26

Be aware that there exist various definitions of a canonical transformation (CT) in the literature: Firstly, Refs. 1 and 2 define a CT as a transformation$^1$ $$ (q^i,p_i)~~\mapsto~~ \left(Q^i(q,p,t),P_i(q,p,t)\right)\tag{1}$$ [together with a choice of a Hamiltonian $H(q,p,t)$ and a Kamiltonian $K(Q,P,t)$; and where $t$ is the time parameter] that ...


24

I understand the definition of a Hilbert space. But I do not understand why non-commutativity compels us to use Hilbert spaces. It doesn't, but that's not what Scrinzi is saying. The reason is doesn't is because we could work, for example, in Wigner quasiprobability representation: $$\rho\mapsto W(x,p) = \frac{1}{\pi\hbar}\int_{-\infty}^\infty\langle x+x'|\...


21

It's subtle. The theorem is not there: quantum flows are compressible (Moyal, 1949). I'll follow Ch. 0.12 of our book, Concise Treatise of Quantum Mechanics in Phase Space, 2014. The analog of the Liouville density of classical mechanics is the Wigner function in phase space quantum mechanics. Its evolution equation (generalizing Liouville's) is $$ {\...


15

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 ...


14

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 ...


14

So, the short answer is that you're quite correct: if the dynamics of a system is subject to Liouville's theorem, then phase space volume is conserved, so the entropy associated to a given probability distribution remains constant as it evolves under those dynamics. This is actually just one instance of a much more general puzzle: how do we reconcile the ...


14

In brief: phase space is not made into a vector space because that additional structure provides no benefit; quantum mechanics uses a Hilbert space because that additional structure does provide benefits. Any time you relate a mathematical structure to a physical concept, you need to ask how useful that relation is. The mathematical structure will have ...


14

Because you do equilibrium statistical mechanics. In the usual ensemble theory we associate to a system (a macrostate) a big number of corresponding microstates, each microstate is a point in phase space, that point is called representative point. Now you want to study how these points move in phase space. First of all the situation of equilibrium for the ...


13

Cool question! Thanks to user lionelbrits for his answer that prompted me to pull out my mechanics books and check the definitions of "canonical transformation" given by different authors. If you look in Goldstein's classical mechanics texts in the section on canonical transformations, then you'll find that canonical transformations are essentially defined ...


13

Let me actually collect most of my comments in an answer attempting to be more coherent than they, or your labile question. In fact, you are piling up three different questions, logically distinct, but with strong and natural connections, so it might be worth splitting them apart, before bringing them back together in the final coda. First, there is plain ...


13

Yes. A phase space trajectory of a smooth system$^1$ has to be a continuous curve. For it to be called "periodic", the movement has to repeat itself, both velocity and position: i.e., it must come back to the same spot in phase space. For a deterministic system, the current condition determines uniquely its future evolution, and there can not be "...


13

In classical physics, the definition of an observable is that it is a (reasonably smooth) function of position and momentum. Therefore, quantum systems obtained purely by quantization of a classical system also have that all their observables are functions of position and momentum. In quantum physics, an observable is just an operator designate to belong to ...


12

For a general and brief overview of the mathematical framework of Quantum Mechanics, see this answer. In a nutshell, Hilbert spaces arise from the representation theory of C*-algebras, which are postulated to be the relevant mathematical object that describes a quantum theory (because it contains observables in its self-adjoint part, and states as special ...


12

"Used in anger" or "killer ap"? To my knowledge, no problem has been solved in the phase-space quantization language that was not solvable in the other two formulations/pictures (Hilbert space or path integrals). This is in contrast to, e.g., path integrals (whose gauge fixing, Faddeev-Popov, and Fujikawa anomaly applications to gauge theories are virtually ...


11

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 ...


11

Liouville's theorem says the accessible volume in phase space does not increase, but it tends to become narrow filaments that "fill up" a much larger volume. If you think of a particle in a reflecting box, you might start it with a known position $\pm 1$ mm in all three axes and a known velocity $\pm 1$ mm/sec in all three axes. This is a phase space ...


11

There is no "wavefunction in phase space" because the wavefunctions $\psi(x)$ and $\psi(p)$ are obtained from the abstract state vector $\lvert \psi\rangle$ by $\langle x\vert \psi\rangle$ and $\langle p\vert \psi\rangle$, respectively. Since position and momentum don't commute, there are no $\lvert x,p\rangle$ to get a naive $\psi(x,p)$. However, from any ...


