Deformation Quantization
What is quantization?
Let us build a strong connection between classical and quantum.
Weyl-Wigner transform
After quantum mechanics was discovered to be a fundamental law of nature, mathematicians pondered the difference between 'quantum' and 'classical'. The first step is the Weyl-wigner transform, which deforms functions in phase space into quantum operators.
Let coordinate of phase space is $(q,p)$, function on phase space is $f$, and qunatum operator of position and momentum are $P$ and $Q$ with $[P, Q]=i\hbar$.
$\Phi [f] = \frac{1}{(2\pi)^2} \int \int \int \int f(q, p) (e^{i(a(Q-q) + b(P-p))})dq dp da db$
Integrating $p$ and $q$ became Fourier transform which goes $f(q,p)$ to $\tilde{f}(a,b)$.
$\Phi [f] = \frac{1}{(2\pi)^2} \int \int \tilde{f} (a, b) e^{iaQ + ibP} da db$
Let's take an example.
$(cq + dp)^n \longmapsto (cQ + dP)^n$
$pq \longmapsto \frac{1}{2} (PQ + QP)$
$p^m q^n \longmapsto \frac{1}{2^n} \sum^n_{r=0} {n \choose r} Q^r P^m Q^{n-r}$
These shows noncommutative of $P$ and $Q$, which is different from $p$ and $q$.
How do operations between functions work? Basically, all we have to do is do the weyl-wigner transformation.
$\Phi[f_1 \star f_2] = \Phi[f_1] \Phi[f_2]$
How to compute this? First define $f \overleftarrow{\partial}_x g = \frac{\partial f}{\partial x} \cdot g$ and $f \overrightarrow{\partial}_x g = f \cdot \frac{\partial g}{\partial x}$.
Then, $f \star g = f \exp (\frac{i\hbar}{2} (\overleftarrow{\partial}_x \overrightarrow{\partial}_p - \overleftarrow{\partial}_p \overrightarrow{\partial}_x))g$.
We can check
$(f \star g) (x, p) = \frac{1}{\pi^2 \hbar^2} \int f (x + x', p + p') g(x + x'', p + p'') \exp (\frac{2i}{\hbar} (x' p'' - x'' p')) dx' dp' dx'' dp''$.
Functions in phase space have different operational rules than Poisson brackets before quantization. The newly defined operation is known as Kontsevich quantization formula.
$f_1 \star f_2 = \sum^\infty_{n=0} \frac{1}{n!} \left( \frac{i\hbar}{2} \right)^n \Pi^n (f_1, f_2)$
For low order,
$\Pi^0 (f_1, f_2) = f_1 f_2$
$\Pi^1(f_1, f_2) = \{f_1, f_2\}$
So,
$[f, g] = f \star g - g \star f = i\hbar \{f, g\} + O(\hbar^2)$
It looks complicated, but it's easy if you look at the eigenfunctions.
$\delta(q) \star \delta(p) = \frac{2}{\hbar} \exp (2i \frac{qp}{\hbar})$
Let's take a look at their roles in phase space.
The Wigner distribution corresponds to the Weyl quantization of the density matrix.
The wigner distribution is a probability distribution over phase space defined as follows.
$P[a \le X \le b] = \int^b_a \int^\infty_{-\infty} W(x, p) dp dx$
It can also be defined as the wigner transform of the density matrix.
$W(x, p) = \frac{1}{\pi\hbar} \int^\infty_{-\infty} \left\langle x-y | \hat{\rho} | x+y \right\rangle e^{2ipu/\hbar} dy$
Then the expected value is expressed as follows:
$\left\langle \hat{G} \right\rangle = \int dx dp W(x, p) g(x, p)$
This naturally satisfies the normalization conditions.
$\int^\infty_{-\infty} dx \int^\infty_{-\infty} dp W(x, p) = Tr(\hat{\rho})$
Time evolution
First, we need to define the eigen(distribution)functions for the Hamiltonian.
$H \star W = E \cdot W$
Since $\frac{\partial f}{\partial t} = -\frac{1}{i\hbar} (f \star H - H \star f)$,
$\frac{\partial W}{\partial t} = - [W, H] = - \{W, H \} + O(\hbar^2)$
In the classical limit, the Liuville theorem is obtained.
Example: Simple Harmonic Oscillator
$H = \frac{1}{2} m\omega^2 x^2 + \frac{p^2}{2m}$
$H \star W = (\frac{1}{2} m\omega^2 x^2 + \frac{p^2}{2m}) \star W = (\frac{1}{2}m\omega^2 (x+\frac{i\hbar}{2}\overrightarrow{\partial}_p)^2 + \frac{1}{2m} (p - \frac{i\hbar}{2} \overrightarrow{\partial}_x)^2) W$
$=(\frac{1}{2} m\omega^2 (x^2 - \frac{\hbar^2}{4} \overrightarrow{\partial})^2_p) + \frac{1}{2m} (p^2 - \frac{\hbar^2}{4} \overrightarrow{\partial}^2_x))W + \frac{i\hbar}{2}(m\omega^2 x \overrightarrow{\partial}_x)W = E \cdot W$
First condition from imaginary part: $\frac{\hbar}{2} (m\omega^2 x \overrightarrow{\partial}_p - \frac{p}{m} \overrightarrow{\partial}_x) \cdot W = 0$
This make eigenfunction form: $W(x, p) = F(\frac{1}{2} m\omega^2 x^2 + \frac{p^2}{2m}) \equiv F(u)$
Which involves the Laguerre polynomials as
$F_u(u) = \frac{(-1)^n}{\pi\hbar} L_n (4\frac{u}{\hbar \omega}) e^{-2u/\hbar\omega}$
Also eigenvalue is $E_n = \hbar \omega (n + \frac{1}{2})$.
Epilogue
However, the above method does not always work. Groenwold's theorem tells us that there is no perfect quantization. The field of focusing on the structure of quantum, giving up the current clear method and finding a general method is called geometric quantization.
Reference
Wikipedia (above links)
Brian C. Hall - Quantum Theory for Mathematicians