In chapter 11-2-2, I&Z discuss Goldstone's theorem. They start by claiming that if an operator $A$ exists, such that $$ \delta a(t) \equiv \langle 0| [Q(t),A]|0\rangle \neq 0 \tag{11-30} $$ the symmetry is spontaneously broken. In Eq. (11-31), they insert a completeness relation as $1=\sum_n |n\rangle\langle n|$: $$\begin{align} \delta a(t) &= \sum_n \int d^3\vec x [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i P\cdot x}-c.c. ] \tag{11-31} \\ &= \sum_n (2\pi)^3 \delta^3(\vec P) [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i E t}-c.c. ] \end{align}$$ After taking the time derivative in Eq. (11-33), which brings down a $E(\vec p)$, they conclude that $\delta^3(\vec p) E(\vec p)$ must be zero, which leads to massless states.

However, if we were to use the following completeness relation, $$1=\sum_n \int \frac{\text{d}^3 \vec p}{(2\pi)^3 2E(\vec p)} |n,\vec p \rangle\langle n,\vec p, |$$ then the energy in the denominator and the energy coming from the time derivative of the exponential function would cancel and the conclusion from above wouldn't work anymore because there's no $E(\vec p)$ anymore!

Edit 1: I found that Ryder's QFT textbook (p. 292) as well as Nair's QFT textbook (p. 246) use the same completeness relation $1=\sum_n |n\rangle\langle n|$, i.e. the proof goes along the same lines as the one in I&Z. But why do they choose this completeness relation?

Edit 2: Maybe the answer is to take the following (off-shell) completeness relation: $$1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$$ since this would not give an $E(\vec p)$ in the denominator...?

Edit 3: Another reference that uses $1=\sum_n|n\rangle\langle n|$: arXiv link (page 5) and one that uses $1=\int\frac{d^3 p}{(2\pi)^3}|p\rangle\langle p|$: arxiv link (page 19).

Edit 4: Found a reference (warning, large PDF file size) that uses the completeness relation $1=\sum_n \int \frac{\text{d}^3 p}{(2\pi)^3} |\vec p \rangle\langle \vec p |$ in eq. (3.2) as well (Thanks to @ChiralAnomaly for pointing this out!).


1 Answer 1


The relation $$ 1=\sum_n\int\frac{d^3p}{(2\pi)^32E(p)} |n,p\rangle\langle n,p| \tag{1} $$ assumes $$ \langle n,p|n',p'\rangle \propto 2E(p)\delta_{nn'}\delta^3(p'-p). \tag{2} $$ In other words, it assumes that massless zero-momentum states are normalized to zero. But to have SSB, the quantity (11-30) $=$ (11-31) is supposed to be nonzero and time-independent. In order to be time-independent, terms with non-zero $E$ must cancel in (11-31), which means that the nonzero constant must come from terms with $E=0$... and that's a problem, because those terms are undefined due to a zero-divided-by-zero issue when we use (1) and (2), so we can't draw any conclusions about what happens when $E=0$. This problem doesn't arise when we use $$ 1=\sum_n\int\frac{d^3p}{(2\pi)^3} |n,p\rangle\langle n,p| \tag{3} $$ instead, because then $$ \langle n,p|n',p'\rangle \propto \delta_{nn'}\delta^3(p'-p). \tag{4} $$

  • $\begingroup$ Thank you very much, this makes sense. Most QFT textbooks I can think of use the normalization (2), since it's nicely Lorentz invariant. Do we use the normalization (4) only to prove the Goldstone theorem and then go back to the covariant normalization (2)? In other words: when working with NG modes later on, do we use (2) or (4)? $\endgroup$
    – ersbygre1
    Jun 17, 2021 at 23:41
  • $\begingroup$ 2nd question: The zero-divided-by-zero issue you mentioned, is this the energy coming from the time derivative and the energy in the denominator of your eq. (1)? $\endgroup$
    – ersbygre1
    Jun 17, 2021 at 23:42
  • $\begingroup$ Thank you very much for your detailed replies, this helps a lot! Concerning your 3rd comment, so this is why I&Z (and others) use a limit of volume -> infinity when talking about the Goldstone theorem. And just to be sure, the fact that they can't be normalized is due to the $\delta^{(3)}(0)$, right? $\endgroup$
    – ersbygre1
    Jun 18, 2021 at 2:05
  • 1
    $\begingroup$ @ersbygre1 Yes, that's correct. $\endgroup$ Jun 18, 2021 at 2:25

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