This post was originally written to ask about transformation properties of fields in the interaction picture of QFT under the Poincare transformations. Arnold Neumaier has pointed out that the question is vacuous, because there is no such thing as interaction picture in relativistic QFT. As far as my (not very educated, I must admit - so please correct me if I'm wrong) understanding goes, the problem is that interacting and non-interacting theories are in a way "infinitely different" and one cannot treat both representations of Poincare group and both field operators as well-defined operators acting in the same Hilbert space. This problem is perhaps one of the reasons why QFT is plagued with mathemathical problems. The approach applied by physicists who actually calculate outcomes of experiments is to neglect this problem, treating it as a technical difficulty and then cure resulting inconsistencies and infinities by renormalizaton procedure, leading to consistent results on the level of perturbation theory. In the remainder of this post I will play the game of assuming that interaction picture actually exists, and all calculations presented are purely formal manipulations.

First let's explain notations. Following development in Weinber's book vol. 1, chapters 2 and 3 I expect two representations of Poincare group to exist. $U_0(\Lambda,a)$ transforms asymptotic (in and out) states. $U(\Lambda, a)$ is representation of Poincare group of interacting theory. The relation between the generators of two representations is $H=H_0+V$, $\vec P = \vec P_0$, $\vec J = \vec J_0$, $\vec K = \vec K_0 + \vec W$, where $\vec W$ is correction term which needs to satisfy certain conditions to yield a Lorentz covariant theory (in the sense that $U_0$ commutes with the $S$ operator) - details are described in Weinberg. $\phi(x)$ is some quantum field in the Heisenberg picture. It can be a tensor, spinor or whatever and I neglect relevant indices. It transforms according to $$ U(\Lambda, a) \phi(x) U(\Lambda,a)^{\dagger}= D(\Lambda ^{-1}) \phi (\Lambda x +a), $$ where $D$ is some finite dimensional representation of Lorentz group. This is the matrix which acts on Lorentz or spinor indices of $\phi$. Therefore saying what is $D$ is equivalent to specifying the algebraic type of the field $\phi$ - for example $D(\Lambda)=1$ corresponds to a scalar field, $D(\Lambda)=\Lambda$ is a vector field etc.

Now I choose an inertial frame and attempt to define fields in the interaction picture $\phi_I(x)$ as $$ \phi_I(x) = U_0(x) \phi (0) U_0(x)^{\dagger}. $$ From the definition, $\phi_I(x)=\phi(x)$ if $x^0=0$ but not otherwise. This definition can be rewritten as $$ \phi_I(x) = U_I(x) \phi(x) U_I(x)^{\dagger}, $$ where $U_I(a)=U_0(a)U(a)^{\dagger}$. It follows easily from the definitions so far that if $R$ is a rotation and $a$ and arbitrary vector then $$ U_0(R,a) \phi_I(x) U_0(R,a) = D(R^{-1}) \phi_I(R x +a), $$ and in this sense interaction picture fields $\phi_I(x)$ transform as free fields under the Poincare group. However, I fail to reproduce the same result if rotation $R$ is replaced by boost or some general Lorentz transformation. The best I can get is $$ U_0(\Lambda, a) \phi_I(x) U_0(\Lambda,a)^{\dagger}=D_0(\Lambda^{-1}) \phi_I(\Lambda x +a), $$ where $D_0$ is some, but not necessarily the same representation of Lorentz group. This seems odd, because introduction of interactions shouldn't change the algebraic character of the fields.

My question can be put as follows: Is there a way to argue that (at least under some assumptions about interactions) the representations $D$ and $D_0$ are equal?

This post imported from StackExchange Physics at 2016-08-31 10:21 (UTC), posted by SE-user Blazej