In standard textbooks in QFT while discussing e.g. the Kallen-Lehmann formula (see e.g. Section 7.1 in the Peskin-Schroeder book) it is always assumed that bound states of two or more particles have higher mass than the one particle states. Why this should be true?

Let us compare this for example with the classical two particles interacting according to the Coulomb law.They may rotate around each other with fixed distance $R$ such that the center of mass is at rest, thus forming a bound state. The total energy goes to $-\infty$ when $R$ goes to 0.

Analogous problem arises in the case of two non-relativistic quantum particles interacting again according to Coulomb law. While the total energy of the bound state cannot be arbitrarily small as in the classical case, the discrete energy levels (corresponding to bound states) are negative. At the same time if the particles are at rest and are far apart from each other and thus do not interact, their energy vanishes.

Am I wrong?

  • 1
    $\begingroup$ related: Why does the branch cut of the self-energy begin at $2m$?. $\endgroup$ Oct 30, 2018 at 14:26
  • $\begingroup$ I think it's just that one is implicitly assuming that he/she is looking at the lightest particles in the spectrum. If the bound states were becoming lighter than that, you would not longer call them bound states and you would start looking at their fields two-point function instead. $\endgroup$
    – TwoBs
    Oct 30, 2018 at 19:09
  • $\begingroup$ @TwoBs: I do not think so. There are elementary fields entering the Lagrangian, and Kallen-Lehmann theorem says something on propagator of such fields; it is necessary eventually to justify Feynmann’s rules. But if there is a bound state lighter than any one particle state corresponding to an elementary field, it is not clear how to construct new field corresponding to it and rewrite the Lagrangian in its terms. $\endgroup$
    – MKO
    Oct 30, 2018 at 19:34
  • $\begingroup$ @MKO kallen-Lehmann works for any type of field, composite or not. And from KL I don't think you can exclude simple poles in $0< \mu^2 <m^2$. As I said, it's implicit associating to the lightest particle a certain field whose KL decomposition reveals that it can generate other, heavier, poles too. $\endgroup$
    – TwoBs
    Oct 30, 2018 at 22:41
  • $\begingroup$ @TwoBs: Well, consider the case of QFT with one scalar field $\phi$ of physical mass $m>0$ with a self interaction say $\phi^4$. Is it true that all bound states in the theory have mass greater than $m$? $\endgroup$
    – MKO
    Oct 31, 2018 at 1:34

1 Answer 1


The two(or more)-particle states you are referring to are not bound states. They are scattering states of multiple particles. Recall that irreducible representations $[m,s]$ of the Lorentz group are labeled by two Casimir operators, the invariant mass:

$$ M^2 = - P_\mu P^\mu \ ,$$

where $P_\mu$ is the energy-momentum vector and the eigenvalues will be denoted by $m^2$; and

$$ - W^\mu W_\mu \ ,$$

where $W_\mu$ is the Pauli-Lubanski pseudovector, but this is just for completeness (this has eigenvalues $s(s+1)$) .

Now the one-particle states that the QFT books consider are simply the representation with just one $[m,s]$. The two-particle states are states $[m,s] \otimes [m',s']$ etc. If we have such a many-particle representation, we can decompose it into irreducibles, but we will get a continuous family of representations from $M = m+m'$ to $\infty$.

Bound states, on the other hand, are one-particle representations with some definite $m$. That such a bound states may be formed from other particles simply means that there is a transition matrix element.

In particular, to get an unitary S-matrix, one has to assume that all bound states are included as particle states in the asymptotic Hilbert spaces.

  • $\begingroup$ I think I do refer to bound states rather than to scattering states of multiple particles - see Section 7.1 in Peskin-Schroeder book. $\endgroup$
    – MKO
    Oct 30, 2018 at 14:49
  • $\begingroup$ Once a bound state has formed, it is a single-particle state (it is a one-particle representation of the Poincare group). And for their mass i can't find any assumption in P&S. $\endgroup$ Oct 31, 2018 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.