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2

Hint: 1. $\phi(x,t)$ at different times are not independent. 2. $\int{d^4p\delta(p^2-m^2)}=\int{d^4p\frac{\delta(p^0-E_p)}{2p^0}}$. The left side of this equation is Lorentz invariant. This time your question is much clearer. If $\phi(x)$ is an arbitrary function of $x$, there's nothing confusing. If $\phi(x)$ is constrained by the Klein-Gordon ...


1

Usually, what you are taking the Laplace transform of, is the system response, $h(t)$, e.g. system's response to a delta input, $\delta(t)$. If system's delta response is nonzero at $t<0$ it would mean the system had been anticipating your delta input at $t=0$ and already started responding, which is non-causal. Therefore a causal system must have $h(t)$ ...


3

It is because, the continuous Fourier transform is equivalent to evaluating the bilateral Laplace transform with imaginary argument $s = iω$. This relationship between the Laplace and Fourier transforms is often used to determine the frequency spectrum of a signal or dynamical system. I assume you do not have any intuition problem with the Fourier transform. ...


1

I don't know the specific context that you may be looking for, but it's quite common to use a four-point correlation function of the quantum dipole moment operator to retrieve non-linear relaxation dynamics. These correlation functions make up the nonlinear response function of a material system to an electric field which, when convolved with the excitation ...


1

It turns out that this form is just a fancy way of writing a normal Fourier transform. Use the definition of the intensity correlation function $$g^{(2)}(\tau) = \langle n(\tau) n(0) \rangle, $$ where $n(t)$ denotes the (Hermitian) intensity at the position of the detector, evaluated at time $t$ in the Heisenberg picture. The angle brackets $\langle ...


0

I think I understand it a little better now so I decided to post my thought and obviously won't accept my own answer. My understanding is that A(k) is the space of wave vectors, however we know these only differ by direction not magnitude for the diffracted wavepacket (Rayleigh scattering). Therefore the function A(k) indicates the "spread" of the ...


0

A problem formulated in the time domain and its equivalent formulation in the frequency domain contain essentially the same information. They just have different mathematical forms. One is easier to solve than the other, that is why we use transformations. Many problems attempted through the equations of motion obtained from Newton's equations are really ...


-2

There are no differences. Frequency domain is used as a mathematics transformation tool (Fourier, Laplace) in order to resolve too complex differential equations in time domain spectra.


0

A single oscillator will make the rope tied between the two walls, to display standing waves, i.e. any point of the rope will rise and get down according to a law f(t) = Asin(ωt), where A is the maximal amplitude of oscillation at that point. But you have two oscillators. The oscillation of a point in time is (1) f(t) = A_1 sin(ω_1 t) + A_2 sin(ω_2 t). ...


1

In your third equation, I think you have $$I=\int \bar{a}(k)\bigl(-\Delta_k\bigr) a(k) dk\; ,$$ where I have rewritten the second derivative as the Laplace operator and corrected tha missing complex conjugate and the fact that you have only one variable. Said that, this can be rewrittenas the scalar product in $L^2$: $$I= \langle a, -\Delta_k a\rangle ...


1

$$ \sum _nf(n)\int \mathrm{d}p\, |\left< p|n\right> |^2=\sum _nf(n)\int \mathrm{d}p\int \mathrm{d}q\, \left< n|p\right> \left< p|n\right> =2\pi \sum _nf(n)\left< n|n\right> =2\pi \cdot \sum _nf(n), $$ where I used the fact that $$ \int \mathrm{d}p\, \left| p\right> \left< p\right| =2\pi, $$ which follows from $$ \left< ...


0

The very definition of a frequency is the rate at which something occurs over time. Therefore by shortening the time interval in which something occurs you are increasing its frequency, and by increasing the time interval you decrease the frequency. Hence, Frequency is inversely proportional to time.


1

First, the context is a function of time that is periodic which means that it is repetitive with repetition period $T$. $$g(t) = g(t + T)$$ So, if one sampled the function every $T$ seconds, one would get the same value each time. Now, we have the period of time $T$ which tells how long it takes for the signal to go through one cycle. The inverse ...


3

To answer the question "same wavelength, different frequencies, which arrives first?": naturally, the one with biggest speed, which is proportional to frequency AND wavelength according to the formula: $$v = \lambda f$$ So, for the same wavelength $\lambda$, the one with bigger frequency $f$ will have bigger speed $v$, thus arriving earlier. For the ...


2

I find the whole notation here a bit confusing since we are talking about momenta modes while using real space notation (maybe I'm the only one...). To clarify what is going on we can switch to momentum space instead. Consider the quadratic term in the exponential: \begin{align} \int d^4x \phi ^2 & = \int d^4x \int d^4k d^4k' \phi _k \phi ^\ast _{ k ' ...


4

I'm not being precise, but morally: Imagine you were integrating out all modes above a frequency $b\Lambda$. Consider $\omega < b\Lambda < 3 \omega$. A mode $\phi$ with frequency $\omega$ when cubed, will have some part of it as mode of frequency $3 \omega$, since: $\sin (3x) = 3 \sin (x) - 4 \sin^3 (x)$. (Easier to see that ${(e^{i \omega t})}^3 = ...


3

A different angle on this that I DON'T believe is in conflict with Terry Bollinger's answer: whether you express a wavefunction in position co-ordinates, or, as its Fourier transform, i.e. in momentum co-ordindates, the two models are precisely the same. So neither the expression of a position co-ordinate wavefunction (such as you find from the solution of ...


1

A pure plane wave is not a physically realizable state for a real particle because it would require the particle's wave function to extend over infinite space. Even if the universe is open and infinite in size, it has only existed for a finite time, so no particle wave function would have had enough time to grow to infinity.


0

Re: comments to the accepted answer. There is a very natural interpretation: For a linear problem the Fourier transform is the same as a plane wave Ansatz. That is, guess that $f(x) = c e^{-ikx}$ for some $k$, and put this into your equation. What you are doing is looking for solutions that are sinusoidal oscillations with a well-defined wavevector $k$. ...


0

A Fourier transform can be done over spatial or temporal dimensions, but the end result is that $f(\mathbf{x},t)$ $\rightarrow$ $F(\mathbf{k},\omega)$. The assumption is that the amplitude, $A$, of the signal is not dependent upon $\mathbf{k}$ or $\omega$, which allows one to say that $F(\mathbf{k},\omega)$ ~ $A \ e^{i (\mathbf{k} \cdot \mathbf{x} - \omega ...



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