# Vanishing of terms in Wigner transform of operators

I have been given the following proof for the fact that the Moyal bracket is the Wigner-Weyl transform of the quantum commutator. However, I am unsure about the vanishing of some boundary terms in an intermediate step of the proof.

Consider three operators $$\hat{A}$$, $$\hat{B}$$, and $$\hat{C}$$ on a Hilbert space $$\mathcal{H}$$, related to each other via $$\hat{C}=\hat{A}\hat{B}$$

The Wigner transform to phase space of the above expression is given by the Moyal product $$C_c(x,p_x)=A_c(x,p_x)\star B_c(x,p_x)$$ where the subscript $$c$$ labels c-number quantities. An integral representation of the above equation is $$C_c(x,p_x)=\int_{-\infty}^\infty db\int_{-\infty}^\infty dc \int_{-\infty}^\infty \frac{d\alpha}{2\pi}\int_{-\infty}^\infty \frac{d\beta}{2\pi}e^{i\alpha b}e^{i\beta c}A_c(x+\frac{c}{2},p_x+\hbar\alpha)B_c(x-\frac{b}{2},p_x+\hbar\beta)$$

To evaluate the integral, we first Taylor expand the integrand as $$A_c(x+\frac{c}{2},p_x+\hbar\alpha)B_c(x-\frac{b}{2},p_x+\hbar\beta)=A_c(x+\frac{c}{2},p_x)B_c(x-\frac{b}{2},p_x)+\hbar\left(\underbrace{\alpha B_c(x-\frac{b}{2},p_x)\left.\frac{\partial A_c}{\partial p_x}\right|_{(x+\frac{c}{2},p_x)}}_{\text{This!}}+\beta A_c(x+\frac{c}{2},p_x)\left.\frac{\partial B_c}{\partial p_x}\right|_{(x-\frac{b}{2},p_x)}\right)+\mathcal{O}(\hbar^2)$$

and consider each term individually. For example, the contribution from the first half (which I denoted "This!" in the expression above) of the term linear in $$\hbar$$, is \begin{align}\hbar\int_{-\infty}^\infty db\int_{-\infty}^\infty dc \int_{-\infty}^\infty \frac{d\alpha}{2\pi}\alpha e^{i\alpha b}\underbrace{\int_{-\infty}^\infty \frac{d\beta}{2\pi}e^{i\beta c}}_{=\delta(c)}B_c(x-\frac{b}{2},p_x)\left.\frac{\partial A_c}{\partial p_x}\right|_{(x+\frac{c}{2},p_x)}=\\=\hbar\int_{-\infty}^\infty db\int_{-\infty}^\infty \frac{d\alpha}{2\pi}\underbrace{\alpha e^{i\alpha b}}_{=-i\frac{\partial}{\partial b}e^{i\alpha b}}B_c(x-\frac{b}{2},p_x)\left.\frac{\partial A_c}{\partial p_x}\right|_{(x,p_x)}=\end{align}

At this point, integration by parts is performed to give $$\begin{split}=i\hbar\int_{-\infty}^\infty db\int_{-\infty}^\infty \frac{d\alpha}{2\pi}e^{i\alpha b}\left.\frac{\partial A_c}{\partial p_x}\right|_{(x,p_x)}\underbrace{\left.\frac{\partial B_c}{\partial b}\right|_{(x-\frac{b}{2},p_x)}}_{=-\frac{1}{2}\left.\frac{\partial B_c}{\partial x}\right|_{(x-\frac{b}{2},p_x)}}=\\=-\frac{i\hbar}{2}\int_{-\infty}^\infty db\underbrace{\int_{-\infty}^\infty \frac{d\alpha}{2\pi}e^{i\alpha b}}_{=\delta(b)}\left.\frac{\partial A_c}{\partial p_x}\right|_{(x,p_x)}\left.\frac{\partial B_c}{\partial x}\right|_{(x-\frac{b}{2},p_x)}=\\=-\frac{i\hbar}{2}\left.\frac{\partial A_c}{\partial p_x}\right|_{(x,p_x)}\left.\frac{\partial B_c}{\partial x}\right|_{(x,p_x)}\end{split}$$

Acting similarly on the other terms gives the final result, $$C_c(x,p_x)=A_c(x,p_x)B_c(x,p_x)+\frac{i\hbar}{2}\left\{A_c(x,p_x),B_c(x,p_x)\right\}+\mathcal{O}(\hbar^2)$$

My question is: why do the boundary terms in the integration by parts vanish? Specifically to the term examined above, why is it that $$\lim_{t\to\infty}\left[B_c(x-\frac{b}{2},p_x)e^{i\alpha b}\right]_{b=-t}^{b=t}=0\quad?$$

I overheard someone saying that the answer lies in the integration contour considered, but I couldn't really hear the answer in full.

• Cool? One gathers so. . – Cosmas Zachos Feb 18 at 14:25

Specifically to the term examined above, why is it that $$\lim_{t\to\infty}\left[B_c(x-\frac{b}{2},p_x)e^{i\alpha b}\right]_{b=-t}^{b=t}=0\quad?$$
Crudely, because for any x you may take a $$t\gg 2x$$, and all these phase space functions are taken to vanish at infinity in Wigner map physics applications.
Consider a phase-space Gaussian $$B(x,p)\propto \exp (-(x^2+p^2)/a)$$. The rapid oscillations of the phase factor $$e^{\pm i\alpha t}$$ are not terribly stable either: in QM they are routinely the mark of 0.