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These notes are a great introduction to deformation quantization but I failed to check the validity of the statement p.9, right before (5.18).

Context: let $(\mathcal{A},+,\mu)$ be an algebra. $\mu:\mathcal{A}\times \mathcal{A} \to \mathcal{A}$ standing for multiplication. Deformation consists in considering a family (paramatrized by $\nu$ in a yet to be chosen space) of product on $\mathcal{A}[[\nu]]$ (formal power series with coefficients in $\mathcal{A}$) generically given by $$ \forall\ f,g\in \mathcal{A},\quad \mu_{\nu}(f,g) := \mu(f,g) + \sum_{k=1}^{+\infty} \nu^k \mu_k (f,g) \label{1}\tag{1}$$ i.e. by a family of bilinear maps $\mu_k:\mathcal{A}\times \mathcal{A} \to \mathcal{A}$ satisfying some conditions and extended to elements $F, G\in \mathcal{A}[[\nu]]$ of the form $F=\sum_{k=1}^{+\infty} \nu^k f_{k},\ f_k \in \mathcal{A}$ by $\mathbb{K}[[\nu]]$-bilinearity. (the idea behind formal power series, as far as I understand, is to ignore convergence issues but still have a structure where one can compare terms of the same degree in $\nu$").

Two of these star-product are equivalent if there exists an invertible algebra isomorphism (transition map) $T:(\mathcal{A}[[\nu]],+,\mu_{\nu}) \longrightarrow (\mathcal{A}[[\nu]],+,\rho_{\nu})$, i.e. a map such that $$ \forall\ F,G \in \mathcal{A}[[\nu]], \quad T\big(\mu_{\nu}(F,G)\big)= \rho_{\nu}(T(F),T(G))$$

Question: Let $\mathcal{A}=\mathcal{C}^{\infty}(\mathbb{R}^2)$ and denote $(a, \overline{a})$ or $(b, \overline{b}$) the variables of the functions. I want to check that the normal product ( (5.4) p.8; with the more usual notation for products) $$f \ast_N g := \sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \frac{\partial^k f}{\partial a^k} \frac{\partial^k g}{\partial \overline{a}^k} = f\, e^{\hbar \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}}}\, g \label{2}\tag{2}$$ is equivalent to the Moyal product ((5.15) p.9, one can consider $\hbar$ as the deformation parameter... although there is usually the factor as in (\ref{5})) $$ f \ast_M g := \sum_{k=0}^{+\infty} \left(\frac{\hbar}{2} \right)^k \frac{1}{k!} \left. \left( \frac{\partial }{\partial a} \frac{\partial }{\partial \overline{b}} - \frac{\partial }{\partial \overline{a}} \frac{\partial }{\partial b}\right)^k f\big(a, \overline{a}\big) g\big(b, \overline{b}\big) \right|_{\genfrac{}{}{0pt}{1}{a=b}{\overline{a}=\overline{b}}} = f\, e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)}\, g \label{3}\tag{3}$$ i.e. $$ T\big(f \ast_N g \big) = T(f) \ast_M T(g)\quad \text{with}\quad T = \genfrac{}{}{0pt}{0}{"}{}\!\!\exp\left(-\frac{\hbar}{2} \frac{\partial^2}{\partial a\, \partial \overline{a}} \right)\genfrac{}{}{0pt}{0}{"}{} \label{4}\tag{4}$$

Remarks:

