# Weyl symbol of product

Are there explicit formulas for the Weyl symbol of $$-f(x)D_x^2$$ where $$D_x:=-i\partial_x$$ and $$\partial_x$$ is the derivative and $$f$$ some sufficiently smooth function?

In the standard quantization the symbol would of course just be $$w(x,\xi)=f(x) \xi^2$$ but in Weyl quantization this seems to be no longer true.

## 1 Answer

By quantization, one is usually more trying to find the operator corresponding to a given symbol than the symbol corresponding to an operator. In your case, it seems the Weyl symbol you are talking about is the Wigner transform (the inverse of the Weyl trasnform). For an operator $$A$$ of kernel $$a(x,y)$$ it is defined by $$w_A(x,\xi) = ∫_{\mathbb{R}} e^{-i\,y\,\xi}\,a(x+y/2,x-y/2)\,\mathrm{d}y.$$ If your operator is $$A = -f(x)\,D^2_x = f(x)\,\partial_x^2$$, then its kernel is $$a(x,y) = f(x)\,\delta_0''(x-y)$$, and so since $$(x+y/2)-(x-y/2) = y$$, we find (replacing the integral by a duality bracket) \begin{align*} w_A(x,\xi) &= \left\langle \delta_0'', e^{-i\,\xi\,\cdot}f(x+\cdot/2)\right\rangle \\ &= \left.\partial^2_y\left(e^{-i\,y\,\xi}f(x+y/2)\right)\right|_{y=0} \\ &= -\xi^2 f(x) - i\,\xi\, f'(x) + \frac{1}{4}\, f''(x). \end{align*}

• Yes I did the correction thanks (without the $1/2$ since it is counted twice) Apr 15, 2020 at 5:44