# Goldstone theorem in Weinberg vol 2

I was reading the proof of Goldstone's theorem (the operator proof starting on page 170) in Weinberg's book on QFT (Volume II) and got confused. I am able to follow each line of the proof, but as a whole, something seems missing in terms of my understanding.

Questions:

1. Suppose we tried to prove the existence of gapless modes by computing the anit-commutator $$\langle \{J^\lambda(y),\phi_n(x)\}\rangle=(2\pi)^{-3}\int d^4 p \bigg[\rho^\lambda_n(p) e^{ip(y-x)}-{\tilde\rho}_n^\lambda(p) e^{ip(y-x)}\bigg],$$ where $$\{,\}$$ is the anti-commutator, instead of starting with the commutator equation $$\langle [J^\lambda(y),\phi_n(x)]\rangle=(2\pi)^{-3}\int d^4 p \bigg[\rho^\lambda_n(p) e^{ip(y-x)}+{\tilde\rho(p)}_n^\lambda e^{ip(y-x)}\bigg]$$ in (eq. 19.2.18). Then, if we assume that $$\{Q,\phi_n(x)\}\neq 0$$, it seems that, going through the exact same proof as in Weinberg, but replacing commutators with anti-commutators we would arrive at the result $$\rho_n(\mu^2)\propto \delta(\mu^2)$$, indicating the presence of a gapless mode. Is this in fact the case? If not, can you show me exactly where the logic breaks down?

2. I am also confused about why $$\rho_n(\mu^2)\propto \delta(\mu^2)$$ proves the existence of a gapless mode. Intuitively, it makes sense that this should be the case, but usually a gapless particle is defined in terms of poles in the S-matrix. How exactly is $$\rho_n(\mu^2)$$ is related to the S-matrix? If we calculated the spectral function $$\rho$$ of some random assortment of operators from our theory and found it to be proportional to a delta-function, would this imply the existence of gapless modes? Or is there something particular about this spectral function defined by (19.2.19) and (19.2.20) that allows us to make statements about gapless modes?

To your second point: Recall that the spectral density function is defined, in equation (10.7.4), as

$$(2\pi)^{-3} \theta(p^0) \rho(-p^2) = \sum_{n \in \text{States}} \delta^{4}(p-p_n) | \langle 0| \Phi(0) |n \rangle |^2$$

If there are a one-particle state of mass $$m$$, they will come in a whole family, indexed by the three-momentum $$\vec{p}$$, and the set $$p^2 = - m^2$$ defines a hyperboloid in (four-vector) $$p$$-space. Correspondingly, there will by one part of the sum over the states which collapses to the hyperboloid:

$$\sum_{n \in \text{States of mass m}} \delta^{4}(p-p_n) | \langle 0| \Phi(0) |n \rangle |^2 =\int d^3 \vec{q} \delta^{4}(p-p(\vec{q})) | \langle 0| \Phi(0) |\vec{q} \rangle |^2 \ .$$

Here $$\vec{p}(\vec{q}) = \vec{q}$$ and $$p^0(\vec{q}) = \sqrt{\vec{q}^2 + m^2}$$. Hence the above becomes

$$\delta\!\left(p^0 - \sqrt{\vec{p}^2 + m^2}\right) | \langle 0| \Phi(0) |\vec{p} \rangle |^2 \ .$$

Now we use the identity

$$\theta(p^0)\delta(p^2 + m^2) = \frac{\delta\!\left(p^0 - \sqrt{\vec{p}^2 + m^2}\right)}{2p^0}$$

and that, as Weinberg points out in equation (10.7.19),

$$| \langle 0| \Phi(0) |\vec{p} \rangle |^2 = (2\pi)^{-3} (2 p^0(\vec{p}))^{-1} Z \ ,$$

with $$Z$$ the wave function renormalization constant. Note that it is here that the definition of a particle as a pole in the S-matrix enters. Thus the above gives

$$\rho(-p^2) = Z \delta(p^2 + m^2) + \cdots$$

We can read this argument backwards: If $$\rho(-p^2)$$ contains a term $$\delta(p^2 + m^2)$$, then we know that the operator $$\Phi(x)$$ creates and annihilates particles of mass $$m$$. For this argument it is completely irrelevant how we acquired the operator $$\Phi(x)$$. It can be a composite operator.

To your first point i can't give a definite answer. I want to point out however that while i can imagine giving a rigorous definition of the operator $$[Q,\Phi(y)]$$, my imagination is tested to much larger extent for $$\{ Q, \Phi(x) \}$$. The point is that assuming a continuous symmetry acts means there is a one-parameter family $$U_t$$ acting on field operators by conjugation, and they are generated by the charges, which then act by the commutator, and that they are well-defined is guaranted by the existence of the symmetry. For the anti-commutator, there is no such argument. Finally, there is also spontaneous breaking of supersymmetry, where this might be different, but that's a completely different chapter.