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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\... 1 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 ... 0 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}\... 1 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 ... 0 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 ... 1 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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