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Spin-0, spin-1/2, and spin-1 particles have an associated quality expressing the spin structure of the field.

Spin-0 fields obviously have no extra quality. Spin-1/2 fields are expressed by spinors and gamma matrix algebra (the gamma matrices can be expressed in spin matrices). Spin-1 fields have a polarization vector.

If we extend this to massless spin-2 fields, in particular, the graviton, then what extra quality the field carries? I will be a tensor, but what will be the components of this tensor? Is it a metric tensor? Or, like the polarization vector for, say, photons, does it give the direction of momentum (=energy, in the case of photons) transfer?

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    $\begingroup$ Your last statement is inaccurate – the polarization vector of a photon does not encode momentum or energy. $\endgroup$
    – Emilio Pisanty
    Commented Aug 22, 2022 at 20:01
  • $\begingroup$ @EmilioPisanty The direction of energy-momentum transfer? Not the E and p themselves. $\endgroup$
    – Gerald
    Commented Aug 22, 2022 at 20:17
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    $\begingroup$ Precisely. Energy has no direction. Momentum does, and it's along the Poynting vector (proportional to the wavevector for a plane wave), not along the polarization. $\endgroup$
    – Emilio Pisanty
    Commented Aug 22, 2022 at 22:14
  • $\begingroup$ @EmilioPisanty If the polarization has two transversal components (like the electric and magnetic field), then, say, an electron, gets a longitudinal push? Don't the electric and magnetic fields combine? Say the electric field quantum gives a transversal push, and then the magnetic quantum gives a longitudinal push? $\endgroup$
    – Gerald
    Commented Aug 22, 2022 at 22:35
  • $\begingroup$ @EmilioPisanty The photon has energy and momentum. The polarization is perpendicular to the momentum. What means the polarization then? That the Poynting vector is in the same direction as the momentum? How is polarizition important? How can photons make an electron oscillate? Because the polarization oscillates? Ain't a photon just absorbed? Or better, its energy and momentum? $\endgroup$
    – Gerald
    Commented Aug 22, 2022 at 22:46

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Spin-0 fields obviously have no extra quality.

When $\ell = 0$ we have $2\ell + 1 = 1$. This is why we say that spin-0 fields transform in the (trivial) 1-dimensional representation of the rotation group.

Spin-1/2 fields are expressed by spinors and gamma matrix algebra (the gamma matrices can be expressed in spin matrices).

When $\ell = 1/2$ we have $2\ell + 1 = 2$. This is why we say that spin-1/2 fields transform according to the two-dimensional irreducible representation of the rotation group.

Often we show the explicit representation using Pauli matrices. For electrons/positrons we actually package up two separate two-dimensional representations together into the four-dimensional gamma matrices.

Spin-1 fields have a polarization vector.

When $\ell = 1$ we have $2\ell + 1 = 3$. This is why we say that spin-1 fields transform according to the three-dimensional irreducible representation of the rotation group. Photons also have a bit of funny business due to the fact that they are massless, so we only end up with transverse polarization (only two polarizations to worry about).

If we extend this to massless spin-2 fields,

You're jumping over spin 3/2, but ok. Just want to point out real quick that we would expect there to be some spin 3/2 particles that would transform according to the irreducible 4-dimensional representation.

in particular, the graviton, then what extra quality the field carries?

It should transform according to a $2\ell + 1 = 5$ dimensional irreducible representation of the rotation group. So, yes, it will be more tensor-like than vector-like. The reason the 3-dimensional irreducible representation is vector-like is because we can repackage up the three $Y_{1,m}$ into something that looks like a vector. For the 5-dimensional representation, we will have components transforming like the five $Y_{2m}$.

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    $\begingroup$ The graviton, being massless, has the same "funny business" as the photon (though the technical details differ especially in arbitrary dimension, see physics.stackexchange.com/q/134197/50583) and should also have only two polarizations in 4D (usually denoted by "+" and "x"), no? $\endgroup$
    – ACuriousMind ♦
    Commented Aug 22, 2022 at 20:08
  • $\begingroup$ But how the actual "polarization" tensor for a graviton looks like? $\endgroup$
    – Gerald
    Commented Aug 22, 2022 at 21:05
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    $\begingroup$ @Gerald Sounds like you are interested in the graviton's degrees of freedom. ACuriousMind's linked answer and this answer may help: physics.stackexchange.com/questions/74307/… $\endgroup$
    – hft
    Commented Aug 22, 2022 at 21:24
  • $\begingroup$ I don't know how to prove that the graviton only has two polarization off the top of my head. Perhaps someone else will be able to. $\endgroup$
    – hft
    Commented Aug 22, 2022 at 21:26
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    $\begingroup$ You will have to wait for someone else to fill in the details, sorry. $\endgroup$
    – hft
    Commented Aug 22, 2022 at 21:34

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