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 '15 at 0:51
  • $\begingroup$ To 1) : They are not different since 1-particle states don't scatter . $\endgroup$ – jjcale Jul 2 '16 at 12:16

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 '15 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 '15 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 '15 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 '15 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 '16 at 12:11

I am thinking about this too. I decided on the following. The one particle states we are talking about are always one particle states of the Full interacting theory. Therefore, the in field operators are not operators of the free theory. They are just the field operators at t -> inf. They are still very much operators of the full theory, acting on states in the Hilbert space of the full theory. However, the infield operators have the property of satisfying the Klein-Gordon equation. So actually the single particle states are created by them. When you hit a single particle state with an infield operator you just create or destroy 1 particle. Bear in mind this is all happening in the full interaction theory Hilbert space. We are not going back and forth between free and interacting theory!

What happens when you use a field operator (not infield, just plain old phi) to act on single particle states? You get a mess, the state becomes a superposition of many particles. Even though that happens, eqn 16.36 still becomes zero, because it just relies on the fact that the single particle states are momentum states on the mass shell $(p^{2} - m^2 = 0)$, and some properties of the P operator (of the full theory).

So in summary, all the states, operators and fields (including infields) are of the Full theory. There is no contradiction!

One thing Im struggling to see is why in the interacting theory, you need to rescale the fields. Its easy to get lost in the long chain of arguments. Is there an example or analogy from perturbation theory of single particle QM?


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