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The Hamiltonian $H(\theta,p_\theta)$ needs to be formulated in terms of the coordinate $\theta$ and its canonically conjugate momentum $p_\theta = \frac{\partial L}{\partial \dot{\theta}} = R^2 \dot\theta$. The correct expression for the Hamiltonian is \begin{align} H(\theta,p_\theta) & = p_\theta \dot{\theta}(\theta,p_\theta) - ...

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The quantization prescription $$[\hat{x},\hat{y}] := \mathrm{i}\hbar\widehat{\{x,y\}}\tag{1}$$ for $x,y$ two classical phase space coordinates does have its subtleties. In particular, as the answer in the linked question says, it leads to inconsistent results when applied to e.g. polar coordinates. The reason for this is two-fold: For the radial ...

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There are at least two generalizations of Noether's theorem. 1) Assume that the Hamiltonian system with Hamiltonian $H(z),\quad z=(p,q)$ has a one-parameter symmetry group $\{g^s_F(z)\}$ which is generated by a Hamiltonian system with Hamiltonian $F$. Then $F$ is a first integral for $H:\quad \{F,H\}=0$, moreover if $dF\ne 0$ then there are local ...

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The problem is that you are equating too many things to $\dot{q_k}$. Usually $\dot{q_k} = \frac{dq_k}{dt}$, a total derivative, as opposed to a partial derivative. If $q_k$ has no explicit time-dependence, i.e. it does not depend directly on $t$ itself, then $\frac{\partial q_k}{\partial t} = 0.$ In this case, the Poisson bracket reduces to: $... 1 In relativistic QFT for instance, the Hamiltonian does not respect Lorenz invariance. That is not to say that the symmetry is not present, in the Hamiltonian. Of course it is, the Lagrangian and Hamiltonian are equivalent ways of analysing the same thing. But the Hamiltonian does not have Lorenz invariance autocratically built into it, as it singles out ... 1 As I see, maybe the problem is energy. So, What is energy? The formal classical definition of energy is: Energy is a dynamical invariant of a system that came from time-translation symmetry. There is also a question here about it. If you want more references about it, let me know. So.. when Bob write,$E = T + V$in dissipative systems (damped OHS for ... 1 Actually the form you are looking for is $$H = X^\dagger M X = \left[\begin{array}{cccc}b^\dagger & d^\dagger & b & d\end{array}\right]\left(\begin{array}{cccc}\alpha_1 & \beta_1 & \gamma_1 & \delta_1 \\ \alpha_2 & \beta_2 & \gamma_2 & \delta_2 \\ \alpha_3 & \beta_3 & \gamma_3 & \delta_3 \\ \alpha_4 & ... 1 You need to use vectors. Since L \neq r \times p, you need to use \vec L = \vec r \times \vec p instead, where the \times is the vector cross product of vectors, not the scalar multiplication of scalars. So you have$$\vec L= \left[(v_o t \cos \theta) \hat x+ (v_o t \sin \theta - \frac{1}{2}gt^2)\hat y\right] \times m\left[(v_o \cos \theta)\hat x+ ... 1 I) It seems OP's main question was spurred by a typo below eq. (4.2) in Ref. 1 in the formula for the unit normal vector $$\tag{1} {\bf n}({\bf x})~:=~ \frac{{\bf N}({\bf x})}{|| {\bf N}({\bf x})||}, \qquad {\bf N}({\bf x})~:=~\frac{\partial f({\bf x})}{\partial {\bf x}},\qquad || {\bf N}({\bf x})||~:=~\sqrt{{\bf N}({\bf x})\cdot {\bf N}({\bf x})}.\qquad$$ ... 1 Also, you can write Hamilton's equations of motion in sympletic form: $$\dot\xi_i = \omega_{ij}\frac{\partial H}{\partial\xi_j}$$ Where$\xi_i$are the coordinates in the phase space, that is,$\xi = (\mathbf q, \mathbf p)$. And,$\omega$is the sympletic matrix:$\$ \omega = \begin{bmatrix} 0 && -I_{n\times n} \\ I_{n\times n} && 0 \\ ...

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