# Weyl Quantization

We won’t dive deep into the analysis problem and will rather present the basic idea behind Weyl quantization, then we will mention some further developments such as Moyal product and deformation quantization.

## Idea

On a flat phase space $\mathbb{R}^{2n}$ with the standard symplectic form $\omega = \sum dq^i \wedge dp_i$, to any reasonable function $a(q,p)\in C^\infty(\mathbb{R}^{2n})$, the Weyl calculus assigns an operator $$A = \hat{a} = \text{Op}(a(q,p)) = q(q,-ih\partial_q),$$ that is defined on functions $u(q) \in S(\mathbb{R}^n)$ by an oscillatory integral $$$$Au (q) = (2\pi)^{-n} \int_{\mathbb{R}^{2n}} \exp(\frac{i}{h}p(q-q’)) a(\frac{q+q’}{2},p) u(q’) dq’ dp. \label{weyl}$$$$ The function $a(q,p)$ is called the Weyl symbol of the operator $A$. $h \in (0,1]$ is a parameter.

### Physical motivation

See Stein’s book for more.

On a flact phase space every classical observable should be a function $\sigma(q,p)$. The quantum observable corresponding to $\sigma(Q,P)$ where $Q$ is the position operator and $P=-i\partial_q$. However we have a ordering problem arising from $[Q,P]\neq 0$, so we need to be more careful on the definition of $\sigma(Q,P)$.

When $\sigma(q,p) = \sigma(q)$, i.e. no dependence on $p$, then we should have the multiplication operator $f\mapsto \sigma(\cdot)f(\cdot)$. Similarly for $\sigma(q,p)=\sigma(p)$ we should have the multiplier operator $\hat{f}\mapsto \sigma(\cdot)\hat{f}(\cdot)$.

We can define $\sigma(Q,P) = A$ where $A$ is the pseudodifferential operator with symbol $\sigma$. This assignment has the effect that $A = A_1 A_2$ if $\sigma(q,p) = \sigma_1(q)\sigma_1(p)$, i.e. this amounts to putting the $P$ operations to the right of the $Q$ operations.

We can formally treat the $P$ and $Q$ operations on a more symmetric footing and with a Fourier transformation put $$Op(\sigma) = \iint \exp(i x\cdot Q + y\cdot P) \hat{\sigma}(x,y) dx dy,$$ this is done a lot in physics. Some manipulations on $\hat{\sigma}$ gives $$Op(\sigma) u (q)= \iint \sigma(\frac{q+q’}{2}) \exp(i(q-q’)\cdot \xi) u(q’) dq’ d\xi.$$

### Mathematical motivation

So far this is motivated by quantum mechanics, but it can also be motivated in a purely mathematical point of view, if we have already motivated the definition of pseudodifferential operators. See 10.11/cpa.3160320304 for more. In fact they are quite similar: both of them arise from the ordering of operators.

Recall from the entry on Fourier integral operators that a pseudodifferential operator on $\mathbb{R}^n$ can be defined through $$Au (x) = \iint e^{i(x-x’)\cdot\xi}\sigma_A(x,x’,\xi) u(x’) dx’ d\xi$$ now suppose that $\sigma$ doesn’t depend on $x’$ and integrate $x’$ out $$$$Au (x) = (2\pi)^{-n}\int e^{ix\cdot\xi}\sigma_A(x,\xi) \hat{u}(\xi) d \xi. \label{def_pdo}$$$$ where $\hat{\cdot}$ denotes the Fourier transform. If $\sigma(x,\xi)$ is a polynomial in $\xi$, then $A$ can be seen as a linear differential operator $\sigma(x,D)$ with $D=-i\partial_x$ put to the right of the coefficients. However one could equally well have decided to put all differentiation to the left of the coefficients.

The two definitions agree if $\sigma$ is a real linear function of $(x,\xi)\in \mathbb{R}^{2n}$. Any two such functions can be transformed into each other by a linear symplectic transformation $\chi$ with the symplectic form given by $\omega = \sum dx^i \wedge d\xi_i$.

Segal proved that corresponding to $\chi$ there is a unitary transformation $U$ in $L^2(\mathbb{R}^n)$ such that $$$$U^{-1} \sigma(x,D) U = (\sigma\circ\chi) (x,D) \label{cov}$$$$ for all linear $\sigma$ and $U$ is unique up to a constant factor.

Remark. The pair $(\sigma(x,D),U)$ is similar to a covariant representation, with $\chi$ seen as a group action on an element $\alpha_g$ of a group $G$, of a dynamical system $(Q,G,\alpha)$ in $B$.
• The group $G$ should be seen as a subgroup of $\text{Sp}(2n,\mathbb{R})$, the symplectomorphism group of $\mathbb{R}^{2n}$.
• $B$ should be the algebra of pseudodifferential operators (is this a $C^\ast$-algebra?)
• $Q$ should be (some completion of) the algebra of the coordinates $(x,\xi)$ hence a subset of $C^\infty(\mathbb{R}^{2n})$.

There is a unique way to modify the definition given by $\eqref{def_pdo}$ so that $\eqref{cov}$ remains valid for general $\sigma$, and $\sigma(x,D)$ is a multilication by $\sigma(x,\xi)$ (i.e. a multiplier) if $\sigma(x,\xi)=\sigma(x)$ id independent of $\xi$. This can be found, by considering when $\sigma$ is bounded exponential by Fourier decomposition of a general $\sigma\in L(\mathbb{R}^{2n})$, to be $$Op(\sigma) u (x) = (2\pi)^{-n}\iint \sigma(\frac{(x+y)\cdot\xi}{2})\exp(i(x-y)\cdot\xi)u(y) dyd\xi,$$ which agrees with $\eqref{weyl}$.

## Moyal bracket and the star product

There is a inverse formula (or the formula for recovering the symbol) for Weyl quantization $\sigma\mapsto Op(\sigma)$ given by a kind of trace formula. Moyal, when studying the interpretation of the symbol $u$ of agiven a quantum operator $Op(u)$, found what is now called the Moyal bracket: $$M(u,v) = \frac{2}{i\hbar} \sinh(\frac{i\hbar}{2}P)(u,v) = P(u,v) + \sum_{r=1}^{\infty}(\frac{i\hbar}{2})^{2r} P^{2r+1}(u,v)$$ where $$P^r(u,v) = \Lambda^{i_1 j_1} … \Lambda^{i_r j_r} (\partial_{i_1\ldots i_r} u) (\partial_{j_1\ldots j_r}v)$$ the $r$-th power of the Poisson bracket $P$. $i_k,j_k = 1,\ldots,2n$ with $k=1,\ldots,r$, $\Lambda = \begin{pmatrix}0 & -I \\ I & 0 \end{pmatrix}$.

We might assume $u,v\in C^\infty(\mathbb{R}^{2n})$ and the sum can be taken as a formal series.

A similar formula for the symbol of a product $Op(u)Op(v)$ is called star product: $$u \star_M v = \exp(\frac{i\hbar}{2}P)(u,v) = uv + \sum_{r=1}^\infty (\frac{i\hbar}{2})^r P^r(u,v).$$