41 votes

What is the "secret " behind canonical quantization?

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 ...
Valter Moretti's user avatar
31 votes
Accepted

Does the poisson bracket $\{f,g\}$ have any meaning if neither of $f$ or $g$ is the system's Hamiltonian?

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 $f$ (...
Valter Moretti's user avatar
22 votes
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The quantum analogue of Liouville's theorem

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 ...
Cosmas Zachos's user avatar
22 votes

Why Equal-time commutation relation?

The problem is that once you define your operators at a fixed time $t_0$, your operators at all later times are immediately determined by the Heisenberg equations of motion. So you are not free to ...
Zack's user avatar
  • 2,828
21 votes
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Hamilton equations from Poisson bracket's formulation

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 ...
J. Murray's user avatar
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17 votes
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Is a canonical transformation equivalent to a transformation that preserves volume and orientation?

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 ...
Valter Moretti's user avatar
16 votes
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The geometrical interpretation of the Poisson bracket

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 \...
ZeroTheHero's user avatar
14 votes

The geometrical interpretation of the Poisson bracket

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_{\...
Robin Ekman's user avatar
  • 14.6k
12 votes

Dequantizing Dirac's quantization rule

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 ...
Cosmas Zachos's user avatar
12 votes
Accepted

Dequantizing Dirac's quantization rule

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....
OkThen's user avatar
  • 804
12 votes

Why Equal-time commutation relation?

OP's proposal $$\begin{align} [\hat{\phi}({\bf x}_1,t_1),\hat{\phi}({\bf x}_2,t_2)]~=~&0,\cr [\hat{\phi}({\bf x}_1,t_1),\hat{\pi}({\bf x}_2,t_2)]~=~&i\hbar\hat{\bf 1}~\delta^3({\bf x}_1-{\bf ...
Qmechanic's user avatar
  • 196k
11 votes
Accepted

Non-Euclidean Phase Space?

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 ...
Qmechanic's user avatar
  • 196k
10 votes
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Is there any relation between Poisson Brackets and the Jacobian Matrix?

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} \...
Qmechanic's user avatar
  • 196k
10 votes
Accepted

What is the physical meaning of the minus sign in the Hamilton equation of momenta?

Consider the Hamiltonian for a single particle in some potential $$H(p,q) = \frac{p^2}{2m} + V(q)$$ where the first term is the kinetic energy and the second term is the potential energy. In this case,...
Lenard Kasselmann's user avatar
9 votes
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Canonical Commutation Relations in arbitrary Canonical Coordinates

The quantization prescription $$ [\hat{x},\hat{y}] := \mathrm{i}\hbar\widehat{\{x,y\}}\tag{1}$$ for $x,y$ two classical phase space coordinates does have its subtleties. In particular, as the answer ...
ACuriousMind's user avatar
  • 122k
9 votes

Is a canonical transformation equivalent to a transformation that preserves volume and orientation?

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.$^1$ ...
Qmechanic's user avatar
  • 196k
8 votes

Motivation for covariant phase space

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 ...
Arnold Neumaier's user avatar
8 votes

Proof of constructing action-angle coordinates on Hamiltonian system

OP's actual question follows directly from Theorem 5.3 in Ref. 2, but that leaves the obvious question: How to prove Theorem 5.3? It seems the only really satisfying answer would be to outline a ...
Qmechanic's user avatar
  • 196k
8 votes

Does the Heisenberg equation for fields and canonical momentums hold as well for the hamiltonian density operator instead of the Hamiltonian operator?

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'...
Qmechanic's user avatar
  • 196k
8 votes

Why do we care only about canonical transformations?

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 ...
Qmechanic's user avatar
  • 196k
8 votes

What is the "secret " behind canonical quantization?

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 ...
Cosmas Zachos's user avatar
8 votes
Accepted

Why are first class constraints harder to quantize than second class constraints?

First-class constraints generate gauge transformations (assuming the Dirac conjecture), i.e. map physically equivalent states onto each other. Even if you do not assume the Dirac conjecture, then ...
ACuriousMind's user avatar
  • 122k
8 votes
Accepted

$ℏ$ in the canonical commutation relation

Note that while the commutator $[\cdot,\cdot]$ is dimensionless, the Poisson bracket $\{\cdot,\cdot\}$ carries dimension of inverse angular momentum. So a quantity of dimension of angular momentum is ...
Qmechanic's user avatar
  • 196k
7 votes
Accepted

Clever way to show a property of Lie transformation

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$...
Qmechanic's user avatar
  • 196k
7 votes

What is the physical meaning of the minus sign in the Hamilton equation of momenta?

More broadly speaking, the asymmetric minus sign in Hamilton's equations is intimately related to the antisymmetry of the symplectic/Poisson structure, which in turn has several consequences, e.g. ...
Qmechanic's user avatar
  • 196k
6 votes
Accepted

Field theory: equivalence between Hamiltonian and Lagrangian formulation

First, note that the EOM are the Euler-Lagrange equations: $$ \frac{\delta S}{\delta \phi}=\partial_\mu\left(\frac{\partial \mathscr L}{\partial\phi_{,\mu}}\right)-\frac{\partial \mathscr L}{\partial ...
AccidentalFourierTransform's user avatar
6 votes

Dequantizing Dirac's quantization rule

The statement is true by the very definition of quantization, i.e. there is nothing to show. So let's talk about the definition of quantization, which is a map from classical observables to quantum ...
ACuriousMind's user avatar
  • 122k
6 votes
Accepted

Whence the $i$ in QM Poisson bracket definition?

The imaginary unit $i$ is there to turn quantum observables/selfadjoint operators into anti-selfadjoint operators, so that they form a Lie algebra wrt. the commutator. Or equivalently, consider the ...
Qmechanic's user avatar
  • 196k
6 votes

Hamilton equations from Poisson bracket's formulation

The "partial" time derivative $\frac{\partial}{\partial t}$ means in this context an explicit time differentiation. A function $$(q,p,t)~\mapsto~ f(q,p,t)$$ of phase space and time is said to have an ...
Qmechanic's user avatar
  • 196k

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