canonical quantization | Physics Library

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 correspondence principle 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 . Then we may obtain from each coordinate function an operator on the Hilbert space $\mathcal{H} = L^2(X, \mu)$ , consisting of functions on that are square-integrable with respect to some measure $\mu$ , by the operator substitution rule:

$\displaystyle x^i \mapsto \hat{x}^i$	$\displaystyle = x^i \cdot,$	(1)
$\displaystyle p_i \mapsto \hat{p}_i$	$\displaystyle = -i \hbar \frac{\partial }{\partial x^i},$	(2)

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,

$x^i x^j \mapsto \hat{x}^i \hat{x}^j = x^i x^j \cdot$ ,
$p_i p_j \mapsto \hat{p}_i \hat{p}_j = -\hbar^2 \frac{\partial ^2}{\partial x^i x^j}$ ,
if $i \neq j$ then $x^i p_j \mapsto \hat{x}^i \hat{p}_j = -i \hbar x^i \frac{\partial }{\partial x^j}$ .

Remark 1 The substitution rule creates an ambiguity for the function $x^i p_j$

when

, since

, 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

$\displaystyle x^i p_j \mapsto \frac{1}{2}\left(\hat{x}^i \hat{p}_j + \hat{p}_j \hat{x}^i\right),$

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 (

)-(

) to a map

$\displaystyle C^\infty(T^*X)$	$\displaystyle \to {\mathrm{Op}}(\mathcal{H})$
$\displaystyle f$	$\displaystyle \mapsto \hat{f}.$

Remark 2 This procedure is called “canonical” because it preserves the canonical Poisson brackets. In particular, we have that

$\displaystyle \frac{-i}{\hbar}[\hat{x}^i, \hat{p}_j] := \frac{-i}{\hbar}\left(\hat{x}^i\hat{p}_j - \hat{p}_j\hat{x}^i\right) = \delta^i_j,$

which agrees with the Poisson bracket $\{ x^i, p_j \} = \delta^i_j$ .

Example 1 Let $X = \mathbb{R}$ . The hamiltonian function for a one-dimensional point particle with mass $m$

$\displaystyle H = \frac{p^2}{2m} + V(x),$

where

is the potential energy. Then, by operator substitution, we obtain the Hamiltonian operator

$\displaystyle \hat{H} = \frac{-\hbar^2}{2m} \frac{d^2}{dx^2} + V(x).$

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