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after looking through a couple QFT texts it seems that all the spin-1/2 fields come associated with a charge of some sort. I was wondering if it's possible to write down a classical lagrangian (with Lorentz invariance) such that after quantization, the excitations of the field are merely spin-1/2 particles with no associated charges (electric, color) of any kind? That is,operators like $a^{\dagger}$ would only make single particle states with helicity $\pm \frac{1}{2}$ and no associated $U(n)$ or $SU(n)$ charges. Thanks.

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Well, yes, neutrinos are examples, aren't they? –  Luboš Motl Jul 28 '12 at 19:20
    
Don't neutrinos carry weak charge? –  John Rennie Jul 28 '12 at 19:24
    
@LubošMotl I thought they were a prime example too. Could you then clarify something for me then please? Given the Lagrangian $i\psi^{\dagger}\bar{\sigma}^\mu \partial_\mu \psi$, this has a conserved U(1) charge $\psi^{\dagger}\psi$. Then do the creation operators associated with this field theory create spin-1/2 particles with charge $\pm 1$? Forgive my naivete. –  kηives Jul 28 '12 at 19:29
    
@knives: This charge is broken when you give neutrinos mass. –  Ron Maimon Jul 28 '12 at 21:09
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The Lagrangian you are looking for is the two-component Weyl spinor form for a massive fermion with Majorana mass, this is believed to describe neutrinos at low energies (below the scale where you see that they are paired with the leptons into electroweak doublets):

$$ S = \int \bar{\psi}^i \sigma^\mu_{ij} \partial_\mu \psi^j + {M\over 2}(\epsilon_{ij}\psi^i\psi^j + \epsilon_{ij}\bar{\psi}^i\bar{\psi}^j) d^4x $$

Where the $\sigma$ is the Pauli matrix 4-vector. This is a 2 component spinor field with no charge. The mass term breaks the phase invariance of the massless field. This often appears in books by combining the $\psi$ and $\bar{\psi}$ into a 4-spinor, but then the conjugate spinor is not independent. The two-component form doesn't appear very often, but it is the most insightful 4d form in my opinion.

The reason we don't see fundamental fields of this form is because such uncharged fermions have a mass which is tunable, and so generically is at the Planck mass. This isn't as bad as scalar masses, because the $M=0$ point has an extra U(1) global symmetry, although global symmetries are never fundamental.

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Thanks Ron, except I should have put mass-less in the title too... sorry... I was wondering if you can have a mass-less, charge-less, spin-1/2 field... It was one of those cases where I thought it in my head, but didn't write it down cause I thought everyone else knew what I was thinking... :( –  kηives Jul 28 '12 at 21:22
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@kηives: You can't in 4d--- in the massless limit you always have a global symmetry. You can do it with a Majorana Weyl spinor in 2d or 10d. –  Ron Maimon Jul 28 '12 at 21:33
    
Thanks Ron. some text. –  kηives Jul 28 '12 at 21:45
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