nLab
star product

Context

Algebra

Geometric quantization

Contents

Idea

In deformation quantization of Poisson manifolds the commutative product \cdot of the commutative algebra of functions is replaced by a noncommutative associative product. This is often called a star product and denoted “\star”.

An archetypal example is the Moyal star product (example 2 below) that deforms the function algebra on a Poisson vector space, and often “star product” is by default understood to be a Moyal star product. Indeed every star product induced from a constant rank-2 tensor on a vector space is isomorphic to a Moyal star product (prop. 2 below).

More recently also nonassociative “star products” have been proposed to be of interest.

Definition

On finite-dimensional vector spaces

Let VV be a finite dimensional vector space and let πVV\pi \in V \otimes V be an element of the tensor product (not necessarily skew symmetric at the moment).

We may canonically regard VV as a smooth manifold, in which case π\pi is canonically regarded as a constant rank-2 tensor. As such it has a canonical action by forming derivatives on the tensor product of the space of smooth functions:

π:C (V)C (V)C (V)C (V). \pi \;\colon\; C^\infty(V) \otimes C^\infty(V) \longrightarrow C^\infty(V) \otimes C^\infty(V) \,.

If { i}\{\partial_i\} is a linear basis for VV, identified, as before, with a basis for Γ(TV)\Gamma(T V), then in this basis this operation reads

π(fg)=π ij( if)( jg), \pi(f \otimes g) \;=\; \pi^{i j} (\partial_i f) \otimes (\partial_j g) \,,

where iffx i\partial_i f \coloneqq \frac{\partial f}{\partial x^i} denotes the partial derivative of the smooth function ff along the iith coordinate, and where we use the Einstein summation convention.

For emphasis we write

C (V)C (V) prod C (V) fg fg \array{ C^\infty(V) \otimes C^\infty(V) &\overset{prod}{\longrightarrow}& C^\infty(V) \\ f \otimes g &\mapsto& f \cdot g }

for the pointwise product of smooth functions.

Definition

(star product induced by constant rank-2 tensor)

Given (V,π)(V,\pi) as above, then the star product induced by π\pi on the formal power series algebra C (V)[[]]C^\infty(V) [ [\hbar] ] in a formal variable \hbar (“Planck's constant”) with coefficients in the smooth functions on VV is the linear map

() π():C (V)[[]]C (V)[[]]C (V)[[]] (-) \star_\pi (-) \;\colon\; C^\infty(V)[ [ \hbar ] ] \otimes C^\infty(V)[ [ \hbar ] ] \longrightarrow C^\infty(V)[ [\hbar] ]

given by

() π()prodexp(π ijx ix j) (-) \star_\pi (-) \;\coloneqq\; prod \circ \exp\left( \hbar \pi^{i j} \frac{\partial}{\partial x^i} \otimes \frac{\partial}{\partial x^j} \right)

Hence

f πg1+π ijfx igx j+ 212π ijπ kl 2fx ix k 2gx jx l+. f \star_\pi g \;\coloneqq\; 1 + \hbar \pi^{i j} \frac{\partial f}{\partial x^i} \cdot \frac{\partial g}{\partial x^j} + \hbar^2 \tfrac{ 1 }{2} \pi^{i j} \pi^{k l} \frac{\partial^2 f}{\partial x^{i} \partial x^{k}} \cdot \frac{\partial^2 g}{\partial x^{j} \partial x^{l}} + \cdots \,.
Proposition

(star product is associative and unital)

Given (V,π)(V,\pi) as above, then the star product () π()(-) \star_\pi (-) from def. 1 is associative and unital with unit the constant function 1C (V)C (V)[[]]1 \in C^\infty(V) \hookrightarrow C^\infty(V)[ [ \hbar ] ].

Hence the vector space C (V)C^\infty(V) equipped with the star product π\pi is a unital associative algebra.

Proof

Observe that the product rule of differentiation says that

iprod=prod( iid+id i). \partial_i \circ prod = prod \circ ( \partial_i \otimes id \;+\; id \otimes \partial_i ) \,.

