New answers tagged hamiltonian-formalism
0
Assuming you're talking about a hamiltonian for a two-dimensional quantum system, any self-adjoint (aka hermitian) $2\times2$ matrix with complex entries will do. Any such matrix must equal its adjoint (conjugate-transpose), and this restricts the matrix to have the following form:
$$
H = \begin{pmatrix}
a & c+id\\
c-id & b
\end{pmatrix}
$$
...
0
What I suspect is stopping OP is different notations for the functional/variational derivative. Let there be given a functional
$$\tag{1} F[u]~=~\int \! dx ~f(x), $$
where the integrand notation $f(x)$ is a short-hand notation for the following function
$$\tag{2} f(x)~=~f(u(x), u^{\prime}(x),u^{\prime\prime}(x),\ldots;x) .$$
If the underlying variational ...
1
Actually, the Liouville theorem is more general - it is valid even if the distribution function depends on time, and even if the Hamiltonian depends on time.
http://en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian)
-> phase space volume preservation but no energy conservation: any Hamiltonian which depends on time, but you already know that. For ...
1
The key inside to OP's question has already been provided by Ikiperu in above comments. Here we just want to show that the problem becomes very simple to study in the corresponding Lagrangian formalism.
The Hamiltonian reads
$$\tag{1} H(p,q) ~:=~ \frac{p^2}{2m} + \lambda pq + \frac{m\lambda^2}{2}\frac{q^6}{q^4+\alpha^4}. $$
Since there is no explicit time ...
1
Liouville's theorem not only depends on the form of Hamilton's equations but also on the fact that $\partial\rho/\partial t = 0$, where $\rho$ is the statistical distribution function of the system. This is strictly true only for closed systems and is approximately true for quasi-closed systems when not observed for too long a time.
Energy of a system is ...
0
If we consider for simplicity a 2d phase space (q,p), then we can interpretate the poisson bracket between two functions f(q,p) and g(q,p) as the vector product of their gradients, which are vector fields in this plane:
$[f,g]=(\nabla f\times \nabla g)\cdot \mathbf{e}_z$
where $e_z$ is a unit vector perpendicular to the plane.
From that definition all the ...
2
Let's assume no explicit time dependence and that our Poisson bracket $\{,\}$ - I prefer curly brackets so square ones $[,]$ can be used to denote the commutator of vector fields - is non-singular, ie there's a corresponding symplectic product $\omega$.
The time derivative
$$
\frac{\mathrm d}{\mathrm dt}=\{\,\cdot\,,H\}
$$
is actually the Lie derivative ...
1
The physical interpretation is integrability conditions being satisfied on the manifold. From the first equation, if you would take A not depending on 't' explicitly then dA/dt = [A,H]. The Poisson bracket contains in it the dynamics involved in canonically conjugate variables and in classical mechanics, we can measure them simultaneously. Apart from this, ...
6
Your interpretation is not quite right. One sharp interpretation one can give to this "cutting" of phase space into cubes of size $h^{2N}$ (here $N$ is the dimension of the system's configuration space), is that it allows one to use classical phase space to count the number of energy eigenstates of the corresponding quantum hamiltonian. Instead of trying ...
1
I would say no, the symplectic form isn't a physical quantity. It's rather a quantity specific to the phase space formulation of a physical system. If you choose not to formulate things in phase space, the symplectic form is absent.
Moreover, even if you do work in phase space, in a system with constraints, the physical quantities are really only defined ...
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