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I think you can see that the a linear dynamical system does not have limit cycle in this way: For a linear dynamical system, the Fourier modes are the eigenvectors of the linear operator, and the solution of the system is some system specific amplitudes (constant in time) multiply the Fourier modes. So the phase portrait ($\dot{x}$ vs $x$) must be cyclic. ...


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The answer to your question is not so difficult. In your notes, the author defines a limit cycle as an isolated periodic trajectory, i.e. nearby there can not be any periodic trajectories. Of course, it is easy to construct periodic trajectories for linear systems, just look at $$ x'(t) = y(t), \quad y'(t) = -x(t), \quad x(0) = x_0, \quad y(0) = 0.$$ ...


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I googled the LSZ reduction formula, but many of the are almost the same as Srednicki's except this one http://porthos.ist.utl.pt/~romao/homepage/publications/Lectures/Lectures-TCA-c2.pdf by Jorge Romao. He didn't insert a T directly in the expression of S matrix element like (5.13), so his final result contains a "disconnected term", plus (5.15). He stated ...


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Fleshing out @Peter Morgan's answer a bit, to the effect that $x$ and $p$ are not bounded operators, so their commutator need not be bounded. First note that $$[x^n,p] = i\hbar nx^{n-1} ~,$$ hence the operator norms of both sides satisfy $$ 2\| p\| ~ \|x\|^n \geq n \hbar \|x\|^{n-1}$$ so that, for any $n$, $$2\|x\|~\|p\|\geq n \hbar~. $$ Since $n$ ...


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In answer to the title question: no, you can't always decompose an $L_2$ function in terms of only the bound spectrum of hydrogen. This is because there are orthogonal functions to all bound states, which naturally represent the free states of the electron. The quickest examples are of course the Coulomb-wave eigenfunctions $|\chi_{E,l,m}⟩$ of the ...


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You may want to refer to Jackson's 'Classical Electrodynamics' for several examples of solutions using Green functions. I also found chapter 7 of 'Mathematics for classical and quantum physics' by Byron and Fuller quite helpful. Its title itself is 'Green functions'.


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While a funny-looking coincidence, this is not a valid alternative expression for entropy in general, since the entropy of a probability distribution (which are what rigorously hides behind the strange word "macrostate") is more generally given by $$ S = - k_B \sum_i p_i\ln(p_i) \tag{1}$$ and becomes only $$S = k_B \ln(\Omega) \tag{2}$$ in the case of a ...


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From a mathematical perspective this means that it is not differentiable. The problem is that you need the discreteness to be able to count states. If you replace the discreteness by something smooth you get something differentiable, but your definition of entropy no longer makes sense. This is just one of the points where mathematicians cringe, but it works ...


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Noting the simple relation $$ \frac{\Delta E_{i,j}}{\Delta E_{i,j}-\hbar\omega-i\eta}-\frac{\Delta E_{i,j}-\hbar\omega-i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta}=\frac{\hbar\omega+i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta} $$ and by Sokhotski-Plemelj theorem $$ \lim_{\eta\rightarrow 0^+} \frac{i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta}=0 $$ because, if the limit ...


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I don't think we need Sokhotski-Plemelj for this. Think of $E_j - E_i$ as a fixed value $E$. Then the formula is re-written as $$\frac{\hbar \omega}{E - \hbar \omega - i \eta}\, .$$ Now let $x \equiv \hbar \omega$ and you get $$\frac{x}{E -x - i \eta} \, .$$ This integral is dominated by the part where $x \approx E$ so let's try shifting the variables $y ...


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This subtlety is related to the fact that the momentum operator $\hat{P}$ (unlike the Hamiltonian $\hat{H}=\frac{\hat{P}^2}{2m}$) has no eigenfunctions compatible with the Dirichlet boundary conditions, and $\hat{P}$ is not a self-adjoint operator. This is essentially Example 4 in F. Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics, ...


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The space of semi-infinite forms is basically the name used by mathematicians for the fermionic Fock space please see for example: Friedrich Wagemann lecture, page 8. Given an infinite dimensional vector space with spanned by: $\{ e_i, i\in \mathbb{Z}\}$, let its dual space be spanned by $\{ f_i, i\in \mathbb{Z}\} (\langle f_j, e_i \rangle = \delta_{ij}$) ...


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Ok, I do not know the Sage algorithm but I am going to offer a conjecture of what is happening. You have to verify the conjecture by further numerical investigations. I assume that the Sage algorithm works optimally for bifurcations of a single equilibrium and can run into problems such as we see here when dealing with equilibria (AKA fixed points) ...


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Define $D=\{(x,y)\in\mathbb R^2:\|(x,y)\|\leq R\}$ as the disk of interest. There are two spaces of interest here: the space of square-integrable functions on $D$, $L_2(D)$, and the space of such functions with Dirichlet boundary conditions, $\mathcal H=\{\psi\in L_2(D):\psi(p)=0\:\forall p\in \partial D\}$. You're interested in the hamiltonian ...


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Disclaimer The following answer is largely taken from the arXiv paper with e-print number 1512.04015 by Treumann and Baumjohann (from here on I abbreviate references to this paper as TB15). Background It is well known that the Maxwell-Jüttner distribution works well for a momentum/energy distribution with an isotropic, scalar temperature, $T$. This ...


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This is typically said of solutions to differential equations which make no sense. For example when calculating skin effect you find that the solution to the field in the conductor is a sum of an exponentially growing term and an exponentially attenuating term. You throw away the first one because it is unphysical*. * Or because it does not satisfy ...


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Causality requires that $\chi(t) \propto \Theta(t)$, where $\Theta(t)$ is the Heaviside step function. In other words, $\chi(t-t') = 0$ for $t'>t$, so that only past influences from times $t'\leq t$ affect the system response at time $t$. This leads to constraints on $\chi(\omega)$ viewed as a function of complex frequency: it must be analytic in the ...



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