Why is the spectrum of the momentum operator continuous for states containing 2 or more particles?

Question

Consider a free theory and let $P^\mu = (H, \mathbf{P})$ be the 4-momentum operator. Since $P_\mu P^\mu = m^2$ is a Lorentz scalar, we get the relation $H^2 - |\mathbf{P}|^2 = m^2$. Here $H$ must be an eigenvalue $E$ of the Hamiltonian so that the momentum is on shell. With this requirement we get that the spectrum of the one particle state is all combinations of scalars $E$ and vectors $\mathbf{P}$ such that $E^2 - |\mathbf{P}|^2 = m^2$, which is a hyperboloid. We also have that the vacuum is the point at the origin. This describes the spectrum of $P^\mu$ for systems consisting of $n = 0, 1$ particles.

Assuming the above is all correct, I am unclear on why states consisting of two or more particles form a continuum in a rest frame with zero momenta. This is stated in Peskin & Schroeder and visualized in Figure 7.1 (page 213). A similar question was asked in another question (Why do the eigenvalues of the 4-momentum operator organize themselves into hyperboloids?), from which I know this somehow follows from boosts and Lorentz invariance, but I still do not fully understand why there is a continuum.

Can anyone explain why there is a continuum for $n \geq 2$ states and why there is a "bound state" between the $n = 1$ and $n = 2$ states in the figure? In a rest frame with two particle, why does $E^2 - |\mathbf{P}|^2 = 4m^2$?

Answer 1

To see why there is a continuous spectrum for multiparticle states look at a n-particle state with zero total momentum, i.e. \begin{equation*} \mathbf{P} = \sum_i \mathbf{p}_i = 0. \end{equation*} The energy of this state is then just \begin{equation*} E = \sum_i \sqrt{m^2 + \mathbf{p}_i^2}. \end{equation*} Since the $\mathbf{p}_i^2$ can be chosen arbitrarily as long as they still sum to 0, this gives you a continuum. This energy has a lower bound of $E = 2\sqrt{m}$ corresponding to two particles at rest so the contiuum starts there.