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6

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 ...


2

Here is the answer (I will not consider the constants on the denominator of your Fourier transform for simplicity, however they are there ;-) ). When you write the operator $\hat{\phi}$ you have to be careful. I will drop the hats, because it will be clearer I think (maybe here the hat stands for an operator and not for the fourier transform). Your operator ...


7

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 ...


0

Using the poisson representation of bessel function $J_n(z) = \frac{(z/2)^n}{\sqrt{\pi} \Gamma(n+1)} \int_0^{\pi} \cos{(z\cos{\theta})} \sin^{2n}{\theta} d\theta$, you can determine the constant.. In 3+1 d, for example, for a space-like distance, we can make a lorentz transformation such that $r = x-y$ is purely spatial, the amplitude is then $D(x-y) \sim ...


1

You found $$x = e^{ik} + e^{-ik} - 2 = 2(\cos(k) - 1) \leq 0 . $$ If the question is whether $g\ge 0$ then the answer is "no" in the sense that it's eigenvalues are non-positive as we have just showed. A bit of intuition We recognize $$\phi(n+1) + \phi(n-1) -2\phi(n)$$ as some kind of second derivative. Consider the first derivative of a function $f$ ...


0

Regarding your third question. If you want to see the state vector $|\psi\rangle$ you considered explicitly, you must choose a basis to expand it. For example, the position eigenkets $|x\rangle$. Then you can expand your state kets with respect to $|x\rangle$. That is $|\psi\rangle=\int dx |x\rangle\langle x|\psi\rangle$. Where $\langle ...


0

See the standard text by Peter J. Brockwell & Richard A. Davis "Time Series:Theory and Methods" 2 ed, p.331, where authors define $\omega_j=2\pi j/n\in(-\pi,\pi]$ (integer multiples of the fundamental frequency $2\pi /n$) as Fourier frequencies. Later on p.335, they say "if $\omega$ is not a Fourier frequency, the analysis is a little more ...



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