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

Question

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?

Answer 1

I'm going to provide an outline for an argument based upon some basic measure theory, along with some assumptions about the Kallen-Lehmann density which hold in many QFTs, that agrees with Weinberg's intuition. First note that

$$\int_0^\infty \rho(\mu^2)d\mu^2=1$$

and hence $\rho $ can be written as a measure $\rho(\mu^2)d\mu^2=d\nu(\mu^2)$. Also recall that the Wightman function has a closed form for spacelike separations given by a Bessel function

$$\Delta_+(x-y,\mu^2)=\frac{\mu}{4\pi^2r}K_1(\mu r)~,~r:=\sqrt{(x-y)^2} $$

For spacelike separations in the mostly minus metric, $(x-y)^2>0$, and therefore the Wightman function is real and rapidly decaying and we can rewrite

$$\frac{1}{4\pi^2 r^2}\int(d\nu-d\bar\nu)\mu r K_1(\mu r)=0~~~, ~~\forall~ r>0$$

At this point we could simply expand the kernel in powers of $r$ and equate all the moments of the two measures, which would show that the measures are equal (under some assumptions). However, this is not immediately possible, since the series expansion of the Wightman function around $r=0$ is somewhat singular. As a matter of fact, we find that

$$xK_1(x)=(1+A_1x+A_2x^3+...)+(B_1 x+B_3 x^3+...)\log x$$

So we assume that the following integrals exist and are finite for all $\alpha> 0$

$$M(\alpha)=\int x^\alpha d\nu\\\bar M(\alpha)=\int x^\alpha d\bar\nu$$ while also asserting that $M,\bar M$ are $C^1$ functions in that domain. The reason why we need the function to be differentiable is clear: we need the logarithmic moments to exist, and those are produced by first derivatives of $M,\bar M$ at odd positive integer values of $\alpha$.

Expanding the kernel, we deduce after some simple algebra that the odd moments vanish:

$$M(2n-1)=\bar{M}(2n-1)~,~\frac{dM}{d\alpha}(2n-1)=\frac{d\bar M}{d\alpha}(2n-1)$$

Now define the measures given by $d\kappa, d\bar\kappa=\mu d\nu, \mu d\bar\nu$ which now have equal even moments, and extend the measures evenly to all of $\mathcal{R}$, so that their odd moments vanish identically. This technical step is required to bring it to the form of the Hamburger moment problem, which states that a measure with given moments exists and is unique if the moments satisfy the Carleman condition

$$|M_\kappa (n)|\leq CD^n n!$$

which shouldn't be too constraining for most QFTs.

PS I don't know for a fact whether the assumption that the log-moments exist is redundant or not in the light of the result above, but better be safe than sorry!