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The neutrino its own antiparticle?

  1. Oct 17, 2009 #1
    I recently discovered that there is a possibility that neutrinos are their own antiparticles. However, I cannot see how lepton number could be conserved if this were so, can anyone help explain it to me?

  2. jcsd
  3. Oct 17, 2009 #2
    The lepton number should be equal to zero in this case.
  4. Oct 17, 2009 #3

    take the example of beta- decay:

    neutron >>> proton + electron + anti electron neutrino

    usually we say that lepton number is conserved as there is a lepton number of 0 on the left and a lepton number of zero on the right (+1 -1=0) But if the neutrino is its own antiparticle, then how can an anti electron neutrino have a negative lepton number if it's actually the same as an electron neutrino?
  5. Oct 17, 2009 #4
    You see, if the neutrino is its own antiparticle, the ends do not meet.
  6. Oct 17, 2009 #5
    What do you mean? "the ends"?
  7. Oct 17, 2009 #6
    I mean if you suppose the neutrino to be truly neutral (particle=antiparticle), it does not match the experimental data.
    Last edited: Oct 17, 2009
  8. Oct 17, 2009 #7
    so it can't be its own antiparticle?
  9. Oct 17, 2009 #8
    No, it can't, unfortunately.

    Despite its electric charge is equal to zero, the neutrino has some other non-zero "charges" (lepton number, for example).
  10. Oct 17, 2009 #9

    Vanadium 50

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    I'm afraid that nothing that Bob for Short posted is accurate.

    A neutrino can be a particle with a distinct antiparticle. Every other lepton we have discovered acts this way, and we call them Dirac particles. However, it's also possible for the anti-particle of the neutrino to be itself: the left-handed neutrino's antiparticle is the right-handed neutrino. These are called Majorana particles, and the only reason this possibility is admitted is because neutrinos are neutral.

    If this were the case, a left handed chiral (massless) neutrino would have lepton number +1, and a right handed one would have lepton number -1. A physical neutrino would be a mix of these two states, and you would have violation of lepton number. However, since the neutrino is almost massless, the mixing is small: it will go something like (m/E)2, where m and E are the mass and energy of the neutrino.

    Again, if this were the case, the violation of lepton number would allow certain radioactive decays to take place, albeit at a very slow pace. For example 130Te could decay to 130Xe plus two electrons. By slowly, I mean slowly - we're talking about half-lives of the order 1025 years.

    People are searching for such decays, without success. These are not simple measurements: we're we're talking about 1 decay per month or year per kilogram of the isotope in question. Background radiation is tens or hundreds of millions of times larger.
  11. Oct 17, 2009 #10
    I am sure that everything that Vanadium_50 posted is accurate, unlike my posts.
  12. Oct 17, 2009 #11
    http://en.wikipedia.org/wiki/Ettore_Majorana" [Broken] :confused:
    Last edited by a moderator: May 4, 2017
  13. Oct 17, 2009 #12
    You are both accurate, it's just a point of view. Left-handed neutrino can't be an antiparticle of left-handed antineutrino, because that would mean blatant nonconservation of lepton number ... however, left-handed neutrino can be an antiparticle of right-handed antineutrino, because neutrinos have no electromagnetic or strong charges, and the only force via which they interact is the weak, which only couples to one chirality.

    Further more, if neutrino were truly massless, it could be its own antiparticle (in this sense of the term) and we'd be none the wiser. You can't really tell the difference between a SM Dirac neutrino, which has four components, two of which are "sterile" and never interact with any other known fields, and a massless Majorana neutrino, which has only two "physical" components.

    Because the anti electron neutrino produced this decay has right handed chirality, and electron neutrinos are left handed. If not for Majorana mass term, those wouldn't couple to each other and you wouldn't be able to observe any nonconservation of anything.
  14. Oct 18, 2009 #13
    To clarify some of this discussion, it's probably worth stating the following explicitly. The standard model contains a left handed neutrino and its CPT conjugate, the right handed antineutrino. In order for neutrinos to have mass, there must also be a right handed neutrino and a left handed antineutrino. If neutrinos get mass the same way that all the other standard model fermions do, the right handed neutrino and left handed antineutrino come from a different fundamental field than the SM neutrino.

