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A common procedure to determine the spin of the excitations of a quantum field is to first determine the conserved currents arising from quasi-symmetries via Noether's theorem. For example, in the case of the Dirac field, described by the Lagrangian, $$\mathcal{L}=\bar{\psi}(i\gamma^\mu \partial_\mu -m)\psi$$ the associated conserved currents under a ...

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Reposting comment as an answer and expanding. The answer is yes. You can find an exposition in Condensed Matter Field Theory by Altland and Simons, starting on page 134 in the second edition. The troubles come from that spin can't be described with a Hamiltonian that is a function of $q$:s and their conjugate $p$:s. However the more general formulation of ...

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If you linearise the theory such that $$g^{\mu \nu}(x) = \eta^{\mu \nu} + h^{\mu \nu}(x)$$ say, you will find that your quantum of gravitation is this tensor $h^{\mu \nu}(x)$. Then clearly it has two free indices, and is what we call a 'spin-2 particle'. The maths to do the linearisation and prove that it transforms as a spin-2 particle would under ...

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I am not sure that I get your question right, but let me try to answer according to my understanding. The spin part of the two electron wave function of the singlet state $|0,0\rangle=|l=0,m=0\rangle$ is $|0,0\rangle=(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)/\sqrt{2}$ The three triplet states look like this: ...

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linearly - then there's exactly the same number of both spins. This is incorrect. If you have a collection of photons in which half are left hand circularly polarized ($L$) and half are right ($R$), then you have unpolarized light (not linearly polarized). If you have linear polarized light, then each photon is in a (quantum) superposition of R and L at ...

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