# Tag Info

34

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 ...

30

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...

18

The statement "$A$ happens given that $B$" is equivalent to "If $B$, then $A$", which is symbolically represented as an implication $$B \Rightarrow A$$ or, if you want to preserve the order of $A$ and $B$ in the original statement $$A \Leftarrow B$$

15

Surely there is a story behind the print on Feynman's tee, these are the CM-1/CM-2 T-shirts. Quoting Tamiko Thiel: The geometric boxes and their 'hard' connections represent the 12-dimensional 'cube of cubes' that forms the internal hardware network connecting all processor chips with each other in a maximum of 12 steps. Feynman is the one who suggested ...

14

Bra ket notation has nothing to do with integrals. If I have two vectors in 3D space, $\vec{v}$ and $\vec{w}$ , I can write them as $|v\rangle$ and $|w\rangle$ if I want. In that case, I would write their dot product (a.k.a. inner product) as $\langle v | w \rangle$. This dot product has nothing to do with integration. When your vectors are functions then ...

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. ... 11 The notation \lvert \rangle is meant to imply that \lvert \text{anything here you want to put here} \rangle is a vector in a Hilbert space. If you have got some wavefunction \psi(x), then you often denote the abstract vector (instead of the concrete realisation in a basis like \psi(x)) it represents by \lvert \psi \rangle. If you have got only a ... 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. 9 If you want to make statements over (discrete) time, linear temporal logic may be worth looking at. For instance, \qquad\displaystyle \Box (A \implies B) means that whenever A holds, B has to hold at the same time. \qquad\displaystyle \Box (A \implies \Diamond B) means that whenever A holds, B will hold at some point in the future. Or yet ... 9 In physicist jargon, we talk about group representations of \mathrm{SU}(2) and \mathrm{SU}(3) by denoting an irreducible representation whose representation vector space has dimension N by \mathbf{N}. Hence, the statement \mathbf{3} \otimes \bar{\mathbf{3}} = \mathbf{1} \oplus \mathbf{8} is the statement that the tensor product of the ... 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. ... 7 (Q\cdot Q)_{ij}=Q_{im}Q_{mj} (Q^T\cdot Q)_{ij}=(Q^T)_{im}Q_{mj}=Q_{mi}Q_{mj} where we use that (Q^T)_{im}=Q_{mi} 7 Those Greek letters are indices indexing the components of g. Generally if one expresses a rank-2 tensor like g as a matrix, the first index indexes the rows, the second the columns. In your example, we have g_{rr} \equiv g_{11} = 1, g_{\theta\theta} \equiv g_{22} = r^2, g_{r\theta} \equiv g_{12} = 0, etc. As you can see, we sometimes use numbers ... 6 Given the way that you've presented your table, I would personally put a "-" rather than a 1 in the units column. This to me would signify that units such as "g, km, s, A" etc. do not apply here. In terms of your symbols, in many branches of physics it is common to use a "hat", "tilde" or "star" notation above a symbol to indicate that it is a ... 6 The Lane-Emden is really a non-dimensional form of the Poisson's equation with spherical symmetry:$$ \nabla^2f(r)\equiv\frac1{r^2}\frac{d}{dr}\left(r^2\frac{df}{dr}\right) $$This should be clear to all that, since we have two factors of d/dr on the right hand side of the above, it must be a 2nd order differential equation. This 2nd-order derivative ... 6 It is a mnemonic notation that indicates that \mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu is the object whose square root is to be used as the infinitesimal line element, traditonally denoted \mathrm{d}s, when determining the lengths of worldlines x : [a,b] \to \mathcal{M} by integrating the line element along them as$$ \begin{align*} ...

6

What they're saying is that $|3\rangle$ represents the third energy eigenstate of the oscillator. So, it replaces something like $\psi_3$. Writing $|3\rangle$ requires context - you would have to explain that you were going to number the nth energy eigenstate of the harmonic oscillator as $|n\rangle$ before using that notation. It's not an abuse of ...

6

I will try to answer this in the more general case where the configuration space is $\mathbb R^n$. In this case the Hilbert space of the quantum theory is $L^2(\mathbb R^n)$ with Lebesgue measure, and the inner product has the representation $$(f,g) = \int_{\mathbb R^n}\overline{f(x)}g(x)\ \text d\lambda(x)$$ where $\lambda$ is the Lebesgue measure. Let $U$ ...

6

Force is indeed a vector. Technically you should write $|\overrightarrow{F}| = 30N$, however there is usually context given that let you omit this. If you are working in one dimension, then the vector-like direction is all encapsulated in the sign once you've defined your coordinate system (e.g. -30N is 30N downwards.) Beyond that, it is typically just a ...

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 ...

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 $$\tag{A} H~=~\frac{p^2}{2m} ... 5 I think the answer is no. It generally precedes some approximation method with a bounded error, but there are so many approximations methods in physics -- some rigorous, some nonrigorous -- that it's way too presumptuous to give it a rigorous definition. Generally, it means one of several things: If a\ll b, expanding in powers of \frac{a}{b} is ... 5 The answer is already on page 2 of your link above: "Among the large number of radionuclides of medical interest, Sc-44 is promising for PET imaging. Either the ground-state Sc-44g or the metastable-state Sc-44m can be used for such applications, depending on the moleculeused as vector." So the metastable state Sc-44m decays to the ground state Sc-44g. 5 Either. It's context dependent. Chemists generally mean the whole atoms, nuclear physicists usually mean the nucleus, and people not in those categories could mean either. And there are exception to all those rules or thumb. And the distinctions is important when people start throwing masses around because the mass of an electron is almost on the same ... 5 Two conventions. First - use a capital M - make sure you make it big and pointy, so it cannot be confused with lower case: When it is right next to the lower case 'm', the difference should stand out clearly. Second - some people use the "computer short hand" E6: 1.7E6 m This is generally understood to mean (but quicker to write than) 1.7\cdot ... 5 Use:$$ u_{\alpha}= g_{\alpha\beta}u^{\beta} $$where g_{\alpha\beta} is the metric tensor. 5 You're getting tripped up by summation notation. Whenever you have a repeated index, this means that that index is to be summed from 1 to 3:$$ \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 & := ...

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