Issue with understanding the Weinberg's proof of the Goldstone boson theorem

Question

To prove Goldstone's theorem Weinberg (in volume II, page 170, equation 19.2.18) starts with the expectation value over the vacuum of the commutator between one field $\phi_n(x)$ and the conserved current associated with the symmetry $J^{\lambda}(y)$, he writes:

Summing over intermediate states this is: $$ \left<[J^{\lambda}(y),\phi_n(x)]\right>_{VAC} = (2\pi)^{-3} \int d^4 p [\rho^{\lambda}_n(p)e^{ip(x-y)} - \tilde{\rho}^{\lambda}_n(p)e^{ip(x-y)}].\tag{19.2.18} $$

I would have written the same formula with a $(2\pi)^{-4}$ instead of a $(2\pi)^{-3}$ my understanding being that the $p$ exponential term and $2\pi$ come from Fourier transforming the left side and thus I would expect a $2\pi$ for each of the coordinates. What am I doing wrong? Any help is greatly appreciated.

Answer 1

$$ \langle \Omega|[J^\mu(y),\phi_n(x)]|\Omega\rangle=\langle \Omega|J^\mu(y)\phi_n(x)-\phi_n(x)J^\mu(y)|\Omega\rangle \\=\int d\Pi\ \langle\Omega|J^\mu(y)|p,\lambda\rangle \langle p,\lambda|\phi_n(x)|\Omega\rangle-\langle\Omega|\phi_n(x)|p,\lambda\rangle \langle p,\lambda|J^\mu(y)|\Omega\rangle \\d\Pi=\frac{d^3 p}{(2\pi)^3}\frac{1}{2E_p(\lambda)} $$ where the integral is also formally over all intermediate single-particle and multi-particle states, which ensures that $\int d\Pi\ |p,\lambda\rangle\langle p,\lambda|$ obeys the completeness relation. This has the proper normalisation for a completeness relation (which you might recognise as a simple extrapolation of the integration measure for the scalar field mode expansion), and is equivalent to the Lorentz-invariant phase space up to the momentum-conserving delta function and factors of $2\pi$. Now, translating to the origin: $$ =\int d\Pi\ \langle\Omega|J^\mu(0)|p,\lambda\rangle \langle p,\lambda|\phi_n(0)|\Omega\rangle e^{ip(x-y)}-\langle\Omega|\phi_n(0)|p,\lambda\rangle \langle p,\lambda|J^\mu(0)|\Omega\rangle e^{-ip(x-y)} $$ Consider only the first term for simplicity: $$ \int d\Pi\ \langle\Omega|J^\mu(0)|p,\lambda\rangle \langle p,\lambda|\phi_n(0)|\Omega\rangle e^{ip(x-y)} \\=\int \frac{d^4 p}{(2\pi)^4}e^{ip(x-y)}\int d\Pi\ (2\pi)^4\delta(p-\tilde p)\langle\Omega|J^\mu(0)|p,\lambda\rangle \langle p,\lambda|\phi_n(0)|\Omega\rangle $$ where $\tilde p$ is the four-momentum of the current intermediate state in the sum. $$ 2\pi\rho_+^\mu(p)\equiv\int d\Pi\ (2\pi)^4\delta(p-\tilde p)\langle\Omega|J^\mu(0)|p,\lambda\rangle \langle p,\lambda|\phi_n(0)|\Omega\rangle \\\Rightarrow \langle \Omega|[J^\mu(y),\phi_n(x)]|\Omega\rangle = \int \frac{d^4 p}{(2\pi)^3} \left(\rho_+^\mu(p)e^{ip(x-y)}-\rho_-^\mu(p)e^{-ip(x-y)}\right) $$ Alternatively, the insertion of the complete set of states brings $d^3p/(2\pi)^3$, and in converting to the spectral representation $\rho_\pm^\mu(m^2)$ which has a delta function $\delta(m^2-\tilde m^2)$ (as all intermediate states in the Hilbert-space are on-shell. Even the spectral density for multiparticle states contains the delta function, although $\rho(m^2)$ is not "spiked" for them), we must add a $\int d(m^2)$ while integrating over all states, which turns the integration measure into $d^4p/(2\pi)^3$.