Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider 2 different Majorana fermions $\Psi_{L}, \Psi_{R}$ (physically, neutrinos). In general case I can write the massive part of lagrangian of these fermions in the form $$ L_{m} = (\bar {\Psi}_{1} \bar {\Psi}_{2})\begin{pmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{pmatrix}\begin{pmatrix} \Psi_{1} \\ \Psi_{2} \end{pmatrix} = \bar {\Psi}^{a}\hat {m}_{ab}\Psi^{b}. \qquad (1) $$ Then let's introduce matrix of some gauge interaction $$ \hat {A}_{\mu}^{ab} = A_{\mu} = \begin{pmatrix} A^{11}_{\mu} & A^{12}_{\mu} \\ A^{21}_{\mu} & A^{22}_{\mu} \end{pmatrix}_{ab}. \qquad (2) $$

and corresponding fermion's charge $g$.

The full lagrangian with interaction and mass term takes the form

$$ L = \bar {\Psi}(i\gamma^{\mu}\partial_{\mu} + g\gamma^{\mu}\hat {A}_{\mu})\Psi - \bar {\Psi}\hat {m}\Psi . $$

I want to represent the lagrangian in terms of states with definite masses. Let's call them $\Psi_{L}, \Psi_{R}$. So by introducing mixing unitary matrix

$$ \begin{pmatrix} \Psi_{L} \\ \Psi_{R} \end{pmatrix} = \begin{pmatrix} cos(\theta ) & -sin (\theta ) \\ sin(\theta ) & cos(\theta ) \end{pmatrix}\begin{pmatrix} \Psi_{1} \\ \Psi_{2} \end{pmatrix} $$ I can rewrite the mass term in a form of $$ L_{m} = (\bar {\Psi}_{L} \bar {\Psi}_{R})\begin{pmatrix} m_{1} & 0 \\ 0 & m_{2} \end{pmatrix}\begin{pmatrix} \Psi_{L} \\ \Psi_{R} \end{pmatrix}. $$ But I can't diagonalize the $A$-matrix in general case in this basis. So the $A$-matrix and $m$-matrix don't commute with each other. So there is the question: what is the physical meaning of this result in the case of Majorana fermions?

share|improve this question
    
Why would you expect that the interaction and mass matrix commute? This is not generically the case. –  JeffDror Feb 3 at 21:25
    
@JeffDror : in this case this fact (non-commuting) is somehow connected with violating some conservation law (in my opinion). I want to know, firstly, which conservation law it violates and, secondly, how exactly this violation is explained by non-commuting. –  Andrew McAddams Feb 3 at 21:35
1  
@AndrewMcAddams, can you edit your post to elaborate on your question as pertaining to conservation? As JeffDror pointed out, we shouldn't in general expect the mass eignestates to be eigenstates of the interaction. An example is the quarks under electroweak interactions; the mass states are non-diagonal under the EW mediators, which gives rise to the flavor-mixing charged current interactions (i.e. the Wqq vertices). –  chase Feb 3 at 23:25
    
@chase : your words are very similar to the answer on my question (maybe, some lepton number doesn't conserve or something like that). So can you write the answer about my special case of two Majorana fermions? –  Andrew McAddams Feb 3 at 23:31
    
@chase : and why for the Dirac massive neutrinos interaction and mass basises are equal? –  Andrew McAddams Feb 3 at 23:50
add comment

1 Answer 1

up vote 1 down vote accepted

We have no reason to expect, in general, that mass eigenstates and interaction eigenstates should be the same. In your example, the Lagrangian written the mass-diagonal basis $\Psi_L,\Psi_R$ will contain additional mixing terms: $$ \bar\Psi_R \hat A_{RL} \Psi_L + \bar\Psi_L \hat A_{LR} \Psi_R $$

If $\hat A$ is a gauge field, physically, these terms represent a vertex:

An example of this in the Standard Model is the flavor-changing charged-current interactions which couple up-type and down-type quarks at vertex with the W boson.

If the $\hat A$ is just some non-diagonalizable potential, you can interpret this as a mass insertion or kinetic mixing term, where the species $\Psi_L$ may spontaneously oscillate into $\Psi_R$ and visa-versa. Physically, this is similar to the oscillation of neutrino species as governed by the elements of the PMNS matrix.

share|improve this answer
    
Thank you! But how to show that non-diagonalized $\hat {A}$ ($\hat {A}$ isn't gauge potential because lagrangian doesn't have SU(2) symmetry) refers to the oscillation? Then what will change if it is diagonalized in mass basis? –  Andrew McAddams Feb 4 at 0:07
    
Is it connected with the connection between the lagrangian of interaction and the amplitude of possibility of "scattering" for transformations $\Psi_{R} \to \Psi_{L}$ and vice versa? –  Andrew McAddams Feb 4 at 0:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.