PlanetPhysics/Canonical Quantization

Canonical quantization is a method of relating, or associating, a classical system of the form $$(T^*X, \omega, H)$$, where $$X$$ is a manifold, $$\omega$$ is the canonical symplectic form on $$T^*X$$, with a (more complex) quantum system represented by $$H \in C^\infty(X)$$, where $$H$$ is the Hamiltonian operator. Some of the early formulations of quantum mechanics used such quantization methods under the umbrella of the \htmladdnormallink{correspondence principle {http://planetphysics.us/encyclopedia/PrincipleOfCorrespondingStates.html} or postulate}. The latter states that a correspondence exists between certain classical and quantum operators, (such as the Hamiltonian operators) or algebras (such as Lie or Poisson (brackets)), with the classical ones being in the real ($$\mathbb{R}$$) domain, and the quantum ones being in the complex ($$\mathbb{C}$$) domain. Whereas all classical Observables and States are specified only by real numbers, the 'wave' amplitudes in quantum theories are represented by complex functions.

Let $$(x^i, p_i)$$ be a set of Darboux coordinates on $$T^*X$$. Then we may obtain from each coordinate function an operator on the Hilbert space $$\mathcal{H} = L^2(X, \mu)$$, consisting of functions on $$X$$ that are square-integrable with respect to some measure $$\mu$$, by the operator substitution rule: $$\begin{matrix} x^i \mapsto \hat{x}^i &= x^i \cdot, (1)\\ p_i \mapsto \hat{p}_i &= -i \hbar \pdiff{}{x^i} (2), \end{matrix}$$ where $$x^i \cdot$$ is the "multiplication by $$x^i$$" operator. Using this rule, we may obtain operators from a larger class of functions. For example,


 * 1) $$x^i x^j \mapsto \hat{x}^i \hat{x}^j = x^i x^j \cdot$$,
 * 2) $$p_i p_j \mapsto \hat{p}_i \hat{p}_j = -\hbar^2 \pdiff{^2}{x^i x^j}$$,
 * 3) if $$i \neq j$$ then $$x^i p_j \mapsto \hat{x}^i \hat{p}_j = -i \hbar x^i \pdiff{}{x^j}$$.

\begin{rmk} The substitution rule creates an ambiguity for the function $$x^i p_j$$ when $$i=j$$, since $$x^i p_j = p_j x^i$$, whereas $$\hat{x}^i \hat{p}_j \neq \hat{p}_j \hat{x}^i$$. This is the operator ordering problem. One possible solution is to choose "$x^i p_j \mapsto \frac{1}{2}\left(\hat{x}^i \hat{p}_j + \hat{p}_j \hat{x}^i\right),$" since this choice produces an operator that is self-adjoint and therefore corresponds to a physical observable. More generally, there is a construction known as Weyl quantization that uses Fourier transforms to extend the substitution rules (2)-(3) to a map $$\begin{matrix} C^\infty(T^*X) &\to \Op (\mathcal{H}) \\ f &\mapsto \hat{f}. \end{matrix}$$ \end{rmk}

\begin{rmk} This procedure is called "canonical" because it preserves the canonical Poisson brackets. In particular, we have that $$\frac{-i}{\hbar}[\hat{x}^i, \hat{p}_j] := \frac{-i}{\hbar}\left(\hat{x}^i\hat{p}_j - \hat{p}_j\hat{x}^i\right) = \delta^i_j, $$ which agrees with the Poisson bracket $$\{ x^i, p_j \} = \delta^i_j$$. \end{rmk}

\begin{ex} Let $$X = \reals$$. The Hamiltonian function for a one-dimensional point particle with mass $$m$$ is $$ H = \frac{p^2}{2m} + V(x), $$ where $$V(x)$$ is the potential energy. Then, by operator substitution, we obtain the Hamiltonian operator $$ \hat{H} = \frac{-\hbar^2}{2m} \frac{d^2}{dx^2} + V(x). $$ \end{ex}