# Tag Info

## Hot answers tagged terminology

11

It's c for constant or celeritas, which means speed in Latin. Everyone uses it because it's convention. You could use $\xi$ or $\zeta$ or $\gamma$ or any other symbol you wanted, but then you'd have to explain what it meant, and people would have to go through the trouble to remember this every time they read your papers. Better to go with convention and ...

3

Confirmation bias might be what you're after. From wiki: Confirmation bias [...] is the tendency of people to favor information that confirms their beliefs or hypotheses. http://en.wikipedia.org/wiki/Confirmation_bias

2

One has to be careful with the given potential. To start with it must be shown that $$h=-(d/dx)^2+V(x),$$ defines a unique self-adjoint operator $H$, i.e., is essentially self-adjoint. In case $$V(x)=ax^2+bx^3+cx^4$$ with $c>0$ this is indeed the case. In fact the resolvent of $H$ is compact (these matters are discussed in the books by Reed and Simon), ...

2

The magnitude of acceleration is simply a measurement of change in speed per unit time. As an example, say you are in a car starting from rest and you begin to speed up. Say that you reach a speed of $20 {m \over s}$ in $2$ seconds. This means the magnitude of your acceleration is: $$a = {20 {m \over s} \over 2s} = 10 {m \over s^2}$$ That is, your speed ...

2

Your question is kind of vague but I will try to respond. Acceleration is defined as the time rate of change of velocity. Since velocity has both magnitude and direction, so does acceleration. In other words, acceleration is a vector. The length of the vector is its magnitude. Its direction is the direction of the vector. So the magnitude of ...

2

I assume you're thinking about Minkowski space, i.e. the metric $\eta_{\mu\nu}=\text{diag}(c^2,-1,-1,-1)$. You should be aware that the dot notation is purely a notational shorthand, and has no other information contained in it. In particular, by definition we have $$\dot{A}\equiv\partial_0A=\frac{1}{c}\frac{\partial A}{\partial t}$$ Thus, there is no ...

2

The set of all possible elements of the form $e^{i\alpha}D(x,p)$ with $\alpha, x,p \in \mathbb R$ verifying the commutation relations you wrote in addition to: $$D(x,p)^* = D(-x,-p)\:,\quad D(0,0)=I$$ is a group and it is called Heisenberg group, it is homeomorphic (diffeomorphic) to $U(1) \times \mathbb R^2$ but not isomorphic as a Lie group. It is a real ...

1

We can realize the displacement operator as $$\tag{1}\hat{D}(x,p)~=~e^{x\hat{P}+p\hat{X}},$$ where the elements $\hat{X}$, $\hat{P}$ and ${\bf 1}$ generates the Heisenberg algebra $$\tag{2} [\hat{X},\hat{P}]=i{\bf 1}.$$ These elements can be realizes as differential operators in the Schrödinger representation. (See also the Stone-von Neumann theorem.) ...

1

Line integrals of the magnetic field strength are magnetic voltage drops. Just google for "magnetic voltage drop" (including the double-quotes). In the quasi-static case ($\dot{\vec{D}}=\vec{0}$) the $\vec{H}$-field within a simply path-connected domain with zero current density has a magnetic potential. In this case you can calculate the magnetic voltage ...

1

I've always treated anharmonic oscillators to mean the potential has the form $$V(x)=\gamma x^2 + \beta_ix^i$$ with $i$ being any value except 2, including negative values as well. Anharmonicity then follows as the deviation of the eigenvalue of $V(x)$ above from the harmonic solution. For example, the paper you link above, Case 1 has an energy eigenvalue ...

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