Using this we compute as follows:

(f πg) πh =prodexp(π ij i j)((prodexp(π kl k l))id)(fgg) =prodexp(π ij i j)(prodid)(exp(π kl k l)id)(fgg) =prod(prodid)exp(π ij( iid j+id i j)exp(π kl k l)id(fgg) =prod(prodid)exp(π ij iid j)exp(π ijid i j)exp(π kl k lid)(fgg) =prod 3exp(π ij( i jid+ iid j+id i j)) \begin{aligned} (f \star_\pi g) \star_\pi h & = prod \circ \exp( \pi^{i j} \partial_i \otimes \partial_j ) \circ \left( \left( prod \circ \exp( \pi^{k l} \partial_k \otimes \partial_l ) \right) \otimes id \right) (f \otimes g \otimes g) \\ & = prod \circ \exp( \pi^{i j} \partial_i \otimes \partial_j ) \circ (prod \otimes id) \circ \left( \exp( \pi^{k l} \partial_k \otimes \partial_l ) \otimes id \right) (f \otimes g \otimes g) \\ & = prod \circ (prod \otimes id) \circ \exp( \pi^{i j} ( \partial_i \otimes id \otimes \partial_j +id \otimes \partial_i \otimes \partial_j ) \circ \exp( \pi^{k l} \partial_k \otimes \partial_l ) \otimes id (f \otimes g \otimes g) \\ & = prod \circ (prod \otimes id) \circ \exp( \pi^{i j} \partial_i \otimes id \otimes \partial_j ) \circ \exp( \pi^{i j} id \otimes \partial_i \otimes \partial_j ) \circ \exp( \pi^{k l} \partial_k \otimes \partial_l \otimes id ) (f \otimes g \otimes g) \\ & = prod_3 \circ \exp( \pi^{i j} ( \partial_i \otimes \partial_j \otimes id + \partial_i \otimes id \otimes \partial_j + id \otimes \partial_i \otimes \partial_j) ) \end{aligned}

In the last line we used that the ordinary pointwise product of functions is associative, and wrote prod 3:C (V)C (V)C (V)C (V)prod_3 \colon C^\infty(V) \otimes C^\infty(V) \otimes C^\infty(V) \to C^\infty(V) for the unique pointwise product of three functions.

The last expression above is manifestly independent of the choice of order of the arguments in the triple star product, and hence it is clear that an analogous computation yields

=f π(g πh). \cdots = f \star_\pi (g \star_\pi h) \,.

On polynomial observables in field theory

Definition

(star products on regular polynomial observables induced from propagators)

Let (E,L)(E,\mathbf{L}) be a free Lagrangian field theory with field bundle EfbΣE \overset{fb}{\to} \Sigma, and let ΔΓ Σ((EE) *)\Delta \in \Gamma'_\Sigma((E \boxtimes E)^\ast) be a distribution of two variables on field histories.

On the off-shell regular polynomial observables with a formal paramater \hbar adjoined consider the bilinear map

PolyObs(E) reg[[]]PolyObs(E) reg[[]] ΔPolyObs(E) reg[[]] PolyObs(E)_{reg}[ [\hbar] ] \otimes PolyObs(E)_{reg} [ [ \hbar ] ] \overset{\star_{\Delta}}{\longrightarrow} PolyObs(E)_{reg}[ [\hbar] ]

given as in def. 1, but with partial derivatives replaced by functional derivatives

A 1 ΔA 2(()())exp( ΣΔ ab(x,y)δδΦ a(x)δδΦ b(y))(A 1A 2) A_1 \star_{\Delta} A_2 \;\coloneqq\; ((-)\cdot(-)) \circ \exp\left( \int_\Sigma \Delta^{a b}(x,y) \frac{\delta}{\delta \mathbf{\Phi}^a(x)} \otimes \frac{\delta}{\delta \mathbf{\Phi}^b(y)} \right) (A_1 \otimes A_2)

As in prop. 1 this defines a unital and associative algebra structure.

If the Euler-Lagrange equations of motion induced by the Lagrangian density L\mathbf{L} are Green hyperbolic differential equations and if Δ\Delta is a propagator for these differential equations, then this star product algebra descends to the on-shell regular polynomial observables

PolyObs(E,L) reg[[]]PolyObs(E,L) reg[[]] ΔPolyObs(E,L) reg[[]]. PolyObs(E,\mathbf{L})_{reg}[ [\hbar] ] \otimes PolyObs(E, \mathbf{L})_{reg} [ [ \hbar ] ] \overset{\star_{\Delta}}{\longrightarrow} PolyObs(E, \mathbf{L})_{reg}[ [\hbar] ] \,.

(Dito 90, Dütsch-Fredenhagen 00 Dütsch-Fredenhagen 01, Hirshfeld-Henselder 02)

Properties

Equivalences of star products

Proposition

(shift by symmetric contribution is isomorphism of star products)

Let VV be a vector space, πVV\pi \in V \otimes V a rank-2 tensor and αSym(VV)\alpha \in Sym(V \otimes V) a symmetric rank-2 tensor.

