New answers tagged

-1

-First we know that $\psi(\overrightarrow{x},t)$ is the density probability of being in $\overrightarrow{x}$ at time $t$ . Then we can say: $$ P_{total}=\int d^3x |\psi(\overrightarrow{x},t)|^2=1 $$ For this reason we have square-integrable condition on $\psi(x,t)$ $$ \int_{-\infty}^\infty |\psi(x,t)|^2 d^3 \overrightarrow{x} <\infty $$ We know that:$$ \...


2

The Fourier transform is a unitary map $L^2[{\mathbb R}^n]\to L^2[{\mathbb R}^n]$, although the integrals involved are not necessarily proper, and limits may have to be taken in the $L^2[{\mathbb R}^n]$ sense. See here and here. A longer, less rigorous, but perhaps more accessible account can be found in chapter 2 of my book with Paul Goldbart, a draft ...


2

See the question Which transform most closely mimics the human auditory system? on the Signal Processing SE. In short, the ear has a filter bank of gammatone-like filters tuned to different frequencies: The impulse response function of an individual filter looks something like this: Notice the impulse response decays rapidly after some finite time.


-2

Actually, your question is related to the precise details of how the body works, and has nothing to do with physics. The ear does not perform a Fourier transform. It is a long time since I studied it, but the biomechanics is quite complex. Certainly it has nothing to do with FTs. The ear contains multiple sensors sensitive to different frequencies. Iirc ...


2

If we can observe a signal for a finite time, we can estimate its frequency approximately, with an uncertainty inversely proportional to the duration of the observationg. As I understand it, the ear works by the sound waves vibrating certain small structures within the ear, which are resonant at different frequencies. These structures excite nerves that ...


2

Both do the job, as long as one can freely integrate by parts on ${\mathscr I}$. At large $|u|$ the fields on ${\mathscr I}$ are finite, i.e. $$ \lim_{u\to\pm\infty}\phi(u,z,{\bar z})= \phi_\pm(z,{\bar z}). $$ You can check that if $C = \phi_+ + \phi_- = 0$, then we can freely integrate by parts (one would have to be careful about the phases $e^{i\omega u}$ ...


3

A function $f(x,t)$ which satisfies the wave equation can be expressed generally as a function of a single argument $f(x-ct)$, where $c=\frac{\omega}{k}$. This is because you can express this function as an integral, according to Fourier Analysis: $$f(x-ct)=\int^\infty_{-\infty}C(r)e^{ir(x-ct)}\mathrm dr \tag{1}$$ This isn't really right. The function you'...


7

Your interpretations all make two basic mistakes. You assume the recorded data was mathematically accurate, and the FFT algorithm used somehow produces "exact" results. Some of the "broad spectrum" at low frequencies is most likely just environmental background noise. The signal to noise ratio compared with the peak amplitude is around 40 ...


1

The 2D Fourier transform of the fct $g(x,y)$ is defined as $$ G(f_x, f_y) \propto \int_{-\infty}^{\infty}dx \int_{-\infty}^{\infty}dy\; g(x,y) \, e^{i 2\pi(f_x x + f_y y)} $$ where $\{f_x, f_y\}$ are the spatial frequencies. The exponentials are the phasors of plane waves. The direction of a plane wave is given by considering its phase (exponent) $$ 2\pi(f_x ...


0

The method you describe of writing the wave function in the basis of momentum eigenstates and then applying the time dependence to each term in that basis, will indeed only work for a free particle. The reason for this is that what you really want to do when you study the time dependence of a system is to write it in the basis of energy eigenstates, because ...


1

They characterize the response in one of the circuits triggered by the signal in the other. E.g., if $F(\omega)$ has only one component: $$F(\omega) = (f_1(\omega),0),$$ Then the responses in the two circuits will be given by $$x_1(\omega) = \Theta_{11}(\omega)f_1(\omega),\\ x_2(\omega) = \Theta_{21}(\omega)f_1(\omega),$$ where $\Theta(\omega)$ is the matrix ...


3

In essence, the answer is yes. You do it by going to the "energy-space", i.e., you use the energy eigenbasis to span your Hilbert space. Let's say the kets forming the energy eigenbasis are $\vert E \rangle$ then the wavefunction of a state $\vert \psi\rangle$ in this basis would be given by $\psi(E) \equiv \langle E \vert \psi \rangle$. Just like $...


3

The basic idea of the Fourier transform is that of a transformation between two different bases in a vector space: In the position representation, the basis is that of states with well-defined positions $\phi_x$ (i.e. Dirac delta functions), and the state is written as $\psi = \int \psi(x)\phi_x \mathrm dx$, i.e. as a linear combination of the basis ...


