Groenewold's derivation of the star product (On the principles of elementary Quantum Mechanics) [closed]

Question

$\newcommand{\dd}{{\rm d}}$ In the paper "On the principles of elementary Quantum Mechanics" am trying to get from equation EQN 4.25 to EQN 4.27. I need help on exponential identities and integration by parts. Basically I need to get from part (9) to (10) of my calculations (Full details after the "--- ---"):

\begin{align} &=\frac{1}{h^2}\int \dd\eta \dd\xi \dd\sigma \dd\tau\,e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\\ &\qquad\times\left(a(\sigma ,\tau )e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \tau }\frac{\delta }{\delta \sigma }-\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \sigma }\right)\frac{\delta }{\delta \tau }\right)}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\right)\\ &\qquad\times\left(e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{\frac{1}{4}\left(-\frac{\delta (2 \hbar )}{i \delta \sigma }\frac{\partial }{\partial \tau }-\frac{\delta (-(2 \hbar ))}{i \delta \tau }\frac{\partial }{\partial \sigma }\right)}b(\sigma ,\tau )\right) \tag{9}\end{align}

Then somehow by partial integrations?

\begin{align} &=\frac{1}{h^2}\int \dd\eta \dd\xi \dd\sigma \dd\tau e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\\ &\qquad\times\left(a(\sigma ,\tau )e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \tau }\frac{\delta }{\delta \sigma }-\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \sigma }\right)\frac{\delta }{\delta \tau }\right)}b(\sigma ,\tau )\right) \end{align}

In summary, how did $$\left(a(\sigma ,\tau )e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \tau }\frac{\delta }{\delta \sigma }-\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \sigma }\right)\frac{\delta }{\delta \tau }\right)}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\right)\left(e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{\frac{1}{4}\left(-\frac{\delta (2 \hbar )}{i \delta \sigma }\frac{\partial }{\partial \tau }-\frac{\delta (-(2 \hbar ))}{i \delta \tau }\frac{\partial }{\partial \sigma }\right)}b(\sigma ,\tau )\right) $$ "become" $$e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\left(a(\sigma ,\tau )e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \tau }\frac{\delta }{\delta \sigma }-\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \sigma }\right)\frac{\delta }{\delta \tau }\right)}b(\sigma ,\tau )\right)?$$
It seems $e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}$ in (9) has "slipped" from the differential actions to arrive at (10)

Relevant Information :

1. $\frac{\delta}{ \delta \sigma }$, or $\frac{\delta}{ \delta \tau }$ acts to the right, for example :

$$f(\sigma)\frac{\delta}{ \delta \sigma }\sigma = (\frac{\partial }{\partial \sigma } f(\sigma) )* \sigma $$ 2. Also the operators $\overset{\rightharpoonup }{p}$ and $\overset{\rightharpoonup }{q}$ don't act on anything ; It does nothing on $\sigma$ , $\tau$ , $\xi$ , $\eta$ , a($\sigma$,$\tau$) or b($\sigma$,$\tau$).

"--- ---"

Full Details :

$$ \overset{\rightharpoonup }{a} \overset{\rightharpoonup }{b} =\frac{1}{h^4}\int\dd\eta \dd\xi \dd\sigma \dd\tau \dd\eta '\dd\xi '\dd\sigma '\dd\tau ' e^{\frac{i \left(\eta \xi '-\xi \eta '\right)}{2 \hbar }} e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }} \exp \left(-\frac{i \left(\eta ' \tau '+\eta \tau +\xi ' \sigma '+\xi \sigma \right)}{\hbar }\right) a\left(\frac{\sigma '}{2}+\sigma ,\tau -\frac{\tau '}{2}\right) b\left(\sigma -\frac{\sigma '}{2},\frac{\tau '}{2}+\tau \right)$$

$$= \frac{1}{h^4}\int\dd\eta \dd\xi \dd\sigma \dd\tau \dd\sigma '\dd\tau ' e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\int\dd\eta '\dd\xi ' \exp \left(-\frac{i \left(\eta ' \left(\frac{\xi }{2}+\tau '\right)+\xi ' \left(\sigma '-\frac{\eta }{2}\right)\right)}{\hbar }\right) a\left(\frac{\sigma '}{2}+\sigma ,\tau -\frac{\tau '}{2}\right) b\left(\sigma -\frac{\sigma '}{2},\frac{\tau '}{2}+\tau \right)\tag{1}$$

