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If black is absolute black, the top bar completely masks bars underneath. If the second layer also uses absolute black, it looks just like the top layer. So you wouldn't be able to discriminate. If you have translucent grey bars, you cannot tell the difference between white-over-black and black-over-white. If you had two layers where one was vertical and ...

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So you're hoping that by measuring the gray level you can tell which bar belongs to which layer? That's potentially possible, but the problem is that a black bar on the top level completely masks any black or white bar on levels below, so there is no way to read all the barcodes. you might be able to reject the lower layers (read just the top) on the basis ...

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The discrepancy comes from the fact that there are different ways to define a quantum Fourier transform and the direct equivalence to the Hadamard transform holds for only one such definition. See for instance this Introduction to Quantum Computing. Consider the general case of $N$ qubits, let the $2^N$ computational basis states be $$|{\bf x}\rangle ... 1 This is mostly because we're usually more interested in the spatial part of a plane wave than in the temporal part, so that plane waves are most convenient when written as$$ e^{i(\mathbf k\cdot\mathbf r-\omega t)}. \tag 1 $$The normalization follows from this choice. In general terms, it's hard to call which factor has more weight. There are plenty of ... 0 "W_{ik} is translationally invariant" means$$ W_{ik} = \frac{1}{N}\sum_{\bf{q}}{W(\bf{q})e^{-i\bf{q}\cdot\left(\bf{R}_i - \bf{R}_k \right)}} $$possibly with$$ \sum_{\bf{q}}{W(\bf{q})} = 0 $$if W_{ii}=0 is explicitly defined. Substitute in the sum defining G(E;\bf{k}, \bf{k}'),$$ \frac{1}{N^2}\sum_{ijk\neq i}{\sum_{\bf{q}}W({\bf ...

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It is perfectly possible to use wavelets to analyse quantum mechanical situations. The wavelets are localised in both time and frequency but they are themselves subject to the uncertainty principle - if you want a better time resolution, you need to pay for it with a coarser frequency resolution. The uncertainty principle is a universal wave phenomenon and ...

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It tells you what is the spectral content of the motion. Some examples of when this might be interesting: The object being measured is a point on a guitar string---then it would tell you what note is being played The object being measured is a planet. Then the peak frequency of the motion is the inverse of the planet's year. The object being measured is ...

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In this case, the Fourier transform will be a dirac pointing the unique frequency of vibration $f$ (or indeed 2, at $-f$ and $f$), with amplitude $A$ and phase $0$.

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I see now that your question is about the interpretation. Well, the interpretation is that you now integrate over the space of all fields in momentum space. Of course, mathematically the region of integration is still the space of functions $\mathbb{R}^4\to\mathbb{R}$ (or whatever kind of field applies) and so the meaning of $\mathcal{D}\phi$ is more or less ...

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You want to use $$\hat x= i\hbar\frac{\partial}{\partial p}$$ in the momentum basis. This means that $$<p|\hat x|\psi>= i\hbar\frac{\partial}{\partial p} <p|\psi>$$ Thus, by hermiticity of $\hat x$, we evaluate $$<x|\hat x|p> = (<p|\hat x|x>)^*$$ $$=(i\hbar\frac{\partial}{\partial p} <p|x>)^*$$ $$... 1 In the following calculation, I ignore some coefficients. According to J(x)=\int d^4 k_1 e^{ik_1 x} , J(y)=\int d^4 k_2 e^{ik_2 y} and D(x-y)=\int d^4k \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon} We have$$W(J) = \int d^4x d^4y d^4 k d^4 k_1 d^4 k_1 J(k_1)e^{ik_1 x} J(k_2)e^{ik_2 x} \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon} W(J)=\int d^4x d^4y d^4 k ...

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Notation: $x=(t,\boldsymbol x)$; $k=(k_0,\boldsymbol k)$; $kx=k_0t-\boldsymbol k\cdot\boldsymbol x$; $\mathrm dx=\mathrm dt\;\mathrm d^3\boldsymbol x$; etc. You can in principle perform the Fourier decomposition on both space and time variables, but to do so you'll need several properties of the Dirac's delta funciton: The first one is: let \$\xi\in\mathbb ...

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