Memorandum on one-variable *-product
I wrote this article with the following intentions:
- To avoid the ordering problem inherent in operator formulations.
- To emphasize that nontrivial phenomena already appear at the level of deformations of function spaces.
- To avoid cumbersome notations such as normal ordering $:A:$.
- To present the theory in a form consistent with operator methods, but written in a more direct way.
- To describe nontrivial structures within a unified and consistent notation.
- To write the theory in a straightforward manner, without unnecessary formal complexity.
Quantum group and its representation
We begin by introducing $ U_\hbar(\mathfrak{sl}_2)$ and its representation on a polynomial algebra.
Let $\hbar$ be a formal variable and put $q=e^{\hbar}$. Define $U_\hbar(\mathfrak{sl}_2)$ to be the $\mathbb C[[\hbar]]$-algebra generated by $E,F,H$ with defining relations
$$
[H,E]=2E,\qquad [H,F]=-2F,\qquad [E,F]=\frac{e^{\hbar H}-e^{-\hbar H}}{e^{\hbar}-e^{-\hbar}}.
$$
Put$
K=e^{\hbar H}.
$
Equip $U_\hbar(\mathfrak{sl}_2)$ with Hopf algebra structure
$$
\Delta(H)=H\otimes 1+1\otimes H,\qquad \Delta(K)=K\otimes K,
$$
$$
\Delta(E)=K\otimes E+E\otimes 1,\qquad \Delta(F)=1\otimes F+F\otimes K^{-1},
$$
and counit
$$
\varepsilon(E)=\varepsilon(F)=\varepsilon(H)=0,\qquad \varepsilon(K)=1.
$$
Let$
A:=\mathbb C[[\hbar]][x,y].
$
Define
$$
\theta_x:=x\partial_x,\qquad \theta_y:=y\partial_y,\qquad [z]:=\frac{q^z-q^{-z}}{q-q^{-1}}.
$$
Define a representation$
\rho:U_\hbar(\mathfrak{sl}_2)\to \operatorname{End}_{\mathbb C[[\hbar]]}(A)
$by
$$
\rho(E)=\frac{x}{y}[\theta_y],\qquad \rho(F)=\frac{y}{x}[\theta_x],\qquad \rho(H)=\theta_x-\theta_y,\qquad \rho(K)=q^{\theta_x-\theta_y}.
$$
Hence, for $a,b\ge 0$,
$$
\rho(K)\cdot x^ay^b=q^{a-b}x^ay^b,
$$
$$
\rho(E)\cdot x^ay^b=[b]x^{a+1}y^{b-1},\qquad \rho(F)\cdot x^ay^b=[a]x^{a-1}y^{b+1}.
$$
Module-algebra structure
We put an algebra structure on the module which is compatible with the coproduct structure of the acting Hopf algebra.
For $\beta=(\beta_1,\beta_2),\gamma=(\gamma_1,\gamma_2)\in \mathbb Z_{\ge 0}^2$, write
$$
X^\beta=x^{\beta_1}y^{\beta_2},\qquad X^\gamma=x^{\gamma_1}y^{\gamma_2}.
$$
Let $g:\mathbb Z^2\times\mathbb Z^2\to \mathbb Z$ be bilinear. Define a product $*$ on $A$ by
$$
X^\beta * X^\gamma=q^{g(\beta,\gamma)}X^{\beta+\gamma},
$$
and extend it $\mathbb C[[\hbar]]$-bilinearly to all of $A$.
$(A,*)$ is an associative algebra.
