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26

Your ear is an effective Fourier transformer. An ear contains many small hair cells. The hair cells differ in length, tension, and thickness, and therefore respond to different frequencies. Different hair cells are mechanically linked to ion channels in different neurons, so different neurons in the brain get activated depending on the Fourier transform ...


20

I see that two examples in optics have been mentioned, a diffraction grating by Mark Eichenlaub, and a lens by sigoldberg1. I would like to elaborate a bit, because there is a subtle difference between the two. On the one hand, a diffraction grating separates out light of different frequencies, i.e. colors, transforming them into different positions. This ...


13

The reason of a more modest version of your statement (your big claim is not right) is that the sum $$\sum_{n=-\infty}^{\infty} |a_n|^2 $$ has to converge. That's because this sum is proportional to $$ \int_0^{2\pi} |f(x)|^2 dx $$ which converges for bounded functions (a basic insight about Fourier expansions and Hilbert spaces of periodic functions). ...


13

Before answering the question more or less directly, I'd like to point out that this is a good question that provides an object lesson and opens a foray into the topics of singular integral equations, analytic continuation and dispersion relations. Here are some references of these more advanced topics: Muskhelishvili, Singular Integral Equations; Courant ...


12

Sine and cosine waves are, physically, the most common. They are definitely the best description to what comes out of a wall socket, not because we like them mathematically, but because it's what comes out; electromotive force is generated in the power plant as a sinusoidal pattern with frequency 50/60 Hz. In the usual kind of generator, this is because in ...


11

This doesn't make much sense: light year is in any case a unit of distance. What is common is to use "reduced units", for examples units where $c=1$ (speed of light) or $h=2\pi$. But in these cases the opposite would happen: you would say "year" to mean a distance. Or for example you say "has a mass of xyz MeV" instead of "$MeV/c^2$". About the Fourier ...


11

In many cases our systems are described by linear differential equations, and these have the property that any linear combination of solutions to the differential equation is also a solution to the differential equation. This is useful because usually any arbitrary solution can be Fourier transformed to express it as a sum of plane waves. So if we can find ...


10

No. Consider any state with a momentum wavefunction symmetric about zero. It's position-space and momentum-space norm-squared probability distributions are not changed by time-reversal, even though the wavefunction clearly is. Here is an explicit example. Take the four Gaussian wavepacket of mean positions $x_0$ or $-x_0$, mean momenta $p_0$ or $-p_0$, ...


10

Dear user1602, yes, $\psi(x)$ and $\tilde\psi(p)$ are Fourier transforms of one another. This answers the only real question you have asked. So if one knows the exact wave function as a function of position, one also knows the wave function as a function of momentum, and vice versa. In particular, there is no "wave function" that would depend both on $x$ ...


10

Yes, it happens in reality too, nicely demonstrating that the Fourier analysis predictions are confirmed. An easy way to see this is to take an electrical sine wave signal, which is nice and monochromatic, and pulse it on and off. If you examine the spectrum of the pulsed wave on a spectrum analyser, you will see the spread of frequencies about the centre ...


9

Remember the double slit experiment? The interference pattern is the Fourier transform of the hole(s). This boggled my mind when I first learned it. In the limit where the screen is far from the mask, the rays of light actually physically compute the Fourier transform (see Fraunhofer diffraction).


9

Your running into circles will stop once you commit yourself to a choice. What to regard as postulate is always a matter of choice (by you or by whoever writes an exposition of the basics). One starts from a point where the development is in some sense simplest. And one may motivate the postulates by analogies or whatever. The CCR are a simple ...


9

Those things are surely not enough to find the inner product $\langle q|p\rangle$ uniquely. For example, starting with the conventional $Q,P$, you may redefine them by a canonical transformation, for example by $$ Q\to Q'=Q, \quad P\to P'= P + Q^3 $$ Then $P', Q'$ obey all the four conditions in the same way as $P,Q$. They also have eigenstates and ...


8

I) Here we will work in the Heisenberg picture with time-dependent self-adjoint operators $\hat{Q}(t)$ and $\hat{P}(t)$ that satisfy the canonical equal-time commutation relation $$\tag{1} [\hat{Q}(t),\hat{P}(t)]~=~ i\hbar~{\bf 1}, $$ and two complete sets of instantaneous eigenstates$^1$ $\mid q,t \rangle $ and $\mid p,t \rangle $, which satisfy ...


8

This has been extensively studied in linguistics and acoustics. Humans and other primates predict speaker gender through a combination of fundamental frequency $F_0$ ("pitch") and Vocal-Tract-Length estimates ($VTL$) which are a proxy for body size. Sometimes "formant dispersion" is used for $VTL$. It is usually defined as ...


