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1

You are correct in assuming that there should be no more than $L$ Fourier modes. In book you specified considered two different sets of boundary conditions. For periodic boundary conditions it is the usual problem and you get multiples of $2\pi/L$ which they wrote as $q=j\pi/L$ with $j=-L,\dots,-2,0,2,\dots,L$ (note that only even values of $j$ appear). ...


0

Turns out that due to orthogonality relations of Hermite-Gauss poly's, Hermite-Gauss modes are orthonormal, so $$ \int \int u_{n,m} \left(u_{n',m'}\right){}^*dxdy=\delta _{m,m'} \delta _{n,n'} $$ Then a's can be found by multiplying both sides by conjugate of u and integrating ...


0

In general, you will run into some problems with "functions" like this one. Consider, for example, U(0), which is $$ U(0)=\frac{1}{(2\pi)^3}\int d^3 p e^{i\vec p \cdot \vec {\Delta x}} $$ $$ =\delta(\vec {\Delta x}) $$ I.e., a three dimension "delta function". This is not really a "function" in the conventional sense, but rather a "functional". So, okay, ...


1

In general, no you cannot. If you're told that the three source signals are all sinusoidal (for example), then Fourier analysis will give you the answer. But if, e.g., the three source signals are each a combination of various waveforms such as sawtooth or square, then there's no way to separate them unambiguously. I would like to warn you that there's no ...


1

You can express the dirac-delta-function as: $$\delta(x-x')=\frac{1}{2 \pi}\int dp e^{i p (x-x')}$$ (simply fourier-transform the dirac-function) compare it with your expression and you get the factor. p.s Your last line from intermediate step is wrong.


1

You're close, but you seem to be saying that $\exp(ikx) = \delta(x)$, which is not true, and you're missing a $2\pi$. The correct identity is $$\int dk\ e^{ikx} = 2\pi \delta(x)$$ Therefore, with a change of variables $p=\hbar k$: $$\int dp\ e^{ipx/\hbar} = \hbar \int dk\ e^{ikx} = 2\pi\hbar$$


1

What is color? Color has two possibilities. It is what our eye retina perceives as red, blue, yellow ... and its study belongs to biology. For example mixing blue paint and yellow paint gives the green color identified by our retina. In the rainbow each frequency is displayed according to the strength coming from the white light source, and our retina ...


5

Consider a single value of $m$. The Fourier series for just that $m$ gives $$a_m \cos(2\pi m t / T) + b_m \sin(2\pi m t / T) \, .$$ This can be rewritten as $$M_m \cos (2\pi m t / T + \phi_m)$$ where $$M_m = \sqrt{a_m^2 + b_m^2} \qquad \text{and} \qquad \phi_m = \tan^{-1}(-b_m/a_m) \, .$$ So, you can see that $a_m$ and $b_m$ are just the cartesian coordinate ...


2

I assume from your question, you are concerned with the Fourier transform of a scalar function of time (no space dependence): $$\tilde f(\omega) = \int_{-\infty}^\infty dt\, e^{-i\omega t} f(t)$$ With the inverse: $$f(t) = \frac{1}{2\pi} \int_{\infty}^\infty d\omega\, e^{i\omega t} \tilde f(\omega).$$ This symmetry is due to the fact, that your signal is ...


0

Expanding on the answer by @AndreaDiBiagio about the use of Fourier transforms to solve differential equations, specific applications I encounter occur in simulations of fluid dynamics, magneto-fluid dynamics (MFD), and plasma physics. Specifically, I have used Fast Fourier Transforms (FFTs) to significantly speed up the solution of a poisson problem for ...


0

By Fourier transforming a signal you indeed obtain frequency magnitude and phase information, and that can be very useful in a analysing experimental results. Another, related, extremely useful application of the Fourier Transform is in terms of solving differential equations. It turns out that many systems can be analysed pretty with differential equations ...


0

The Fourier Transform has many applications and you can find it in many places of maths and physics. I will show a pair of well-know examples; in Quantum Mechanics, the position space and the momentum space are linked by a Fourier Transform: $$\Psi(\vec{r},t)=\frac{1}{(2\pi \hbar)^{3/2}}\int_{\mathbb{R}^3} ...


4

There is an integration formula (see "Table of integrals, series and products" 7ed, p337 section3.324 1st integral) $$\int_0^\infty d\beta \exp\left[-\frac{A}{4\beta}-B\beta\right]=\sqrt{\frac{A}{B}}K_1\left(\sqrt{AB}\right)\qquad [\mathrm{Re}A\ge0, \mathrm{Re}B>0].$$ If $\mathrm{Re}A\ge0, \mathrm{Re}B>0$ is violated, the integral will be divergence. ...


2

The poles lay on the light cone, i.e. $G(x,y) \rightarrow \infty$ for $x\rightarrow y$. To see this, try calculating the integral via residue calculus. First you make a lorentz transformation of the integration variable, so that x has only one entry left (this is either the temporal entry for timelike x or one of the spatial for spacelike). Now for $s = ...


1

For any arbitrary collection of such travelling waves will always be a wave envelope that retains the same shape as the collection of waves propagate? No, it will not. For example, a Gaussian wave-packet will spread out in time. Wave packets are used to represent localization of particles in Quantum Mechanics.Group velocity will give the physical velocity of ...


2

You might see it from the notion of Fourier transformation. For example, you could express an arbitrary quantum state of your Hilbert space in momentum representation by applying a Fourier transformation on your position representation. More explicitly, for $\left|\psi\right>\in\mathcal{H}$, you might define $$\left< \mathbf{q} |\psi\right> = ...


1

since $\phi(x)$ is real, $\phi^*(q) = \phi(-q)$ implies Re$(\phi(q))$ is even and Im$(\phi(q))$ is odd, but nonetheless independent unlike the real and imaginary parts of $\phi(x)$. So in the new measure it must be understood that you are summing only over the even/odd functions. There for ...


1

This is actually a cool proof that doesn't show up in a lot of intro quantum books. It involves infinitesimal translation operators. It's kinda lengthy, but essentially, you can show that infinitesimal translation is a unitary operator, and so can be written as $\hat{T}_{\epsilon}=\hat{1}-i\epsilon\hat{p}$, up to first order in the small translation ...


4

Observe that $$ p \langle x \vert p \rangle = \langle x \vert \hat{p} \vert p \rangle =-\mathrm{i} \partial_x \langle x \vert p \rangle$$ since, with $\hbar = 1$, $\hat p = -\mathrm{i}\partial_x$ when acting on wavefunctions $\langle x \vert \psi \rangle$. The solution of this differential equation is then the inner product $$ \langle x \vert p \rangle = ...


0

Since there is no interaction between the parts of your lattice, the response of the total system is sum of responses of the individual parts. These parts are harmonic oscillators, whose response function is well-known. If these have many different frequencies, the total system will have response function with peaks at these frequencies. No need for boson ...



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