According to the Kallen-Lehmann representation of the propagator, the propagator has a branch cut beginning around $p^2\approx4m^2$. This appears to invalidate the use of the usual $\frac{i}{p^2-m^2+i\epsilon}$ for intermediate propagators in computing an amplitude, as the branch cut would dominate it. To cite an example, the tree level graph for two-to-two scattering in $\phi^3$ theory is generally calculated whilst ignoring the branch cut region of the propagator.

# [Physics] Kallen-Lehmann representation versus perturbation theory

perturbation-theorypropagatorquantum-field-theory

#### Related Solutions

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.

What you found is a very simple example of an **infra-red divergence**, which plague all physical theories with massless particles.

This kind of divergences are already present in the classical case (see for example ref.1), and they usually signal that you are asking an unphysical question, not that the theory itself is unphysical. For example, if you have massless particles it becomes meaningless to ask about the total number of them in a certain physical configuration, while asking about the total energy is a well-posed question. This is reflected in the mathematics of the theory through divergences: the first question results in a divergent expression, while the second does not.

These divergences are inherited to the quantum mechanical case. For example, IR divergences are ubiquitous in QED, but it can be proven that these divergences always cancel out for "meaningful cross-sections", i.e., for predictions that are well-posed in the sense of the previous paragraph. This is sometimes known as the Bloch-Nordsieck cancellation, or in the more general case of the Standard model, as the Kinoshita-Lee-Nauenberg theorem. See for example refs.2,3,4.

In your particular case, as you are dealing with a scalar theory as opposed to a gauge boson, the analysis is slightly simpler (although one must in general abandon on-shell renormalisation schemes if we are to have massless particles). This is discussed in ref.5.

For a further discussion of infra-red divergences, see refs. 6-8.

**References**

Itzykson C., Zuber J.-B. Quantum field theory, section 4-1-2.

Peskin, Schroesder. An introduction To Quantum Field Theory, section 6.5.

Itzykson C., Zuber J.-B. Quantum field theory, section 8-3-1.

Schwartz M.D. Quantum Field Theory and the Standard Model, chapter 20.

Srednicki M. Quantum Field Theory, chapters 26 and 27.

Ticciati R. Quantum Field Theory for Mathematicians, section 19.9.

Pokorski S. Gauge Field Theories, sections 5.5 and 8.7.

Weinberg S. Quantum theory of fields, Vol.1. Foundations, chapter 13.

## Best Answer

Kallen-Lehmann representation originates from the decomposition of $\hat{\phi}(x)|\Omega\rangle$ into the Hamiltonian eigenstates $|\lambda_{\mathbf{p}}\rangle$. If $|\langle\Omega|\hat{\phi}(x)|\lambda_{\mathbf{p}}\rangle=0$ that state doesn't give contribution into the propagator spectral representation.

In the free theory $\hat{\phi}(x)|0\rangle$ is simply a superposition of single particle states. Even though there are other free Hamiltonian eigenstates - many-particle states, they have zero overlap with $\hat{\phi}(x)|0\rangle$ and you get just one pole in the free propagator.

However for interacting theory $\hat{\phi}(x)|\Omega\rangle$ decomposes into all kinds of states including many-particle states that give you branch cut.

The main point of my answer to your question is that Feynman's perturbation theory uses the propagators of the free theory for the lines of the diagrams and they as I said don't have branch cut. The full propagator is given by the sum of all diagrams and you will see that it has branch cuts when you compute loop-level contributions.