# Tag Info

## New answers tagged fourier-transform

2

My first question is related perhaps to notation choice or a typo. Should we use the same function used for the filed to write the Fourier transform? Or we should put Φ(p,t) instead of ϕ(p,t)? This is a standard awful physicist convention. Clearly the functional forms of the two "phis" are different, but the symbol being used is the same. This ...

2

I would say that an answer is that length of wave packet and width of spectrum are related by: $\Delta\omega \Delta t\approx 1$ "Width of spectrum" here is characteristic range of frequencies that signal contains, that is width of Fourier transform of the signal. Infinite sine wave contains only 1 frequency, that is its spectrum/Fourier transform is ...

5

In Scalar Field Theory one defines creation and annihilation operators as: $$a_p = E(p)\phi(p) + i\pi(p)$$ and $$a^\dagger_p = E(−p)\phi(−p) − i\pi(−p)$$ combining together to give the momentum space representation of the quantum field: $$\tilde{\phi}(p) = \frac{a_p + a^\dagger_{-p}}{2E(p)}$$ By taking the Fourier transform of this momentum space ...

1

The size of the image in frequency domain $f_{max}$ is inversely proportional to the grid spacing in real space $\Delta x$. (i.e. the finer step, the hight frequency you can sample). And grid spacing step in frequency domain $\Delta f$ is inversely proportional to size of real space image $x_{max}$ ( i.e. the longer interval of data you have, the more ...

1

Momentum is be conserved iff the Hamiltonian has translational symmetry. Usual boundary conditions such as homogeneous Dirichlet or Neumann conditions don't allow for such symmetry. But there still are specific conditions, which do allow the Hamiltonian to have translational symmetry on the bounded domain: Born—von Karman boundary conditions. Thus in the ...

4

This would mean that the eigenbasis of a physical observable is not orthogonal. Is there an error in my derivation, and if not, how can this be understood physically? The set of eigenfunctions of $\hat p$ in the sense $$\hat{p}\phi = p\phi$$ is sure to be orthogonal if they belong to a subset of $L^2((0,1))$ on which the operator $\hat{p}$ is ...

2

Let's drop the scaling constants and say that the wavefunction in momentum co-ordinates $\psi_p(\vec{p})$ is the FT of that $\psi_x(\vec{x})$ in position co-ordinates , i.e. $$\psi_p(\vec{p}) = \mathfrak{F}_{\vec{p}}(\psi_x(\vec{x}))$$ where $\mathfrak{F}$ is the Fourier transform. Now you propose to take the inverse Fourier transform of $\psi_x$. The ...

0

How to prove that the position operator in momentum is iℏ∂/∂p Apply the following useful result. If $$F(k) = \int_{-\infty}^{\infty} f(x)e^{-ikx}dx$$ then $$\int_{-\infty}^{\infty} xf(x)e^{-ikx}dx = \int_{-\infty}^{\infty} f(x)\left(i\frac{\partial}{\partial k} e^{-ikx}\right)dx = i\frac{\partial}{\partial k}\int_{-\infty}^{\infty} f(x)e^{-ikx}dx = ... 2 From this step,$$U_G=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x U(x)e^{-iGx}=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx}$$note that the summation is an impulse train with spacing a. Since the integral is from -\frac{a}{2} to \frac{a}{2}, just the impulse at x = 0 is integrated over so only the n=0 term ... 2 (\hbar omitted in the following.) That is not weird, it is one of the crucial properties of the Fourier transform F(\bar{}) that$$ F(\partial_x f) = \mathrm{i}p F(f) i.e. differentiation by one variable becomes multiplication with the Fourier conjugate variable and vice versa. Because of this, Fourier transformation is a powerful tool to solve ...

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