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When proving Goldstone's theorem in Vol. 2, p. 171, Weinberg takes the following: $$\langle \{J^\lambda(y),\phi_n(x)\}\rangle= \frac{\partial}{\partial y_\lambda}\int d\mu^2 \rho_n(\mu^2) \left[ \Delta_+(y-x;\mu^2) - \Delta_+(x-y;\mu^2)\right]$$ with $$ \Delta_+(z;\mu^2)=(2\pi)^{-3}\int d^4p\theta(p_0)\delta(p^2+\mu^2)e^{ipz}$$ and he computes the case for $\lambda=0, x^0=y^0=t$ finding: $$\langle \{J^\lambda(y),\phi_n(x)\}\rangle=2i(w\pi)^{-3}\int d\mu^2\rho_n(\mu^2)\int d^4p\sqrt{{\bf p}^2+\mu^2}e^{i{\bf p}({\bf y}-{\bf x})}\delta(p^2+\mu^2)=i\delta^3(y-x)\int d \mu^2\rho_n(\mu^2)$$

Why the Fourier transform of the derivative is $\sqrt{\mu^2+{\bf p}^2}$ instead of $p$ and why in the last passage when integrating over the delta the result is not 0. Thanks in advance

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He considered $\lambda=0$, i.e. time derivative. The time derivative gives you $p^0$. Integration over the delta function $\delta(p^2+\mu^2)$ forces $p^0=\sqrt{\mu^2+\mathbf{p}^2}$, hence the result.

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  • $\begingroup$ Thanks for your answer but then shouldn't he cancel both the integral and the delta function? $\endgroup$
    – Ringo_00
    Commented May 10, 2021 at 20:38
  • $\begingroup$ The delta function is $\delta(p^2+\mu^2)$ (single variable), not $\delta^{(4)}(p^2+\mu^2)$, it only cancel the $dp^0$ integral, you are still left with $d^3\mathbf{p}$. $\endgroup$
    – JF132
    Commented May 10, 2021 at 20:43
  • $\begingroup$ To do the $dp^0$ integral, you will need an identity of the delta function: $\delta(x^2-\alpha^2)=\frac{1}{2|\alpha|}\big(\delta(x+\alpha)+\delta(x-\alpha)\big)$ $\endgroup$
    – JF132
    Commented May 10, 2021 at 21:06

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