Then the linear map

C (V) exp(12α) C (V) f exp(12α ij i j)f \array{ C^\infty(V) &\overset{\exp\left(\tfrac{1}{2}\alpha \right)}{\longrightarrow}& C^\infty(V) \\ f &\mapsto& \exp\left( \tfrac{1}{2}\hbar \alpha^{i j} \partial_i \partial_j \right) f }

constitutes an isomorphism of star product algebras (prop. 1) of the form

exp(12α):(C (V)[[]], π)(C (V))[[]], π+α), \exp\left(\hbar\tfrac{1}{2}\hbar\alpha \right) \;\colon\; (C^\infty(V)[ [\hbar] ], \star_{\pi}) \overset{\simeq}{\longrightarrow} (C^\infty(V))[ [\hbar] ], \star_{\pi + \alpha}) \,,

hence identifying the star product induced from π\pi with that induced from π+α\pi + \alpha.

In particular every star product algebra (C (V)[[]], π)(C^\infty(V)[ [\hbar] ],\star_\pi) is isomorphic to a Moyal star product algebra 12π\star_{\tfrac{1}{2}\pi} (example 2) with 12π skew ij=12(π ijπ ji)\tfrac{1}{2}\pi_{skew}^{i j} = \tfrac{1}{2}(\pi^{i j} - \pi^{j i}) the skew-symmetric part of π\pi, this isomorphism being exhibited by the symmetric part 2α ij=12(π ij+π ji)2\alpha^{i j} = \tfrac{1}{2}(\pi^{i j} + \pi^{j i}).

Proof

We need to show that

C (V)[[]]C (V)[[]] exp(12α)exp(12α) C (V)[[]]C (V)[[]] π π+α C (V)[[]] exp(12α) C (V)[[]] \array{ C^\infty(V)[ [\hbar] ] \otimes C^\infty(V)[ [\hbar] ] & \overset{ \exp\left( \tfrac{1}{2}\hbar \alpha \right) \otimes \exp\left( \tfrac{1}{2}\hbar \alpha \right) }{\longrightarrow}& C^\infty(V)[ [\hbar] ] \otimes C^\infty(V)[ [\hbar] ] \\ {}^{\mathllap{\star_{\pi}}}\downarrow && \downarrow^{\mathrlap{\star_{\pi + \alpha}}} \\ C^\infty(V)[ [\hbar] ] &\underset{\exp\left( \tfrac{1}{2} \alpha \right) }{\longrightarrow}& C^\infty(V)[ [\hbar] ] }

hence that

prodexp((π+α))(exp(12α)exp(12α))=exp(12α)prodexp(π). prod \circ \exp( \hbar(\pi + \alpha) ) \circ \left( \exp\left( \tfrac{1}{2}\alpha\right) \otimes \exp\left( \tfrac{1}{2}\alpha \right) \right) \;=\; \exp\left( \tfrac{1}{2}\alpha \right) \circ prod \circ \exp( \pi ) \,.

To this end, observe that the product rule of differentiation applied twice in a row implies that

i jprod=prod(( i j)id+id( i j)+ i j+ j i). \partial_i \partial_j \circ prod \;=\; prod \circ \left( (\partial_i \partial_j) \otimes id + id \otimes (\partial_i \partial_j) + \partial_i \otimes \partial_j + \partial_j \otimes \partial_i \right) \,.

Using this we compute

exp(12α ij i j)prodexp(π ij i j) =prodexp(12α ij(( i j)id+id( i j)+ i j+ j i))exp(π ij k l) =prodexp((π ij+α ij) i j)exp(12α ij( i j)id12α ijid( i j)) =prodexp((π ij+α ij) i j)(exp(12α)exp(12α)) \begin{aligned} \exp\left( \hbar\tfrac{1}{2}\alpha^{i j} \partial_i \partial_j \right) \circ prod \circ \exp( \hbar \pi^{i j} \partial_{i} \otimes \partial_j ) & = prod \circ \exp\left( \hbar \tfrac{1}{2}\alpha^{i j} \left( (\partial_i \partial_j) \otimes id + id \otimes (\partial_i \partial_j) + \partial_i \otimes \partial_j + \partial_j \otimes \partial_i \right) \right) \circ \exp( \hbar \pi^{i j} \partial_{k} \otimes \partial_l ) \\ & = prod \circ \exp\left( \hbar (\pi^{i j} + \alpha^{i j}) \partial_i \otimes \partial_j \right) \circ \exp\left( \hbar \tfrac{1}{2} \alpha^{i j} (\partial_i \partial_j) \otimes id \hbar \tfrac{1}{2} \alpha^{i j} id \otimes (\partial_i \partial_j) \right) \\ & = prod \circ \exp\left( \hbar (\pi^{i j} + \alpha^{i j}) \partial_i \otimes \partial_j \right) \circ \left( \exp\left( \tfrac{1}{2} \hbar \alpha \right) \otimes \exp\left( \tfrac{1}{2} \hbar \alpha \right) \right) \end{aligned}