  • In fact I already checked (\ref{4}) up to second order in $\hbar$ but it did not work at order 3 (although I'm not sure as the calculations were quite tedious...). It was not a priori clear that (\ref{4}) hold, one could have the other way round $ T\big(f \ast_M g \big) = T(f) \ast_N T(g)$ instead but this seems to fail at order 1. I only want to check the first few orders, but I would gladly take a proof for all order. I will soon write what I have done, but as I mentionned, it's tedious.
  • The Moyal product is first defined in the text (3.5-3.6) p.5 by $$ f \ast_M g := \sum_{k=0}^{+\infty} \frac{\nu^k}{k!} \underbrace{\left(\frac{\partial }{\partial q_1} \frac{\partial }{\partial p_{2}} - \frac{\partial }{\partial p_{1}} \frac{\partial }{\partial q_2} \right)^k f(q_1,p_1)\, g(q_2,p_2)}_{\mu_k(f,g)}\left.\vphantom{\frac{T}{T}}\right|_{\genfrac{}{}{0pt}{1}{q_1=q_2}{p_{1}=p_{2}}}\quad \text{with}\quad \nu = \frac{i\hbar}{2} \label{5}\tag{5}$$ and it does coincide with (\ref{3}) via (these are the correct $\sqrt{2}$ factors...) $$ \left\lbrace \begin{aligned} a & := \frac{1}{\sqrt{2}} \left(q + i\hspace{.5pt} p \right) \\ \overline{a} & := \frac{1}{\sqrt{2}} \left( q - i\hspace{.5pt} p \right) \end{aligned} \right. \enspace \Longrightarrow\quad \left\lbrace \begin{aligned} \frac{\partial}{\partial\hspace{.7pt} q} & = \frac{\partial\hspace{.7pt} a}{\partial\hspace{.7pt} q} \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial\hspace{.7pt} \overline{a}}{\partial \hspace{.7pt} q} \frac{\partial}{\partial\hspace{.7pt} \overline{a}} =\frac{1}{\sqrt{2}} \left( \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial}{\partial\hspace{.7pt} \overline{a}} \right) \\ \frac{\partial}{\partial\hspace{.7pt} p} & = \frac{\partial\hspace{.7pt} a}{\partial\hspace{.7pt} p} \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial\hspace{.7pt} \overline{a}}{\partial \hspace{.7pt} p} \frac{\partial}{\partial\hspace{.7pt} \overline{a}} = \frac{i}{\sqrt{2}} \left( \frac{\partial}{\partial\hspace{.7pt} a} - i\, \frac{\partial}{\partial\hspace{.7pt} \overline{a}}\right) \end{aligned} \right. $$
  • To make (\ref{3}) (same for (\ref{5})) more explicit, let me write the $k=2$ term: (notation $\displaystyle \partial_a=\frac{\partial}{\partial a},\ \partial_{ab}= \frac{\partial^2}{\partial a \partial b}$ etc.) $$\begin{split} \mu_2(f,g) &= \Big(\partial_{aa\overline{b}\overline{b}} - 2 \partial_{a\overline{a}b\overline{b}} + \partial_{\overline{a}\overline{a}bb} \Big) f(a,\overline{a})g(b,\overline{b})\left.\vphantom{\frac{T}{T}}\right|_{\genfrac{}{}{0pt}{1}{a=b}{\overline{a}=\overline{b}}}\\ &= (\partial_{aa}f)(\partial_{\overline{a}\overline{a}}g) - 2 (\partial_{a\overline{a}}f)(\partial_{b\overline{b}}g) + (\partial_{\overline{a}\overline{a}}f) (\partial_{aa}g) \end{split} \label{6}\tag{6}$$ One can also use the $\overleftarrow{\partial}$ or $\overrightarrow{\partial}$ notations or a tensorial notation.
Noix07
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2 Answers2

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Your reference has this explained in its ref [26], namely Zachos (2000) J Math Phys 41, 5129–5134, hep-th/9912238.

In any case, it is straightforward to prove your (4) through elementary Fourier analysis. That is, use test/sample functions $$ f=\exp (ma+n\bar a), ~~~~ g=\exp (ka+s\bar a), $$ so (4) presents as $$ \exp (\hbar ms -\hbar(m+k)(n+s)/2) ~ \overset{?}{=} ~ \exp (-\hbar mn/2-\hbar ks/2 +\hbar(ms-nk)/2), $$ indeed, an identity.

You might, or might not, appreciate the geometrical features associated with it.

For the mainstream review of all such moves, see this booklet.

