Tag Info

0

I do not think probability distributions are preserved by the Hamiltonian flow...consider a probability distribution that is a $\delta$-function on the phase space at initial time (you have just one point with probability one), so it is a particle with fixed coordinate and momentum. If you evolve in time by the Hamiltonian flow, you will find yourself at the ...

0

As you state in the comments, $$\frac{dF}{dt}=\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t}$$ So popping this into the Lagrangian, $$L'=L+\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t}$$ The Hamiltonian $H=p\dot q-L$ implies $$H'=p'\dot{q}-L'=p\dot q+something\tag{1}$$ where $something$ is for you to work out. ...

4

The Hamiltonian is so useful because it is actually the operator providing translation in time (in autonomous systems). We know that any physical quantity on the phase space in the Hamiltonian formalism is evolved like $$\frac{df}{dt} = \{f,H \}$$ Where $\{\}$ is the Poisson bracket. It is thus natural to say $$\frac{d}{dt} = \{\cdot,H\}$$ and a small ...

6

I agree with Qmechanic but just to put a different perspective. While one may write down formulae for the Lagrangian, like $$L = \frac{p^2}{2m} - U(x)$$ which only differs from the Hamiltonian by the minus sign, and while it's possible to simply put hats above all the operators, unlike the Hamiltonian, the Lagrangian isn't a natural operator in any sense. ...

2

Comment to the question (v1): Unlike the Hamiltonian $H$ (which is a constant of motion if there is no explicit time dependence), the Lagrangian $L$, as an observable, is typically not conserved in time. Think e.g. of a harmonic oscillator.

5

Explicitly proving non-integrability of an arbitrary Hamiltonian system is an open problem. For some classes of Hamiltonian systems (e.g systems on a plane) is possible to prove explicitly the non-integrability of the system, using theorems of Poincare, Burns, Ziglin and Yoshida (and generalizations). For example there is a theorem of Poincare: For a ...

4

Consider a non-relativistic massless particle with charge $q$ on a 2D torus $$\tag{1} x ~\sim~ x + L_x , \qquad y ~\sim~ y + L_y,$$ in a constant non-zero magnetic field $B$ along the $z$-axis. Locally, we can choose a magnetic vector potential $$\tag{2} A_x ~=~ \partial_x\Lambda, \qquad A_y ~=~ Bx +\partial_y\Lambda,$$ where $\Lambda(x,y)$ is ...

4

$U(1)$ Chern-Simons theory with (physical) space a 2-torus is such an example. Its phase space is the gauge equivalence classes of flat connections on the 2-torus. These are specified by the holonomies around two 1-cycles forming a basis of $H_1(T^2)$. This is of course a 2-torus $U(1) \times U(1)$. Because of the form of the Chern-Simons action, these ...

2

In solid state physics, the bulk of a crystal is usually given periodic boundary conditions to avoid the sticky problem of what to do at the termination of the crystal. So the crystal is all bulk, no surface. This turns out to be a very good approximation to the bulk of a real crystal. It also gives the solid the topology of a 3-torus.

1

John R. Taylor's Classical Mechanics has a couple of chapters on Lagrange and Hamilton that I found very helpful.

0

The weak equality $f \approx 0$ means, that we first must evaluate all of the Poisson brackets of the theory (the equations of motion etc.) and only after that we may set $f$ to zero. It's because the hamiltonian doesn't contain info about primary constraints (the good example is electrodynamics), and so it doesn't contain info about the secondary ...

1

Comments to the question (v2): The canonical expansion of the two-fermion wave function seems more related to the canonical form of antisymmetric real matrices in the framework of vector spaces and linear algebra. The Darboux' theorem in the framework of manifolds and differential geometry (which is a surprisingly potent result) is overkill for the ...

0

Not sure to answer properly, but if I remember, using the Dirac bracket allowing you to get rid off the second class constraints and deal at the end only with first class constraints. And still at the end, you consider only weak equality, no =.

0

Comment to the question (v2): According to Ref. 1, the weak equality symbol $\approx$ usually means equality modulo all constraints: primary, secondary, tertiary, $\ldots$, constraints. (or in Dirac's classification) first and second class constraints. References: M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994; p. 13.

7

The phase space is a symplectic manifold, so any manifold $\mathcal{M}$ that admits a closed nondegenerate 2-form is a possible phase space. Now, what is necessary (or sufficent) for admitting such a form? First, as you mention, $\mathcal{M}$ must be even-dimensional. Second, $\mathcal{M}$ must be orientable. Why? Because orientability is equivalent to ...

10

The bible for the mathematical formulation of classical Mechanics, namely Foundations of Mechanics by Abraham and Marsden, defines a hamiltonian system as a triple $(M, \omega, X_H)$ where $(M, \omega)$ is a symplectic manifold, and $X_H$ is the Hamiltonian vector field corresponding to a hamiltonian function $H:M\to\mathbb R$. Now, are there typically any ...

0

The answer to the second question is actually quite straightforward: by computing $\partial_t E$ and using what given in the Gross-Pitaevskii for $\dot{\psi}$ and $\dot{\psi^*}$ one can check that all the terms cancel out so that $\partial_tE=0$. To derive the expression for the energy one could also start considering a lagrangian giving the ...

0

I think that the answer to your question is a cautious and hand-wavy "most of the time" in the direction of information conservation and unitarity. But also "no" in the sense of a well defined "holonomy". The 3+1 decomposition is based on the assumption of global hyperbolicity of the space-time. The cases where 3+1 decomposition breaks down without ...

4

The first thing you should realize is the fact that while $\phi$ has an equation of motion with second time derivatives, it is not the wave function, and therefore there is no problem with QM. The field is just an operator (more or less), not a state. Acting with the fields on the vacuum state you generate the other states which do evolve with an hamiltonian ...

Top 50 recent answers are included