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1

A rotation is of the form $$\begin{bmatrix} \cos(\theta) & - \sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$ A reflection is of the form $$\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & - \cos(2\theta) \end{bmatrix}$$ If we want to find a $2-$dimensional representation of a $3-$dimensional rotation then we can ...


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As a matter of fact, there is a much easier way to derive those matrices. What you are really after is the matrix representation of the $\mathfrak{su}(2)$ Lie algebra generators $J_x$, $J_y$ and $J_z$ in the irreducible $\mathrm{SU}(2)$ representation with spin $j$, given by $$ J_z|j,m\rangle = m |j,m\rangle,\quad J_\pm|j,m\rangle = C_\pm(j,m) |j,m\pm 1\...


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For $SU(2)$ the spinor representation has dimension 2. Your questions is not clear, but for rotation groups (or more precisely, their associated spin groups), we have: For $so(2n)$ (with $n\ge 2$), there are two spinor representations of dimension is $2^{n-1}$. E.g., for $so(2\times 5)$, the spinor is 16. For odd dimensions, $so(2n+1)$, the spinors have ...


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This might be easier to understand in terms of the Pauli matrices: $\sigma_z=\pmatrix{0 &1\\1 &0}$, $\sigma_y=\pmatrix{0 &-i\\i &0}$, $\sigma_x=\pmatrix{1 &0\\0 &-1}$, where the operators for $z, y$, and $x$, are given by $\hat{S_z}=\tfrac{\hbar}{2}\sigma_z$, $\hat{S_y}=\tfrac{\hbar}{2}\sigma_y$, and $\hat{S_x}=\tfrac{\hbar}{2}\...


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The spin-statistic theorem holds in all dimensions of spacetime greater than two. The proper definitions of "half-integer" or "integer" spin in general dimension is simply how the rotation operator of a full rotation, $R(2\pi)$ is represented - "integer spin" or "bosonic" representations will have it as the identity, while "half-integer" or "fermionic ...


3

For massive spinors "right-handed" and "left-handed" chirality isn't tied so much to true rotations, as to the casting of Lorentz transformations as "space-time rotations". In this case, a very popular short answer to the conceptual question is that Lorentz transformations "rotate" $(1/2, 0)$-spinors one way in space-time, and $(0, 1/2)$-spinors the opposite ...


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In general you are getting it right: Non-commuting operators do not share eigenstates, thus measuring $S_x$ on an eigenstate of $S_z$ will result in a state that is not an eigenstate of $S_z$ anymore. The spin operators do not commute because they are defined via the Lie-algebra relation $[S_i, S_j] = i \hbar \varepsilon_{ijk} S_k$. Next, if the particle ...


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Mutually non-commuting operators cannot have simultaneous eigenstates, namely the eigenstates of the former must by all means be expressed as a linear combinations of (all) the eigenstates of the latter. In the case at hand, given ${|+\rangle}_z$ as eigenstate of the operator $S_z$, the following must hold: $$ {|+\rangle}_z = c_1 {|+\rangle}_x + c_2 {|-\...


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The two papers talk about very different things. In Kapustin's paper, he considered non-orientable space-time manifold to classify SPT (i.e. the partition function of the phase on these manifolds). To do that, one has to first Wick rotate to Euclidean space-time, where time-reversal becomes a mirror reflection, but with a sign change. In Watanabe et. al. ...



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