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There is no "Fourier optics" in superconductors in general, only in Josephson junction, and it's due to the Aharonov-Bohm effect. In bulk superconductor, the Aharonov-Bohm gives the so-called Little-Parks effect : the oscillation of the critical temperature with respect to the magnetic flux enclosed in a non-connected superconductor (= a superconductor with ...


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Here is some documentation for Matlab's fft: https://www.google.nl/url?sa=t&source=web&rct=j&ei=d50VVcCiFumz7gbl54G4Aw&url=http://www.mathworks.com/help/matlab/ref/fft.html&ved=0CBwQFjAA&usg=AFQjCNEW3G7KD5j1-T99VPbdeCb80SHHLg Octave is supposed to be very similar.


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I think you have the answer for your second question. For your first question let me clarify: $$\langle x\ |\ \hat{x}\ |\ p\rangle \overset{(1)}{=} \int dx'\langle x\ |\ \hat{x}\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(2)}{=} \int dx'x'\langle x\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(3)}{=} x\langle x\ |\ p\rangle ...


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Second, if I try to do the same thing by using the $p$-basis representation of $\hat{x}$, I get into more trouble: $$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dp'\langle x\ |\ \hat{x}\ |\ p'\rangle\langle p'\ |\ p\rangle=\int dp'i\hbar\frac{\partial}{\partial p'}\langle x\ |\ p'\rangle\langle p'\ |\ p\rangle$$ $$=\int dp' ...


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This question (v6) [concerning the overall minus sign in OP's calculation] is essentially a Fourier transformed version of e.g. this Phys.SE post, see Emilio Pisanty's answer and my answer. The main point is again that the derivative in the momentum Schrödinger representation $$\hat{x}~=~i\hbar\frac{\partial}{\partial p}, \qquad \hat{p}~=~p,$$ acts on the ...


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$$\frac{\partial}{\partial t}\Psi(x,t) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) e^{i\hbar kx}\left(\frac{\partial}{\partial t}e^{-\frac{1}{2}\frac{\hbar^2k^2}{m}t/\hbar}\right) dk$$ $$\frac{\partial}{\partial x}\Psi(x,t) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) \left(\frac{\partial}{\partial x}e^{i\hbar ...


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Yes it is the (square root of) amplitude versus frequency. It presumably breaks the audio into short intervals and then DFT's each interval individually. The DFT of the entire sound would not be very insightful.


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I don't believe it means a great deal. The reason is that the the Fraunhofer approximation or other farfield approximations assume that the point where you're calculating the EM field is at a distance from the source that is (i) big compared with the extent of the source and (ii) big compared with the wavelength. These two assumptions are needed to make e.g. ...


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A spread of momentum is sometimes used literally (as in, the momentum is definitely bigger than $a$ and definitely less than $b$ so the spread of momentum of is $b-a$), and sometimes it is used colloquially just to say that most of the time the momentum is between an $a$ and a $b$ such that $b-a$ equals your spread of momentum. The reason having a spread of ...


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"I just want to know, before entering the slit, the particles had no vertical momentum. But after that, they got vertical momentums. Now, how it proved the Uncertainty Principle? Feynman wrote that as the pattern diffracted, the uncertainty of vertical position i.e. vertical momentum increased. Can you tell me how?" Let's take the things one by one. ...


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You are correct that time and frequency domains are just Fourier transforms of each other. However you only have full information if you have amplitude and phase information (as opposed to a power spectral density which is only amplitude information in the frequency domain). Frequency spectra might tell you that you have multiple modes that exist, but it ...


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Missing from these answers is that CCDs like your camera can't, in principle, see a Fourier transform. Even if you stick a lens at the focal plain you won't see the Fourier Transform! Your camera is not an interferometer! Borrowing the equation from Colin K's answer we see that the integral, can be negative, positive or even imaginary. \begin{equation} ...


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Taking as given the state of your particle, the result of a momentum measurement is a random variable. $\Delta p$ is the standard deviation of that random variable.


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As you didn't quote the site with Feynman discourse, I make a guess what was the issue. The single-slit experiment illustrates diffraction. The wave-front that passes through the slit generates, according to Huygens' principle, a new wave-front, and so on. We obtain in all, beyond the slit, a deviation of the direction of the light. i.e. the linear momentum ...


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My first question is related perhaps to notation choice or a typo. Should we use the same function used for the filed to write the Fourier transform? Or we should put Φ(p,t) instead of ϕ(p,t)? This is a standard awful physicist convention. Clearly the functional forms of the two "phis" are different, but the symbol being used is the same. This ...


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I would say that an answer is that length of wave packet and width of spectrum are related by: $\Delta\omega \Delta t\approx 1$ "Width of spectrum" here is characteristic range of frequencies that signal contains, that is width of Fourier transform of the signal. Infinite sine wave contains only 1 frequency, that is its spectrum/Fourier transform is ...



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