30

Well, $\{f, \cdot \}$, similarly to $\{H,\cdot\}$, computes the derivative of the argument $\cdot$ with respect to the action of the one-parameter group of canonical transformations generated by $f$ (see the note below for the complete definition) $$\phi_a^{(f)} : F \to F\:,\quad a \in \mathbb R\:,$$ satisfying $$\phi_a^{(f)} \circ \phi_b^{(f)}= \phi_{a+b}^{...


27

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 ...


27

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, ...


26

Indeed, canonical quantization works just when it works. It is in my view wrong and dangerous to think that this is the way to construct quantum theories even if it sometimes works: it produced astonishing results as the theoretical explanation of the hydrogen spectrum. However, after all the world is quantum and classical physics is an approximation: the ...


23

There are three easy tests to check if a transformation is canonical. Note that some multiplicative constants might pop up in certain textbooks, depending on the exact definition of canonical transformation. Notation Let $x = (p, q)$ be the $2n$ variables, and the transformed variables be $\tilde{x}(x) = (\tilde{p}(p, q), \tilde{q}(p, q))$. The method ...


22

According to the topic of deformation quantization, the first few entries in the dictionary between $$ \text{Quantum Mechanics}\quad\longleftrightarrow\quad\text{Classical Mechanics}\tag{0}$$ read $$ \text{Operator}\quad\hat{f}\quad\longleftrightarrow\quad\text{Function/Symbol}\quad f,\tag{1}$$ $$ \text{Composition}\quad\hat{f}\circ\hat{g} \quad\...


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 $$ {\...


21

The confusion stems from a bit of mathematical sloppiness. It's necessary useful sloppiness, because as you'll see in a moment, the full machinery is a pain, but it's helpful to see and keep in the back of your mind when confusions like this arise. Phase Space Trajectories I won't go through the trouble of constructing the cotangent bundle to the ...


17

I) OP is given a problem of the form $$\tag{1} \dot{q}~=~f(q,p), \qquad \dot{p}~=~g(q,p), $$ where $f$ and $g$ are two given smooth functions. OP is asked to derive the integrability condition for the eqs. (1) to be Hamilton's eqs. $$\tag{2} \dot{q}~=~\frac{\partial H}{\partial p}, \qquad \dot{p}~=~-\frac{\partial H}{\partial q}.$$ OP correctly ...


16

In dimension $2n>2$ they are not equivalent since (for time-independent transformations) canonical is equivalent to $$\sum_{k=1}^n dq^k\wedge dp_k = \sum_{k=1}^n dQ^k\wedge dP_k\tag{1}$$ whereas conservation of oriented volume means $$dq^1\wedge \cdots \wedge dq^n \wedge dp_1\cdots \wedge dp_n = dQ^1\wedge \cdots \wedge dQ^n \wedge dP_1\cdots \wedge dP_n\:...


15

I) It is worthwhile mentioning that there exists a basic approach well-suited to physics applications (where we usually assume locality) that avoids multiplying two distributions together. The idea is that the two inputs $F$ and $G$ in the Poisson bracket (PB) $$\tag{1}\{F,G\} ~=~ \int_M \!dx \left( \frac{\delta F}{\delta \phi(x)}\frac{\delta G}{\delta \pi(...


13

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

I do not know about deep questions. And people seem to give pretty deep answers here. My contribution is to show $$ \lim_{\hbar \to \infty} \frac{1}{i\hbar}[ F(p,x) , G(p,x)] = \{F(p,x), G(p,x)\}_{P.B.} $$ where $ [ F, G] = F G - G F $ and $$ \{ F(p,x), G(p,x) \} = \frac{\partial F}{\partial x} \frac{\partial G}{\partial p} - \frac{\partial G}{\partial ...


10

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 ...


10

Let me rearrange the logic of the Moyal Bracket that @ACuriousMind discussed neatly, by visiting a notional planet where people somehow discovered classical mechanics and quantum mechanics independently; but suffered a terrible mental block that prevented them from appreciating there was a connection between the two, at first. Then, one day, their ...


10

The Poisson bracket is the bracket of a Lie algebra defined by the symplectic 2-form. That's a lot to unpack so let's go through it slowly. A 2-form $\omega$ is an anti-symmetric two-tensor $\omega_{\mu\nu}$. If $\omega_{\mu\nu}x^\nu \neq 0$ at points where $x^\nu \neq 0$, then $\omega$ is said to be non-degenerate. So $\omega$ is like the metric tensor, ...


9

The main difference between Hamilton and Lagrange/Newton mechanics is, that it happens directly on the phase space, i.e. any point on your manifold already fully determines the state of your system. Intuitively, you realize this by specifying position and momentum coordinates. On a mathematical level, the world we see is some smooth manifold (a priori not ...


