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In Eq.10.7.11 of Weinberg's textbook on QFT vol.I, it is claimed that equality of spectral densities $\rho(\mu^2)$ for particles and $\bar{\rho}(\mu^2)$ for antiparticles (obtained by iserting complete set of states in $\langle \phi \phi^\dagger\rangle_0$ and $\langle \phi^\dagger \phi\rangle_0$ respectively) is necessary condition to have the commutator on the vacuum $\langle[\phi(x),\phi^\dagger(y)]\rangle_0$ vanishing at spacelike separation.

I don't think I understand the argument presented by Weinberg, it looks more a non-sequitur to me. (Although it is eventually true, because of CPT. I am interested in understanding if Weinberg's argument, works as well or if it must be amended, completed...).

The argument he gives (as I understand it) is that $$ \forall\, (x-y)_{\mbox{(spacelike)}}\qquad \langle[\phi(x),\phi^\dagger(y)]\rangle_0\big|_{spacelike}= \int d\mu^2 \Delta_{+}(x-y,\mu^2)\left(\rho(\mu^2)-\bar{\rho}(\mu^2)\right)=0\,\qquad \Rightarrow \rho(\mu^2)=\bar{\rho}(\mu^2) $$ where $\Delta_+ (x-y,\mu^2)$ is given n Eq. 10.7.7 that I find useful to rewrite it by integrating the $p^0$ against the delta function as $$ \Delta_+(x-y,\mu^2)\propto \int \frac{d^3 p}{2\omega_{\mu^2}(p^2)} e^{ip(x-y)}\,,\qquad \omega_{\mu^2}(p^2)=\sqrt{\vec{p}^2+\mu^2}\,. $$ For spacelike separations there is always a Lorentz boost where x and y are simultaneous, $x^0=y^0$ and by Lorentz invariance we can rewrite the commutator as a $$ \langle[\phi(x),\phi^\dagger(y)]\rangle_0\big|_{spacelike} = \int d\mu^2 \int \frac{d^3 p}{2\omega_{\mu^2}(p^2)} e^{i\vec{p}(\vec{x}-\vec{y})}\left(\rho(\mu^2)-\bar{\rho}(\mu^2)\right)\,. $$ This would imply, in my understanding, that $$ \int d\mu^2 \frac{1}{2\omega_{\mu^2}(p^2)} \left(\rho(\mu^2)-\bar{\rho}(\mu^2)\right)=0 \quad \forall p^2 $$ Now, rescaling the integration variable $z=\mu^2/p^2$, this implies $$ \int dz \frac{1}{\sqrt{1+z^2}} \left(\rho(z p^2)-\bar{\rho}(z p^2)\right)=0 \quad \forall p^2\,. $$ At this point it looks like there could be solutions of this constraint with $\rho\neq \bar{\rho}$, in contrast to Weinberg's claim.

For example, if I try taking $\rho$ and $\bar{\rho}$ to be homogeneous functions of degree $n$, that is $\rho(z p^2)=p^{2n}\rho(z)$ and $\bar{\rho}(z p^2)=p^{2n}\bar{\rho}(z)$, I can pull out the $p^2$ dependence so that the constraint becomes $$ \mbox{If }\quad \rho(z p^2)=p^{2n}\rho(z)\,,\bar{\rho}(z p^2)=p^{2n}\bar{\rho}(z) \Rightarrow \int dz \frac{1}{\sqrt{1+z^2}} \left(\rho(z)-\bar{\rho}(z)\right)=0 $$ which does not require vanishing of the integrand. This would-be counter example does not really work in the end because by the homogenity assumption I could have equally pulled out the $z$-dependence, $$ \mbox{If }\quad \rho(z p^2)=p^{2n}\rho(z)\,,\bar{\rho}(z p^2)=p^{2n}\bar{\rho}(z) \Rightarrow \left(\rho(p^2)-\bar{\rho}(p^2)\right) \int dz \frac{z^{2n}}{\sqrt{1+z^2}} =0\Rightarrow \rho(p^2)-\bar{\rho}(p^2)=0 $$ and the constraint would have indeed implied equality of spectral densities (in this class of homogeneous spectral densities, assuming $n$ sufficiently negative to ensure convergence.

Here is my question: is there a missing (or implicit) assumption Weinberg is making or his argument can be carried until the end? Am I somehow missing some assumption? For example, I dont' think I have used the positivity of the spectral densities. This should play a role in the argument?

As example of extra assumption I could think of adding is that at large $\mu^2$ the spectral density becomes the one of a CFT (inspired by thinking of QFT as the RG evolution of an UV CFT deformed by relevant deformations). I can reach that region by taking large $p^2$ (in the first intergral version in the z-variable above), then the spectral density would be homogeneous (from the CFT assumption), the degree being fixed by the scaling dimension of $\phi$ (assumed to be primary). From there I would conclude that indeed the spectral density must be equal but only at large $p^2$. Somehow even this does not look enough.

Any Ideas?

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  • $\begingroup$ I'm not sure why Weinberg considers these spectral densities to be different quantities in the first place, since in quantum mechanics generally $\langle n |\Phi^\dagger(0)|0\rangle=\langle 0|\Phi(0)|n\rangle^*$... Also he has already used this equality to derive line (10.7.3), so I can't fathom why this whole argument is necessary. $\endgroup$ 1 hour ago

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