Context for the following questions: two widely stated claims hinge on what appears to be an inconsistent argument. The claims are that (1) an interacting field can produce, in addition to 1-particle states, continuum states as well and (2) imposing a strong asymptotic condition--$\lim_{x^0\rightarrow -\infty} \varphi(x) \rightarrow \sqrt Z \varphi_{in}(x)$--leads to a contradiction, and that one needs to use a weaker asymptotic condition, instead. I am adding this preamble in an attempt to counter the impression that this is about some obscure technicality.

  1. Are the 1-particle momentum eigenstates of an interacting field (say $\phi^4$ theory) different from the 1-particle momentum eigenstates of the corresponding in-field? I am assuming that the 1-particle states of the interacting field are the eigenstates of the full interacting Hamiltonian, while the 1-particle states of the in-field are the eigenstates of the Hamiltonian of the free theory whose mass parameter is the renormalized mass of the interacting field.

  2. If the answer to (1) is yes, as I suspect it is, then I am confused by the ambiguous interpretation of 1-particle kets in equations 16.36 and 16.38 in Bjorken and Drell. In eq. 16.36 it appears that one interprets the 1-particle ket as the momentum eigenstate of the interacting field, and in 16.38 one interprets the same ket as the momentum eigenstate of the corresponding in-field. Am I missing something?

Here are the relevant equations:

$(\Box + m^2) \varphi(x) = j(x)$

where $j(x) := \lambda \varphi^3(x) + (m^2-m_0^2) \varphi(x)$ for the $\varphi^4$ theory with $m_0$ being the mass parameter and $m$, the renormalized mass.

$\varphi_{in}$ is defined by the equation

$\sqrt{Z} \varphi_{in}(x) = \varphi(x) - \int d^4 y \ \Delta_{ret} (x-y;m) j(y)$

where $\Delta_{ret}$ is the retarded Green's function (vanishes for $x^0 < y^0$) that satisfies the equation

$(\Box_x + m^2) \Delta_{ret}(x-y;m) = \delta^4(x-y)$.

Consider the matrix element,

$\langle 0 | \varphi(x) | p\rangle = \sqrt Z \langle 0 | \varphi_{in}(x) | p\rangle + \int d^4y \ \Delta_{ret}(x-y;m) \langle 0 | j(y) | p\rangle$

Eq. 16.36 (Bjorken, Drell):

$\langle 0 | j(y) | p\rangle = (\Box + m^2) \langle 0 | \varphi(y) | p \rangle = (\Box + m^2) e^{-ip.y} \langle 0 | \varphi(0) | p\rangle = (p^2-m^2) \langle 0 | \varphi(y) | p \rangle =0$

The above equation uses the translation invariance $\varphi(y) = e^{i \hat P^\mu y_\mu} \varphi(0) e^{-i \hat P^\mu y_\mu}$, where $\hat P^\mu$ is the 4-momentum operator for the interacting field theory.

Equation 16.38:

$\langle 0 | \varphi_{in}(x) | p\rangle = \int d^3 k \frac{e^{-ik.x}}{\sqrt{(2\pi)^3 2 \omega_k}} \langle 0 | a_{in}(k) | p\rangle = \frac{e^{-ip.x}}{\sqrt{(2\pi)^3 2\omega_p}}$

The same $|p\rangle$ appears to be used as the eigenket of both $\hat P^\mu$ of the interacting field, as well as the ket generated by acting the creation operator of the in-field on the in-field vacuum. Am I missing something?

  • $\begingroup$ Welcome to Physics! Added to the above comment is that this site has MathJax enabled, which means you can use Latex-like syntax to add in equations for readability. $\endgroup$
    – Kyle Kanos
    Jul 24, 2015 at 0:51
  • $\begingroup$ To 1) : They are not different since 1-particle states don't scatter . $\endgroup$
    – jjcale
    Jul 2, 2016 at 12:16

2 Answers 2


The "1-particle" momentum eigenstates of the free theory are certainly not those of the interacting theory! For one, if the field is not free, we do not have access to its mode expansion in the usual way, and it becomes unclear what a "1-particle state" is supposed to be.

Additionally, Haag's theorem states that the interacting Hilbert space is unitarily inequivalent to the free Hilbert space, so the states of the free theory and those of the interacting theory should be thought of as lying in completely different Hilbert spaces.

Whenever you see something like "n-particle momentum state", it should mean a free state, possibly in the asymptotic past/future as would be usual in the LSZ-formalism for scattering.

  • $\begingroup$ In that case there appears to be a problem in the argument that shows that the interacting field can produce continuum states, unlike the free field (that is, the argument that Z < 1, where phi(x) -> \sqrt{Z} phi_{in}(x) in the asymptotic past). The 1-particle states appear to be used inconsistently in 16.36 and 16.38 of Bjorken and Drell's book, unless I'm missing something. $\endgroup$
    – Coriolis
    Jul 24, 2015 at 0:35
  • $\begingroup$ Regarding your comment "it becomes unclear what a "1-particle state" is supposed to be", can't one reasonably assume that the eigenstates of the full Hamiltonian (Hermitian) operator exist and the subspace corresponding to P^2 = m^2 constitute 1-particle states, even if it is difficult to construct them explicitly? $\endgroup$
    – Coriolis
    Jul 24, 2015 at 0:39
  • $\begingroup$ Do we have any actual example of the difference between the two, say in any exactly solvable interacting QFT model like the Thirring model? $\endgroup$
    – Slereah
    Jul 24, 2015 at 8:52
  • $\begingroup$ @Slereah To see the non-trivial effects of interaction you need to consider models where the renormalization is necessary. There are few where it has been done non-perturbatively, and are quite difficult to investigate. Nevertheless the Haag's theorem has been proved to hold in those models, and the interacting fields have been constructed. As a (very technical) reference you should see e.g. the works of Glimm and Jaffe on the $(\varphi^4)_3$ model $\endgroup$
    – yuggib
    Jul 24, 2015 at 9:39
  • $\begingroup$ 1-particle states correspond to irreducible representations of the Poincare group which are associated to the positive discrete part of the mass spectrum. So if the discrete part of the mass spectrum contains a positive value then there are 1-particle states. $\endgroup$
    – jjcale
    Jul 2, 2016 at 12:11

You proved: $\langle 0|\phi(x)|p\rangle=\sqrt{Z}\langle 0|\phi_{in}(x)|p\rangle$

Then you try to imply that because $\vert p\rangle$ is the one particle ket of $a_{in}(p)$, so $\vert p\rangle$ must also be the one particle ket of $a(p,t)$ (note how weird this gets given the time dependence of $a(p,t)$). Anyway, this logic is wrong, because that would require the property:

$\langle k|\phi(x)|p\rangle=\sqrt{Z}\langle k|\phi_{in}(x)|p\rangle$ for all k and p.


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