Momentum eigenstates in an interacting quantum field theory

Question

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?

Answer 1

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.

Answer 2

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.