6

Your solution $A_n$ isn't zero for every $n$. Looking at your initial conditions you might expect there to be only one non-zero coefficient since the IC already have the form of a sine wave. When $n=1$ you get $\frac 0 0$ so you have to be extra careful. Evaluating your solution for $A_n$ as a limit: $$\lim_{n\rightarrow 1}\frac{2\sin n\pi}{\pi(n^2-1)}=-1$$ ...


4

Wolframalpha is likely the source of confusion here: the initial condition is the term with $n=1$, i.e. $A_1 = 1$, whereas all others are $A_n=0$. The equation for the initial condition is one equation with the infinite number of unknowns in the rhs - no surprise that an automatic routine gives an irrelevant answer there.


0

Note that $y$ is a deterministic signal and thus independent of $Az(t)$. The answer should just be the sum of the two autocorrelations. More precisely, you want to compute the following expectation: $$\mathbb{E}_{\tau}[(Az(\tau)+B\cos(\omega \tau))(Az(\tau+t)+B\cos(\omega(\tau +t)))]$$ By linearity of the expectation you get the two autocorrelations that you ...


2

Fock space description and second quantization are not specific to harmonic oscillators - this is simply counting how many particles are in each state, whatever is the nature of the states. Creation/annihilation operators serve here to increase or reduce the number of particle in a state. What often serves as a source of confusion is that for a one-particle ...


5

Two quantum mechanical systems which are defined on the same Hilbert space (e.g. the free particle on a line and the harmonic oscillator, which are both defined on $L^2(\mathbb R)$) are distinguished only by which operator is chosen to be the Hamiltonian. It follows immediately that any procedure which gives you information about the set of energy ...


0

The momentum transfer $q$ has to be an even integral multiple of $\pi/L$. There is no need for it be in the allowed momenum lattice.


0

$e^{i\omega t} = \cos(\omega t) + i\sin(\omega t)$. Now when $t=T i$ very large both $\cos$ and $\sin$ go over $-1$ to $+1$. Integral is the sum of these values. It can be $-1$ or $+1$. The average over a long time will be $-1$ or $1/T$ which becomes very small. Therefore this does not contribute.


3

The time average of a quantity between $t=0$ and $t=T$ is $$\overline f = \frac{1}{T}\int_0^T f(t) dt$$ if $f(t) = Ae^{i\omega t}$, then $$\overline f = \frac{A}{T}\int_0^T e^{i\omega t} dt = \frac{A(e^{i\omega T}-1)}{i\omega T}$$ which goes to zero as $T\rightarrow \infty$. It's not that $f$ doesn't contribute at all, but if you integrate it over a long ...


3

$e^{i\omega t}$ makes a cycle. Average it over the whole cycle and it averages out. Average it over many cycles and it averages out except the fraction of the last cycle. So it mostly averages out. If it isn't $e^{i\omega t}$ but something more complicated like $|\sin(i\omega t)e^{i\omega t}|$ then it might not average out. You have to check.


1

The reason is simply this $$A\int_{0}^{\frac{2\pi}{\omega}} e^{i \omega t}=A\frac{1}{\omega}e^{i \omega t} \bigg \vert_0^{\frac{2\pi}{\omega}}=A\frac{1}{\omega}(e^{2 \pi i}-1)=A\frac{1}{\omega}(1-1)=0.$$ If one picks a longer time interval, it can be split into many smaller intervals that contribute nothing.


0

The Fourier transform of a rectangle is the sinc function. Since the Fourier transform is simply a superposition of different sine-waves (a cosine-wave is a sine-wave, which is shifted by $\pi/2$) this is an example of your problem. Please go ahead an check out the derivate of the above mentioned Fourier transform.


3

You’re exactly right: $|\phi(p)|^2$ gives the probability of measuring momentum $p$ at time $t=0$. An analogous relation holds for the time-dependent case: $$\Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}dx e^{-i px/\hbar}\Psi(x,t)$$ This is simply due to the fact that one independently transforms between position and momentum space and between ...


1

Consider the two point function $$\langle \tilde{\mathcal{O}}(-k) \tilde{\mathcal{O}}(k) \rangle = \int d^d x e^{i k \cdot (x_1-x_2) }\left\langle\mathcal{O}(x_1) \mathcal{O}(x_2)\right\rangle,$$ where the integral is over the differences $x=x_1-x_2$. Performing the rescaling $k\rightarrow \lambda k$ we have $$\left\langle \tilde{\mathcal{O}}\left(-k \...


3

If $\Lambda \subset \mathbb{R}^d$ is a lattice, the reciprocal lattice $\Lambda^* \subset \mathbb{R}^d$ may be defined as the lattice of vectors $a$ such that for all $v \in \Lambda$, $v \cdot a \in \mathbb{Z}$. The point of the reciprocal lattice is that if we take the Fourier transform of a function $f(v)$ valued on $\Lambda$, meaning we take $$\tilde f(k) ...


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