Integral representations

Proposition

(integral representation of star product)

If π\pi skew-symmetric and invertible, in that there exists ωV *V *\omega \in V^\ast \otimes V^\ast with π ijω jk=δ k i\pi^{i j}\omega_{j k} = \delta^i_k, and if the functions f,gf,g admit Fourier analysis (are functions with rapidly decreasing partial derivatives), then their star product (def. 1) is equivalently given by the following integral expression:

(f πg)(x) =(det(ω) 2n)(2π) 2ne 1iω((xy˜),(xy))f(y)g(y˜)d 2nyd 2ny˜ \begin{aligned} \left(f \star_\pi g\right)(x) &= \frac{(det(\omega)^{2n})}{(2 \pi \hbar)^{2n} } \int e^{ \tfrac{1}{i \hbar} \omega((x - \tilde y),(x-y))} f(y) g(\tilde y) \, d^{2 n} y \, d^{2 n} \tilde y \end{aligned}

(Baker 58)

Proof

We compute as follows:

(f πg)(x) prodexp(π ijx ix j)(f,g) =1(2π) 2n1(2π) 2ne iπ(k,q)e ik(xy)f(y)e iq(xy˜)g(y˜)d 2nkd 2nqd 2nyd 2ny˜ =1(2π) 2nδ(xy˜+πk)e ik(xy)f(y)g(y˜)d 2nkd 2nyd 2ny˜ =1(2π) 2nδ(xy˜+z)e iω(z,(xy))f(y)g(y˜)d 2nzd 2nyd 2ny˜ =(det(π) 2n)(2π) 2ne 1iω((xy˜),(xy))f(y)g(y˜)d 2nyd 2ny˜ \begin{aligned} \left(f \star_\pi g\right)(x) & \coloneqq prod \circ \exp\left( \hbar \pi^{i j} \frac{\partial}{\partial x^i} \otimes \frac{\partial}{\partial x^j} \right)(f, g) \\ & = \frac{1}{(2 \pi)^{2n}} \frac{1}{(2 \pi)^{2n}} \int \int \underbrace{ e^{ i \hbar \pi(k,q) } } \underbrace{ e^{i k \cdot (x-y)} f(y) } \underbrace{ e^{i q \cdot (x- \tilde y)} g(\tilde y) } \, d^{2 n} k \, d^{2 n} q \, d^{2 n} y \, d^{2 n} \tilde y \\ & = \frac{1}{(2 \pi)^{2n}} \int \delta\left( x - \tilde y + \hbar \pi \cdot k \right) e^{i k \cdot (x-y)} f(y) g(\tilde y) \, d^{2 n} k \, d^{2 n} y \, d^{2 n} \tilde y \\ & = \frac{1}{(2 \pi)^{2n}} \int \delta\left( x - \tilde y + z \right) e^{ \tfrac{i}{\hbar} \omega(z, (x-y))} f(y) g(\tilde y) \, d^{2 n} z \, d^{2 n} y \, d^{2 n} \tilde y \\ & = \frac{(det(\pi)^{2n})}{(2 \pi \hbar)^{2n} } \int e^{\tfrac{1}{i \hbar}\omega((x - \tilde y),(x-y))} f(y) g(\tilde y) \, d^{2 n} y \, d^{2 n} \tilde y \end{aligned}

Here in the first step we expressed ff and gg both by their Fourier transform (inserting the Fourier expression of the delta distribution from this example) and used that under this transformation the partial derivative π abϕ aϕ b\pi^{a b} \frac{\partial}{\partial\phi^a}{\frac{\partial}{\phi^b}} turns into the product with iπ ijk ik ji \pi^{i j} k_i k_j (this prop.). Then we identified again the Fourier-expansion of a delta distribution and finally we applied the change of integration variables k=1ωzk = \tfrac{1}{\hbar}\omega \cdot z and then evaluated the delta distribution.

Next we express this as the groupoid convolution product of polarized sections of the symplectic groupoid. To this end, we first need the following definnition:

Definition

(symplectic groupoid of symplectic vector space)

Assume that π\pi is the inverse of a symplectic form ω\omega on 2n\mathbb{R}^{2n}. Then the Cartesian product

2n× 2n pr 1 pr 2 2n 2n \array{ && \mathbb{R}^{2n} \times \mathbb{R}^{2n} \\ & {}^{\mathllap{pr_1}}\swarrow && \searrow^{\mathrlap{pr_2}} \\ \mathbb{R}^{2n} && && \mathbb{R}^{2n} }

inherits the symplectic structure

Ω(pr 1 *ωpr 2 *ω) \Omega \;\coloneqq\; \left( pr_1^\ast \omega - pr_2^\ast \omega \right)

given by

Ω =ω ijdx idx jω ijdy idy j =ω ij(dx idy i)(dx j+dy j). \begin{aligned} \Omega & = \omega_{i j} d x^i \wedge d x^j - \omega_{i j} d y^i \wedge d y^j \\ & = \omega_{i j} ( d x^i - d y^i ) \wedge ( d x^j + d y^j ) \end{aligned} \,.

