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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 ...

9

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 ...

2

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 ...

1

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: ...

0

The tides would be shorter, as the speed of the moon relative to the earths surface would change. Currently the tides change from furthest out to in every 6hrs and 12mins. This is due to the fact that the earth rotates once every 24 hours, and the tide changes every time the moon gets 1 quater of the way around the earth. 24/4 = 6 hours, plus 12 minutes ...

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The problem here is that you're looking at the magnetic field $\textit{at the proton}$. Using this approach you can't derive the spin-orbit coupling that you want. Because of the symmetry of the system you would expect the magnetic fields at each particle to have the same magnitude, but the energy of spin-orbit coupling comes from the $\textit{electron's}$ ...

1

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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Let's speak about 1D particles for simplicity. What should be understood first of all is that for indistinguishable particles configuration space isn't the same as for distinguishable ones. For two distinguishable spinless 1D particles configuration space is a square: one side is for $x_1$, another for $x_2$. But if the particles appear indistinguishable, ...

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Could you justify that Hamiltonian? Why do you neglect the $p_z^2/2m$ term. I feel like it's unnecessary to have the kinetic energy term at all. Most of the spin-1/2 problems I've done have simply had a Hamiltonian of the form $H = - \gamma \cdot B$. Furthermore I'm not sure what you mean by equations of motion of the z direction. It seems like you are ...

2

It's funny how hard this question turns out to be. I wrote a blogpost about it last year which you can read here. But here's the short version. There are six possible states you can create in terms of total spin (l) and z-axis spin (m), which are: (3/2, 3/2) (3/2, 1/2) (1/2, 1/2) (1/2, -1/2) (3/2, -1/2) (3/2, -3/2) It's easy to make states 1 and 6 from ...

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No, it is not equivalent. The Spin $s = S = 3/2$ particle will have spin projections between $S_3 = 3/2$ and $-3/2$, as you have worked out. That is it, it will just be a multiplet with 5 members. The three particles with spin $s = 1/2$ can also have a combined spin with $S = 3/2$ which will form the same 5-multiplet. However, they can also form a $S = 1/2$ ...

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