For $\beta,\gamma,\delta\in \mathbb Z_{\ge 0}^2$,
$$
(X^\beta*X^\gamma)*X^\delta
=q^{g(\beta,\gamma)+g(\beta+\gamma,\delta)}X^{\beta+\gamma+\delta},
$$
while
$$
X^\beta*(X^\gamma*X^\delta)
=q^{g(\beta,\gamma+\delta)+g(\gamma,\delta)}X^{\beta+\gamma+\delta}.
$$
Since $g$ is bilinear,
$$
g(\beta,\gamma)+g(\beta+\gamma,\delta)=g(\beta,\gamma+\delta)+g(\gamma,\delta).
$$
Therefore
$$
(X^\beta*X^\gamma)*X^\delta=X^\beta*(X^\gamma*X^\delta).
$$
Let $H$ be a Hopf algebra and $A$ an algebra. Assume that $A$ is a left $H$-module via $\rho:H\to \operatorname{End}(A)$. Then $A$ is called an $H$-module-algebra if
$$
\rho(h)\cdot (f*g)=\sum \bigl(\rho(h_{(1)})\cdot f\bigr)*\bigl(\rho(h_{(2)})\cdot g\bigr)
$$
for all $h\in H$ and $f,g\in A$, where
$$
\Delta(h)=\sum h_{(1)}\otimes h_{(2)},
$$
and moreover
$$
\rho(h)\cdot 1_A=\varepsilon(h)1_A
$$
for all $h\in H$.
With the product $*$ and the representation $\rho$ defined above, $A=\mathbb C[[\hbar]][x,y]$ is a $U_\hbar(\mathfrak{sl}_2)$-module-algebra.
We consider$A$ as a polynomial ring over $\mathbb C[[\hbar]]$, and equip it with an additional product $*$.
Note that the usual product does not give a module-algebra structure.
click
contentIt is enough to verify the defining identity for the generators $K,E,F$.
First, since
$$ \Delta(K)=K\otimes K, $$
we must show
$$ \rho(K)\cdot(f*g)=\bigl(\rho(K)\cdot f\bigr)*\bigl(\rho(K)\cdot g\bigr). $$
For monomials $X^\beta,X^\gamma$,
$$ \rho(K)\cdot(X^\beta*X^\gamma) =\rho(K)\cdot\bigl(q^{g(\beta,\gamma)}X^{\beta+\gamma}\bigr) =q^{g(\beta,\gamma)}q^{(\alpha,\beta+\gamma)}X^{\beta+\gamma}, $$
where $\alpha=(1,-1)$. On the other hand,
$$ \bigl(\rho(K)\cdot X^\beta\bigr)*\bigl(\rho(K)\cdot X^\gamma\bigr) =q^{(\alpha,\beta)}X^\beta * q^{(\alpha,\gamma)}X^\gamma =q^{g(\beta,\gamma)}q^{(\alpha,\beta)+(\alpha,\gamma)}X^{\beta+\gamma}. $$
Since
$$ (\alpha,\beta+\gamma)=(\alpha,\beta)+(\alpha,\gamma), $$
the two sides are equal.
Next, since
$$ \Delta(E)=K\otimes E+E\otimes 1, $$
we must show
$$ \rho(E)\cdot(f*g)=\bigl(\rho(K)\cdot f\bigr)*\bigl(\rho(E)\cdot g\bigr)+\bigl(\rho(E)\cdot f\bigr)*g. $$
For monomials $X^\beta,X^\gamma$, using
$$ \rho(E)\cdot X^\beta=[\beta_2]X^{\beta+\alpha}, $$
we obtain exactly this identity from the definition of $*$.
Similarly, since
$$ \Delta(F)=1\otimes F+F\otimes K^{-1}, $$
we must show
$$ \rho(F)\cdot(f*g)=f*\bigl(\rho(F)\cdot g\bigr)+\bigl(\rho(F)\cdot f\bigr)*\bigl(\rho(K^{-1})\cdot g\bigr). $$
For monomials $X^\beta,X^\gamma$, using
$$ \rho(F)\cdot X^\beta=[\beta_1]X^{\beta-\alpha}, $$
this again follows from the definition of $*$.
Finally, the unit condition
$$ \rho(h)\cdot 1=\varepsilon(h)1 $$
holds for all $h\in U_\hbar(\mathfrak{sl}_2)$.
