Ron Maimon wrote "*In the Dyson-style interaction picture, there's a turning on-and-off function \(f(t)\) that describes the interaction strength. This turning on and off is supposed to be adiabatic, so that the bare vacuum slowly relaxes into the interacting vacuum, the scattering happens, and then you turn off the interaction again.*"

OK let us see it on a toy model. Let us consider a particle in a 1D box and with some regular time-independent potential \(V(z)\) in it. For simplicity we can choose a linear potential. Then the exact eigenfunctions \(\psi_n(z)\) will be combinations of Airy functions. It can physically be a neutron jumping in the gravitational field between two mirrors, so \(V(z)=g\cdot z\) and \(\psi_n(z)=N_n\left[Ai(z-z_n(L))-Bi(z-z_n(L))\frac{Ai(-z_n(L))}{Bi(-z_n(L)}\right]\), where \(z_n(L)\) are the roots of the equation \(\psi_n(L)=0\).

The "bare" eigenfunctions are just \(\varphi_n (z)\propto\sin\frac{\pi (n+1)z}{L};\;n\ge0\). Let us compare the ground (vacuum) states in case when \(L\) is sufficiently big to have only one Airy function in \(\psi_0\). Qualitatively the ground state functions are given in Fig. 1.

They are different.

We can represent \(\psi_0 (z)\) as a spectral sum \(\psi_0(z)=\sum p_n \varphi_n (z)\), so what? Does that mean the interacting vacuum \(\psi_0\) "contains" bare states or bare particles? No. No, Ron, no!

There is no way to observe physically those \(\varphi_n\) in \(\psi_0\), nor exists a mathematical way to "extract" \(\varphi_n\) from \(\psi_0\). Only their sum makes sense, namely, it represents the ground state. This statement about dumb functions \(\varphi_n\) is valid for any function \(F(z)\) expanded in the Fourier series. Saying that interaction populates the vacuum state with bare particles ("condensat", according to Lubosh Motl or "constant density" particles, according to you) is a very sloppy and misleading way of expressing this simple Fourier decomposition. This point should be clear.

Now, Dyson interaction picture. Although some people use the adiabatic function \(f(t)\) in the time evolution calculations, in fact it serves just to start perturbative calculations from the zeroth-order approximation (or "bare" states in the above sense). If you analyze carefully what is going on with such a calculation, you will see nothing, but dressing. It means even in the initial state we have a dressed state rather than a bare one. Let us see it. For that, let us calculate the "time evolution" of a bare vacuum state \(\varphi_0(z) e^{-iE_0^{(0)} t}\) due to turning the potential \(V(z)\) on with help of \(f(t)\) (no time-dependent scattering potential is considered for instance). The Dyson's solution can be represented as a time ordered exponential with the interaction potential \(V(z)\) in it. If you make the first order calculation, you will see that the potential \(V(z)\) transforms \(\varphi_0(z) e^{-iE_0^{(0)} t}\) into \(\left[\varphi_0 (z) +\sum_{m>0}\frac{V_{m0}\varphi_m(z)}{E_0^{(0)}-E_m^{(0)}}\right]e^{-i(E_0^{(0)}+V_{00}) t}=\psi_0^{(1)} e^{-iE_0^{(1)} t}\). In the *k*-th order, the Dyson's solution gives \(\psi_0^{(k)} e^{-iE_0^{(k)} t}\) where \(\psi_0^{(k)}\) and \(E_0^{(k)}\) are just the usual *stationary perturbation theory solutions* of the *k*-th order. So the Dyson's solution is the dressed vacuum state \(\psi_0 e^{-iE_0 t}\) at **any **time \(t\), Ron. At any time means at the initial time too. There is no trace of \(f(t)\) here because it is always on.

If we add now a scattering potential, we will factually start from the dressed states, we will scatter them, and we will finish with dressed states, and no bare states will be involved, strictly speaking (Fig. 2). Representing the dressed states as spectral decompositions over bare states does not involve the latter physically. This is the true picture in QFT. The scattering phenomenon (or, better, an additional time-dependent external potential) makes the initial dressed state \(\psi_0 e^{-iE_0 t}\) evolve into a *physical* superposition \(\psi_0 (z)e^{-iE_0 t} \to\sum C_n\psi_n(z)e^{-iE_n t}\) with observable stated \(\psi_n(z)\), \(C_n(t)\) being the occupation number amplitudes. The external potential turns on and off itself, without our help.

Unfortunately, our perturbative spectral representation of solutions \(\psi_n(z)\) in course of occupation amplitude calculations confuses two perturbative calculations - dressing calculation (=time-independent thing) and scattering calculation (=time evolution of occupation number amplitudes \(C_n(t)\)). That's the reason of your opinion about bare particles being present in the initial/final states and in the physical vacuum.