According to the the definition of anti-particles, they are particles with same mass but opposite charge. Neutrinos by definition have no charge. So, how can it have an anti-particle?
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9$\begingroup$ the opposite of zero is zero $\endgroup$– anna vMay 5, 2014 at 17:44
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2$\begingroup$ @annav Then how do you differentiate whether it is a particle or an anti-particle? $\endgroup$– YashbhattMay 5, 2014 at 17:48
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4$\begingroup$ @Yashbhatt: see Identification of particles and anti-particles. It is possible that neitrionos are Majorana particles though I get the impression this idea is going out of fashion. $\endgroup$– John RennieMay 5, 2014 at 17:51
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3$\begingroup$ From quantum number rules. In any case the neutrino is found from energy conservation and momentum conservation rules, and its quantum numbers are also defined by conservation of flavor,spin. $\endgroup$– anna vMay 5, 2014 at 17:57
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2$\begingroup$ This question is also related, and close to being a duplicate. $\endgroup$– David ZMay 5, 2014 at 18:00
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2 Answers
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There are other neutral particles with antiparticles, such as the neutron and the $K^0$ meson. In those cases we have a microscopic theory that says those particles are made of quarks: for instance, the $K^0$ is made of a down quark and an anti-strange quark, while its antiparticle the $\bar K^0$ is made of a strange quark and an anti-down.
The neutrino is different from these because we have no evidence that it has any composite structure. While the neutrino doesn't have any electric charge, it does have a quantum number that appears to be conserved in the same way as electric charge: lepton number. We find in experiments that neutrinos are never created alone. A neutrino is always produced in conjunction with a positive lepton ($e$, $\mu$, or $\tau$), and an antineutrino is always produced in conjunction with a negative lepton.
There is another key property of neutrinos that's important when thinking about their antiparticles, which is their spin. Weak decays break mirror symmetry (or "parity symmetry"). If you have a beta-decay source that doesn't have any spin to it at all, and you measure the spins of the decay electrons that come out, you'll find that they are strongly polarized: beta-decay electrons prefer to be "left-handed", or traveling so that their south poles point forwards and their north poles point backwards. Beta-decay antielectrons, by contrast, prefer to be right-handed. The neutrinos follow the same rule: neutrinos have left-handed spins, and antineutrinos have right-handed spins.
If a neutrino had exactly zero mass, this polarization would be complete. However, we now have convincing evidence that at least two flavors of neutrino have finite mass. This means that it's possible, in theory, for an relativistic observer to "outrun" a left-handed neutrino, in which reference frame its north pole would be pointing along its momentum — that observer would consider it a right-handed neutrino. Would a right-handed neutrino act like an antineutrino? That would imply that the neutrino is actually its own antiparticle (an idea credited to Majorana). Would the right-handed neutrino simply refuse to participate in the weak interaction? That would make them good candidates for dark matter (though I think there is other evidence against this).
It's an open experimental question whether there is really a difference between neutrinos and antineutrinos, apart from their spin, and there are several active searches, e.g. for forbidden double-beta decays.
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$\begingroup$ I am confused here:"——that observer would consider it a right-handed neutrino. Would a right-handed neutrino act like an anti-neutrino?". Ok then that observer would consider similarly the right-handed antineutrino in some frame as left-handed antineutrino in his frame. So what is the justification of "Would a right-handed neutrino act like an antineutrino?". $\endgroup$ Aug 5, 2014 at 5:14
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$\begingroup$ @user22180 It depends on whether neutrinos are Dirac fermions (like the charged leptons and quarks) or Majorana fermions. In that hypothesis the neutrino and antineutrino are the same, and it's the $W$ boson that selects between the two by handedness. Don't underestimate how strange neutrinos are — essentially every guess we've made has been wrong in several ways before we got it right. $\endgroup$– rob ♦Aug 5, 2014 at 12:37
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1$\begingroup$ @rob: a word of caution when you wrote "[…] that observer would consider it a right-handed neutrino. Would a right-handed neutrino act like an antineutrino?" The boost revert the helicity of the neutrino, not its chirality. Weak interaction is only sensitive to the chirality not the helicity. A anti-neutrino is chiral right-handed (and a neutrino chiral left-handed). So you cannot transform a neutrino into a anti-neutrino thanks to a boost. $\endgroup$– PaganiniOct 18, 2015 at 13:34
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$\begingroup$ @Paganini I think I disagree. If I remember correctly, a massive particle in its rest frame has equal parts left- and right-chiral, and the correlation between chirality and helicity appears as you boost enough that the mass becomes negligible. $\endgroup$– rob ♦Oct 18, 2015 at 14:50
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$\begingroup$ For example, the decay $\pi\to\rm e\nu$ is suppressed relative to $\pi\to\mu\nu$ even though the former would liberate more kinetic energy. Because the $\pi$ is spinless, so the charged lepton in both decays must come out with helicity and chirality opposed; this is easier for the heavier/less-boosted $\mu$. However the charged lepton from those decays is not a "pure chiral" lepton in its rest frame. $\endgroup$– rob ♦Oct 18, 2015 at 14:50
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Whilst the neutrino is electrically neutral, electric charge $Q$ can be expressed as a combination of (the 3rd component of) weak isospin $T_3$ and weak hypercharge $Y_W$
$$Q = T_3 + \frac{Y_W}{2}$$
For the (left handed) neutrino, $T_3 = \frac{1}{2}$ and $Y_W = -1$ thus, the electric charge of the neutrino is
$$Q = \frac{1}{2} - \frac{1}{2} = 0$$
For the (right handed) antineutrino, the charges are opposite: $T_3 = -\frac{1}{2}$ and $Y_W = 1$ thus, the electric charge of the antineutrino is
$$Q = -\frac{1}{2} + \frac{1}{2} = 0$$
answered May 5, 2014 at 19:24