I always thought that a particle is an eigenvector of $P^2=H^2-\boldsymbol P^2$ with an isolated eigenvalue. In other words, a necessary condition for $\varphi$ to be a particle is that $$ P^2|\varphi\rangle=m^2|\varphi\rangle\tag{1} $$ and $$ \mathrm dE(\mu^2)=\delta (\mu^2-m^2)\mathrm d\mu^2+\mathrm d\sigma(\mu^2)\tag{2} $$

My problem is that the answer Particle/Pole correspondence in QFT Green's functions seems to suggest that a particle is an isolated eigenvalue of $H$, not $P^2$. An isolated eigenvalue of $H$ cannot be an isolated eigenvalue of $P^2$, because $$ E(\boldsymbol p)^2=\boldsymbol p^2+m^2\tag{3} $$ is continuously connected to $E(0)^2=m^2$.

Therefore, I can frame my question as follows: in order for $\varphi$ to be a particle, should it be an isolated eigenvalue of $P^2$, or $H$? or neither?

  • 2
    $\begingroup$ of $P^2$. The vectors spanned by the one particle states form a unitary irreducible representation of Poincaré group where $P^2 = m^2I$... $\endgroup$ Jan 20, 2017 at 17:38
  • $\begingroup$ @ValterMoretti thanks, that's what I thought... but then how can I make sense out of Arnold's answer? $\endgroup$ Jan 20, 2017 at 17:41
  • $\begingroup$ Arnolds answer assumes that the spatial momentum vanishes, see my answer. $\endgroup$
    – Moe
    Jan 20, 2017 at 18:00
  • $\begingroup$ Notice that in free space $P^2 \propto H$. So perhaps the question is "What is a non-free particle ?". $\endgroup$ Jan 20, 2017 at 19:01
  • $\begingroup$ Actually, the squared masses are indeed isolated eigenvalues of $H^2-P^2$, but with an infinite multiplicity, due to degeneracy generated by the Poincare group symmetry. $\endgroup$ Oct 1, 2020 at 12:30

2 Answers 2


The most rigorous way of defining a particle is as a irreducible representation of the Poincaré algebra, i.e. you classify the object "particle" by its behaviour under all possible symmetry transformation of Minkowskian spacetime.

The representation of this algebra is infinitedimensional as the Poincaré group is noncompact and therefore its representations are infinitedimensional but you can study its representations via induced representation.

We start with the momentum operators $P^\mu$ which form an invariant subalgebra (an invariant subalgebra is a set of mutually commuting generators of the Lie algebra). You can diagonalize all $P^\mu$ simultaneously (as one does with the $L_z$ and $L^2$ for the representations of SO(3), although $L^2$ is somewhat exceptional, see below) and denote their eigenstate by $|p\rangle$ with $$ P^\mu|p\rangle = p^\mu |p\rangle. $$ Obviously such a state is a eigenstate of the translation operator (i.e. a momentum eigenstate) but it transforms under a Lorentz transformation $J$ as $$ |p\rangle \rightarrow |Jp\rangle $$ as can be deduced from the commutator $[J^{\mu \nu}, P^\rho]$, just as we naively expect from the momentum of a particle.

Now the operator $P^2 = P^\mu P_\mu$ commutes with all generators of the algebra (it is called a Casimir invariant) and therefore its eigenvalues label a specific representation, i.e. are a "quantum" number, just as the eigenvalues of the $L^2$ operator labels its SO(3) via the spin quantum number. We have $$ P^2 |p\rangle = p^2 |p\rangle = m^2 |p\rangle $$ and now define the single particle state $|p\rangle$ to describe an elementary particle with mass $m$ moving with 4-momentum $p$. In this sense the mass $m$ is a "conserved quantum number" and a particle is an eigenvalue of $P^2$ classified by its mass (and its spin).

Your problem seems to be an odd framing in the answer to the pole/particle question. If you look at the LSZ reduction formula for Green's function in QFT, you will notice that the particle/pole correspondence actually arises from a pole in $$ \frac{1}{p^2 -m^2}, $$ so for $\boldsymbol{p}=0$, the eigenvalues of $H$ and $P^2$ coincide and exactly this is assumed in the linked answer:

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.

For a much better explanation of the "isolatedness" of the eigenvalue solution see Arnold Neumaiers answer.

What is missing in the above classification of a particle as an irrep. of the Poincaré algebra is the spin, which of course is also carried by elementary particles. Indeed, the algebra contains another Casimir invariant with the associated quantum number spin. With this additional quantum number, the representations can be distinguished according to their mass being greater, smaller or equal to zero. This interesting work is done via the little group.

For more information on the extremely interesting representation theoretical aspect of particles, see the Wigner's classification or the first chapter in Weinberg I.


''in order for $\varphi$ to be a particle, should it be an isolated eigenvalue of $P^2$, or $H$? or neither?''


A particle in quantum field theory is defined to have an isolated mass shell.

The reason is that, since the four components of the momentum commute, the discrete eigenspaces only appear upon considering joint eigenvalues of all components of the momentum. The joint spectrum of these is a union of mass shells, and if a mass shell is isolated, the joint eigenspace corresponding to any point on this mass shell is finite-dimensional.

Note that one has a very similar situation in nonrelativistic field theory and in translation invariant $N$-particle quantum mechanics. The only difference is that in the nonrelativistic case the mass shells are hyperplanes in momentum space rather than hyperboloids.

The criterion with isolated eigenvalues of the Hamiltonian applies only for systems where translation invariance has been removed by considering the system in its rest frame. This is equivalent of intersecting the spectrum with the line defined by zero 3-momentum, which turns an isolated mass shell into an isolated eigenvalue of $H$.


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