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 Apr 14 awarded Nice Answer Nov 26 awarded general-relativity Nov 23 awarded Yearling Jul 17 awarded Notable Question Feb 26 awarded Favorite Question Nov 23 awarded Yearling Oct 23 comment Differences between symmetric, Hermitian, self-adjoint, and essentially self-adjoint operators @Ruslan : Corrected Oct 23 revised Differences between symmetric, Hermitian, self-adjoint, and essentially self-adjoint operators Correction Oct 11 comment Could you help me understand this paper (PRL 106:136806)? The function $f_−(\vec r)$ is odd under the reflection relatively to the line defect. So $f_−(\vec r)=0$ for $\vec r$ being on the line defect. Maybe in my answer, I should have use $\vec r_\perp$ instead of $\vec r$. I have updated the end of the answer. Oct 11 revised Could you help me understand this paper (PRL 106:136806)? Clarification Oct 9 comment Could you help me understand this paper (PRL 106:136806)? $\langle\vec{r}|-\rangle$ is not equals to $0$. Replace the notation $|-\rangle$ by $|f_-\rangle$. The complex number $\langle\vec{r}|f_-\rangle$ means simply $f_-(\vec{r})$. It is not possible that $f_-(\vec{r})=0$ (for all $\vec r$) because it would mean $|f_-\rangle=0$. As explained in my answer, it is only $f_-(\vec 0)$ which is equals to zero, because the function $f_-(\vec{r})$ is odd under reflection. Oct 4 comment Relation between the curvature of a manifold and the number of covariantly constant vector fields that it admits If the four $K^a$ are linearly independent, they are a basis for vectors (because the tangent space of a $4$- dimensional manifold has dimension $4$) So any vector $X$ will be covariantly constant : $D_\mu X=0$. So $0 = (D_\mu D_\nu - D_\nu D_\mu) X= \mathcal R_{\mu\nu}X$, the latter equality being an equality between vectors, $\mathcal R_{\mu\nu}$ being the "matrix" of elements ${(\mathcal R_{\mu\nu})}_\lambda^\rho = {R_{\mu\nu}}_\lambda^\rho$, the vector $X$ having coordinates $X^\lambda$. This must be true for all $X$, so ${R_{\mu\nu}}_\lambda^\rho=0$, and the space-time is flat. Oct 4 comment Significance of total divergence anomaly term @ACuriousMind : In the Wikipedia article (baryonic charge violation), since $K_\mu$ is the Hodge dual of the Chern-Simons 3-form, then the anomaly could be considered as "topological". No ? Oct 4 comment Tetrad choice for Pauli-Lubanski in the massless case Your link is dead. Oct 4 comment The criterion to obtain covariant spinor derivatives in superspace @sbthesy : I begin to think that there is an error in the text (beginning of the page $40$): I would have written : $i\partial_\alpha (\delta_\epsilon S) =i\partial_{\alpha}[S, \epsilon Q+\bar{\epsilon}\bar{Q}] = i\partial_\alpha(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})S \neq i(\epsilon\mathcal{Q}+\bar{\epsilon}\bar{\mathcal{Q}})(\partial_\alpha S) = \delta_\epsilon (\partial_\alpha S)$ Oct 4 comment Could you help me understand this paper (PRL 106:136806)? @Hai-YaoDeng : For the last part of your question, I think that one may see that like a kind of closed path. Your are starting from one side of the line defect, you get a $\Gamma$ to go to the other side, a propagator $G$ to move along the line defect, an other $\Gamma$ to come back to the initial side of the line defect, then an other propagator $G^\dagger$ coming back to the initial point. Oct 4 comment Could you help me understand this paper (PRL 106:136806)? It is said in the text : "As the reflection oper- ator maps A sites onto B sites, and vice versa", so $|-\rangle \sim (|A\rangle - |B\rangle)$ is odd under reflection. Oct 3 answered Could you help me understand this paper (PRL 106:136806)? Oct 3 answered The criterion to obtain covariant spinor derivatives in superspace Oct 3 answered Bell State vs. Bell Measurement