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10

Spin is best understood as an intrinsic angular momentum. It is probably easier to understand the concept for a charged particle. A classical charged particle moving along a circle has an angular momentum and the "circuit" has a magnetic moment. Further, the two are proportional to each other. It is experimentally found that a charged particle like an ...

3

Spin arises from the need to represent the rotation group $\mathrm{SO}(3)$ upon our Hilbert space of states. We need such a representation because the rotations (together with space translations) correspond to the non-relativistic changes of reference frames. Since states are only determined up to rays in the Hilbert space, the true space of states on which ...

2

Analyzing the spectrum of the strings, one finds that it contains $N^2$ massless vector states, which is precisely the number of gauge fields corresponding to a $U(N)$ group. Note that this is only true for massless oriented open strings; the unoriented case yields $SO(N)$ or $Sp(N)$. As is described in the same chapter of the book, open string states can ...

0

The defining relation for the Clifford algebra, $Cl(1,d)$ is $$\{\gamma_\mu,\gamma_\nu\}=2 \eta_{\mu\nu}\ \mathbf{1}\ ,$$ For simplicity, I will assume that $\eta_{\mu\nu}=\text{Diag}(1,-1,\ldots,-1)$ with $\mu,\nu=0,1,\ldots,d$. Other signatures can easily be incorporated. It is easy to see that $\gamma_0^2=-\gamma_i^2=\mathbf{1}$ for $i=1,\ldots,d$. ...

2

Your understanding of reducible and irreducible representations is a little bit muddled. Let me try to clarify this a bit: A reducible representation $D:G\to \text{GL}(V)$ is one that has a nontrivial invariant subspace $W$. That is, there exists a nonzero $W<V$ such that for all $g\in G$ and all $w\in W$, the action $D(g)w\in W$ remains in the ...

2

1) Why don't we consider finite dimensional representations of this group? As you said, we ask (anti)unitarity, so it is impossible to find finite-dimensional representation. 2) Why associate the Lorentz group to fields? The essence of the answer is what Trimok already said in his comment: the "translational part" of the Poincarè group is already ...

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