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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 ...


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 ...


0

See some basic QM textbook as mention above, @Todd R, Briefly, $S$ which in $3$ dimensional case can be represented in $x, y,z$ basis as \begin{align} \vec{S} = S_x \hat{x} + S_y \hat{y} + S_z \hat{z} \end{align} If we choose $z$ direction as (consider $S_z$ and state as a eigenvalue problems : Usually many textbook choose $z$, actually choice of ...


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.


1

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 ...


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 ...


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+ ...


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 & := ...


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, ...


0

I think the other answers are more easily summarized as: A continuous function is easier to integrate (or differentiate) and provides an answer which is more than accurate enough for any use. This is similar to replacing binomial distributions with continuous distributions for large populations.


0

The problem is that in order to work with distributions of velocities / densities etc, we need to consider "every possible value". There may well not be, at any given moment, any molecules with that precise value - in fact, if you specify the value to enough precision, there will never be a single molecule that has that value. And so we move from the realm ...


0

We can consider this to be the mean number of molecules, or the expectation value. There are almost always random fluctations to the numbers of molecules in any given interval -- say if 50% of the time we expect 2, and the other 50% 3, the expectation value or mean value would be 2.5 -- even though we can never see 2.5 molecules. In this way, the ...


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 ...


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 ...


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. ...


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. ...


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} ...


0

As shown in the diagram $|dr|$ represents the magnitude of the vector difference(that involves the laws of vector addition/subtraction) between $\vec{r_2}\quad \& \quad \vec{r_1}$ while $d|r|$ represents the difference between magnitudes of two vectors which is simply the difference in their lengths.


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

In response to your edit: The real part of $(A^*(S)A(T))$ is equal to the real part of $(A(S)A^*(T))$, since they are just complex conjugates of each other. So he could have written either one. Concrete example: let $(A^*(S)A(T))=a+ib$ for some $a,b$. Then $(A(S)A^*(T))=a-ib$, since it's just the complex conjugate. Then $(A^*(S)A(T))+(A(S)A^*(T))=2a$. We ...


9

$\mathfrak{R}e$: real part. $A^*$: complex conjugate of probability amplitude $A$.



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