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32

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 ...


31

Part of it is that Newtonian mechanics is described in terms of calculus. When we consider vibrational motions, we're talking about some particle that tends to not be displaced from some equilibrium position. That is, the force on the particle, at displacement $x$, $F(x)$, is equal to some function of displacement $x$, $g(x)$. There are two ways calculus ...


21

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 ...


18

In this case it's probably best to be pragmatic. A pulse can be described as a superposition of sine waves that extend infinitely into space and time. But it's just that: a mathematical description that is useful for your purposes. There is not necessarily a physical meaning connected to it. Nevertheless, in quantum mechanics the wave-description of ...


15

One of the big reasons not discussed above is Fourier theory -- any function $f(x)$ can be expressed in the form $f(x) = \int dk\, A(k)e^{ikx}$, which basically means that any function can be decomposed into an infinite sum of sines and cosines. Since this is the case and dealing with sine and cosine is mathematically simpler than the general case of ...


14

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 $$\tag{...


14

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 ...


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). That'...


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

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 $|p'\...


12

First of all, why are Fourier transforms useful? The Fourier transform is special because the complex exponential functions are eigenfunctions of he translation. In any linear problem with translation invariance, the Fourier transform turns a differential equation into an algebraic one. A simple example is a driven damped harmonic oscillator. The equation ...


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

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.


11

This issue is a bit confused in textbooks, however the statement of the professor is physically wrong (mathematically all the procedure can be rigorously justified using the theory of distributions). The point is that the claimed position operator is not the position operator because it is not even self-adjoint (nor Hermitian) in the relevant Hilbert space ...


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

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 ...


10

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 coordinate-...


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

Because cycles and oscillations and things with periodicity, are all intimately related to the circle. And $sin$ and $cos$ are defined based on the circle.


9

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 ...


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

I think Crawford is an incredible book - full of insight and clearly written by someone who loves the material. I used it for my waves-course sophomore year, and I think it's too bad it's out of print now. If you want something more theoretical, though, try Howard Georgi's book. Also, I'll second French. Also, David Morin has a set of drafted chapters of ...


9

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 $$\frac{\sum_{i=1}^n(F_{i+1}-F_i)}{...


9

First of all, the Gibbs phenomenon is a mathematical effect – it is the appearance of narrow but intense oscillations around the right value whenever a function with a discontinuity is approximated by its Fourier expansion that is truncated. This phenomenon doesn't occur when the relevant function is a wave function $\psi(x,y,z)$ in quantum mechanics for a ...


8

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 ...


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 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 }...



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