Cosmas Zachos
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  • In had in fact downloaded this book a while a go... I should have time to read it during the new confinement!! I also stumbled upon the survey on deformation quantization by Martin Bordemann, also a great text which gave me the idea to test the equality for polynomial functions. But in fact exponentials will also do. – Noix07 Oct 29 '20 at 14:00
  • Yes, it is a theorem that one-parameter ($\hbar$) associative generalizations of Poisson Brackets lead to antisymmetrizations of equivalent star products. In the 1990s, people had not appreciated the theorem, and were "inventing" "new" star products by the dozens; and only providing the explicit T-dressing maps would slow them down, if that. It's so wonderful they've calmed down since... – Cosmas Zachos Oct 30 '20 at 16:06
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It is in fact possible to prove
$$\begin{split} T\big(f \ast_N g \big) = T(f) &\ast_M T(g)\quad \text{with}\quad T = \genfrac{}{}{0pt}{0}{"}{}\!\!\exp\left(-\frac{\hbar}{2} \frac{\partial^2}{\partial_{a}\, \partial_{\overline{a}} } \right)\genfrac{}{}{0pt}{0}{"}{} \\ \Longleftrightarrow \quad e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left(f\, e^{\hbar \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} }\, g \right) & = \left(e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} f\right) e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)} \left(e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} g\right) \end{split}\tag{4}$$ by brute force. The idea is to reckognize the simplification that occurs for the exponents of exponential functions suggested by Cosmas but in terms of series: $$\begin{split} \text{L.h.s.} &= e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left(f\, e^{\hbar m s }\, g \right) = e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left( e^{\big(ma + n\overline{a} \big) + \hbar m k + \big(ka + s\overline{a} \big)} \right)\\ & = e^{-\frac{\hbar}{2} (m+k)(n+s)} \left(f\, e^{\hbar m s }\, g \right)\\ &= \left( e^{-\frac{\hbar}{2} m(n+s)} f\right) e^{\hbar m s } \left(\,e^{-\frac{\hbar}{2} k(s+n)} g \right) \end{split} \label{4L}\tag{4L}$$

$$ \begin{split} \text{R.h.s.} &= \left(e^{-\frac{\hbar}{2} mn\ + \big(ma + n\overline{a}\big)} \right) e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)} \left(e^{-\frac{\hbar}{2} ks\ + \big(ka + s\overline{a}\big)}\right) \\ &= \left(e^{-\frac{\hbar}{2} mn\ + \big(ma + n\overline{a}\big)} \right) e^{\frac{\hbar}{2}\big( ms - nk \big)} \left(e^{-\frac{\hbar}{2} ks\ + \big(ka + s\overline{a}\big)}\right) \end{split} \label{4R}\tag{4R}$$ The simplification for the exponents to go from L.h.s. to R.h.s. $-\frac{\hbar}{2} ms + \hbar m s -\frac{\hbar}{2} kn = \frac{\hbar}{2} (ms-nk) $.

Remark: One could probably exploit this argument by decomposing a certain class of functions in terms of exponentials, probably with a Laplace transform in analogy with the case of complex exponentials: a tempered distribution can be written $u =\mathcal{F}^{-1}(\hat{u})=\genfrac{}{}{0pt}{0}{"}{}\!\!\frac{1}{(2\pi)^2}\int_{\mathbb{R}^2}\hat{u}(k,\overline{k})\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k}\, \genfrac{}{}{0pt}{0}{"}{}$.

