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1) Stretch your arm out. 2) Rotate your hand as in the plate trick. Start at time $t=0$, complete one rotation at time $t=1$, and complete the second rotation at time $t=2$. 3) Think of your arm as a copy of the unit interval, stretching from $s=0$ at your shoulder to $s=1$ at the tip of your hand. 4) Imagine a copy of the standard basis for ${\... 6 There are two ways to have absolutely forbidden transitions, one rigorous, the other only valid in the thermodynamic limit. One class of forbidden transitions is those that violate conservation of a conserved quantity. In practice, the only absolutely conserved quantities are the (locally conserved) gauge charges of the electromagnetic, strong, and ... 4 "The same rule" is associativity, but everything else is quite different. That is, the group commutator is, in some convention, $$[R_{x}(\alpha),R_{z}(\beta)]= R_{x}(\alpha)R_{z}(\beta) R_{x}(-\alpha)R_{z}(-\beta),$$ where$R_{x}(\alpha)= e^{-i\alpha J_x}$, etc... For notational simplicity, let's call$-i\alpha J_x =A$and$-i\beta J_z =B$, antihermitean, ... 3 Your orthogonal matrix $$R(\phi,\theta)= \begin{bmatrix} \cos(\theta) &\sin(\theta) & 0 \\ \sin(\theta)\cos(\phi) & -\cos(\theta)\cos(\phi) & \sin(\phi)\\ \sin(\theta)\sin(\phi) & -\cos(\theta)\sin(\phi) &-\cos(\phi) \end{bmatrix}$$ must have antisymmetric generators. To find ... 3 Both$p^2$and the sign of$p^0$are invariant under infinitesimal Poincaré transformations (we're really looking at representations of the Poincare algebra or rather projective representations of the connected component of the Poincaré group here). Hence any representation that involves more than one value of$p^2$and$\mathrm{sgn}(p^0)$is not ... 1 Yes classically massless$\phi^4$has a scaling symmetry. The reason why people don't mention this is because it is not a quantum symmetry i.e. it doesn't survive the quantization, the technical term for it is that it is anomalous. People write it as$e^\lambda$because there is an implicit assumption of the transformation having a definite sign. So you can ... 1 Following the answer by @ACuriousMind I believe I got the point and decided to post a version which is a little bit more expanded with what I understood. Let$f(p)$be some function defined in momentum space. We can define the function acting on the momentum operators$P$by the usual procedure using the eigenstate basis of the commuting set$\{P^\mu\}$:$$f(... 1 The parametrization you've given just isn't of the exponential form you're after. The matrix you've written down is parametrized by the location of the$x$axis after the rotation, which has polar angle$\theta$and azimuthal angle$\phi$in polar spherical coordinates drawn around the old$x$axis. It is not a rotation by angle$\phi\$ about an axis at ...