Kallen–Lehmann spectral representation for an arbitrary spin

Question

Let's have Kallen–Lehmann spectral representation for the scalar theory: $$ \tag 1 D(p) = \int \limits_{0}^{\infty} d(\mu^{2})\frac{\rho (\mu^{2})}{p^{2} - \mu^{2} + i\varepsilon}. $$ We can represent $(1)$ in a form $$ D (p) = D_{free}(p, m)Z + \int \limits_{m^{2}}^{\infty}d(\mu^{2})\rho (\mu^{2})D(p, \mu), $$ where $\mu$ refers to the mass of the multiparticle states and $m$ refers to the mass of one-particle free state. From here there is interesting result: it can be showed that $0 \leqslant Z \leqslant 1$.

How to get analogous result for the arbitrary spin field (or at least for fermions with spin $\frac{1}{2}$ and for photons)?

I know the general structure of the propagator of the free theory (for simplicity I assume massive case): for field $\Psi $ with spin $s$ $$ \hat {\Psi}_{l} (x)= \sum_{\sigma = -s}^{s}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi)^{3}2E_{\mathbf p}}}\left( u^{\sigma}_{l}(\mathbf p )e^{-ipx}\hat {a}_{\sigma}(\mathbf p ) + v^{\sigma}_{l}(\mathbf p )e^{ipx}\hat {b}^{\dagger}_{\sigma}(\mathbf p ) \right) $$ it will be

$$ D_{lm}(p) = \frac{F_{lm}(p)}{p^{2} - m^{2} + i\varepsilon}, \quad F_{lm} = \sum_{\sigma}u^{\sigma}_{l}(\mathbf p )\left(u^{\sigma}_{m}(\mathbf p) \right)^{\dagger}. $$

For the derivation of spectral representation for the scalar field see the reference above. Here also you can see the problem with this derivation which appears for the case of field with nonzero spin: it is impossible to introduce the scalar density function $\rho (\mu^{2})$.

Also there is bigger problem: there was using relation $\langle |[\hat {\Psi}(\mathbf x , t), \frac{d}{dt}\hat {\Psi}^{\dagger}(\mathbf y, t)]|\rangle = i\delta (\mathbf x - \mathbf y)$ for getting relation $0 \leqslant Z \leqslant 1$ (look to the Weinberg's "QFT" (chapter "Kallen–Lehmann spectral representation" of the section "Nonperturbative methods") or to Greiner's "Field Quantization" (pp. 278-282)). But this relation is correct only if one field is canonical conjugated to another (one field is canonical coordinate while the another one is canonical momentum). But for the arbitrary spins the canonical momentums must be chosen "individually".

Answer 1

Let's call $\Delta^+(x-y;m^2)$ the 2-point function for a free scalar field $\varphi$ of mass $m$: denoting the ground state as $\Psi_0$ $$ \Delta^+(x-y;m^2)\equiv \langle\varphi(x)\varphi(y)\rangle \equiv \left(\Psi_0, \varphi(x)\varphi(y) \Psi_0 \right)= \int\frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_m(\mathbf{k})}e^{-ik\cdot(x-y)} $$ where implicitly $k_0=\omega_m(\mathbf{k})\equiv\sqrt{\mathbf{k}^2+m^2}$. This way $$ \langle[\varphi(x),\varphi(y)]\rangle = \Delta^+(x-y;m^2)-\Delta^+(y-x;m^2)\equiv\Delta(x-y;m^2). $$ As you quoted, the 2-point function for a higher spin field will have the same structure, a part from a factor which allows to preserve the tensor behavior of the function: to be practical consider the free spin $1/2$ field of mass $m$ $$ \langle \psi(x)\overline{\psi}(y)\rangle = \int\frac{d^3k}{(2\pi)^3}\frac{\sum_{s=1}^2u_s(\mathbf{k})\overline{u_s}(\mathbf{k})}{2\omega_m(\mathbf{k})}e^{-ik\cdot(x-y)}= \int\frac{d^3k}{(2\pi)^3}\frac{\hat{k}+m}{2\omega_m(\mathbf{k})}e^{-ik\cdot(x-y)}=\\ =(i\hat\partial+m)\Delta^+(x-y;m^2), $$ where $\hat{k}\equiv\gamma^\mu k_\mu$ and so on.