Therefore $(A,*,\rho)$ is a $U_\hbar(\mathfrak{sl}_2)$-module-algebra.
In $A$-type quantum groups, the coproduct is not unique. For instance, one has a one-parameter family
$$
\Delta(E_i)=K_i^{1-\mu}\otimes E_i+E_i\otimes K_i^{-\mu},\qquad
\Delta(F_i)=K_i^{\mu}\otimes F_i+F_i\otimes K_i^{\mu-1},\qquad
\Delta(K_i)=K_i\otimes K_i.
$$
Accordingly, the module-algebra product changes as
$$
P_\mu=\tilde P\, q^{(1-\mu)\theta_x\otimes\theta_y-\mu\theta_y\otimes\theta_x}.
$$
For $\mu=1,0,\tfrac12$, one obtains the normal ordering, anti-normal ordering, and Weyl ordering, which lead after a change of variables to the DO product formulas and the Moyal product.
In what follows, we do not focus on these differences. The essential point is the reduction to a one-variable product.
Extension of the function ring
We enlarge the algebra from $ A$ to an extended object $A'$, where additional expressions are allowed.
Recall$
A=\mathbb C[[\hbar]][x,y].
$ We now adjoin logarithmic and exponential-type expressions and formal function in $\hbar$ write the enlarged algebra informally as $
A'.
$
Here $A'$ is deliberately left ambiguous: it is an extension of $A$ large enough to contain the expressions used below, but we do not give a rigorous intrinsic definition of $A'$ at this stage. The product on this enlarged object will be denoted
$$
P:A'\otimes A'\to A'.
$$
On $A$ the product behaves regularly, whereas on $A'$ new and somewhat strange phenomena appear.
Define
$$
\varpi_{00}:=2\exp\!\left(\frac{2}{\hbar}\log x\,\log y\right),
\qquad
\bar\varpi_{00}:=2\exp\!\left(-\frac{2}{\hbar}\log x\,\log y\right).
$$
We call $\varpi_{00}$ the vacuum element and $\bar\varpi_{00}$ the conjugate vacuum element. These elements lie in $A'$, not in $A$.
$$
(\log y)*\varpi_{00}=\varpi_{00}*(\log x)=0,
$$
$$
(\log x)*\bar\varpi_{00}=\bar\varpi_{00}*(\log y)=0.
$$
The point of passing from $A$ to $A'$ is that certain natural elements are no longer polynomial in $x,y$ and $1/\hbar$ may appear(breaking the filter structure). The original product is well behaved on $A$, but on $A'$ one encounters additional phenomena.
To isolate the one-variable structure, set
$$
\zeta:=\log(xy)
$$
and let $A_1\subset A'$ be the subspace generated by $\zeta$.
The product $P$, obtained as the specialization of the original $x,y$-product to functions of $\zeta=\log(xy)$, satisfies
$$
P=\tilde P\,\exp\!\left(\hbar\,\partial_\zeta\otimes\partial_\zeta\right):A_1\otimes A_1\to A_1
$$
and we define
$$
f*g:=\tilde P\,\exp\!\left(\hbar\,\partial_\zeta\otimes\partial_\zeta\right)(f\otimes g).\qquad (f,g\in A_1)
$$
If $f,g\in A_1$ depend only on $\zeta$, then the above formula is the specialization of the original two-variable product
$$
P=\tilde P\,q^{\theta_x\otimes\theta_y}:A'\otimes A'\to A'.
$$
click
Since $q=e^{\hbar}$,
$$ P=\tilde P\,e^{\hbar\,\theta_x\otimes\theta_y}. $$
Also,
$$ \theta_x=x\partial_x=\partial_{\log x},\qquad \theta_y=y\partial_y=\partial_{\log y}. $$
For a function of $\zeta=\log x+\log y$ alone,
$$ \partial_{\log x}f(\zeta)=\partial_\zeta f(\zeta),\qquad \partial_{\log y}f(\zeta)=\partial_\zeta f(\zeta), $$
and similarly for $g(\zeta)$. Hence, on such elements,
$$ \theta_x\otimes\theta_y=\partial_\zeta\otimes\partial_\zeta. $$
Therefore the original product specializes to
$$ P=\tilde P\,\exp\!\left(\hbar\,\partial_\zeta\otimes\partial_\zeta\right):A_1\otimes A_1\to A_1. $$
This is exactly the stated one-variable deformation.
