented Apr 7, 2017 at 16:10
  • $\begingroup$ Hey @rwold, looking again at this again, I cannot see how this all works out. Any chance you could be a bit more explicit? Thanks again! $\endgroup$ Commented Oct 16, 2017 at 18:52

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