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2

Note that you can also write $\phi$ and $\pi$ as $$\phi(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\vec{p}}}} \left(a_\vec{p} + a_{-\vec{p}}^\dagger \right) e^{i\vec{p} \cdot \vec{x}}$$ $$\pi(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} (-i) \sqrt{\frac{\omega_{\vec{p}}}{2}} \left(a_\vec{p} - a_{-\vec{p}}^\dagger \right) e^{i\vec{p} \cdot \... 0 1) \sigma^\pm = \tau \pm \sigma so you get something of the form$$ \sum_n e^{-in\tau}\Big(e^{-in\sigma} + e^{in \sigma} \Big) $$which gives you the cosine. 2) The \alpha' (and l) is just a convenient normalisation. 1 Like Anders Sandberg mentioned in his comment, the Fourier transform (fft function in MATLAB) returns a vector of complex numbers. When you try to plot that, you will end up with its representation on the complex plane, where the x-axis is the real part and the y-axis is the imaginary part. If you would like to plot that as magnitude against frequency you ... 2 Periodic functions f(t) (with a time period T) can be approximated by a Fourier series, i.e. by summing harmonic oscillations with the discrete frequencies 0, \frac{2\pi}{T}, 2\frac{2\pi}{T}, 3\frac{2\pi}{T}, \ldots .$$f(t)=\sum_{n=-\infty}^{+\infty} F_n e^{in\frac{2\pi}{T}t}$$But, as you already noticed, with a Fourier series you cannot build ... 1 Such spatial Fourier transforms are commonly used in optics, both in theoretical treatments and in image analysis. An optical field is, of course, an example of an electromagnetic field. The main difference is that much of optics is done in the scalar approximation where one ignores the vector nature of the field. But you can also Fourier analyse vector ... 0 Use spherical coordinates (R,\theta, \phi); in the Fraunhofer far-field zone (R \to \infty) the field, say \mathcal {E}(\theta, \phi), can be written as a 2D Fourier transform of the source distribution F(\xi, \eta) here assumed to be planar whose rectangular coordinates are (\xi, \eta):$$ \mathcal {E}(\theta, \phi) \approx \frac {\mathfrak{j}}{2\...

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For your first question, remember that we want the electromagnetic field to be real. Adding the complex conjugate within the Fourier transform ensures this. Without it, we would generally have a complex vector potential, which is unphysical. As for your second question: you are certainly right that if the integral is zero this does not necessarily mean that ...

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In case anyone else runs into this problem, with some outside help I managed to find a solution. If one makes the substitution $t_r = t-\frac{r}{c}$ in the last integral, the full expression reads, $$\vec{A}(\omega) \sim e^{i\omega\frac{r}{c}} \int_{-\infty}^{\infty} e^{i\omega t_r} e^{-\frac{k}{m}t_r} dt_r.$$ So now the two integrals only differ by a phase ...

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I would like to add some words. You should not mix concepts of particle and wave. In order to understand behavior of a matter wave, you'd better stick with the concept of wave, rather than the particle picture. The Gaussian wave that you gave the expression spreads out and it can be observed far away from $x=0$ after a long time with certain velocity, which ...

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When the field detector is not in the beampath it is not sensitive to the field of the beam in the beampath. It is sensitive to the field in the mode next to the beam path which is zero in this experiment.

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Reviewing the mode expansion The laser being monomode and always turned on, the quantum state inside of its cavity or of the light it radiates can be described using coherent states of frequency $\omega_0$. If photons exist at other frequencies could you write down the interaction that created them? The first issue here is that you're assuming that modes ...

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Consider you have a monochromatic laser of frequency $ω_0$, for example made with hydrogen transition. Consider the beam has a given width $w$. The contradiction begins right here. Even if your device emits light from $t=-\infty$ to $t=\infty$, a finite beam width precludes monochromaticity. Especially so if we consider that your measured field abruptly ...

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This is the answer that I misunderstood the question. If you want to find an answer using Lorentz transformation and quantum fields, then skip this and see below. The interaction you want to have is simply the interaction between electron bound to an atom of blocking material and electromagnetic waves. The monochromatic laser in your experiment drive dipole ...

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Consider you have a monochromatic laser of frequency ω0. Consider it is infinitely small in width (see it as a line). An infinitely small line width source has to be stationary from negative infinity to positive infinity. Yours isn't (you have it change at t=0), so it must contain additional frequencies. But this doesn't make sense. Indeed, I only ...

3

The apparent paradox is analogous to the problem of blind men describing an elephant https://en.wikipedia.org/wiki/Blind_men_and_an_elephant (it's like a rope, a tree, a tent, a snake---). A Fourier transform is only one example of a way to represent a wave form. The same wave form can be represented as a sum of delta functions, Gabor wavepackets, and even ...

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The nonlinearity that you can't find in your monochromatic laser is here (or was in v1 of the question): Then you let the light pass until a time $t_0+T$ when you cut again the light using the plate again. A metal plate is a charge distribution where the lattice of positive ions has a different spatial arrangement and response function that the Fermi gas ...

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I'm not sure I fully understand the question, but I'll try to explain the physical meaning a bit. The parameter $a$ tells us about how "localized" the wavefunction is in the initial configuration. For large $a$, the particle has a large probability of being observed near the origin (at early times). For small $a$, the particle is more likely to be ...

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The Schrödinger's form of the wave equation is an expression of kinetic energy + potential energy = total energy. A wave function is one specific solution to this wave equation. A wave packet is a specific collection of solutions to the wave equation. Think of it as more than one photon or more than one particle behaving as a wave, all traveling in a ...

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Any superposition/linear combination of monochromatic left- and right-moving exponential waves is of course equivalent to a superposition/linear combination of sine and cosine waves, as OP demonstrated above. However from the point of view of a 1D scattering experiment where wavepackets enter from (and leave to) spatial infinity, the travelling exponential ...

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If the spatial part were to be multiplied by (for example) $\cos \left( \dfrac{\hbar k^2}{2m}t \right)$ this would be true. In this case you get a standing wave. But multiplying by $e^{-i\frac{\hbar k^2}{2m}t}$ yields a travelling wave.

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The short answer is yes. Not just a yes, a YES! Optics is one of the subjects that uses Fourier transforms all of the time. (Like a lot of other subjects in physics and engineering) If we'll think about a the light wave (or more precisely, the electric field) in one point in space we we'll see it varying with time. If it has a specific frequency $\omega$, ...

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A glass prism is not a Fourier Transformer, instead it is a gadget that moves the various frequency components to different spatial locations, it is more like an array of almost filters. Almost filters because to be a filter it also needs a rejection mechanism of frequencies not needed, for example an array of slits to remove the unneeded frequencies from ...

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