We next rewrite this one-variable product in a form suitable for calculation.
Define the invertible operator $
T:=\exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\right).
$Then
$$
P=T\,\tilde P\,(T^{-1}\otimes T^{-1}),
$$
$$
f*g=T\bigl((T^{-1}f)(T^{-1}g)\bigr).\qquad(f,g\in A_1)
$$
click
Since $\partial_\zeta$ is primitive,
$$ \Delta(\partial_\zeta)=\partial_\zeta\otimes 1+1\otimes\partial_\zeta. $$
Moreover, the ordinary product $\tilde P:f\otimes g\mapsto fg$ satisfies
$$ \partial_\zeta\,\tilde P=\tilde P\,\Delta(\partial_\zeta). $$
Hence the same relation holds for any function of $\partial_\zeta$, in particular for
$$ T=\exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\right), $$
so that
$$ T\tilde P=\tilde P\,\Delta(T). $$
Therefore
$$ P =T\,\tilde P\,(T^{-1}\otimes T^{-1}) =\tilde P\,\Delta(T)(T^{-1}\otimes 1)(1\otimes T^{-1}). $$
Now
$$ \Delta(\partial_\zeta)^2 =(\partial_\zeta\otimes 1+1\otimes\partial_\zeta)^2 =\partial_\zeta^2\otimes 1+2\partial_\zeta\otimes\partial_\zeta+1\otimes\partial_\zeta^2, $$
hence
$$ \Delta(T) =\exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\otimes 1+\hbar\,\partial_\zeta\otimes\partial_\zeta+\frac{\hbar}{2}1\otimes\partial_\zeta^2\right). $$
Multiplying by $T^{-1}\otimes 1$ and $1\otimes T^{-1}$, the terms
$\frac{\hbar}{2}\partial_\zeta^2\otimes 1$ and $\frac{\hbar}{2}1\otimes\partial_\zeta^2$ cancel, so we obtain
$$ P=\tilde P\,\exp\!\left(\hbar\,\partial_\zeta\otimes\partial_\zeta\right). $$
This proves the claim.
Deformation of special functions
In this section, we show that the $*$-product deforms classical special function theory. The key object is the gauge operator $
T=\exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\right),
$ which coincides with the Weierstrass transform (Gaussian transform), i.e. the heat kernel evolution operator. See for instance
https://en.wikipedia.org/wiki/Weierstrass_transform
and
https://mathlog.info/articles/2834
.
Accordingly, the $*$-product is written as
$$
f*g = T\bigl((T^{-1}f)\cdot (T^{-1}g)\bigr).
$$
Hence any $*$-function $F_*$ is induced from an ordinary function $F$ by
$$
F_*(z)=T\cdot F(T^{-1}\cdot z).
$$
In particular, functions such as
$$
\exp_*(\zeta),\quad \delta_*(\zeta),\quad \zeta_*^{-1}
$$
are obtained from classical functions by this transformation.
