Disclaimer: Nothing in this note is original.

The Hamiltonian formulation of classical mechanics live on phase space, the Lagrangian formulation live on configuration space. Geometrically, the phase space is a symplectic manifold of dimension $\mathbb{R}^{2d}$. The coordinates in the phase space are positions and conjugate momenta. Observables are given by functions on phase space.

The time evolution is controlled by the Hamiltonian $H(p,x)$ through Hamilton’s equation

\[\dot{x}=\partial _ {p}H, \quad \dot{p}=-\partial _ {x}H\]

and the Poisson bracket of two observables $A,B$ are defined to be

\[\left\{ A,B \right\} :=\partial _ {p}A \times \partial _ {x}B - (A\leftrightarrow B).\]

Hamilton’s equation can be written in terms of the Poisson bracket

\[\dot{x}=\left\{ H,x \right\} ,\quad \dot{p}=\left\{ H,p \right\}\]

in fact for any observable the time evolution is given by

\[\dot{A} = \left\{ H,A \right\} .\]

The time evolution in phase space can be regarded as a flow $\phi^{t}$ parametrized by time $t$, if the initial position is given by $p(0),x(0)$ the the flow is denoted by $\phi^{t}(p_ {0},x_ {0})$ where $p_ {0}:=p(0)$. It is indeed a flow since $\phi^{t}$ form a one-parameter group.

Weyl quantization

The question we wanna ask is, given a classical observable as a function of $x$ and $p$, how can be get the quantized version of it? We know how to quantize $x$ and $p$, by making then into operator $\hat{x},\hat{p}$ which satisfy the commutation relation $[\hat{x},\hat{p}]=i$. We have chosen the natural units. What about a general operator $A(x,p)$ then? Since now $\hat{x},\hat{p}$ are not commutative anymore, we need to be careful about the ordering when quantizing $A(x,p)$. A possible (good) choice is to take symmetrised products of $\hat{x}$ and $\hat{p}$,

\[xp \to \frac{1}{2} (\hat{x}\hat{p}+\hat{p}\hat{x}),\]

this is called the Weyl ordering.

In the following we will only consider 1-dimensional case, as it can be trivially generalized to arbitrary dimension $d$.

Let $A(x)$ be a classical operator depending on $x$ alone, not on $p$. The motivation is that, we can use a somehow weird form of “$\delta$ function” to replace $x$ with $\hat{x}$, if we define something like $\delta(x-\hat{x})$, and integrate over $x$,

\[\hat{A}(\hat{x}) := \int dx \, \delta(x-\hat{x})A(x).\]

But what is this delta function between a c-number and a q-number? To make sense of it, we further write it using the Fourier transform,

\[\delta(x-\hat{x}) = \int \frac{dp}{2\pi} \, e^{ ip(x-\hat{x}) }\]

and

\[\hat{A} := \frac{1}{2\pi}\int dx dp \, e^{ ip(x-\hat{x}) } A(x),\]

here the operator $\hat{x}$ appears in the exponent and in principal we know how to deal with it, for example with the help of Baker-Campbell-Hausdorff formula.

If the operator depends on both $x$ and $p$, we can introduce another “delta function”, bring the operator $\hat{p}$ to the exponent using a Fourier transform. As a possible choice, we define the two “delta functions” together are like

\[\delta(x-\hat{x})\delta(p-\hat{p}) := \int \frac{d\xi}{2\pi} \frac{dy}{2\pi} \, e^{ iy(p-\hat{p})+i\xi(x-\hat{x}) }.\]

Note that this is just one of many possible definitions, it is by no means determined from the first principals.

For future convenience, let’s define an operation $\text{Op}$

\[\hat{A} := \text{Op}[A(x,p)] := \int \frac{d\xi}{2\pi} \frac{dy}{2\pi} dx dp \, e^{ iy(p-\hat{p})+i\xi(x-\hat{x}) } A(x,p).\]

The integral over $\xi$ and $y$ are not surprising, however the integral over $p$ and $x$ could appear unexpected to people who see it for the first time. To rewrite the definition of $\text{Op}$ in a more comfortable (for myself) form where $dp$ is always divided by $2\pi$, I prefer

\[\hat{A} := \text{Op}[A(x,p)] := \int \frac{d\xi}{2\pi} \frac{dp}{2\pi} dx dy \, e^{ iy(p-\hat{p})+i\xi(x-\hat{x}) } A(x,p).\]

The legitimacy of this choice mostly come from the fact that it agrees with the Weyl ordering in Eq. (1). To see that, just apply it to $xp$ and you’ll see that they indeed agree to each other.

This is called the Weyl quantization. $A(p,x)$ is called the Weyl symbol of quantum operator $\hat{A}$.

To fully understand the physical meaning of the symbol, we go to the Schrodinger representation and see what the symbol does to wave functions. Notice that

\[\left\langle z \middle\vert e^{ i\xi \hat{x} }\psi \right\rangle = \psi(i\xi z)\]

and

\[\left\langle z \middle\vert e^{ iyp } \right\rangle =\psi(z-y)\]

we can calculate (though the calculation is long and tedious) $\left\langle z \middle\vert \hat{A}\psi \right\rangle$, eventually we could get rid of the hatted operators in the exponent and arrive at

\[\boxed { \left\langle z \middle\vert \hat{A}\psi \right\rangle =\int \frac{dp}{2\pi}dy \, A\left( p,\frac{z+x}{2} \right) e^{ ip(z-x) } \psi(x) }\]

which is usually referred to as the definition of Weyl quantization.