The pair groupoid on 2n\mathbb{R}^{2n} equipped with this symplectic form on its space of morphisms is a symplectic groupoid.

A choice of potential form Θ\Theta for Ω\Omega, hence with Ω=dΘ\Omega = d \Theta, is given by

Θω ij(x i+y i)d(x jy j)) \Theta \coloneqq -\omega_{i j} ( x^i + y^i ) d (x^j - y^j) )

Choosing the real polarization spanned by x i y i\partial_{x^i} - \partial_{y^i} a polarized section is function F=F(x,y)F = F(x,y) such that

ι x j y j(dF1i14ΘF)=0 \iota_{\partial_{x^j} - \partial_{y^j}}(d F - \tfrac{1}{i \hbar} \tfrac{1}{4} \Theta F) = 0

hence

(1)F(x,y)=f(x+y2)e 1iω(xy2,x+y2). F(x,y) = f\left( \tfrac{x + y}{2} \right) e^{ \tfrac{1}{i \hbar} \omega\left( \tfrac{x - y}{2} , \tfrac{x + y}{2} \right)} \,.
Proposition

(polarized symplectic groupoid convolution product of symplectic vector space is given by Moyal star product)

Given a symplectic vector space ( 2n,ω)(\mathbb{R}^{2n}, \omega), then the groupoid convolution product on polarized sections (1) on its symplectic groupoid (def. 3), given by convolution product followed by averaging (integration) over the polarization fiber, is given by the star product (def. 1) for the corresponding Poisson tensor πω 1\pi \coloneqq \omega^{-1}, in that

F(x,t)G(t,y)d 2ntd 2n(xy) =(f πg)((x+y)/2). \begin{aligned} \int \int F(x,t) G(t,y) \, d^{2n} t \, d^{2n} (x-y) & = (f \star_\pi g)((x+y)/2) \end{aligned} \,.

(Weinstein 91, p. 446, Garcia-Bondia & Varilly 94, section V)

Proof

We compute as follows:

F(x,t)G(t,y)d 2ntd 2n(xy) f((x+t)/2)g((t+y)/2)e 1i14ω(xt,x+t)+1i14ω(ty,t+y)d 2ntd 2n(xy) =f(t/2)g((t(xy))/2)e 1i14ω((x+y)+(xy)t,t)+1i14ω(t(x+y),t(xy))d 2ntd 2n(xy) =f(t/2)g(t˜/2)e 1i14ω((x+y)t˜,t)1i14ω((x+y)t,t˜)d 2ntd 2nt˜ =f(t)g(t˜)e 1i14ω((x+y)2t˜,2t)1ii14ω((x+y)2t,2t˜)d 2ntd 2nt˜ =f(t)g(t˜)e 1iω(12(x+y)t˜,12(x+y)t)d 2ntd 2nt˜ =(f ωg)((x+y)/2) \begin{aligned} \int \int F(x,t) G(t,y) \, d^{2n} t \, d^{2n} (x-y) & \coloneqq \int \int f((x + t)/2) g( (t + y)/2 ) e^{ \tfrac{1}{i \hbar} \tfrac{1}{4} \omega( x-t, x+t ) + \tfrac{1}{i \hbar} \tfrac{1}{4} \omega(t-y, t + y) } \, d^{2n} t \, d^{2n} (x-y) \\ & = \int \int f(t/2) g( (t - (x - y))/2 ) e^{ \tfrac{1}{i \hbar} \tfrac{1}{4} \omega( (x+y) + (x - y) - t, t ) + \tfrac{1}{i \hbar} \tfrac{1}{4} \omega(t-(x+y), t - (x-y)) } \, d^{2n} t \, d^{2n} (x-y) \\ & = \int \int f(t/2) g( \tilde t / 2) e^{ \tfrac{1}{i \hbar} \tfrac{1}{4} \omega( (x+y) - \tilde t, t ) - \tfrac{1}{i \hbar} \tfrac{1}{4} \omega((x+y)-t, \tilde t) } \, d^{2n} t \, d^{2n} \tilde t \\ & = \int \int f(t) g( \tilde t ) e^{ \tfrac{1}{i \hbar} \tfrac{1}{4} \omega( (x+y) - 2 \tilde t, 2 t ) - \tfrac{1}{ii \hbar} \tfrac{1}{4} \omega((x+y)- 2 t, 2 \tilde t) } \, d^{2n} t \, d^{2n} \tilde t \\ \\ & = \int \int f(t) g(\tilde t ) e^{ \tfrac{1}{i \hbar} \omega\left( \tfrac{1}{2}(x+y) - \tilde t, \tfrac{1}{2}(x + y) - t \right)} \, d^{2n} t \, d^{2n} \tilde t \\ & = (f \star_\omega g)((x+y)/2) \end{aligned}

The first line just unwinds the definition of polarized sections from def. 3, the following lines each implement a change of integration variables and fnally in the last line we used prop. 3.

