# Tag Info

2

I think the Hamiltonian is not necessarily the energy for the following reason: you can demonstrate that the Lagrangian may be deduced from the D'alembert principle which is linked to the concept of force, etc. but it may be also deduced from the Hamilton's principle which is a pure mathematical concept applied to physics (a certain quantity has to be an ...

4

Problem: Given Newton's second law $$\tag{1} m\ddot{q}^j~=~-\beta\dot{q}^j-\frac{\partial V(q,t)}{\partial q^j}, \qquad j~\in~\{1,\ldots, n\},$$ for a non-relativistic point particle in $n$ dimensions, subjected to a friction force, and also subjected to various forces that have a total potential $V(q,t)$, which may depend explicitly on time. I) ...

1

A single photon mode is described by a creation operator. The problem is not complicated and really generic in bosonic problems. How to include the driving is just a matter of taste. Suppose your circuit is described by the degrees of freedom $x$ and $p$ (say position and momentum). A force $f$ would be added as a term $x\cdot f$ in the Hamiltonian for ...

0

$$i {{\partial}\over{\partial t}}\pi=[\pi,\int d^3x\tfrac{1}{2}\pi^2+\tfrac{1}{2}\phi()\phi]$$ $$=[\pi,\int d^3x\tfrac{1}{2}\phi()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()]\phi+\phi()[\pi,\phi]$$ $$=\tfrac{1}{2}\int ... 0 A symplectic manifold is defined as a smooth manifold, M, which is equipped with a closed non-degenerate differential 2-form, \omega, called the symplectic form. A multisymplectic manifold of degree k is a manifold equipped with a closed nondegenerate k-form. 1 Hints: OP's eq. (1) is the equation for a constant of motion \frac{df}{dt}=\{f,H\}_{PB}+\frac{\partial f}{\partial t}=0 of a harmonic oscillator H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2 x^2. Let us assume for simplicity that m\omega=1, and leave it to the reader to generalize to arbitrary m and \omega. Complexify z=x+ip\in\mathbb{C}. Then the ... 1 The canonical momentum p is just a conjugate variable of position in classical mechanics, for we have the relation p=\frac{\partial L}{\partial \dot{r}}. When making the transition to quantum mechanics: we need substitute p by an operator -ih\nabla in the Hamiltonian; similarly, we need substitute r by i\hbar \nabla_p in momentum representation. ... 3 The classical poisson bracket with the generator of any transformation gives the infinitesimal evolution with respect to that transformation. The familiar$$ \partial_t f = \{H,f\}$$means nothing else than the time evolution of any observable is given by its Poisson bracket with the Hamiltonian, which is the generator of time translation. More generally, ... 1 The condition you wrote, namely that the partial derivative of the phase density with respect to time vanishes, is a standard one placed on phase densities that describe equilibrium systems. See, for example, page 29 of Eric D'Hoker's statistical mechanics lecture notes which can be found here: http://www.pa.ucla.edu/content/eric-dhoker-lecture-notes On ... 1 There is something you should be careful of regarding Liouville's theorem. If there are momentum-dependent forces, then Liouville's theorem changes because phase density is no longer incompressible. Suppose we define f_{s} = f_{s}(\mathbf{x},\mathbf{p},t) \equiv the particle distribution function of species s, which is non-negative, contains a ... 1 Quantization is often treated as a mystery, yet it can be seen to arise naturally from classical Hamiltonian mechanics.1 This yields the quantization prescription$$ \{\dot{},\dot{}\} \mapsto \frac{1}{{\mathrm{i}\hbar}}[\dot{},\dot{}]$$for all classical observables on the phase space that are to be turned into quantum operators.2 Therefore, if we want to ... 0 Quantization, in the Heisenberg picture is very much related to establishing quantum non-trivial commutators between the main observables of the theory. Originally this is linked to the classical Poisson bracket, which in the Hamiltonian formalism includes the cannonical momentum. Hope I gave some insight, as to why we are led to define the conjugade ... 1 Your Hamiltonian has the "potential" U = V + F, just because F is time-dependent doesn't make it not part of the potential. You are correct that the Hamiltonian is not conserved. The Hamiltonian is the total energy of the system if the Lagrangian does not contain terms linear in the velocity or when the generalized coordinates in the Lagrangian do not ... 1 Let me give a further comment (not exactly an answer) The quantity \mathbf{r} \cdot \mathbf{r} represents the magnitude of the radius as such it does not change under rotation (poisson commutator with ang. momenutm L) The quantity \mathbf{p} \cdot \mathbf{p} represents the magnitude of (linear) momentum, as such it also does not change under rotation ... 2 The classical poisson bracket with the generator of any symmetry gives the infinitesimal evolution with respect to that symmetry. The most familiar statement of this is that the time-evolution of any observable f on the phase space is given by$$ \partial_t f = \{H,f\} Similarily, for a rotation around the $i$-th axis with angle parameter $\phi$, the ...

Top 50 recent answers are included