$$= \frac{1}{h^4}\int\dd\eta \dd\xi \dd\sigma \dd\tau \dd\sigma '\dd\tau ' e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\int h * h *d\sigma '\dd\tau ' \delta \left(\sigma '-\frac{\eta }{2}\right) \delta \left(\frac{\xi }{2}+\tau '\right) e^{-\frac{i \left(\eta ' \tau '+\xi ' \sigma '\right)}{\hbar }} a\left(\frac{\sigma '}{2}+\sigma ,\tau -\frac{\tau '}{2}\right) b\left(\sigma -\frac{\sigma '}{2},\frac{\tau '}{2}+\tau \right)\tag{2}$$

$$=\frac{1}{h^2}\int\dd\eta \dd\xi \dd\sigma \dd\tau e^{-\frac{i (\eta \tau +\xi \sigma )}{\hbar }} a\left(\frac{\eta }{4}+\sigma ,\tau -\frac{\xi }{4}\right) b\left(\sigma -\frac{\eta }{4},\frac{\xi }{4}+\tau \right) e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\tag{3}$$

By Taylor theorem :

$$=\frac{1}{h^2}\int\dd\eta \dd\xi \dd\sigma \dd\tau e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\left(a(\sigma ,\tau )+\frac{1}{4}\left(\eta \frac{\partial }{\partial \sigma }-\xi \frac{\partial }{\partial \tau }\right)a(\sigma ,\tau )+\text{...}\right)e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\left(b(\sigma ,\tau )-\frac{1}{4}\left(\eta \frac{\partial }{\partial \sigma }-\xi \frac{\partial }{\partial \tau }\right)b(\sigma ,\tau )+\text{...}\right)\tag{5} $$

Notice that $$e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\frac{1}{4}\left(\eta \frac{\partial }{\partial \sigma }\right)a(\sigma ,\tau )=e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }\right)a(\sigma ,\tau )$ where $\frac{\delta }{\delta \tau }$$ acts on the left.

$$ =\frac{1}{h^2}\int\dd\eta \dd\xi \dd\sigma \dd\tau e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\left(a(\sigma ,\tau )+\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)a(\sigma ,\tau )+\text{...}\right)e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\left(b(\sigma ,\tau )-\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)b(\sigma ,\tau )+\text{...}\right)\tag{6}$$

Another expression for Taylor expansion $$ = \frac{1}{h^2}\int\dd\eta \dd\xi \dd\sigma \dd\tau e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\left(e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)}a(\sigma ,\tau )\right)\left(e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{-\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)}b(\sigma ,\tau )\right) \tag{7}$$

Notice $$e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)}a(\sigma ,\tau )=a(\sigma ,\tau )e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \tau }\frac{\delta }{\delta \sigma }-\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \sigma }\right)\frac{\delta }{\delta \tau }\right)}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}$$

$$ = \frac{1}{h^2}\int\dd\eta \dd\xi \dd\sigma \dd\tau e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\left(a(\sigma ,\tau )e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \tau }\frac{\delta }{\delta \sigma }-\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \sigma }\right)\frac{\delta }{\delta \tau }\right)}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\right)\left(e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{-\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)}b(\sigma ,\tau )\right) \tag{8}$$

Notice that $$e^{-\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)}$ in $e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{-\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)}b(\sigma ,\tau )$ : $e^{-\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }\right)}=e^{-\frac{1}{4}\left(-\frac{\delta (-(2 \hbar ))}{i \delta \sigma }\frac{\partial }{\partial \tau }-\frac{2 \hbar }{i}\frac{\delta }{\delta \tau }\frac{\partial }{\partial \sigma }\right)}=e^{\frac{1}{4}\left(-\frac{\delta (2 \hbar )}{i \delta \sigma }\frac{\partial }{\partial \tau }-\frac{\delta (-(2 \hbar ))}{i \delta \tau }\frac{\partial }{\partial \sigma }\right)}$$

$$ =\frac{1}{h^2}\int\dd\eta \dd\xi \dd\sigma \dd\tau e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\left(a(\sigma ,\tau )e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \tau }\frac{\delta }{\delta \sigma }-\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \sigma }\right)\frac{\delta }{\delta \tau }\right)}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\right)\left(e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{\frac{1}{4}\left(-\frac{\delta (2 \hbar )}{i \delta \sigma }\frac{\partial }{\partial \tau }-\frac{\delta (-(2 \hbar ))}{i \delta \tau }\frac{\partial }{\partial \sigma }\right)}b(\sigma ,\tau )\right)\tag{9}$$

Then somehow by partial integrations? $$ =\frac{1}{h^2}\int\dd\eta \dd\xi \dd\sigma \dd\tau e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}e^{-\frac{i (\eta \tau +\xi \sigma )}{2 \hbar }}\left(a(\sigma ,\tau )e^{\frac{1}{4}\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \tau }\frac{\delta }{\delta \sigma }-\left(-\frac{2 \hbar }{i}\frac{\partial }{\partial \sigma }\right)\frac{\delta }{\delta \tau }\right)}b(\sigma ,\tau )\right)\tag{10} $$