9

Counterexample: The transformation $$Q^1~=~2q^1 ,\qquad P_1~=~p_1,\qquad Q^2~=~\frac{1}{2}q^2 ,\qquad P_2~=~p_2 $$ preserves phase space volume & orientation, but is not a symplectomorphism.$^1$ For 2D phase space, the canonical phase space volume form $$\Omega~=~\frac{1}{n!}\omega^{\wedge n}$$ is the symplectic 2-form $\omega$ itself, so that the ...


9

Instead, we were given a recipe how to quantize a classical theory, which is based on the rule of transforming all quantities to operators, and that Poisson bracket is transformed to a commutator. For me it seems like a big secret remains out there, it just hard to me to believe that this is the way our world behaves without further intuitive explanations. ...


8

Yes, because $\{J_i,J_j\}_{PB}=\sum_{k=1}^3\epsilon_{ijk} J_k$. More generally, the statement that the Poisson bracket of any two constants of motion is again a constant of motion is known as Poisson's Theorem.


8

You are confusing in the index, such calculations must be carried out very carefully. I would start with your difention. $$M_i=\epsilon _{ijk} q_j p_k$$ $$M_p=\epsilon _{pnm} q_n p_m$$ $$\{M_i, M_p\}=\sum_l \left(\frac{\partial M_i}{\partial q_l}\frac{\partial M_p}{\partial p_l}-\frac{\partial M_i}{\partial p_l}\frac{\partial M_p}{\partial q_l}\right)$$ ...


8

Quantization usually means the association of a Hilbert space to the classical phase space (in our case a Poisson manifold). However, in deformation quantization, this task is achieved indirectly, first through the construction of an associative $C^*$ algebra, in this case the deformed algebra of functions equipped with a star product which serves as the ...


8

For a 2-dimensional phase space, they are the same. More generally, for a $2n$-dimensional symplectic manifold $(M;\omega)$ with symplectic two-form $$\tag{1} \omega~=~\frac{1}{2} dz^I ~\omega_{IJ} \wedge dz^J, \qquad \omega_{IJ}~=~-\omega_{JI}, $$ the Poisson bracket is given by $$ \tag{2} \{f,g\}_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ} \frac{\...


8

The answer is No. For starters for dimensional reasons. A density carries dimension $L^{-3}$. In the classical (as opposed to the quantum) case, it is tempting to (at least partially) incorporate OP's suggestion for functionals $$ F~=~\int \! d^3x~f(x), \qquad G~=~\int \! d^3x~g(x), \tag{1} $$ by changing the definition from the standard field-theoretic ...


8

See What's the point of Hamiltonian mechanics? for a closely related question. The answers in that thread mention various benefits with the Hamiltonian formulation, among other things: Analysis of structural patterns, symmetries & conservation laws, separability & integrability, Liouville theorem, Hamilton-Jacobi theory, etc, without solving the ...


7

As with most if not all (ultimately) things in physics, one does NOT derive $(2)$ from $(1)$, one guesses $(2)$ from $(1)$ and then confirms the soundness of the guess through experiment. I'm being a bit flippant with the word "guess" as a slight satirisation of ourselves as physicists when we (myself definitely included) lose site of the fact that we are ...


7

The basic idea is the following one. For the sake of simplicity, I henceforth assume that every function does not depend explicitly on time (with a little effort, everything could be generalized dealing with a suitable fiber bundle over the axis of time whose fibers are spaces of phases at time $t$). On a symplectic $2n$ dimensional manifold (a space of ...


7

It's been some time since you asked this, but I'll give it a shot. The biggest problems in curved spacetime are defining exactly what you mean by a Hamiltonian, and what you mean by a Poisson bracket. Let's say you're dealing with a real scalar field $\phi$ which minimizes some action $$ W = \int_{\mathcal{M}}\mathcal{L}(\phi,\partial_\mu \phi,x) d^4 x $$...


7

Although OP strictly speaking writes something different in the question formulation (v4), it seems OP essentially wants to prove the following Lemma. Lemma. Given a (possibly infinite-dimensional$^1$) Lie algebra $L$ over a field $\mathbb{F}\supseteq\mathbb{R}$ with Lie bracket $[\cdot,\cdot]:L\times L \to L$ and Lie algebra derivation$^2$ $D:L\to L$, ...


7

Given a dynamical system $\dot x=F(x)$ with continuously differentiable $F$, there is a canonical bijection between initial conditions at a fixed time and solution trajectories. If the dynamical system is Hamiltonian, the phase space is the space of initial conditions, hence is canonically isomorphic to the space of solution trajectories. Considering the ...


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