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The Heisenberg Uncertainty Principle has two distinct aspects: One is the identification of matter as a wave and, in particular, the relationship between a particle's momentum $p$ and its wavelength $\lambda$ through de Broglie's relationship $p=h/\lambda$. This is the crucial bit of physical input. The second one is purely mathematical, and it's the ...

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When you have a bunch of interrelated phenomena in physics, trying to figure out which one is the "reason" for the other ones is often just a recipe for confusion. Different people will start from different postulates, so they will disagree on which results are trivial and which aren't, but hopefully everyone agrees on what's true. In a first course on ...

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Yes, this is not the "true" reason. The direct reason is that they are not commuting operators. See the Robertson-Schrodinger Relation. You end up getting that the product of the uncertainties is bounded by 1/2 of the absolute value of the failure to commute. In the case of position and momentum this is $\hbar/2$.

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If you have a system of independently oscillating point charges as radiators and they do not have a coherence among themselves. Then, If you take single dipole it radiate in a dumbbell shape. If you orient these radiators randomly oriented in space the radiation will propagate as a spherical wave. If you let this wave pass through a slit then you will see ...

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$$b_{j} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi pj/n}\beta_{p}\qquad b_{j+1} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi q(j+1)/n}\beta_{q}$$ Then b_{j}^{\dagger}b_{j+1}^{\dagger} = \frac{1}{n}\sum_{p,q}e^{\pi i(p+q)j/n}e^{\pi iq/n}\beta_{q}^{\dagger}\beta_{p}^{\dagger} = \sum_{p} e^{-\pi ip/n}\beta_{-p}^{\dagger}\beta_{p}^{...

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A sine wave at $\omega$ is periodic in $\omega /2$, but that does not mean that a sine at $\omega$ has frequency components at $\omega /2$. Another way of looking at it is noting that you cannot synthesize a sine at $\omega$ as the sum of sines of other frequencies.

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To keep things mathematically both precise and simple, let's stick with the discrete Fourier transform. Signals are vectors of $N$ complex points, where $N$ is the dataset's length. The dimension of this vector space is $N$. In this setting, the Fourier transform is simply a resolution of a signal, thought of as a vector, into components with respect to a ...

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I don't know what is happening there. It always worked for me. If you have large number of cycles with smooth variation i.e. large time scale with small time interval you will definitely see the frequency components. First of all matlab stores its frequency components like. 0 to $\omega_ {max}$ then $-\omega_{max}$ to 0. Hence if you want your zero ...

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From yuggib's answer: "…we consider physically meaningful only wavefunctions such that there exist a continuous and differentiable representative in its equivalence class. However, also this requirement is not physical." Not quite. A set of countable pointwise discontinuities may be tolerable, at least at first sight, but there is actually a very good ...

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There is no physical requirement for a wave function to be smooth, or even continuous. At least if we accept the common interpretation that a wavefunction is nothing else than a "probability amplitude". I.e. it represents, when multiplied with its complex conjugate, a probability density. Now a probability density can be discontinuous or even undefined on ...

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Discontinuity in first derivative of wave function implies that the wave function experiences a sudden force that changes its momentum instantly. Thus physically speaking this is not possible as there are no dirac delta potentials. There are potentials very close to dirac delta and thus in the dirac delta approximation the wavefunction will have a first ...

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Your definition in your point 1 is the one that is generally used. Once you make this decision, there is almost no redundancy in the way you can choose $\mathrm{Im}(\tilde{E})$. Recall that, if two square integrable functions have the same Fourier transform, they are equal almost everywhere (in the measure theoretic sense), i.e. they can only differ on a ...

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You are on the right track in thinking about Fourier transforms. Consider an arbitrary scalar function $f(x)$, which we will take to be real; that is $$f^*(x)=f(x).$$ Now we write $f(x)$ in terms of its Fourier transform $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty} dk\; e^{ikx}f(k).$$ Complex ...

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Well I'll give it a try. I guess you have to be careful that after transforming you get two different quantized momenta. You're then summing/integrating over $q',q$. Leaving out the normalization constants: \begin{align} H[\tilde{\phi}] &= \int dx \left[k_1 \left( \partial_x \int dq \tilde{\phi}(q)e^{-iqx}\right)\left(\partial_x \int dq' \tilde{\phi}(q'...

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To solve this first you take two spherical waves emanating from two pin holes $\frac{1}{\lambda z} exp(ikr)$. The intensity from individual wave at any point (x,z) will be mod squared sum of the two waves. Here $r$ will be $\sqrt{z^2+(x\pm d/2)^2}$ expand the bracket and take out $z$, expand binomially keep only first term. Now add the two expressions with +...

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