Examples

General examples

Some examples of star products as in def. 1:

Example

If π=0\pi = 0 in def. 1, then the star product 0=\star_0 = \cdot is the plain pointwise product.

Example

(Moyal star product)

If the tensor π\pi in def. 1 is skew-symmetric, it may be regarded as a constant Poisson tensor on the smooth manifold VV. In this case 12π\star_{\tfrac{1}{2}\pi} is called a Moyal star product and the star-product algebra C (V)[[]], π)C^\infty(V)[ [\hbar] ], \star_\pi) is called the Moyal deformation quantization of the Poisson manifold (V,π)(V,\pi).

Wick algebras of normal ordered products

Finite dimensional

Definition

(Kähler vector space)

An Kähler vector space is a real vector space VV equipped with a linear complex structure JJ as well as two bilinear forms ω,g:V V\omega, g \;\colon\; V \otimes_{\mathbb{R}} V \longrightarrow \mathbb{R} such that the following equivalent conditions hold:

  1. ω(Jv,Jw)=ω(v,w)\omega(J v, J w) = \omega(v,w) and g(v,w)=ω(v,Jw)g(v,w) = \omega(v,J w);

  2. with VV regarded as a smooth manifold and with ω,g\omega, g regarded as constant tensors, then (V,ω,g)(V, \omega, g) is an almost Kähler manifold.

Example

(standard Kähler vector spaces)

Let V 2V \coloneqq \mathbb{R}^2 equipped with the complex structure JJ which is given by the canonical identification 2\mathbb{R}^2 \simeq \mathbb{C}, hence, in terms of the canonical linear basis (e i)(e_i) of 2\mathbb{R}^2, this is

J=(J i j)(0 1 1 0). J = (J^i{}_j) \coloneqq \left( \array{ 0 & -1 \\ 1 & 0 } \right) \,.

Moreover let

ω=(ω ij)(0 1 1 0) \omega = (\omega_{i j}) \coloneqq \left( \array{0 & 1 \\ -1 & 0} \right)

and

g=(g ij)(1 0 0 1). g = (g_{i j}) \coloneqq \left( \array{ 1 & 0 \\ 0 & 1} \right) \,.

Then (V,J,ω,g)(V, J, \omega, g) is a Kähler vector space (def. 4).

The corresponding Kähler manifold is 2\mathbb{R}^2 regarded as a smooth manifold in the standard way and equipped with the bilinear forms J,ωgJ, \omega g extended as constant rank-2 tensors over this manifold.

If we write

x,y: 2 x,y \;\colon\; \mathbb{R}^2 \longrightarrow \mathbb{R}

for the standard coordinate functions on 2\mathbb{R}^2 with

zx+iy 2 z \coloneqq x + i y \;\coloneqq\; \mathbb{R}^2 \to \mathbb{C}

and

z¯xiy 2 \overline{z} \coloneqq x - i y \;\coloneqq\; \mathbb{R}^2 \to \mathbb{C}

for the corresponding complex coordinates, then this translates to

ωΩ 2( 2) \omega \in \Omega^2(\mathbb{R}^2)

being the differential 2-form given by

ω =dxdy =12idzdz¯ \begin{aligned} \omega & = d x \wedge d y \\ & = \tfrac{1}{2i} d z \wedge d \overline{z} \end{aligned}

and with Riemannian metric tensor given by

g=dxdx+dydy. g = d x \otimes d x + d y \otimes d y \,.

The Hermitian form is given by

h =giω =dzdz¯ \begin{aligned} h & = g - i \omega \\ & = d z \otimes d \overline{z} \end{aligned}

(for more see at Kähler vector space this example).

