- #1

#### Tio Barnabe

It seems there's not a obvious way to make a product between a spinor and a three-vector.

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- Thread starter Tio Barnabe
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- #1

It seems there's not a obvious way to make a product between a spinor and a three-vector.

- #2

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There is an obvious way. The tensor product.

- #3

Even that spinors are not tensors?The tensor product.

- #4

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Spinors belong to a vector space.Even that spinors are not tensors?

- #5

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Even that spinors are not tensors?

One can show that, if ##V^{\mu}## are components of a 4-vector, i.e. an object transforming under the fundamental representation of the restricted Lorentz group, then

[tex] \psi_{\alpha\dot{\beta}} =: \eta_{\mu\nu}V^{\mu}\left(\sigma^{\nu}\right)_{\alpha\dot{\beta}} [/tex]

are the components of a spinor tensor with respect to the ##\mbox{SL}(2,\mathbb{C})## group.

- #6

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As commented above there are finite dimensional spinorial representations of the Lorentz group that relates those objects without any contradiction. Maybe what makes you uneasy and don't know how to pinpoint mathematically is that these finite dimensional representations of the Lorentz group when irreducible are not unitary since the group is not compact. This is why one has to turn to infinite dimensional representations. Still, QFT makes use of the finite dimensional representations in the rest frame(since in the neighbourhood of the identity unitarity is retained) of massive particles and it constructs their invariant mass and spin with its Lie algebra Casimir invariants.

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- #7

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- #8

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Are you referring to this part of my post?:

I saw that it was misleadingly worded when it was late to edit it.Still, QFT makes use of the finite dimensional representations in the rest frame(since in the neighbourhood of the identity unitarity is retained) of massive particles and it constructs their invariant mass and spin with its Lie algebra Casimir invariants.

Of course the only unitary irrep used is the infinite dimensional one and the Casimir invariants belong to this infinite dimensional unitary representation. When talking about the finite dimensional rep I was thinking about the rotational group that is the subgroup of the Lorentz group that leaves the subspace rep of vanishing three-momentum invariant. The sentence is not correct as written since I didn't specify I was referring to the rotations subgroup. Apologies.

- #9

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(So far), physically relevant are the cases with ##m^2>0## (massive particles) and ##m^2=0## (massless particles). For ##m^2>0## the natural choice of the standard momentum is ##\vec{p}=0##, and the little group is the rotation group SO(3), which is substituted by its covering group SU(2), i.e., importantly you include the possibility of half-integer spin.

The case ##m^2=0## is more subtle. Here the standard choice for the standard momenum is ##k^0=\pm |\vec{k}|##, ##\vec{k}=k \vec{e}_3##. The little group is ISO(2), which has no proper finite-dimensional irreps. It's like the group of a non-relativistic particle in a plane, i.e., you have two translation operators with a continuous spectrum, but continuous intrinsic degrees of freedom have not been observed (yet?). Thus one restricts the irreps. to such for which the translations in this abstract ISO(2) are represented trivially, and then the non-trivial subgroup is O(2) (the rotations around the three-axis) or rather its covering group U(1). Since for the full Lorentz group you get also the full rotations as a subgroup the corresponding helicities (angular momentum component in direction of the momentum of the particle) are restricted to the set ##h \in \{0,\pm 1/2,\pm1,\ldots\}##. Thus, massless particles with spin ##\geq 1/2## all have two physical polarization-degrees of freedom (here using the basis of good helicity to construct the irreps).

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