Notice that Leibniz' rule can be written: $\partial^j_{a}(fg) =f \left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j g$ and similarly with two variables: $\partial^j_{a\overline{a}}(fg)=f \left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j g$. Hence $$\begin{split} \text{L.h.s.} &=\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2} \partial_{ a\, \overline{a}}\right)^j \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \partial^k_a f\, \partial^k_{\overline{a}} g\right)\\ &= \sum_{j=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^j \frac{1}{j!} f\left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right) g \\ &= f \left( \sum_{j=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^j \frac{1}{j!} \left(\sum_{0\leq i\leq j} {j\choose i} \overleftarrow{\partial}_a^{j-i} \overrightarrow{\partial}_a^i \right) \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right) \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j \right) g \\ &= f \left( \sum_{\genfrac{}{}{0pt}{1}{j,k=0}{i+l=j}}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!} \overleftarrow{\partial}_a^{i}\ \left(-\frac{\hbar}{2}\right)^l\!\! \frac{1}{l!} \overrightarrow{\partial}_a^l \left( \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right) \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^{i+l} \right) g \\ &=f \underbrace{\left( \sum_{i=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!} \overleftarrow{\partial}_a^{i} \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^i \right)}_{"e^{-\frac{\hbar}{2} m(n+s)}"} \underbrace{\left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)}_{"e^{\hbar ms}"} \underbrace{\left( \sum_{l=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^l\!\! \frac{1}{l!} \overrightarrow{\partial}_a^l \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^{l} \right)}_{"e^{-\frac{\hbar}{2} k(s+n)}"} g \end{split}$$ Now it suffices to check the equivalent of $e^{-\frac{\hbar}{2} m(n+s)} = e^{-\frac{\hbar}{2} mn} e^{-\frac{\hbar}{2} ms},\ e^{-\frac{\hbar}{2} ms} e^{\hbar ms}= e^{\frac{\hbar}{2} ms}$ and $e^{\frac{\hbar}{2} ms} e^{-\frac{\hbar}{2} nk}= e^{\frac{\hbar}{2} (ms-nk)}$. Since they are all similar, let us just write the first one: \begin{equation} \begin{split} & f \left( \sum_{i=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!} \overleftarrow{\partial}_a^{i} \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^i \right) g\\ &= f \left( \sum_{\genfrac{}{}{0pt}{1}{i=0}{0\leq h\leq i}}^{+\infty} \left(-\frac{\hbar}{2}\right)^{i-h+h} \! \overleftarrow{\partial}_a^{i-h+h} \, \frac{1}{(i-h)!} \overleftarrow{\partial}^{i-h}_{\overline{a}}\, \frac{1}{h!} \overrightarrow{\partial}_{\overline{a}}^h \right) g\\ &= f \left( \sum_{\genfrac{}{}{0pt}{1}{i=0}{h +l =i}}^{+\infty} \left(-\frac{\hbar}{2}\right)^{l} \! \frac{1}{l!} \overleftarrow{\partial}^{l}_{a\overline{a}}\ \left(-\frac{\hbar}{2}\right)^{h} \! \frac{1}{h!} \overleftrightarrow{\partial}^h_{\overline{a}} \right) g\\ &= f \left( \sum_{l=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^{l} \! \frac{1}{l!} \overleftarrow{\partial}^{l}_{a\overline{a}}\right) \left(\sum_{h=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^{h} \! \frac{1}{h!} \overleftarrow{\partial}^h_a \overrightarrow{\partial}_{\overline{a}}^h \right) g \end{split} \end{equation}

I also thought about the fact that a linear map from certain classes of functions of $\mathbb{R}^n$ (different versions, e.g. $\mathcal{C}^{\infty}_c(\mathbb{R}^n)$) which commutes with translations is a convolution operator (there is also a continuity condition, so one has to specify the topology on the space of functions, and this will constrain the convolution operator: convolution with what kind of function or distribution). Unfortunately, a quick computation shows that it would be a convolution with the Fourier transform of $e^{\frac{\hbar}{2}k^2}$ which does not make sense... $$ \begin{split} T(f) &= e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \hat{f}(k,\overline{k})\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k} = \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \hat{f}(k,\overline{k})\,e^{\frac{\hbar}{2} k\overline{k}}\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k} \\ &= \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \left(\int_ {\mathbb{R}^2} f(b,\overline{b})\,e^{-ikb -i\overline{k}\overline{b}} db\, d\overline{b}\right) e^{\frac{\hbar}{2} k\overline{k}}\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k}\\ &= \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} f(b,\overline{b}) \left(\int_ {\mathbb{R}^2} e^{\frac{\hbar}{2} k\overline{k}}\, e^{ik(a-b) +i\overline{k}(\overline{a}-\overline{b})}\, dk\, d\overline{k}\right) db\, d\overline{b} \end{split}$$ Writing things differently also leads to the same result: $$\begin{split} T(f) &= \mathcal{F}^{-1}\circ\mathcal{F}\left(\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2} \partial_{ a\, \overline{a}}\right)^j f\right) \\ &=\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2}\right)^j (ik)^j (i\overline{k})^j f \end{split}$$

Way out: $T$ seems to be unbounded but one can always imagine that its "inverse" (we didn't define its domain and target space...) could be convolution by a Gaussian. Another possibility would be to take a modified Fourier transform, $ \hat{f}(k,\overline{k}) := \int_ {\mathbb{R}^2} f(a,\overline{a})\, e^{- ika +i\overline{k}(\overline{a}-\overline{b})}\, da\, d\overline{a}$ or as remarked, take a Laplace transform. cf. also Inverse Weierstrass transform

Noix07
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