The following facts hold for interacting fields: for the scalar field $$ \langle\varphi(x)\varphi(y)\rangle = \int d\rho(\mu^2)\Delta^+(x-y;\mu^2), $$ and for the spinor field $$ \langle\psi(x)\overline{\psi}(y)\rangle=(i\hat\partial+m)\int d\sigma(\mu^2)\Delta^+(x-y;\mu^2)=\int d\sigma(\mu^2) \int\frac{d^3k}{(2\pi)^3}\frac{\hat{k}+\mu}{2\omega_\mu(\mathbf{k})}e^{-ik\cdot(x-y)}, $$ where the two measures $d\rho(\mu^2),\ d\sigma(\mu^2)$ are scalar. This general form follows essentially from covariance and from the classification of all Lorentz-invariant positive measures by Garding and Lions (see Strocchi, An introduction to the non perturbative foundations of quantum field theory, page 50, page 98). From this form, we can extend the argument for scalar fields: Let's insert a complete set of (improper) eigenstates of momentum (and spin) for intermediate states of all possible masses $\mu$: $|\mathbf{k};\mu^2\rangle$ in the following way: $$ \langle\psi_\alpha(x)\overline{\psi}_\beta(y)\rangle= \int d\mu^2\int \frac{d^3k}{(2\pi)^2}\frac{1}{2\omega_\mu(\mathbf{k})}\langle \psi_\alpha(x) |\mathbf{k};\mu^2\rangle \langle\mathbf{k};\mu^2|\overline{\psi}_\beta(y)\rangle=\\ =\int d\mu^2\int \frac{d^3k}{(2\pi)^2}\frac{1}{2\omega_\mu(\mathbf{k})}\langle \psi_\alpha(0) |\mathbf{k};\mu^2\rangle \langle\mathbf{k};\mu^2|\overline{\psi}_\beta(0)\rangle e^{-ik\cdot(x-y)}. $$ Consider now a boost that brings the particle of momentum $\mathbf{k}$ at rest: $U(\mathbf{k})|\mathbf{k};\mu^2\rangle = |\mathbf{0};\mu^2\rangle$ and recall that for the spinor representation $S$ of the Lorentz boost $\Lambda^{-1}$ we have $$ U(\mathbf{k})\psi_\alpha(x)U(\mathbf{k})^\ast = S_{\alpha\beta}(\mathbf{k})\psi(\Lambda(\mathbf{k})x). $$ So by considering such a boost $$ \langle \psi_\alpha(0) |\mathbf{k};\mu^2\rangle \langle\mathbf{k};\mu^2|\overline{\psi}_\beta(0)\rangle=S_{\alpha\tau}S^{-1}_{\beta\xi}\langle \psi_\tau(0) |\mathbf{0};\mu^2\rangle \langle\mathbf{0};\mu^2|\overline{\psi}_\xi(0)\rangle $$ where all the dependence on $\mathbf{k}$ has been absorbed by the spinor transformation matrices. To sum up: $$ \langle\psi_\alpha(x)\overline{\psi}_\beta(y)\rangle= \int d\mu^2 \langle \psi_\tau(0) |\mathbf{0};\mu^2\rangle \langle\mathbf{0};\mu^2|\overline{\psi}_\xi(0)\rangle \int \frac{d^3k}{(2\pi)^2}\frac{1}{2\omega_\mu(\mathbf{k})}S_{\alpha\tau}S^{-1}_{\beta\xi}e^{-ik\cdot(x-y)}. $$ Now we can compare with the general form given above by tracing both expressions over the spinor indices; we get, since the gamma matrices are traceless and the two spinor trasformation matrices give a delta $$ \langle\psi_\alpha(x)\overline{\psi}_\alpha(y)\rangle=\int d\mu^2 \langle \psi_\tau(0) |\mathbf{0};\mu^2\rangle \langle\mathbf{0};\mu^2|\overline{\psi}_\tau(0)\rangle\Delta^+(x-y;\mu^2)=\\ =\int d\sigma(\mu^2)4\mu \Delta^+(x-y;\mu^2). $$ So, by comparison, we get an expression for the explicit form of the spectral measure: $$ \sigma(\mu^2)=\frac{1}{4\mu}\langle \psi_\alpha(0) |\mathbf{0};\mu^2\rangle \langle\mathbf{0};\mu^2|\overline{\psi}_\alpha(0)\rangle. $$ A similar result hold for a vector field, where instead of the spinor matrices you get the boost itself, which in turn preserves the metric and thus gives a four-dimensional invariant trace between matrix elements. You can extend these results to commutators by using the $\Delta(x-y;\mu^2)$ and to the Feynman propagator by using the corresponding free field Green function.