# Weinberg Sec 5.1 - Why is the transformation of fields in QFT independent of position?

In Weinberg's Quantum Theory of Fields Vol. 1, it is claimed that annihilation and creation fields should transform as $$U_0(\Lambda, a)\psi_l^+(x)U_0^{-1}(\Lambda, a) = \sum_\bar{l}D_{l\bar l}(\Lambda^{-1})\psi_{\bar l}^+ (\Lambda x+a).\tag{5.1.6}$$

This is equation (5.1.6). But I do not see why this is the only way the creation and annihilation operators can transform in order to build up a $$H(x)$$ that satisfies the cluster decomposition principle. In particular, why should $$D(\Lambda^{-1})$$ be position independent? Presumably we could write down a field with a transformation that is different at each point in position, and then contract the indices so that we have a local $$H(x)$$?

• $D$ is a representation of the Lorentz group. You can use this to build a representation of the Poincare group by taking a field at the origin $x=0$ and then defining its Lorentz representation at that point. Then you use the translation generators to translate it to a generic point $x$. Following this construction it is clear that the $x$ dependence of $D$ would be completely fixed by symmetry arguments. The fact that it is independent of $x$ is a consequence of the calculation. Feb 4, 2022 at 14:27

For simplicity just consider a rotation $$R$$ at the reference momentum $$k$$ (the momentum chosen to fix the little group) $$U(R) a(k,\sigma) U^{-1}(R)= \sum_{\bar{\sigma}}D^{(j)}_{\bar{\sigma\sigma}}(R)a(k,\bar{\sigma})$$ This $$D^{(j)}$$ matrix acts on the spin index in the quantum mechanical operators, and it is not position dependent. The $$D_{l\bar{l}}$$ matrix in your question acts on the field index, and it ultimately will come from the $$D^{(j)}$$ matrix given suitable chosen coefficient functions $$u,v$$. This is the content of eq (5.1.23) and figuring out the coefficient functions from this relation makes up the bulk of chapter 5.
• I think I understand what you're saying. I have one last question though - what about the independence of $b$? i.e. why do the states transform in a representation of the Lorentz group and not the Poincare group? Feb 5, 2022 at 12:07
• The states do transform under the full Poincare group, just for simplicity I decided to only consider rotations to highlight the main issue. If you have a non-zero translation $b$ then all that happens according to (5.1.11) is the ladder operators pick up an exponential factor since they generate eigenstates of 4-momentum. So that means that in order to have the x coordinate of the field transform appropriately, all the x dependence must also be given by an exponential factor as in (5.1.15) Feb 5, 2022 at 16:14
• Let me clarify what I mean - I see that the fields do transform in a rep of the Poincare group, but why do the components of the fields not mix under translations? i.e. why can we not have $D_{l \bar{l}}(\Lambda^{-1}, b)$ in (5.1.6). Is it not an ad hoc restriction to ignore this possibility? Feb 6, 2022 at 10:08
• What my answer is saying is that Weinberg in a sense explained it out of order. He is not assuming (5.1.6), this form is dictated by transformation properties of the ladder operators. Start with the unnumbered equations just below eq (5.1.11 and 12) on the top of page 194. The transformation is already done, the only thing you are free to choose is the form of the c-number functions $u_l(x)$ and $v_l(x)$. We want to choose them so that a under a translation $b$, x changes to $x+b$. The only possible choice that works is (5.1.15 and 16), after this we have completely fixed our $x$ dependence. Feb 6, 2022 at 20:40
• In this case the "problem" is solved neatly, but I think in general, if you wanted to add translations to the finite-dimensional representation (that is, have $D_{l \bar{l}}(Λ^{−1},b)$), it can do only trivially. One of the Wightman axioms expresses this as well, in general what we'd expect in a field theory (in flat spacetime). Feb 8, 2022 at 21:29