# Tag Info

35

There is a consistent definition, but it involves a couple of arbitrary thresholds, so I doubt you'd consider it rigorous. The construction $X \gg Y$ means that the ratio $\frac{Y}{X}$ is small enough that subleading terms in the series expansion for $f\bigl(\frac{Y}{X}\bigr) - f(0)$ can be neglected, where $f$ is some relevant function involved in the ...

15

$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction ...

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. ... 12 The notion of tensor product is independent from the Hilbert space structure, it is defined for vector spaces on the field \mathbb K (usually \mathbb R or \mathbb C). A formal definition is given below (there are many equivalent approaches). First, if V is a vector space, V^* denotes its algebraic dual space, namely the vector space of the linear ... 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 ... 9 \mathfrak{R}e: real part. A^*: complex conjugate of probability amplitude A. 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. ... 6 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. 6 Dirac notation is ill-suited for non-self-adjoint operators. Here's why: Let (-,-) be the inner product on our Hilbert space. The expectation value of AB is then$$ \langle AB \rangle_\psi = (\psi,AB\psi)$$by definition, and Dirac notation writes \langle \psi \vert AB \vert \psi \rangle. for this. But, in this notation, it is no longer clear to which ... 6 Here's various things used in index notation : Types of indices : Greek indices for spacetime indices (tensor indices), lower case latin indices for 1) spacelike components 2) local Lorentz components 3) group components (for gauge indexes). Upper case latin indexes for spinor indexes, dotted upper case latin indexes for conjugate spinor indexes. ... 6 Take your first relation for the 3 Pauli matrices individually:$$\sigma_1(t)=U^\dagger\sigma_1(0)U\sigma_1(t)=U^\dagger\sigma_2(0)U\sigma_3(t)=U^\dagger\sigma_3(0)U$$Now you define a "vector" for notational convenience like the OP says in the question. I will choose to rewrite it as a column vector to visually show the relation of the above 3 ... 5 You're getting tripped up by summation notation. Whenever you have a repeated index, this means that that index is to be summed from 1 to 3:$$ \delta_{ij} \delta_{ik} \equiv \sum_{i=1}^3 \delta_{ij} \delta_{ik}. You're right that there are two terms in this sum where i \neq j, and so the contribution to the sum from these terms is zero. But the ... 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 & := ... 5 I think the answer is no. It generally precedes some approximation method with a bounded error, but there are so many approximations methods in physics -- some rigorous, some nonrigorous -- that it's way too presumptuous to give it a rigorous definition. Generally, it means one of several things: If a\ll b, expanding in powers of \frac{a}{b} is ... 5 Comments to the question (v5): In this quantum case the overline/bar notation \bar{A}=\langle A\rangle is borrowed from statistics and it denotes a quantum expectation value of a quantity A. See also Ehrenfest theorem. The problem from Ref. 1 considers a harmonic oscillator with Hamiltonian operator\tag{A} H~=~\frac{p^2}{2m} ...

5

It appears that none of Physical Review Style and Notation Guide, the AIP Style Manual, the IAU Style Manual, or The ACS Style Guide: A Manual for Authors and Editors, weigh in at all on this matter, so I would say it is to some extent up to personal taste. On the other hand, the NIST Manuscript Checklist does take a position: $$\begin{array}{rl} ... 5 As Qmechanic pointed out in the comments, you're mixing Einstein and abstract index notation a bit. To make things absolutely clear, we will use early Latin indices for abstract indices (abc) and Greek indices for component indices (\mu\nu\rho) and will always indicate Einstein summation explicitly. First and foremost, an abstract index is nothing more ... 5 These states represent intermediate coupling schemes that are halfway between the usual LS coupling and the more extreme jj coupling that happens in heavier atoms where relativistic effects mean that the spin-orbit coupling for each individual electron can match or exceed the orbit-orbit coupling between different electrons. The intermediate coupling ... 5 Just a coincidence. There are too many quantities and not enough letters. It probably does make a difference that the fields in which these two equations exist (material science and electromagnetism) are well enough separated that you typically won't see them both in the same papers or textbooks; if that weren't the case, people would start using different ... 5 It's a bit hard to see in this typography, but the two p are supposed to be different. The p on the l.h.s. is the four-momentum p = (p^0,p^1,p^2,p^3)^T, the one on the r.h.s is the three-momentum \vec p = (p^1,p^2,p^3)^T, and then$$ p^2 = E^2 - \vec p ^2$$for p^0 = E is tautologically true just from the definition of p^2. 5 s^2 = x^2 + y^2 + z^2 - (ct)^2, where x, y, z are the usual measures of distance, c is the speed of light, and t is the usual measure of time; then s is the space-time interval. The space-time interval, s, is a relativistic invariant, giving the same measure between two distinct space-time events, without regard to the relative velocities of observers. ... 4 This is a covariant derivative along a world line (if you would not consider a world line the proper time \tau would not make any sense). So you consider a curve in space time parametrized in dependence of the proper time x^\mu(\tau). Then you have:$$\frac{DA^\mu}{d\tau} = \frac{\partial A^{\mu}\big(x(\tau)\big)}{\partial \tau} + ...

4

It is a symbol and an idea used in mathematics too. But the important part is just that $B$ is 'ignorable' relative to $A$. This depends on the level of precision that is being used experimentally. If you're working to a precision of 1 part in 100, then $B$ should not effect the answer to that level of precision. If you're working to 1 part in a million, ...

4

It is common to write $$\partial_i = \frac{\partial}{\partial x^i}$$ for the derivative with respect to the $i$-th coordinate. Since time is customarily written as the $0$-th coordinate, $\partial_0$ is the time derivative.

4

Radioactivity occurs (with the exception of k-capture in which the nucleus captures an orbiting electron) inside the nucleus of atoms, not in the electron shells. Yes, the product has one extra proton, and thus is now a different element, jumping up one in the Periodic Table. No nucleus is electrically neutral. Whether or not an atom is an ion has nothing to ...

4

Your problem is that the thing you want to express is only true in the particular basis you're working in. But equations in the summation convention are true in every basis. Your expression involving an explicit summation is fine and is probably what any reasonable mathematician would do. But if you want to show unreasonable devotion to the summation ...

3

$\frac{DA^\mu}{D\lambda}=\frac{DA^\mu}{d\lambda}$ are two notations for the same object.

3

It's the time independent Schrödinger equation written using Dirac notation.

3

Your last expression is not valid, for two reasons: first, any given index can only occur twice per term in the Einstein convention, once as an upper index and once as a lower index. Remember that when an index is repeated, it means you sum over it with the metric: $$T^a T_a = \sum_{a,b} g_{ab} T^a T^b$$ You have terms like ...

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