# Tag Info

## New answers tagged hamiltonian-formalism

0

After reviewing the problem further, it looks like there is an error in the text. In fact, transforming to polar coordinates should be $P_\rho(x,y,P_x,P_y)=\dfrac{x*P_x+y*P_x}{(x^2+y^2)^{1/2}}$ and NOT $P_\rho(x,y,P_x,P_y)=\dfrac{x*P_x-y*P_x}{(x^2+y^2)^{1/2}}$ which was given.

-1

Hamiltonian can be written as : $dH=\frac{\partial H}{\partial q_i}dq_i+\frac{\partial H}{\partial p_i}dp_i+\frac{\partial H}{\partial t}dt$. or,$\frac{dH}{dt}=\frac{\partial H}{\partial q_i}\dot{q_i}+\frac{\partial H}{\partial p_i}\dot{p_i}+\frac{\partial H}{\partial t}$. We lso know that $\frac{\partial H}{\partial p_i}=\dot{q_i}$ and $\frac{\partial ... 0 I guess what the exercise means is the following: The pendulum is a massless, stiff rod with a mass$m$at one end. The other end of the rod is constrained to be at$\vec r(t) = a \begin{pmatrix} \cos(\omega t) \\ 0 \\ \sin(\omega t) \end{pmatrix}.$If the pendulum can only move in the$x$-$z$-plane, this leaves only one degree of freedom (the angle ... 4 The Hamiltonian$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) - ... 1 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

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 ...

3

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 ...

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: $$\omega = \begin{bmatrix} 0 && -I_{n\times n} \\ I_{n\times n} && 0 \\ ... 1 I) It seems OP's main question was spurred by a typo below eq. (4.2) in Ref. 1 in the formula for the unit normal vector$$\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$$... 1 You need to use vectors. Since L \neq r \times p, you need to use \vec L = \vec r \times \vec p instead, where the \times is the vector cross product of vectors, not the scalar multiplication of scalars. So you have$$\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

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

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 ...

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