# Tag Info

4

There is an integration formula (see "Table of integrals, series and products" 7ed, p337 section3.324 1st integral) $$\int_0^\infty d\beta \exp\left[-\frac{A}{4\beta}-B\beta\right]=\sqrt{\frac{A}{B}}K_1\left(\sqrt{AB}\right)\qquad [\mathrm{Re}A\ge0, \mathrm{Re}B>0].$$ If $\mathrm{Re}A\ge0, \mathrm{Re}B>0$ is violated, the integral will be divergence. ...

2

The poles lay on the light cone, i.e. $G(x,y) \rightarrow \infty$ for $x\rightarrow y$. To see this, try calculating the integral via residue calculus. First you make a lorentz transformation of the integration variable, so that x has only one entry left (this is either the temporal entry for timelike x or one of the spatial for spacelike). Now for $s = ... 1 For any arbitrary collection of such travelling waves will always be a wave envelope that retains the same shape as the collection of waves propagate? No, it will not. For example, a Gaussian wave-packet will spread out in time. Wave packets are used to represent localization of particles in Quantum Mechanics.Group velocity will give the physical velocity of ... 2 You might see it from the notion of Fourier transformation. For example, you could express an arbitrary quantum state of your Hilbert space in momentum representation by applying a Fourier transformation on your position representation. More explicitly, for$\left|\psi\right>\in\mathcal{H}$, you might define $$\left< \mathbf{q} |\psi\right> = ... 1 since \phi(x) is real, \phi^*(q) = \phi(-q) implies Re(\phi(q)) is even and Im(\phi(q)) is odd, but nonetheless independent unlike the real and imaginary parts of \phi(x). So in the new measure it must be understood that you are summing only over the even/odd functions. There for ... 1 This is actually a cool proof that doesn't show up in a lot of intro quantum books. It involves infinitesimal translation operators. It's kinda lengthy, but essentially, you can show that infinitesimal translation is a unitary operator, and so can be written as \hat{T}_{\epsilon}=\hat{1}-i\epsilon\hat{p}, up to first order in the small translation ... 4 Observe that$$ p \langle x \vert p \rangle = \langle x \vert \hat{p} \vert p \rangle =-\mathrm{i} \partial_x \langle x \vert p \rangle$$since, with \hbar = 1, \hat p = -\mathrm{i}\partial_x when acting on wavefunctions \langle x \vert \psi \rangle. The solution of this differential equation is then the inner product$$ \langle x \vert p \rangle = ... 0 Since there is no interaction between the parts of your lattice, the response of the total system is sum of responses of the individual parts. These parts are harmonic oscillators, whose response function is well-known. If these have many different frequencies, the total system will have response function with peaks at these frequencies. No need for boson ... 0 If you combine the two statements In the paraxial approximation for a monochromatic light field, the complex light field in the back focal plane of a lens is the Fourier transform of the complex light field in its front focal plane. The lens in the human eye is a lens that can accomodate its focal length so that the retina lies in the focal plane of the ... 2 No, the (elementary solution for the position representation of the) wavefunction of a free particle,$\psi(x) = \mathrm{e}^{\mathrm{i}px}$is not an "explicit function of both" position and momentum. It is a function of position - the momentum of the plane wave is fixed, and the momentum space wave function of this is its Fourier transform$\tilde{\psi}(k) ...

1

A solution to the Schrodinger equation for a free particle is a plane wave, and because any combination of solutions is also a solution we can construct solutions by summing up plane waves. The equation you quote is constructing a solution by Fourier synthesis. Since the plane wave function $e^{i(kx-\omega t)}$ is a solution we can use Fourier synthesis to ...

3

Line (2) states that the Fourier transform of a function is zero. Then that function is also zero, at least almost everywhere.

0

In an alternating current circuit the electric current changes his direction periodically, by definition. Thus, if you look at the voltage magnitude, i.e. the magnitude of the electric potential $$\mathbf E = \nabla V$$ where $\mathbf E$ is the electric field; it will have a periodic shape. The changing sign of the magnitude states the inversion of the ...

0

Nobody explicitly uses the proper units for amplitudes, the square root of (action divided by the product of all your axes/variables). Instead, one usually expresses this fundamental quantity using two conjugate variables. For example, if you regard a simple mechanical oscillator as a wave by looking at its status over time, the interesting quantity (action ...

1

It's not volts per millisecond, it's volts at particular milliseconds. Think of the wave as displayed on an oscilloscope. Waves can actually be anything as a function of anything. It's just that voltage as a function of time is a really nice example. Of course, when you're talking about Fourier transforms, you get into complex numbers. For example, you can ...

2

The OP is correct in stating that the Fourier transform $$\xi(\omega) = \int\mathrm{d}t\, \mathrm{e}^{\mathrm{i}\omega t} \xi(t),$$ vanishes upon averaging over realisations, $\langle \xi(\omega)\rangle = 0$, so long as we assume that the noise is also zero on average in the time domain, $\langle \xi(t)\rangle = 0$. However, the noise is not only ...

2

The point of restricting the string to length $L$ is that we can then construct a periodic function (with wavelength $L$) by imagining repeated copies of the string connected to each other. In this case we can construct the function as a Fourier series with the lowest frequency sine/cosine having the same wavelength $L$. If we have an infinite string then ...

Top 50 recent answers are included