We first calculate the $*$-exponential. Since $\exp(a\zeta)$is eigenfunction of $\partial_\zeta$ with eigenvalue $a$,
$$
\exp_*(a\zeta)=T\cdot e^{aT^{-1}\cdot \zeta}=\exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\right)\cdot e^{a\zeta}
=\exp\!\left(a\zeta+\frac{\hbar}{2}a^2\right).
$$
Thus the ordinary exponential is deformed by the quadratic correction term $\frac{\hbar}{2}a^2$.
$$ \exp_*(a\zeta)=\exp\!\left(a\zeta+\frac{\hbar}{2}a^2\right). $$
We next turn to the $*$-delta. Since$
T=\exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\right)
$is the Weierstrass transform, it has the Gaussian kernel form
$$
Tf(\zeta)=\frac1{\sqrt{2\pi\hbar}}
\int_{-\infty}^{\infty}
f(u)\exp\!\left(-\frac{(\zeta-u)^2}{2\hbar}\right)\,du.
$$
Applying this to the ordinary Dirac delta at $0$ gives the following proposition.
$$ \delta_*(\zeta):=T(\delta)(\zeta)=\frac1{\sqrt{2\pi\hbar}} \exp\!\left(-\frac{\zeta^2}{2\hbar}\right). $$
Thus the $*$-delta is nothing but the Gaussian heat kernel, namely the image of the point mass under the Weierstrass transform.
Let
$$\operatorname{Daw}(z) := \frac{\sqrt{\pi}}{2} \exp(-z^2)\operatorname{erfi}(z)=e^{-x^2}\int_0^x e^{t^2}\,dt$$
be the Dawson function(with (+) manner).
For $\Re(\hbar)<0$,
$$
\zeta_{\pm}^{-1}=\int_0^{\pm\infty}\exp_*(-t\zeta)\,dt=
\zeta_*^{-1}\mp i\pi\,\delta_*(\zeta),
$$
where $\pm $depends on choice of branch and
$$
\zeta_*^{-1}
:=\frac{\sqrt2}{\sqrt{{\hbar}}}\,\operatorname{Daw}\!\left(\frac{\zeta}{\sqrt{2\hbar}}\right).
$$
proof1
Using the $*$-exponential, one writes a Laplace-type formula for the $*$-inverse of $\zeta$:
$$ \begin{aligned} \zeta_{\pm}^{-1} &=\int_0^{\pm\infty}\exp_*(-t\zeta)\,dt \\ &=\pm\int_0^\infty \exp\!\left(\mp t\zeta+\frac{\hbar}{2}t^2\right)\,dt \\ &=\pm\exp\!\left(-\frac{\zeta^2}{2\hbar}\right) \int_0^\infty \exp\!\left(\frac{\hbar}{2}\left(t\mp\frac{\zeta}{\hbar}\right)^2\right)\,dt \\ &= \pm\frac{\sqrt{\pi}}{\sqrt{-2\hbar}}\, \exp\!\left(-\frac{\zeta^2}{2\hbar}\right)\, \operatorname{erfc}\!\left(\pm\frac{\zeta}{\sqrt{-2\hbar}}\right) \\ &= \mp i\frac{\sqrt{\pi}}{\sqrt{2\hbar}}\, \exp\!\left(-\frac{\zeta^2}{2\hbar}\right)\, \operatorname{erfc}\!\left(\mp i\frac{\zeta}{\sqrt{2\hbar}}\right) \\ &= \mp i\frac{\sqrt{\pi}}{\sqrt{2\hbar}}\, \exp\!\left(-\frac{\zeta^2}{2\hbar}\right)\, \left(1\pm i\,\operatorname{erfi}\!\left(\frac{\zeta}{\sqrt{2\hbar}}\right)\right) \\ &= \mp i\frac{\sqrt{\pi}}{\sqrt{2\hbar}}\, \exp\!\left(-\frac{\zeta^2}{2\hbar}\right) +\frac{\sqrt{\pi}}{\sqrt{2\hbar}}\, \exp\!\left(-\frac{\zeta^2}{2\hbar}\right)\, \operatorname{erfi}\!\left(\frac{\zeta}{\sqrt{2\hbar}}\right) \\ &= \frac{\sqrt2}{\sqrt{\hbar}}\, \operatorname{Daw}\!\left(\frac{\zeta}{\sqrt{2\hbar}}\right) \mp i\pi\,\delta_*(\zeta). \end{aligned} $$
proof2
We can also obtain the same result by Borel summation.
