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4

The answer is no: if you were to write $p_i\propto q_{i+3}$ for $i=1,2,3$, then your quantum Hamiltonian would be $$ H\sim \sum_{i=1}^6 q_i^2 $$ which looks like the Hamiltonian of a free particle in a six dimensional space, but it is not. It is not because the commutation relations are not $[q_i,q_j]=0$ but $$ [q_1,q_2]=0\quad [q_1,q_3]=0\quad ...


4

Hint: Eq. (6) in its current form (v4) is meaningless since the lhs. depends on $x$, while the rhs. is integrated over $x$. The functional derivative $$\tag{A}\frac{\delta F}{\delta u(x^{\prime})}~\stackrel{(B)}{=}~\frac{\delta u(x)}{\delta u(x^{\prime})}~=~\delta(x\!-\!x^{\prime})$$ in eq. (6) for the functional $$\tag{B} F[u]~:=~u(x)~=~\int \! ...


2

Take the orbit equation (using $u=\frac1r$): $$u(\theta)=A\left(1+e\cos\theta\right) $$ Where $A$ is some constant that you can work out from the differential equation if desired, and $e$ is the eccentricity of the orbit. Now, what happens when $\theta$ goes through an angle of $2\pi$? Notice that we would not have returned to the original radius if ...


2

Well, not really. We COULD write hamiltonian as square root - if we know, what is a square root of an operator. Of course we have simple approximation: $$\sqrt{1+x}=1+\frac x2-\frac{x^2}{8}+O(x^3)$$ Using this we could write your hamiltonian as: $$\mathcal H=mc^2\sqrt{1+\frac{p^2}{m^2c^2}}=mc^2+\frac{p^2}{2m}+O(p^4).$$ The problem is that this form of ...


2

Comments to the question (v2): OP is considering the higher-derivative Lagrangian density $$ {\cal L}_1~=~ \frac{1}{2}(\partial\phi)^2 +\frac{g\phi}{2} (\partial^2\phi)^2,\tag{1} $$ where $g$ is a coupling constant. We use Minkowski sign convention $(+,-,-,-)$. Quantum mechanically, the model is not unitary and therefore ill-defined, cf. the Ostrogradsky ...


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Comments to the question (v2): On one hand, the Kuramoto-Sivashinsky (KS) equation is a dissipative differential equation (DE). Each term has an even number of spatial derivatives. It's a non-linear version of the heat equation. Dissipative systems rarely have variational action formulations nor Hamiltonian formulations. On the other hand, in the Korteweg ...


2

Consider a map $$S \ni\phi \mapsto F[\phi] \in \mathbb R$$ defined on a class $S$ of smooth functions $\phi$ defined on the compact set $\Omega \subset \mathbb R^n$ obtained by taking the closure of an open set with regular boundary $\partial \Omega$. Thus the map $F$ associates a real number $F[\phi]$ to each function $\phi\in S$. We say that the ...


1

A general advice: Before trying to understand Hamiltonian field theory, make sure you understand Lagrangian field theory. Before trying to understand Lagrangian field theory, make sure you understand Lagrangian point mechanics. In Lagrangian point mechanics, the functional derivative of the action is $$\tag{1} \frac{\delta S}{\delta q(t)} ...


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The derivation of ideal gas equation from Hamilton's equations will take the same procedure as what you have seen in Wikipedia. Since you said you haven't understood the way in which the equation is derived I will give you a step by step explanation on it. So we have a system of perfect gas molecules. Of course they are non-interacting. We are going to ...


1

Yes. Not only is it possible to find a Hamiltonian density but it is even possible to find a plain Hamiltonian. I gave the construction in this EPL ArXiv1303.6143: Euro Physics Letters, 103 (2013) 28004 This Letter is short but dense, I can only outline the method here. The reason why you can't write an hamiltonian for the electromagnetic field is that ...


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The Lagrangian is in fact an equation of $\vec \Omega$, however, in general it will be a quadratic function of $\vec \Omega$, as the rotational kinetic energy would be given by $$\frac{1}{2}\vec \Omega ^T \mathbf{I}\ \vec \Omega$$ This would give you the desired generalized momenta as a function of the general velocity vectors, as the diagonal entries of the ...



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