11

The importance comes from the equations of motions: $$ \dot{q}=\frac{\partial H}{\partial p}\, ,\qquad \dot{p}=-\frac{\partial H}{\partial q} $$ which can be rewritten as $$ \dot{q}=\{q,H\}\, ,\qquad \dot{p}=\{p,H\} $$ In particular, for an arbitrary function of $f(p,q)$, we have $$ \frac{d}{dt}f(p,q,t)=\{f,H\}+\frac{\partial f}{\partial t}\, . $$ ...


11

You probably need to internalize Ivan Todorov's accessible Quantization is a mystery. Your best bet for addressing your questions is Geometric quantization, not phase space quantization that you appear to be hung up on. Since I did not follow the logic of your conclusion/question after the Dirac 1925 thesis heuristic vision map you wrote, and I have no broad ...


10

First, one must appreciate that the phase space is classically parameterized by $x,p$ and coordinates on an ordinary plane commute with each other, $xp=px$. However, in quantum mechanics, this ain't the case. Instead, we have the Heisenberg commutator $$ xp - px = i\hbar. $$ This means that quantum mechanically, the phase space is not an ordinary plane (or ...


10

I) Let us for simplicity work in 1D with $\hbar=1$. (The generalization to higher dimensions is straightforward.) Moreover, let us for simplicity take an operator $\hat{f}(\hat{X},\hat{P})$ without any ordering ambiguities, i.e., each monomial term in the symbol $f(x,p)$ depends only on either $x$ or $p$, but not on both. Then one possible motivation of ...


10

Hints to the question (v1): We cannot resist the temptation to generalize the background spacetime metric from the Minkowski metric $\eta_{\mu\nu}$ to a general curved spacetime metric $g_{\mu\nu}(x)$. We use the sign convention $(-,+,+,+)$. Let us parametrize the point particle by an arbitrary world-line parameter $\tau$ (which does not have to be the ...


10

The essential idea of a Poincaré map is to boil down the way you represent a dynamical system. For this, the system has to have certain properties, namely to return to some region in its state space from time to time. This is fulfilled if the dynamics is periodic, but it also works with chaotic dynamics. To give a simple example, instead of analysing the ...


10

i will try this one. A Hamiltonian system is (fully) integrable, which means there are $n$ ($n=$ number of dimensions) independent integrals of motion (note that completely integrable hamiltonian systems are very rare, almost all hamiltonian systems are not completely integrable). What this states in essence (and intuitively) is that the hamiltonian system ...


10

The factor of $V$ comes from integration over $\mathrm d\vec x$ (as we assume that the Hamiltonian is position independent). On the other hand the factors of $2\pi$ and $\hbar$ are essentially irrelevant: you only care about the derivatives of the logarithm of the partition function, and therefore any global factor always drops out: you can multiply the ...


9

Let there be given a $2n$-dimenional real symplectic manifold $(M,\omega)$ with a globally defined real function $H:M\times[t_i,t_f] \to \mathbb{R}$, which we will call the Hamiltonian. The time evolution is governed by Hamilton's (or equivalently Liouville's) equations of motion. Here $t\in[t_i,t_f]$ is time. On one hand, there is the notion of complete ...


9

No, it's not a problem. The reason is that, in order for expressions like $$\mu=-T\left(\tfrac{\partial S}{\partial N}\right)_{E,V}.$$ to be meaningful, you have to be using the grand canonical ensemble (or a generalisation thereof), in which particles are able to enter and leave the system. Consequently, $N$ stands not for an integer number of particles, ...


9

Non-equilibrium systems are most often considered in the approximation where local equilibrium is valid, yielding a hydrodynamic or elasticity description. Local equilibrium means that equilibrium is assumed to hold on a scale large compared to the microscopic scale but small compared with the scale where observations are made. In this case, one considers a ...


9

Given a symplectic manifold $(M,\omega)$, it is natural to ponder what tangent bundle connection $$\nabla: \Gamma(TM)\times\Gamma(TM)\to \Gamma(TM) \tag{1}$$ to chose? Generically, it is natural to choose $\nabla$ to be torsionfree $$T~=~0,\tag{2}$$ and compatible $$\nabla \omega~=~0\tag{3}$$ with the symplectic $2$-form $\omega$. One may show (via ...


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