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

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    $\begingroup$ if any row or column squared sums to something less than one (experimentally), this is an indication that we're missing something, e.g. a fourth row/column corresponding to an additional species of neutrinos. $\endgroup$ Nov 7, 2014 at 9:51

3 Answers 3


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 : $⟨\nu_{e}|\nu_{e}⟩=1=(U_{e1}^{*}⟨\nu_{1}|+U^{*}_{e2}⟨\nu_{2}|+U^{*}_{e3}⟨\nu_{3}|)( U_{e1}|\nu_{1}⟩+U_{e2}|\nu_{2}⟩+U_{e3}|\nu_{3}⟩ )$



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.


It really goes deeper than just a theoretical demand on a particular domain. The time-evolution operator for any system must be unitary, because that preserves the total probability at one. And the PMNS matrix appears as a factor in the time evolution operation for neutrino mixing.

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 students often don't take much note of it.

  • $\begingroup$ 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 $\endgroup$
    – agemO
    Apr 6, 2014 at 10:38
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    $\begingroup$ What do you make of this note in the wiki article? en.wikipedia.org/wiki/PMNS_matrix#cite_note-1 $\endgroup$
    – innisfree
    Apr 6, 2014 at 10:49
  • $\begingroup$ 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. $\endgroup$
    – innisfree
    Apr 6, 2014 at 10:52
  • $\begingroup$ Do you really mean this: "The Hamiltonian for any system must be unitary"? $\endgroup$
    – user22180
    Aug 10, 2014 at 13:49
  • $\begingroup$ @user22180 It must be the complete Hamiltonian, with all corrections and the bits we would normally neglect, and I should probably add at least one more weasel word line "isolated" to imply that if we hold some interaction out as "external" this rule no longer applies; but yes or the reason explained. $\endgroup$ Aug 10, 2014 at 15:59

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.

  • $\begingroup$ @innisfree- Neutrino mass matrix is not hermitian. If you are talking about effective mass matrix, after seesaw, even then it is in general complex symmetric but not hermitian. So your second reasoning doesn't apply. However, a complex symmetric matrix M can be diagonalized by a unitary transformation as $D=U^TM U^*$. $\endgroup$
    – SRS
    Sep 21, 2016 at 9:20
  • $\begingroup$ Yes I'll fix this, that's not right is it. $\endgroup$
    – innisfree
    Sep 21, 2016 at 21:55
  • $\begingroup$ I meant $D=M_{diag}=U^TMU$ with U being unitary, and there will be complex conjugation of U on the right. I was careless about that. $\endgroup$
    – SRS
    Sep 22, 2016 at 6:49

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