35

Poisson brackets play more or less the same role in classical mechanics that commutators do in quantum mechanics. For example, Hamilton's equation in classical mechanics is analogous to the Heisenberg equation in quantum mechanics: $$\begin{align}\frac{\mathrm{d}f}{\mathrm{d}t} &= \{f,H\} + \frac{\partial f}{\partial t} & \frac{\mathrm{d}\hat f}{\...


27

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}^{...


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 ...


21

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 ...


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 $$ {\...


20

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\...


20

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, ...


20

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

Both the commutator (of matrices) and the Poisson bracket satisfy the Jacobi identity, $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$. This is why Dirac was inspired by Heisenberg's use of commutators to develop a Hamilton-Jacobi dynamics style of Quantum Mechanics which provided the first real unification of Heisenberg's matrix mechanics with Schroedinger's wave ...


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 $$ \frac{d}{dt}f(p,q,t)=\{f,H\}+\frac{\partial f}{\partial t}\, . $$ ...


10

Regarding the significance of the observables momentum and position there are many similarities between Classical and Quantum mechanics. Some of the algebraic relations have been pointed out. In the end, there is still an important difference, which is obvious by the fact that the function algebra generated by classical quantities is commutative $$q·p=...


10

It seems that OP just wants to trace a sign convention in a specific book (Ref. [JS]). To answer this most convincingly, we should document where we read what. In the case of canonical coordinates (also known as Darboux coordinates), [JS] uses a convention where the positions $q^i$ are ordered before the momenta $p_i$, $$\xi^i~=~(q^1, \ldots, q^n,p_1, \...


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 ...


9

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 ...


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 ...


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

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

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 ...


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

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 ...


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

Standard Hamiltonian mechanics in $N$-particle phase space $R^{6N}$ is inadequate to describe mechanical systems of interest that are not of the $N$ particle form, for example rigid bodies. However, all major techniques in classical mechanics do not depend on the specific structure of $R^{6N}$ but only on the fact that one can define on it a Poisson bracket. ...


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

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

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, ...


7

What one requires for the uncertainty principle to arise is that the relevant observables should not commute, i.e., their commutator is non-zero. The Poisson bracket of two observables is not the same as their commutator. Even if the Poisson bracket of the classical observables is non-zero, they do commute in classical mechanics--they are just real numbers. ...


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