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1

To quantize a classical system, start from the Poisson bracket $$\{x_i, p_j\} = \delta_{ij}.$$ This relation defines $p_i$ as the momentum canonically conjugate to $x_i$ and is equivalent to Hamilton's equations. Quantize by letting $x_i, p_j$ be Hermitian operators on a Hilbert space, with commutator $$[\hat x_i, \hat p_j] = i\delta_{ij} $$ (identity ...


6

What's the mathematical process and physical logic? The Fourier transform of position space ($\vec x$ domain) is wave number space ($\vec k$ domain). This is an unambiguous, well understood mathematical result. By the De Broglie hypothesis, the momentum is $\vec p = \hbar \vec k$. This is physical hypothesis with experimental confirmation. Although ...


0

A Fourier transform is the decomposition of a position space function into a basis of plane waves, each of which has a well defined momentum. $$ f(x) \sim \int \text{d}p\; F(p) e^{\text{i}px} $$ This relies on the quantum mechanical idea that waves can have a well defined momentum.


2

The error in the OP's question comes from the second Kronecker delta, which is not correct (furthermore, the sum over $k$ is not dealt with...). After using the first Kronecker delta $\delta_{k\,k'}$, one has to use the identity $$\frac{1}{N}\sum_{k=0}^{N-1}\exp{[\frac{2\pi i k(j'-j)}{N}]}=\delta_{j'\,j},$$ which directly gives that $F^\dagger F =1$. One ...


0

I don't believe it's true that $\langle k | k' \rangle = \delta_{k',k}$. If I recall correctly, the whole point of the $\frac{1}{\sqrt{N}}$ factor is to normalize the momentum eigenstates, so we should have $\langle k | k' \rangle = N\delta_{k',k}$.


0

When you act $\phi(t,\vec{x})$ on the vacuum, you get a particle at $(\vec{x},t)$, that is $|\vec{x},t>=\phi(t,\vec{x})|0>=\int \frac{d^3 k}{(2\pi)^3}e^{-ikx}a^\dagger _k|0>$. It's a Fourier transformation. And it means that if a particle wants to be in $(\vec{x},t)$ then it should have all kinds of momenta, which is actually the uncertainty ...


0

Ok I think I solved the problem. So to divide the density FFT by k^2 i need actual values of k in k-space for my system. FFTW orders the result of transformation in so called "in-order" output, that means in first quadrant of FFT the first pixel corresponds to DC frequency and next to k/L frequency (k is from 0 to N-1) where L is length of whole system. ...


0

Substitute $$ \varphi(x,y) = \int dx dy \phi(k_x, k_y) e^{i k_x x + i k_y y},~~\varrho(x,y) = \int dx dy \rho(k_x, k_y) e^{i k_x x + i k_y y} $$ in the first equation and this should immediately give the equation you desire. Also, just for potential future purposes, note $$ \int dx e^{i k x } = 2\pi \delta (x) $$


1

This is simply a Fourier transform. The wave equation, Hamiltonian, and Green's functions are all simpler in momentum space than in position space. This is because, for the free, theory, the different momenta decouple, and you can create a particle with momentum $k$ with $a^\dagger_k \left|0\right\rangle$, and that state will evolve in a nice way. This is ...


4

John has answered this partially, however the fundamental mathematical idea is missing: If we think of a function being member of some vectorspace, then basisvectors exist. This concept will surely be familiar to you from Quantum mechanics. But also from there we remember, that problems were alot easier to handle, if we know the Eigenbasis of the Operators ...


2

There are some situations in which the plane wave ansatz is useful. It is a solution to many wave equations. Plane waves are also familiar and we have mathematical techniques to handle them, such as Fourier series. However, in some situations you can quite badly wrong using the planewave ansatz. Notably when the wave interacts with a structure that has ...


11

In many cases our systems are described by linear differential equations, and these have the property that any linear combination of solutions to the differential equation is also a solution to the differential equation. This is useful because usually any arbitrary solution can be Fourier transformed to express it as a sum of plane waves. So if we can find ...


0

What's your problem? How do you pass from a certain vector |Q> expressed according to some base of vectors {|$B_i$>} to the same vector |Q> expressed according to another base of vectors {|$C_j$>}? Let's see it together. Take for example (1) |Q> = $∑_i$ $a_i$ |$B_i$>, where $a_i$ are the amplitudes. For passing to the new base you project |Q> on each ...


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 ...



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