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 "--- ---"):
$
=\frac{1}{h^2}\int \int \int \int d\eta d\xi d\sigma d\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)
$ (9)
Then somehow by partial integrations?
$
=\frac{1}{h^2}\int \int \int \int d\eta d\xi d\sigma d\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)
$ (10)
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 \int \text{...}\int \int d\eta d\xi d\sigma d\tau d\eta ' d\xi ' d\sigma ' d\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)$
$= \frac{1}{h^4}\int \int \text{...}\int \int d\eta d\xi d\sigma d\tau d\sigma ' d\tau ' e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\int \int d\eta ' d\xi ' \exp \left(-\frac{i \left(\eta ' \left(\frac{\xi }{2}+\tau '\right)+\xi ' \left(\sigma '-\frac{\eta }{2}\right)\right)}{\hbar }\right) a\left(\frac{\sigma '}{2}+\sigma ,\tau -\frac{\tau '}{2}\right) b\left(\sigma -\frac{\sigma '}{2},\frac{\tau '}{2}+\tau \right)$ (1)
$= \frac{1}{h^4}\int \int \text{...}\int \int d\eta d\xi d\sigma d\tau d\sigma ' d\tau ' e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}\int \int h * h *d\sigma ' d\tau ' \delta \left(\sigma '-\frac{\eta }{2}\right) \delta \left(\frac{\xi }{2}+\tau '\right) e^{-\frac{i \left(\eta ' \tau '+\xi ' \sigma '\right)}{\hbar }} a\left(\frac{\sigma '}{2}+\sigma ,\tau -\frac{\tau '}{2}\right) b\left(\sigma -\frac{\sigma '}{2},\frac{\tau '}{2}+\tau \right)$ (2)
$=\frac{1}{h^2}\int \int \int \int d\eta d\xi d\sigma d\tau e^{-\frac{i (\eta \tau +\xi \sigma )}{\hbar }} a\left(\frac{\eta }{4}+\sigma ,\tau -\frac{\xi }{4}\right) b\left(\sigma -\frac{\eta }{4},\frac{\xi }{4}+\tau \right) e^{\frac{i \left(\xi \overset{\rightharpoonup }{p}+\eta \overset{\rightharpoonup }{q}\right)}{\hbar }}$ (3)
By Taylor theorem :
$=\frac{1}{h^2}\int \int \int \int d\eta d\xi d\sigma d\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)
$ (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 \int \int \int d\eta d\xi d\sigma d\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)
$ (6)
Another expression for Taylor expansion
$
= \frac{1}{h^2}\int \int \int \int d\eta d\xi d\sigma d\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)
$ (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 \int \int \int d\eta d\xi d\sigma d\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)
$ (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 \int \int \int d\eta d\xi d\sigma d\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)
$ (9)
Then somehow by partial integrations?
$
=\frac{1}{h^2}\int \int \int \int d\eta d\xi d\sigma d\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)
$ (10)
Which by change of variables $(\xi \to x,\eta \to y,\sigma \to p,\tau \to q)$ is equivalent to EQN 4.27
$\overset{\rightharpoonup }{a} \overset{\rightharpoonup }{b}=\frac{1}{h} \int \int dxdy e^{\frac{i \left(x \overset{\rightharpoonup }{p}+y \overset{\rightharpoonup }{q}\right)}{\hbar }} \frac{1}{h} \int \int dpdq e^{-\frac{i (p x+q y)}{\hbar }}\left(a(p,q)e^{\frac{\hbar }{2 i}\left(\frac{\delta }{\text{$\delta $p}}\frac{\partial }{\partial q}-\frac{\delta }{\text{$\delta $q}}\frac{\partial }{\partial p}\right)}b(p,q)\right)$
OK, if you find this derivation fast, who am I to stand in your way?
Clearing away notational clutter of superfluous components, let me take you from the penultimate line of Groenewold’s (4.25) to his (4.27), eschewing changes of notation. The conventions are Groenewold’s and yours, and of course right $\partial$s and left $\delta$s commute with each other.
His historic evaluation is, in essence, much simpler, $$ \int d\sigma d \tau ~e^{-\frac{i}{\hbar} (\xi \sigma +\eta \tau)} a(\sigma +\eta/4, \tau-\xi/4) b(\sigma-\eta/4,\tau +\xi/4)=\int d\sigma d \tau ~e^{-\frac{i}{\hbar} (\xi \sigma +\eta \tau)} \left ( a(\sigma , \tau) e^{\frac{\hbar}{2i}(\delta_\sigma \partial_\tau- \delta_\tau \partial_\sigma)} b(\sigma,\tau )\right ) . $$
I illustrate this simply with $a=e^{A\sigma+\alpha \tau}$ and $b=e^{B\sigma+\beta \tau}$ for you to appreciate the point, and then you may easily insert i’s into the constants and use these exponentials as Fourier modes, so the general expression will hold if it holds for each Fourier mode separately
(This is the essence of (13)-(16), (15), in our CTQMPS .)
The left-hand side is then $$ \int d\sigma d \tau ~e^{-\frac{i}{\hbar} (\xi \sigma +\eta \tau)} e^{A(\sigma +\eta/4)+\alpha (\tau-\xi/4) } e^{B(\sigma -\eta/4)+\alpha (\tau+\xi/4) } . $$
Now you may shift the (dummy) integration variables by $\sigma \to\sigma +\hbar(\beta-\alpha)/4i$ and $\tau\to \tau +\hbar (A-B)/4i$, so the above reduces to ∴ $$ \int d\sigma d \tau ~e^{-\frac{i}{\hbar} (\xi \sigma +\eta \tau) + (A\sigma+\alpha \tau) + \frac{\hbar}{2i}(A\beta-B\alpha) +(B\sigma+\beta \tau) }= \\ \int d\sigma d \tau ~e^{-\frac{i}{\hbar} (\xi \sigma +\eta \tau) } \left(e^{A(\sigma +\frac{\hbar}{2i}\partial_\tau )+\alpha (\tau -\frac{\hbar}{2i}\partial_\sigma)} e^{B\sigma+\beta \tau}\right )= \\ \int d\sigma d \tau ~e^{-\frac{i}{\hbar} (\xi \sigma +\eta \tau) } \left (e^{A\sigma +\alpha \tau } e^{\frac{\hbar}{2i}(\delta_\sigma \partial_\tau -\delta_\tau\partial_\sigma)} e^{B\sigma+\beta \tau}\right )~, $$ as required.