# Kallen-Lehmann + micro-causality do really imply equality of spectral densities for particles and anti-particles?

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?

• 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. 1 hour ago