Which by change of variables $(\xi \to x,\eta \to y,\sigma \to p,\tau \to q)$ is equivalent to EQN 4.27

$$\overset{\rightharpoonup }{a} \overset{\rightharpoonup }{b}=\frac{1}{h} \int dxdy e^{\frac{i \left(x \overset{\rightharpoonup }{p}+y \overset{\rightharpoonup }{q}\right)}{\hbar }} \frac{1}{h} \int dpdq e^{-\frac{i (p x+q y)}{\hbar }}\left(a(p,q)e^{\frac{\hbar }{2 i}\left(\frac{\delta }{\text{$\delta $p}}\frac{\partial }{\partial q}-\frac{\delta }{\text{$\delta $q}}\frac{\partial }{\partial p}\right)}b(p,q)\right)$$

Answer 1

One can see it directly, but Cosmas Zachos' suggestion "to working in multi-Fourier space so there are no derivatives present" is the most systematic & fool-proof approach. Here is a sketched derivation: $$\ldots e^{-\frac{i}{\hbar}(xp+yq)}\left( a(p,q) \exp\left(\frac{\hbar}{2i}\left(\frac{\stackrel{\leftarrow}{\partial}}{\partial p}\frac{\stackrel{\rightarrow}{\partial}}{\partial q}-\frac{\stackrel{\leftarrow}{\partial}}{\partial q}\frac{\stackrel{\rightarrow}{\partial}}{\partial p} \right)\right)b(p,q)\right) \tag{4.27} $$ $$~=~\ldots\left.e^{-\frac{i}{\hbar}(xp+yq)} \exp\left(\frac{\hbar}{2i}\left(\frac{\partial}{\partial p}\frac{\partial}{\partial q^{\prime}}- \frac{\partial}{\partial q}\frac{\partial}{\partial p^{\prime}} \right)\right) a(p,q)b(p^{\prime},q^{\prime})\right|^{p^{\prime}=p}_{q^{\prime}=q} $$ $$~=~\ldots e^{-\frac{i}{\hbar}(xp+yq)}\int \! \mathrm{d}p^{\prime}\mathrm{d}q^{\prime} \delta(p-p^{\prime}) \delta(q-q^{\prime}) \exp\left(\frac{\hbar}{2i}\left(\frac{\partial}{\partial p}\frac{\partial}{\partial q^{\prime}}- \frac{\partial}{\partial q}\frac{\partial}{\partial p^{\prime}} \right)\right) a(p,q)b(p^{\prime},q^{\prime}) $$ $$~=~\ldots e^{-\frac{i}{\hbar}(xp+yq)}\int \! \mathrm{d}p^{\prime}\mathrm{d}q^{\prime} \frac{\mathrm{d}u}{2\pi\hbar}\frac{\mathrm{d}v}{2\pi\hbar} e^{\frac{i}{\hbar}(u(p-p^{\prime})+v(q-q^{\prime}))} $$ $$\exp\left(\frac{\hbar}{2i}\left(\frac{\partial}{\partial p}\frac{\partial}{\partial q^{\prime}}- \frac{\partial}{\partial q}\frac{\partial}{\partial p^{\prime}} \right)\right) a(p,q)b(p^{\prime},q^{\prime}) $$ $$~\stackrel{\text{int. by parts}}{\sim}~\ldots e^{-\frac{i}{\hbar}(xp+yq)}\int \! \mathrm{d}p^{\prime}\mathrm{d}q^{\prime} \frac{\mathrm{d}u}{2\pi\hbar}\frac{\mathrm{d}v}{2\pi\hbar} e^{\frac{i}{\hbar}(u(p-p^{\prime})+v(q-q^{\prime}))} e^{\frac{i}{2\hbar}(xv-yu)} a(p,q)b(p^{\prime},q^{\prime}) $$ $$~=~\ldots e^{-\frac{i}{\hbar}(xp+yq)}\int \! \mathrm{d}p^{\prime}\mathrm{d}q^{\prime} \delta(p-\frac{y}{2}-p^{\prime}) \delta(q+\frac{x}{2}-q^{\prime}) a(p,q)b(p^{\prime},q^{\prime}) $$ $$~=~\ldots a(p,q)b(p-\frac{y}{2},q+\frac{x}{2}) $$ $$~\stackrel{\text{shifting int. var.}}{\sim}~ \ldots a(p+\frac{y}{4},q-\frac{x}{4})b(p-\frac{y}{4},q+\frac{x}{4}). \tag{4.25} $$