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A unitary linear operator which resolves a function on $\mathbb{R}^N$ into a linear superposition of "plane wave functions". Most often used in physics for calculating the response of a time shift invariant linear system as the sum of its response to time harmonic excitation or for transforming a quantum state in position co-ordinates into one in momentum co-ordinates and contrawise. There is also a discrete, fast Fourier transform for discretised data.

7 votes
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Fourier transform of the product

Hint: integrate each and every gradient by parts, $$ \int d^d r ~ e^{-i kr} f(-i \nabla)\tilde g(r) = \int d^d r ~ \tilde g(r) f(i \nabla) e^{-i kr} = \int d^d r ~ f( k) \tilde g(r) e^{-i kr} = f( …
Cosmas Zachos's user avatar
0 votes
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How to prove the equivalence of Wigner distribution function expressions?

Looks like a homework problem. In any case, for $$ G(k) = \int \!\! dx ~ e^{2\pi j xk}~ g(x) \qquad \leadsto G^*(k) = \int \!\! dx ~ e^{-2\pi j xk}~ g^*(x), \leadsto \\ g(x+u/2)= \int \!\! dk' ~ e^{- …
Cosmas Zachos's user avatar
3 votes
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Inversion of scalar field formula

It's a plain inverse 3d Fourier transform, once you reverse your garbling of it: $kx=\omega({\mathbf k}) t- {\mathbf k} \cdot {\mathbf x}$. $$\phi(x) = \int \frac{d^3{\mathbf K}}{(2\pi)^3}\frac{1}{2\o …
Cosmas Zachos's user avatar
1 vote

Fourier transform of matrix element of evolution operator

The "complete orthonormal basis" indicates $$ \langle \theta | \ell\rangle= e^{i\ell \theta}/\sqrt{2\pi} ~, \\ \int_0^{2\pi}\!\!\! d\theta ~~ \langle \ell'| \theta \rangle \langle \theta | \ell\r …
Cosmas Zachos's user avatar
2 votes

Finite-time Fourier transform of a wavefunction

This is but a sum of cardinal sine functions with peaks at $\omega = E_n/\hbar$, for all n. Absorb $\hbar$ into the Es out of respect for sanity. Recall $$ |g\rangle=\sum_n |\psi_n\rangle \langle \ps …
Cosmas Zachos's user avatar
4 votes

How to convert position expectation value in terms of momentum representation?

You just need to heed the definitions of your book, $\langle p|\psi\rangle=\tilde{\psi}(p)$, $\langle \psi|x\rangle= \psi^* (x)$, $\langle x|p\rangle=e^{ixp/\hbar}/\sqrt{2\pi\hbar}$, insert two compl …
Cosmas Zachos's user avatar
0 votes
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Derivation of Wigner Function of cosine with phase

Hint: $$\cos(2\pi f_i t + \theta) = \cos(2\pi f_i (t + \theta/2\pi f_i)= x(\tilde t), \\ \hbox{where}~~\tilde t\equiv t + \theta/2\pi f_i. $$ So, just plug into your expression for x(t), if you t …
Cosmas Zachos's user avatar
2 votes
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Marginal of Wigner Function calculation

It's evident. $$ \psi(q)=\int_{-\infty}^{\infty}\!\! \frac{dp}{\sqrt{2\pi \hbar}}~e^{-ipq/\hbar} \phi(p), $$ so that $$ \int_{-\infty}^{\infty}\!\! dq~ W(q, p) =\frac{1}{(2 \pi \hbar)^2} \int_{-\inf …
Cosmas Zachos's user avatar
6 votes

Group theory, character, representations, delta function, unit element

Yes it is. You are writing down the projection to the origin of the group through its character expansion, indeed a generalization of the Fourier transform, as you guessed. A good book would cover i …
Cosmas Zachos's user avatar
2 votes
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Split-step FFT: Evolving a wave freely makes it spread

Yes, free wave packet solutions of the dispersive Schrödinger equation almost always (*) spread like this, independently of your group velocity $k_0$ of the wavepacket. If you set that to 0, they'd s …
Cosmas Zachos's user avatar
1 vote
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Fourier transform of cross-spectral density space matrix elements

I'll parallel your book's proof in my language of Lemma 3 in our booklet, but now for the off-diagonal Wigner function, setting $\hbar=1$ , easy to reinstate by dimensional analysis. The time depen …
Cosmas Zachos's user avatar
1 vote

Fourier transform of the state?

This is a flip excuse to expose you to Condon's 1937 classic paper about Heisenberg-picture rotations in phase space, leading to the unitary equivalence of $\hat x$ and $\hat p$. In particular, you ma …
Cosmas Zachos's user avatar
0 votes
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Understanding action of position operator on momentum space representation of a position eig...

I am providing a parallel proof/insight to the impeccable answer of @Prahar, as you appear conflicted about dummy variables. Your last equation is not even wrong: it's gobbledygook. A derivative w.r.t …
Cosmas Zachos's user avatar
3 votes
Accepted

Field operators on vacuum

Your definitions/conventions trivially yield $$\psi(\vec{x})|0\rangle = \int \frac{d^3p}{(2\pi)^{3}}\frac{1}{ 2E_p } e^{-i\vec p \cdot \vec x} |\vec p\rangle $$ and $$\pi(\vec{x})|0\rangle = \fr …
Cosmas Zachos's user avatar
6 votes

Infinite square well in momentum space

Let me tweak and discuss a bit the answer of @Prasst, to make sure the student does not simply shrug and walk away because the issue is moot and problematic as per @Valter_Moretti 's answer and @Emili …
Cosmas Zachos's user avatar

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