Let
$$ F(\zeta):=\exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\right)\cdot \frac1\zeta . $$
Its formal expansion is
$$ F(\zeta)\sim \sum_{n=0}^\infty \frac{1}{n!}\left(\frac{\hbar}{2}\right)^n \partial_\zeta^{2n}\!\left(\frac1\zeta\right) =\frac1\zeta\sum_{n=0}^\infty \left(\frac12\right)_n\left(\frac{2\hbar}{\zeta^2}\right)^n . $$
Hence the Borel transform is
$$ \widehat\Phi(\xi)=(1-\xi)^{-1/2}. $$
So the lateral Borel sums along the contours $0\to (1\pm i\epsilon)\infty$ are
$$ \mathcal S_\pm F(\zeta) = \frac1\zeta\int_{0}^{(1\pm i\epsilon)\infty} e^{-t}\left(1-\frac{2\hbar}{\zeta^2}t\right)^{-1/2}dt. $$
Put
$$ u=\frac{2\hbar}{\zeta^2}t,\qquad a=\frac{\zeta^2}{2\hbar}. $$
Then
$$ \mathcal S_\pm F(\zeta) = \frac{\zeta}{2\hbar}\int_{0}^{1\pm i\epsilon\,\infty} e^{-au}(1-u)^{-1/2}du. $$
Now split at $u=1$. For $0< u<1$,
$$ \frac{\zeta}{2\hbar}\int_0^1 e^{-au}(1-u)^{-1/2}du = \frac{\sqrt{2}}{\sqrt{\hbar}}\, \operatorname{Daw}\!\left(\frac{\zeta}{\sqrt{2\hbar}}\right). $$
For $u>1$, along the two lateral contours,
$$ (1-u\pm i0)^{-1/2}=\mp i\,(u-1)^{-1/2}, $$
so
$$ \frac{\zeta}{2\hbar}\int_{1}^{(1\pm i\epsilon)\infty} e^{-au}(1-u)^{-1/2}du = \mp i\sqrt{\frac{\pi}{2\hbar}}\,e^{-a}. $$
Therefore
$$ \zeta_{\pm}^{-1}= \mathcal S_\pm\!\left(\exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\right)\cdot \frac1\zeta\right) = \frac{\sqrt{2}}{\sqrt{\hbar}}\, \operatorname{Daw}\!\left(\frac{\zeta}{\sqrt{2\hbar}}\right) \mp i\sqrt{\frac{\pi}{2\hbar}}\, e^{-\zeta^2/(2\hbar)} $$
So the principle value is
$$ \zeta_{*}^{-1}:= \exp\!\left(\frac{\hbar}{2}\partial_\zeta^2\right)\cdot \operatorname{p.v.}\frac1\zeta = \frac{\sqrt{2}}{\sqrt{\hbar}}\, \operatorname{Daw}\!\left(\frac{\zeta}{\sqrt{2\hbar}}\right) $$
We can rewrite *-delta function like Sato hyperfunction form:
$$ \zeta_+^{-1}-\zeta_-^{-1}=-2\pi i\delta_*(\zeta)$$
Finally, we explain why this produces the singular phenomenon. Since $\partial_\zeta^2\zeta=0$, one has
$$
T^{-1}\zeta=\zeta.
$$
Therefore
$$
\zeta*\delta_*(z)
=T(\zeta\delta(z))=0,
\qquad
\delta_* * \zeta
=T(\delta(z)\zeta)=0.