Definition

(Wick algebra of a Kähler vector space)

Let ( 2n,σ,g)(\mathbb{R}^{2n},\sigma, g) be a Kähler vector space (def. 4). Then its Wick algebra is the formal power series vector space [[ 2n]][[]]\mathbb{C}[ [ \mathbb{R}^{2n} ] ] [ [ \hbar ] ] equipped with the star product (def. 1) which is given by the bilinear form

(2)πi2ω 1+12g 1, \pi \coloneqq \tfrac{i}{2} \omega^{-1} + \tfrac{1}{2} g^{-1} \,,

hence:

A 1 πA 2 (()())exp(k 1,k 2=12nπ ab a b)(A 1A 2) =A 1A 2+k 1,k 2=12nπ k 1k 2( k 1A 1)( k 2A 2)+ \begin{aligned} A_1 \star_\pi A_2 & \coloneqq ((-)\cdot (-)) \circ \exp \left( \hbar\underoverset{k_1, k_2 = 1}{2 n}{\sum}\pi^{a b} \partial_a \otimes \partial_b \right) (A_1 \otimes A_2) \\ & = A_1 \cdot A_2 + \hbar \underoverset{k_1, k_2 = 1}{2n}{\sum}\pi^{k_1 k_2}(\partial_{k_1} A_1) \cdot (\partial_{k_2} A_2) + \cdots \end{aligned}

(e.g. Collini 16, def. 1)

Proposition

(star product algebra of Kähler vector space is star-algebra)

Under complex conjugation the star product π\star_\pi of a Kähler vector space structure (def. 5) is a star algebra in that for all A 1,A 2[[ 2n]][[]]A_1, A_2 \in \mathbb{C}[ [\mathbb{R}^{2n}] ][ [\hbar] ] we have

(A 1 πA 2) *=A 2 * πA 1 * \left( A_1 \star_\pi A_2 \right)^\ast \;=\; A_2^\ast \star_\pi A_1^\ast
Proof

This follows directly from that fact that in π=i2ω 1+12g 1\pi = \tfrac{i}{2} \omega^{-1} + \tfrac{1}{2} g^{-1} the imaginary part coincides with the skew-symmetric part, so that

(π *) ab =i2(ω 1) ab+12(g 1) ab =i2(ω 1) ba+12(g 1) ba =π ba. \begin{aligned} (\pi^\ast)^{a b} & = -\tfrac{i}{2} (\omega^{-1})^{a b} + \tfrac{1}{2} (g^{-1})^{a b} \\ & = \tfrac{i}{2} (\omega^{-1})^{b a} + \tfrac{1}{2} (g^{-1})^{b a} \\ & = \pi^{b a} \,. \end{aligned}
Example

(Wick algebra of a single mode)

Let V 2Span({x,y})V \coloneqq \mathbb{R}^2 \simeq Span(\{x,y\}) be the standard Kähler vector space according to example 3, with canonical coordinates denoted xx and yy. We discuss its Wick algebra according to def. 5 and show that this reproduces the traditional definition of products of “normal ordered” operators as above.

To that end, consider the complex linear combination of the coordinates to the canonical complex coordinates

zx+iyAAAandAAAz¯xiy z \;\coloneqq\; x + i y \phantom{AAA} \text{and} \phantom{AAA} \overline{z} \coloneqq x - i y

which we use in the form

a *12(x+iy)AAAandAAAa12(xiy) a^\ast \;\coloneqq\; \tfrac{1}{\sqrt{2}}(x + i y) \phantom{AAA} \text{and} \phantom{AAA} a \;\coloneqq\; \tfrac{1}{\sqrt{2}}(x - i y)

(with “aa” the traditional symbol for the amplitude of a field mode).

Now

ω 1=yxxy \omega^{-1} = \frac{\partial}{\partial y} \otimes \frac{\partial}{\partial x} - \frac{\partial}{\partial x} \otimes \frac{\partial}{\partial y}
g 1=xx+yy g^{-1} = \frac{\partial}{\partial x} \otimes \frac{\partial}{\partial x} + \frac{\partial}{\partial y} \otimes \frac{\partial}{\partial y}

so that with

z=12(xiy)AAAAz¯=12(x+iy) \frac{\partial}{\partial z} = \tfrac{1}{2} \left( \frac{\partial}{\partial x} -i \frac{\partial}{\partial y} \right) \phantom{AAAA} \frac{\partial}{\partial \overline{z}} = \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right)

we get

i2ω 1+2g 1 =2z¯z =aa * \begin{aligned} \tfrac{i \hbar}{2}\omega^{-1} + \tfrac{\hbar}{2} g^{-1} & = 2 \hbar \frac{\partial}{\partial \overline{z}} \otimes \frac{\partial}{\partial z} \\ & = \hbar \frac{\partial}{\partial a} \otimes \frac{\partial }{\partial a^\ast} \end{aligned}