    On the other hand, since neutrinos carry no conserved gauge charges, it is possible that the left handed neutrino is, in fact, identical to the left handed antineutrino (and similarly for the right handed states). In that case, the neutrino is a Majorana particle and lepton number is explicitly not conserved.

    All this said, if the neutrino is a Majorana particle, its mass violates the SU(2)xU(1) electroweak symmetry, since the mass term violates the conservation of any charge. This suggests that neutrinos should still be massless at scales above electroweak symmetry breaking, meaning that the neutrino mass will depend on the Higgs vev. There are several ways to generate a naturally small neutrino mass at electroweak symmetry breaking, which are collectively referred to a See-Saw mechanisms. For example, in the classic type 1 See-Saw, the SM neutrinos couple to a separated right handed neutrino field via a Yukawa coupling to the Higgs. This term generates a Dirac mass term. But, the right handed neutrinos, being uncharged under all SM gauge groups, can also have a Majorana mass without breaking any symmetry except for lepton number conservation. If the Majorana mass, M, is much larger than the Dirac mass, m, the mass eigenstates are Majorana particles with mass approximately given by M and [itex]\frac{m^2}{M}[/itex].
  15. Oct 19, 2009 #14
    neutrino + antineutrino annihilation...

    What happens in neutrino + antineutrino annihilation?

    The two particles meet at a single point and annihilate each other, producing a virtual Z boson, which is the neutral (i.e. no electric charge) carrier of the weak nuclear force. This Z boson then immediately decays to produce another particle/antiparticle pair, either a new pair of neutrinos, two charged leptons, or a quark/anti-quark pair:
    [tex]\bar \nu + \nu \rightarrow Z^0 \rightarrow \bar e^+ + e^-[/tex]

    Conservation of classical weak leptonic current:
    [tex]\partial_{\mu} J_{\mu}^L = \sum_j \partial_\mu(\bar \nu_j Z_\mu^0 \nu_j) = 0[/tex]

    Conservation of Energy states that what you can produce depends on how much energy there is available from the colliding neutrinos.

    Violations of the lepton number conservation laws:
    In the Standard Model, leptonic family number (LF) would be preserved if neutrinos were massless. Since neutrinos do have a tiny nonzero mass, neutrino oscillation has been observed, and conservation laws for LF are therefore only approximate. This means the conservation laws are violated by chiral anomalies, which results in anomalous leptonic current, although because of the smallness of the neutrino mass they still hold to a very large degree for interactions containing charged leptons. In other words, the classical leptonic current [tex]J_{\mu}^L[/tex] is conserved under the Standard Model:

    [tex]\partial_{\mu} J_{\mu}^L = \sum_j \partial_\mu(\bar e_j^+ \gamma_{\mu} e_j^-) = 0[/tex]

    Thus, it is possible to see rare muon decays such as:

    [tex]\begin{matrix} & \mu^{-} & \rightarrow & e^{-} & + & \nu_e & + & \overline{\nu}_{\mu} \\ L: & 1 & = & 1 & + & 1 & - & 1 \\ L_e: & 0 & \ne & 1 & + & 1 & + & 0 \\ L_{\mu}: & 1 & \ne & 0 & + & 0 & - & 1 \end{matrix}[/tex]

    Because the lepton number conservation law in fact is violated by chiral anomalies, there are problems applying this symmetry universally over all energy scales. However, the quantum number (B − L) is much more likely to work and is seen in different models such as the Pati-Salam model.

    Baryonic charge non-conservation:
    Baryonic charge violation appears through the Adler-Bell-Jackiw quantum chiral anomaly of the [tex]U(1)[/tex] group.