8

The route to the uncertainty principle went something like this: In Heisenberg's brilliant 1925 paper [1], he addresses the problem of line spectra caused by atomic transitions. Starting with the known $$\omega(n, n-\alpha) = \frac{1}{\hbar}\{W(n)-W(n-\alpha) \} $$ where $\omega$ are the angular frequencies, $W$ are the energies and $n, \alpha$ are integer ...


8

The sound that reaches your ear is just air pressure fluctuating over time. You can use a transducer of some sort to convert the value of air pressure to some other form - for example: to the depth of a groove being cut into a helical track on a layer of wax on a rotating drum to the depth of a groove being cut into a spiral track on a circular disc of ...


8

It's sloppy language that is confusing you here. A Jacobian is not a transformation. The Jacobian of a transformation measures by how much the transformation expands or shrinks volume(/area/length/hypervolume/whatever) elements. Example: let $x' = 2x$. Then $dx = dx'/2$. The Jacobian is the $1/2$, meaning nothing more than "a unit of the $x'$ scale has a ...


8

The functions $e^{i \bf p \cdot \bf x}$ as functions of $\bf x$ are linearly independent for different $\bf p$'s, hence every coefficient in the linear superposition (that is, in the integral) must be zero.


8

The reason you can get rid of the integral and the exponential is due to the uniqueness of the Fourier transform. Explicitly we have, \begin{align} \int \frac{ \,d^3p }{ (2\pi)^3 } e ^{ i {\mathbf{p}} \cdot {\mathbf{x}} } \left( \partial _t ^2 + {\mathbf{p}} ^2 + m ^2 \right) \phi ( {\mathbf{p}} , t ) & = 0 \\ \int d ^3 x \frac{ \,d^3p }{ (2\pi)^3 ...


8

The factor of $1/2\pi$ is an artifact of the normalization convention being used for the momentum eigenstates. To begin to see how this is so, let us note that the choice of normalization of a Dirac-orthogonal continuous basis completely determines the form of the resolution of the identity. Writing an arbitrary state $|\psi\rangle$ in a given ...


7

It's not $e$ per se, but rather the exponential function $$ \exp(x) \equiv \mathrm{e}^{x} = \sum_{n \in \mathbb{N}} \frac{x^n}{n!}$$ and the exponential functions appears naturally in various contexts, the most relevant in QM being the Fourier transform and the map between self-adjoint and Hermitian operators, or, in another guise, as the natural map from ...


7

Your visual range includes roughly one octave as compared to roughly twelve in your aural range. Further your visual system uses only four types of light sensors each with limited frequency discrimination, while your hearing has fine frequency discrimination. So while light spectra could have harmonic structure your visual apparatus is ill equipped to ...


7

The wavefunction vector $|\Psi (t) \rangle $ is supposed to be a function of time only. When you write $ | \Psi (t) \rangle $ you are not considering the projection of the wavefunction nor on the position neither on the momentum space, but just the state of the system at time $ t $, which is nothing but a postulate of Quantum Mechanics. You will have the ...


6

Let's look at frequency instead of notes. Let's say the string has a natural frequency of $100 Hz$ and that harmonics are present when you pluck it. Then, the frequency content of the sound will be of the form: $a_1 \cdot 100 Hz + a_2 \cdot 200 Hz + a_3 \cdot 300 Hz + ... $ Now, let's say you fret this string halfway such that the natural frequency ...


6

Momentum is not the Fourier transform of position. In the position representation, position is the operator of multiplication by $x$, whereas momentum is a multiple of differentiation with respect to $x$. These observables (operators) are not Fourier transforms of each other. In the momentum representation, momentum is the operator of multiplication by ...


6

If you integrate $$\int f(x)\textrm{d}x = F$$ then $F$ has the units of $f$ times the units of $x$. Similarly if you differentiate, $$\frac{\textrm{d}f}{\textrm{d}x}$$ has units of $f$ divided by units of $x$. If you look at the simple example of integrating and differentiating with respect to time to go between position, velocity, and acceleration, you'll ...


6

That's equivalent simply to $c\int dx/x$. Switch to the Euclidean spacetime, $k_0=ik_4$ where $(k_1,\dots k_4)$ is $k_E$; i.e. analytically continue in $k_0$ (Wick rotation). The integral is $$\int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(ik\cdot \epsilon)$$ So it's proportional to the Fourier transform of $1/k_E^4$. The original function is ...


6

Yes, there is a very strong interconnection. A particle in q.m. hasn't got a defined position. Instead, there is a function describing the probability amplitude distribution for the position: the wavefunction $u(x)$. This is always told even in books for the general public. However, also the momentum of the particle isn't, in general, well defined: for it ...



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