Groenewold derivation of Moyal Bracket (On the principles of elementary Quantum Mechanics)

Question

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 $q}}\frac{\partial }{\partial p}\right)}b(p,q)\right)$

Answer 1

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.

• The integration by parts of Groenewold amounts to the shift of dummy integration variables here, as, in general, by Lagrange’s translation operator, $$ \int dz ~f(z) g(z+a) = \int dz~ f(z) e^{a\partial_z}g(z+a) = \int dz ~f(z)e^{-a\delta_z} g(z) = \int dz~ f(z-a) g(z)~. $$