Using this, we find the star product

A πB=prodexp(aa *) A \star_\pi B \;=\; prod \circ \exp\left( \hbar \frac{\partial}{\partial a} \otimes \frac{\partial }{\partial a^\ast} \right)

to be as follows (where we write ()()(-)\cdot (-) for the plain commutative product in the formal power series algebra):

a πa =aa a * πa * =a *a * a * πa =a *a a πa * =aa *+ \begin{aligned} a \star_\pi a & = a \cdot a \\ a^\ast \star_\pi a^\ast & = a^\ast \cdot a^\ast \\ a^\ast \star_\pi a & = a^\ast \cdot a \\ a \star_\pi a^\ast & = a \cdot a^\ast + \hbar \end{aligned}

and so forth, for instance

(aa) π(a *a *) =a *a *aa+4a *a+ 2 \array{ (a \cdot a ) \star_\pi (a^\ast \cdot a^\ast) & = a^\ast \cdot a^\ast \cdot a \cdot a + 4 \hbar a^\ast \cdot a + \hbar^2 }

If we instead indicate the commutative pointwise product by colons and the star product by plain juxtaposition

:fg:fgAAAAfgf π :f g: \;\coloneqq\; f \cdot g \phantom{AAAA} f g \;\coloneqq\; f \star_\pi

then this reads

:aa::a *a *: =:a *a *aa:+4:a *a:+ 2 \array{ :a a: \, :a^\ast a^\ast: & = : a^\ast a^\ast a a : + 4 \hbar \, : a^\ast a : + \hbar^2 }

This is the way the Wick algebra with its operator product π\star_\pi and normal-ordered product :::-: is traditionally presented.

Infinite-dimensional

Example

(normal-ordered products and time-ordered products)

In cases where VV is not finite dimensional, but for instance a space of sections of a linear field bundle, the expression involved in the definition of the star product in def. 1 still may be interpreted as products of distributions, but these only exist if the wavefront sets satisfy Hörmander's criterion.

In these situations it happens that the star product for some Poisson tensor 12π\tfrac{1}{2}\pi is not defined due to wavefront set collision (violation of Hörmander's criterion), but the star product 12π+α\star_{\tfrac{1}{2}\pi + \alpha} for the tensor 12π+α\tfrac{1}{2}\pi + \alpha shifted by a symmetric contribition α\alpha as in prop. 2 is defined, and serves as the proper stand-in for the non-existing 12π\star_{\tfrac{1}{2}\pi}.

This is the case notably for Wick algebras (algebras of quantum observables of free fields): here 12π\tfrac{1}{2}\pi is the causal propagator for which π\star_\pi does not exist, and π=12π+α\pi = \tfrac{1}{2}\pi + \alpha is a Wightman propagator, so that π\star_{\pi} does exist on microcausal functionals: it is the “normal-ordered product” of quantum observables.

free field algebra of quantum observablesphysics terminologymaths terminology
1)supercommutative productAA:A 1A 2:\phantom{AA} :A_1 A_2:
normal ordered product
AAA 1A 2\phantom{AA} A_1 \cdot A_2
pointwise product of functionals
2)non-commutative product
(deformation induced by Poisson bracket)
AAA 1A 2\phantom{AA} A_1 A_2
operator product
AAA 1 HA 2\phantom{AA} A_1 \star_H A_2
star product for Wightman propagator
3)AAT(A 1A 2)\phantom{AA} T(A_1 A_2)
time-ordered product
AAA 1 FA 2\phantom{AA} A_1 \star_F A_2
star product for Feynman propagator
perturbative expansion
of 2) via 1)
Wick's lemma
Moyal product for Wightman propagator Δ H\Delta_H
A 1 HA 2= (()())exp((Δ H) ab(x,y)δδΦ a(x)δδΦ b(y))(A 1A 2) \begin{aligned} & A_1 \star_H A_2 = \\ & ((-)\cdot (-)) \circ \exp \left( \hbar \int (\Delta_H)^{ab}(x,y) \frac{\delta}{\delta \mathbf{\Phi}^a(x)} \otimes \frac{\delta}{\delta \mathbf{\Phi}^b(y)} \right)(A_1 \otimes A_2) \end{aligned}
perturbative expansion
of 3) via 1)
Feynman diagrams
Moyal product for Feynman propagator Δ F\Delta_F
A 1 FA 2= (()())exp((Δ F) ab(x,y)δδΦ a(x)δδΦ b(y))(A 1A 2) \begin{aligned} & A_1 \star_F A_2 = \\ & ((-)\cdot (-)) \circ \exp \left( \hbar \int (\Delta_F)^{ab}(x,y) \frac{\delta}{\delta \mathbf{\Phi}^a(x)} \otimes \frac{\delta}{\delta \mathbf{\Phi}^b(y)} \right)(A_1 \otimes A_2) \end{aligned}

References

In perturbative quantum field theory:

Last revised on January 11, 2018 at 06:11:06. See the history of this page for a list of all contributions to it.