    Baryons are not conserved by the usual electroweak interactions due to quantum chiral anomaly. The classic electroweak Lagrangian conserves baryonic charge. Quarks always enter in bilinear combinations [tex]q \bar q[/tex], so that a quark can disappear only in collision with an antiquark. In other words, the classical baryonic current [tex]J_\mu^B[/tex] is conserved under the Standard Model:

    [tex]\partial_\mu J_\mu^B = \sum_j \partial_\mu(\bar q_j \gamma_{\mu} q_j) = 0[/tex]

    However, quantum corrections destroy this conservation law resulting in anomalous baryonic current non-conservation, instead of zero in the right hand side of this equation:
    [tex]\partial_\mu J_\mu^B = \frac{g^2 C}{16\pi^2} G_{\mu\nu} \tilde{G}_{\mu\nu}[/tex]

    High energy physics (B − L):
    In high energy physics, B − L is the difference between the baryon number (B) and the lepton number (L).
    This quantum number is the charge of a global/gauge U(1) symmetry in some GUT models, called [tex]U(1)_{(B-L)}[/tex]. Unlike baryon number alone or lepton number alone, this hypothetical symmetry is not broken by chiral anomalies or gravitational anomalies, as long as this symmetry is global, which is why this symmetry is often invoked. If B − L exists as a symmetry, it has to be spontaneously broken to give the neutrinos a nonzero mass if we assume the seesaw mechanism.

    The chiral anomalies that break baryon number conservation and lepton number conservation individually cancel in such a way that B − L is always conserved.

    One example is proton decay where a proton (B = 1; L = 0) decays into a pion (B = 0, L = 0) and positron (B = 0; L = −1).

    [tex]\begin{matrix} & p^{+} & \rightarrow & \pi^{0} & + & \bar e^+ \\\ B - L: & 1 & = & 0 & + & 1 \end{matrix}[/tex]

    Proton decay:
    [tex]\begin{matrix} & p^{+} & \rightarrow & \bar e^{+} & + & \pi^0 & \rightarrow & \bar e^+ & + & 2 \gamma \\\ B - L: & 1 & = & 1 & + & 0 & \rightarrow & 1 & + & 0 \end{matrix}[/tex]

    Neutron decay:
    [tex]\begin{matrix} & n^{0} & \rightarrow & p^{+} & + & e^- & + & \bar \nu_e \\\ B - L: & 1 & = & 1 & - & 1 & + & 1 \end{matrix}[/tex]

    Muon decay:
    [tex]\begin{matrix} & \mu^{-} & \rightarrow & e^{-} & + & \nu_e & + & \overline{\nu}_{\mu} \\ B - L: & -1 & = & -1 & - & 1 & + & 1 \end{matrix}[/tex]

    Neutrino + antineutrino annihilation:
    [tex]\begin{matrix} & \bar \nu & + & \nu & \rightarrow & Z^0 & \rightarrow & \bar e^{+} & + & e^{-} \\ B - L: & 1 & - & 1 & = & 0 & = & 1 & - & 1 \end{matrix}[/tex]

    Hence proton decay conserves B - L, even though it violates both lepton number and baryon number conservation and neutron decay conserves baryon number B and lepton number L separately, so also the difference B - L is conserved.

    Where [tex]Q[/tex] is the electrical charge in elementary charge units and [tex]T_z[/tex] is the third component of weak isospin. The weak hypercharge can be expressed as:
    [tex]Y_W = 2(Q - T_z)[/tex]

    Weak hypercharge [tex]Y_W[/tex] is related to B − L via:
    [tex]X + 2Y_W = 5(B - L)[/tex]

    Integration via substitution:
    [tex]X + 4(Q - T_z) = 5(B - L)[/tex]

    Where [tex]X[/tex] is the [tex]U(1)_{(B-L)}[/tex] symmetry GUT-associated conserved quantum number.

    http://en.wikipedia.org/wiki/Neutrino" [Broken]
    http://en.wikipedia.org/wiki/Antineutrino" [Broken]
    http://en.wikipedia.org/wiki/Chiral_anomaly" [Broken]
    http://en.wikipedia.org/wiki/Pati%E2%80%93Salam_model" [Broken]
    http://en.wikipedia.org/wiki/B%E2%88%92L" [Broken]
    http://en.wikipedia.org/wiki/Weak_hypercharge" [Broken]
    http://en.wikipedia.org/wiki/Weak_isospin" [Broken]
    http://en.wikipedia.org/wiki/Seesaw_mechanism" [Broken]
    Last edited by a moderator: May 4, 2017
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