# 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?

• There is also a feeble excuse that a field destroys the state with the same labels, but most practitioners shrug it off and simply check the self-consistency of their calculations. – Cosmas Zachos Jun 16 '18 at 14:19
• @CosmasZachos My question is, if the mixing between fields is governed by $U$, will the mixing between states also governed by $U$? – SRS Jun 16 '18 at 14:28
• Of course !! The field just destroys a particle with the same name and creates an antiparticle with the C q-numbers. – Cosmas Zachos Jun 16 '18 at 14:43
• I would guess that it is because, as mentioned already, a neutrino field annihilates neutrinos. To get the phases to match up for the states, you can work with antineutrino fields instead, since they create neutrinos. But the antineutrino fields are related by $U^*$, not $U$. – knzhou Jun 16 '18 at 14:44

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 $U^*$.