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14

IMHO, the notation $\int_a^b\mathrm{d}x\,f(x)$ is much cleaner than $\int_a^b f(x)\,\mathrm{d}x$, because the integration variable ($x$) and its associated integral range $(\int_a^b$) are kept together. This is particularly important in lengthy and multi-dimensional integrals. Consider \Upsilon_{pq}(k)= \int_0^\infty\mathrm{d}x ... 12 It's not just QFT literature. Physicists, especially adult research physicists, find this notation sensible and popular – even though it may be more popular among particle physicists than elsewhere. Formally, dx\,f(x) is a product of two factors and \int is a form of a sum. Because product is commutative, it doesn't hurt when the order is interchanged. ... 10 What exactly does \Delta_{r_e} mean ? Your wave function isn't a field in space, it is a field on configuration space, i.e. it assigns complex numbers to a configuration. If your electron is at (x_e,y_e,z_e) and your proton is at (x_p,y_p,z_p) then the configuration is the point (x_e,y_e,z_e,x_p,y_p,z_p) in a 6d space. A point in that 6d space ... 8 Besides the reasons listed in Lubos Motl's answer, here is another reason for the \int \!dx ~f(x) notation: By writing the integral sign \int_a^b and dx next to each other in multiple nested integrations, it becomes more easy to trace which limits belong to which integration. This becomes particularly handy when changing the orders of integration. ... 5 The index of the Laplacian tells you which of the coordinates it acts on, that is, if you write r = (r_x,r_y,r_z)^T and R = (R_x,R_y,R_z)^T as Cartesian coordinates, then \begin{align} \Delta_r & := \frac{\partial^2}{\partial {r_x}^2} + \frac{\partial^2}{\partial {r_y}^2} + \frac{\partial^2}{\partial {r_z}^2} \\ \Delta_R & := ... 3 The answer is already on page 2 of your link above: "Among the large number of radionuclides of medical interest, Sc-44 is promising for PET imaging. Either the ground-state Sc-44g or the metastable-state Sc-44m can be used for such applications, depending on the moleculeused as vector." So the metastable state Sc-44m decays to the ground state Sc-44g. 3 Your example is an outlier, in my experience (personally, I would have written (F_{\mu\nu})^T instead of F_{\nu\mu}). Almost always, it's the order of the indices that determines row vs. column. If someone writes T^i_j, then while technically there's no way to tell, I would say that it would be far less confusing to make the upper index label the rows ... 2 In my experience, reading the indices left to right and top to bottom, the first index is the row and the second is the column. Your screenshot from Carroll doesn't have to be contradictory (although it's definitely confusing/doesn't make rigorous sense). You can just imagine he omits a little "_{\mu \nu}" on the matrix:F_{\mu \nu}=\Bigg( \cdots ...

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There's a small error here: you say, "...these two vectors $v$ and $\mathbf{v}$" (emphasis mine). The problem there is that $v$ is not a vector. Rather, it's the magnitude of a vector; specifically, it is the magnitude of the velocity vector $\mathbf{v}$. This is actually implicit in your derivation: you created a unit vector in the direction of the motion ...

1

$\Delta_e$ is called the Laplacian operator and it is defined as $$\Delta_e \equiv \frac{\partial^2}{\partial e_x^2} + \frac{\partial^2}{\partial e_y^2}+ \frac{\partial^2}{\partial e_z^2}$$ The Laplacian operator is related to the Del operator $\nabla_e$ $$\nabla_e \equiv \frac{\partial}{\partial e_x}\hat e_x+ \frac{\partial}{\partial e_y}\hat e_y+ ... 1 The left exponential evolves the \langle \alpha_0 \lvert on the left. This is one of the pitfalls of Dirac notation, it would be unambiguous to write$$ (\mathrm{e}^{-\mathrm{i}E_{\alpha_0} t} \lvert \alpha_0 \rangle,B \mathrm{e}^{-\mathrm{i}E_{\alpha_0} t} \lvert \alpha_0 \rangle)$$where (\dot{},\dot{}) denotes the inner product on the Hilbert space, ... 1 This is pretty niche notation, and it is indeed not defined in the paper, but the name "vector-coupled product" does seem to be used by a few people beyond Varga and Suzuki. In essence,$$ [\mathcal Y_{l_1}(\mathbf x_1)\mathcal Y_{l_2}(\mathbf x_2)]_{LM} $$is a coupled wavefunction with total angular momentum L that's made up of the single-particle ... 1 Using polar coordinates it holds  |d{\bf r}| = \sqrt{(d|{\bf r}|)^2 + |{\bf r}|^2(d{\bf \phi})^2}. From this equation you can see, the two expressions you are asking about are actually only equal (in absolute value) for a straight line through the origin, thus otherwise different. For them to be exactly equal, the d|{\bf r}| should be moreover pointing ... 1 If$$ \overrightarrow{r}=r_{x}\widehat{i}+r_{y}\widehat{j} $$then$$ \left | \overrightarrow{r} \right |=\sqrt{r_{x}^{2}+r_{y}^{2}} $$and$$ d\left | \overrightarrow{r} \right |=\frac{r_{x}dr_{x}+r_{y}dr_{y}}{\sqrt{r_{x}^{2}+r_{y}^{2}}} $$on the other hand$$ d\overrightarrow{r}=dr_{x}\widehat{i}+dr_{y}\widehat{j} $$and$$ \left | d\overrightarrow{r} ...

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