$$
So $\delta_*$ is annihilated by $\zeta$ under the $*$-product. As a consequence, once the function space is enlarged so that such singular objects are allowed, the inverse of $\zeta$ is no longer unique:
$$
\zeta_*^{-1}+a\,\delta_*(\zeta)
$$
is again a $*$-inverse for any $a\in\mathbb C$ and it is a linear combination of $\zeta_{\pm}^{-1}$. This is the precise point where the deformed special function theory shows both its beauty and its strange behavior: classical special functions survive in deformed form, but singular terms appear simultaneously and destroy naive uniqueness properties.
$$
\zeta*\delta_*=0,
\qquad
\delta_* * \zeta=0.
$$
Hence
$$
\zeta_*^{-1}+a\,\delta_*(\zeta)
$$
is again a $*$-inverse of $\zeta$ for every $a\in\mathbb{C}$.
The following shows that the extension introduces a failure of associativity once singular terms are allowed.
$(A_1,*)$ is not associative.
For $a,b\in\mathbb{C}$, one has
$$
(\zeta_*^{-1}+a\,\delta_*(\zeta))*\zeta*(\zeta_*^{-1}+b\,\delta_*(\zeta))
=
\zeta_*^{-1}+(a\ \text{or}\ b)\,\delta_*(\zeta),
$$
where the result depends on the bracketing.
The coexistence of $1/z$,$\delta(z)$, that is, an inverse and a nontrivial annihilator in the same algebra forces the breakdown of associativity.
To avoid confusion with the earlier notation $q=e^\hbar$ used for the quantum group, we introduce a new symbol$ Q:=e^{\hbar/2}. $Thus the theta-function parameter is $Q$, not the previous $q$.
Let$
Q=e^{\hbar/2},~~z=e^\zeta.
$Then
$$
\theta_3(z;Q)=\sum_{n\in\mathbb Z} Q^{n^2}z^n,
$$
$$
\theta_4(z;Q)=\sum_{n\in\mathbb Z} (-1)^n Q^{n^2}z^n,
$$
$$
\theta_2(z;Q)=\sum_{n\in\mathbb Z} Q^{(n+\frac12)^2}z^{n+\frac12},
$$
$$
\theta_1(z;Q)=\sum_{n\in\mathbb Z} (-1)^n Q^{(n+\frac12)^2}z^{n+\frac12}.
$$
We now express these functions in terms of the $*$-exponential.
$$
\theta_3(z;Q)=\sum_{n\in\mathbb Z}\exp_*(n\zeta),
$$
$$
\theta_4(z;Q)=\sum_{n\in\mathbb Z}(-1)^n\exp_*(n\zeta),
$$
$$
\theta_2(z;Q)=\sum_{n\in\mathbb Z}\exp_*\!\left(\left(n+\frac12\right)\zeta\right),
$$
$$
\theta_1(z;Q)=\sum_{n\in\mathbb Z}(-1)^n\exp_*\!\left(\left(n+\frac12\right)\zeta\right).
$$
This follows immediately from the definition of the $*$-exponential. Indeed,
$$
\exp_*(a\zeta)=\exp\!\left(a\zeta+\frac{\hbar}{2}a^2\right).
$$
Substituting $a=n$ gives
$$
\exp_*(n\zeta)=\exp(n\zeta)\exp\!\left(\frac{\hbar}{2}n^2\right)
=z^n Q^{n^2}.
$$
Hence
$$
\sum_{n\in\mathbb Z}\exp_*(n\zeta)=\sum_{n\in\mathbb Z}Q^{n^2}z^n=\theta_3(z;Q).
$$
Similarly, inserting the factor $(-1)^n$ gives $\theta_4$. Replacing $n$ by $n+\frac12$ gives
$$
\exp_*\!\left(\left(n+\frac12\right)\zeta\right)
=z^{n+\frac12}Q^{(n+\frac12)^2},
$$
which yields $\theta_2$, and adding the factor $(-1)^n$ gives $\theta_1$.
Thus each theta function is obtained as a lattice sum of $*$-exponentials, and the quadratic weight arises naturally from the deformation term in $\exp_*(a\zeta)$.
