# Unitarity of PMNS matrix

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Why should the neutrino mixing matrix (PMNS matrix) be unitary? Is the unitarity dictated by experiments or is it a theoretical demand?

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user Roopam
asked Apr 5, 2014

## 3 Answers

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It's a theoretical demand : $$\begin{pmatrix} \nu_{e}\\ \nu_{\mu}\\ \nu_{\tau} \end{pmatrix} = \begin{pmatrix} U_{e1} & U_{e2} & U_{e3} \\ U_{\mu1} & U_{\mu2} & U_{\mu3} \\ U_{\tau1} & U_{\tau2} & U_{\tau3} \end{pmatrix} \begin{pmatrix} \nu_{1}\\ \nu_{2}\\ \nu_{3} \end{pmatrix}$$

You know that all states are normalized, for example : $⟨\nu_{e}|\nu_{e}⟩=1=(U_{e1}^{*}⟨\nu_{1}|+U^{*}_{e2}⟨\nu_{2}|+U^{*}_{e3}⟨\nu_{3}|)( U_{e1}|\nu_{1}⟩+U_{e2}|\nu_{2}⟩+U_{e3}|\nu_{3}⟩ )$

so

$U_{e1}^{*}U_{e1}+U_{e2}^{*}U_{e2}+U_{e3}^{*}U_{e3}=1$

You can do the same for the whole matrix and find $U^{+}U=I$

EDIT : as dmckee pointed out it's a general feature in quantum mechanics, the matrix you use to change the basis (here from mass eigenstate to flavour eigenstate) must be unitary.

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user agemO
answered Apr 5, 2014 by (40 points)
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It really goes deeper than just a theoretical demand on a particular domain. The Hamiltonian for any system must be unitary, because that preserves the total probability at one.

This is important because if I start with some state and let it evolve for a while the system must afterwards exist in some state which means that the sum of the probabilities taken across all final states must come to 1. Otherwise things can undergo---in the words of Douglas Adams---"a sudden and gratuitous total existence failure".

Nor is it acceptable to start with a single state and end up with the probability to exist in one of all possible state larger than one. What would that even mean? Sudden and gratuitous extra existence?

This was probably mentioned on the first day you started studying quantum mechanics, but it is so obvious that student often don't take much note of it.

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user dmckee
answered Apr 5, 2014 by (420 points)
The PMNS matrix is not an Hamiltonian, but you're right, it's more general, any change of observable basis (here from mass egeinstate to flavor eigenstate) must be done with a unitary matrix so that probability are conserved

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user agemO
What do you make of this note in the wiki article? en.wikipedia.org/wiki/PMNS_matrix#cite_note-1

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user innisfree
oh i see what it means, in the see-saw model, the $3\times3$ (flavor) PMNS mixing matrix might not be unitary, but the full mixing matrix with all flavors and LH and RH neutrinos must be unitary. a bit misleading, but i suppose it;s true that $3\times3$ PMNS need not be unitary.

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user innisfree
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I will offer two reasons. First, unitarity of mixing matrices insures that probabilities sum to one. The probability of an oscillating neutrino having electron, muon or tau flavour should equal one.

Second, because the neutrino mass matrix is Hermitian it is diagonalised by a unitary matrix.

This post imported from StackExchange Physics at 2014-04-13 14:45 (UCT), posted by SE-user innisfree
answered Apr 5, 2014 by (295 points)

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