particle physics - If $U$ determines the mixing between fields and $U^*$ determines the mixing between states-Why?

In this pdg review, Eq. (14.1), the mixing between the flavour neutrino fields and neutrino fields corresponding to mass eigenstates are denoted as $$\nu_{lL}=\sum\limits_{j}U_{lj}\nu_{jL}\tag{1}$$ where $U$ stands for the neutrino mixing matrix. On the other hand, the mixing between flavour eigenstates and mass eigenstates, in Eq. (14.27), is given by $$|\nu_{lL}\rangle=\sum\limits_{j}U^*_{lj}|\nu_{jL}\rangle\tag{2}.$$

Why did they use $U$ in Eq.(14.1) and its complex conjugate $U^*$ in Eq.(14.27)? I guess this is not a typo because I have seen it at other places as well.

One of the questions tagged by AccidentalFourierTransform in the comment below, asks about conventions. My question is not about the convention. Having fixed $U$ for the mixing between fields, should I use $U^*$ for mixing between states? Or, having fixed $U^*$ for the mixing between fields, should I use $U$ for mixing between states?

Answer

Field operators are defined so that they annihilate states. That is, $$ \langle \Omega | \nu_{i\, L}(x) | \nu_{j,L}(\vec{p})\rangle = \delta_{ij} u(\vec{p}) e^{-ipx}$$

Hence, if the theory is invariant under the transformation $\nu_{i,L}(x) \mapsto U_{ik}\nu_{k,L}(x)$ with $U$ unitary, then we must have $|\nu_j\rangle \mapsto U^†_{lj} |\nu_l \rangle$ so that $$\langle \Omega| \nu'_i(x) |\nu'_j\rangle = U_{ik}U^†_{lj}\langle\Omega |\nu_k|\nu_l\rangle = U_{ik}U^†_{lj}\delta_{kl} = (UU^†)_{ij} = \delta_{ij}$$

Another way of looking at is that it is the conjugate field operator that creates the state, ie $|\nu_i\rangle \propto \nu^†_i(x)|\Omega\rangle$. So if the field transforms as $U$, then the states should transform as $U^*$.