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$\newcommand{\bra}[1]{\left< #1 \right|} \newcommand{\ket}[1]{\left| #1 \right>} \newcommand{\bk}[2]{\left< #1 \middle| #2 \right>} \newcommand{\bke}[3]{\left< #1 \middle| #2 \middle| #3 \right>}$ I shall be assuming that all vector spaces are finite-dimensional for purposes of simplicity. $\ket{f} \in V$ where $V$ is a vector space. $H^\...


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To start: What is $\mathbf{s}(i)$? This is not an operator (not a matrix) but rather meant to denote a vector of operators. \begin{equation}\mathbf{s}(i)=(s_x(i),s_y(i),s_z(i)),\end{equation} where $s_\sigma(i)$ is an operator, a $(2s+1)^N$-dimensional matrix. What does the dot product between them mean? It means to imitate the usual dot product, as in \...


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In short, yes, the convention does apply, but let me make a few remarks before I get to the main point: the tangent space at a point is not an element of the manifold, nor a subset. More specifically, the expression $T_p \in M$ is wrong. $T_p$ can be seen, from an extrinsic point of view, as the hyperplane tangent to the manifold at $p$ the metric is not an ...


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$\renewcommand\mean[1]{\langle #1\rangle}$ $\renewcommand\norm[1]{||#1||}$ $\renewcommand\h{\hbar}$ $\renewcommand\ket[1]{|#1\rangle}$ $\renewcommand\expval[2]{\langle #1|#2|#1\rangle}$ $\renewcommand\braket[2]{\langle #1|#2\rangle}$ $\renewcommand\Braket[3]{\langle #1|#2|#3\rangle}$ OP's original question The step that really confuses me here is that we ...


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Just to elaborate on Cosmas's answer: when one writes $\hat{p} = -i \hbar \ d/dx$, what they really mean is that $\hat{p}$ is given by this expression in the $x$-representation. In other words, $\hat{p}$ has the following matrix elements with respect to the $|x\rangle$ basis: $$ \langle x | \hat{p} | x' \rangle = -i \hbar \frac{d}{dx} \delta(x-x') $$ One ...


2

You are misunderstanding the Dirac notation language. The proper relation you write should be $$\langle x| \hat{p} |0\rangle = \hat{p}_x \langle x|0\rangle, $$ i.e., the right-hand side operator is the realization of the momentum operator acting on functions of x, not vectors (kets), as it does on the left-hand side. It is shorthand for the less confusing ...


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The hat on $x$ represents the fact that it's not the variable but an operator. Sometimes authors use the capital letter $X$ to show that they mean operator (Position operator here). While $x$ is an eigenvalue of operator $X$. Or outside quantum mechanics simply a variable for position. The expectation value of some physical observable $\Omega$ in-state $|\...


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The metric, written invariantly is: $g: XM \otimes_M XM \rightarrow M\mathbb{R}$ Here, $XM$, is the sheaf of tangent fields, that is the sheaf of sections of the tangent bundle, $TM$. The metric, locally, and which means in any chart, is expressed as you have written in it. And this is normally how it's written in the physical literature. And this means ...


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Things become clearer if the symmetric product or anti-commutator is defined directly in the real vector space $V$ (or more precisely: in the non-associative algebra) of traceless self-adjoint (complex) matrices with scalar product $\langle A,B\rangle = \frac{1}{2} Tr(A, B)$, rather than in "component space" with respect to a basis $\{\Lambda_1,\...


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To add to the confusion there we have also the invariant notation for Lie-algebra valued forms: $\alpha [\wedge] \beta := [\alpha \wedge \beta]$ This is usually defined on simple forms and then extended by linearity. More precisely, say $\alpha = \alpha' \otimes X$ and $\beta = \beta' \otimes Y$ where $\alpha, \beta$ are ordinary forms and $X,Y$ are ...


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It depends on the context and it should say somewhere in your textbook. Often it is a shorthand for \begin{align}(\partial \phi)^2&=(\partial_\mu\phi)(\partial^\mu\phi)\\ &=\eta^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi) \end{align} where $\eta_{\mu\nu}$ is the Minkowski metric. Depending on your textbook the Minkowski metric is either $\text{diag}(...


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I believe that $\partial$ has no index because it is being squared. What really is written is $ (\partial \phi)^2 = \partial_\mu \phi \partial^\mu \phi $ which has no indices. Therefore, the author writes $ \partial $ without indices to indicate that the resulting object is a scalar.


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I think they mean a $3\times9$ matrix like $$M=(M_{j,ik})=\left( \begin{matrix} M_{111}&M_{112}&M_{113}&M_{121}&M_{122}&M_{123}&M_{131}&M_{132}&M_{133}\\ M_{211}&M_{212}&M_{213}&M_{221}&M_{222}&M_{223}&M_{231}&M_{232}&M_{233}\\ M_{311}&M_{312}&M_{313}&M_{321}&M_{322}&M_{323}&...


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I also used to also find these sorts of things confusing. It helps me to think of the general case: take a vector of $N$ numbers ${a_i}$ and a vector of $M$ numbers ${b_i}.$ So long as $N,M$ are finite, we have $$ \sum_i \sum_j a_i b_j = \sum_j \sum_i a_i b_j \equiv \sum_{i,j} a_i b_j. $$ Where the double sum notation $\sum_{i,j}$ is used because the order ...


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