For simplicity, take $\hbar=1$ and consider a Hermitian scalar, renormalized field $\phi(x)$; other fields are treated analogously. Then
$$G(E)=i\int_0^\infty dt e^{itE}\langle \phi(0)\phi(t,0)\rangle
=i\int_0^\infty dt e^{itE}\langle \phi(0)e^{-itH}\phi(0)\rangle\\
=\Big\langle\phi(0)\int_0^\infty dt ie^{it(E-H)}\phi(0)\Big\rangle
=\langle \phi(0)(E-H)^{-1}\phi(0)\rangle
=\psi^*(E-H)^{-1}\psi,$$
where $\psi=\phi(0)|vac\rangle$, since the vacuum absorbs the other exponential factors. The spectral theorem for the self-adjoint operator $H$ guarantees a spectral decomposition $\psi=\int d\mu(E')\psi(E')$ into proper or improper eigenvectors $\psi(E')$ of $H$ with eigenvalus $E'$, where $d\mu$ is the spectral measure of $H$. Orthogonality of the eigenvectors gives
$$G(E)=\psi^*(E-H)^{-1}\psi
=\int d\mu(E')\psi(E')^*(E-E')^{-1}\psi(E')
=\int d\mu(E')\frac{|\psi(E')|^2}{E-E'}.$$
Therefore
$$(1)~~~~~~~~~~~~~~~~~~~~~~~~~~G(E)=\int\frac{d\rho(E')}{E-E'},$$
with the measure $d\rho(E')=d\mu(E')|\psi(E')|^2$. For negative $E$, the Greens function is finite; since the measure $\rho$ is positive, this implies that the integral is well-behaved and has no other singularities apart from those explicitly visible in the right hand side of (1).
Note that (1) holds for every self-adjoint Hamiltonian, no matter what its spectrum is. It thus describes the most general singularity structure an arbitrary Greens function can have.
In general, the spectral measure consists of a discrete part corresponding to the bound states and a continuous part corresponding to scattering states. (There might also be a singular spectrum, which is usually absent and does not affect the main conclusion.)
Formula (1) therefore says that (in the absence of a singular spectrum) the only possible singularities of the Greens function are poles and branch cuts. Moreover, (1) implies that $G(E)$ has a pole precisely at those discrete eigenvalues $E'$ of $H$ for which $\psi(E')$ is nonzero, and branch cuts precisely at the part of the continuous spectrum in the support of the measure $d\rho$.
In particular, any pole of $G(E)$ must be a discrete eigenvalue of $H$, whereas the converse only holds under the qualification that $\psi=\phi(0)|vac\rangle$ has a nonzero projection to the corresponding eigenspace of $H$.
The above argument applies to quantum field theories after elimination of the center of mass degrees of freedom. Thus we consider the subspace of states in which the spatial momentum vanishes. On this subspace, the bound states have a finite multiplicity only, as each mass shell is intersected with the line in momentum space defined by vanishing spatial momentum.
In terms of the original question, the direction ''pole $\to$ particle'' holds without qualification, whereas the direction ''particle $\to$ pole'' (claimed by the OP to be the obvious part) is not universally correct but only holds under a nondegeneracy condition.
This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user Arnold Neumaier
Q&A (4872)
