Star Product

The star product is the phase-space product that represents operator multiplication. If $A_W(x,p)$ and $B_W(x,p)$ are Weyl symbols of operators $A$ and $B$, then

$$(AB)_W = A_W\star B_W.$$

The ordinary product $A_WB_W$ is not enough because $A$ and $B$ may not commute. The star product keeps the noncommutative operator algebra while using functions on phase space.

This page is the canonical home for the product itself. The Moyal Bracket page uses the antisymmetric part of the star product to write quantum dynamics.

Why Ordinary Multiplication Fails

The Weyl Transform sends

$$\hat x \longmapsto x, \qquad \hat p \longmapsto p.$$

If ordinary multiplication represented operator multiplication, then one would have

$$(\hat x\hat p)_W=xp.$$

But the Weyl symbols of the ordered products are

$$(\hat x\hat p)_W = xp+\frac{i\hbar}{2}, \qquad (\hat p\hat x)_W = xp-\frac{i\hbar}{2}.$$

Their difference is

$$(\hat x\hat p-\hat p\hat x)_W=i\hbar,$$

matching the canonical commutator. Ordinary multiplication would erase this difference. The star product is the corrected multiplication rule.

Definition

For one degree of freedom, the Weyl-Moyal star product is

$$(A\star B)(x,p) = A(x,p) \exp\left[ \frac{i\hbar}{2} \left( \overleftarrow{\partial_x}\overrightarrow{\partial_p} - \overleftarrow{\partial_p}\overrightarrow{\partial_x} \right) \right] B(x,p).$$

The arrows indicate which factor the derivative acts on. For example, $\overleftarrow{\partial_x}$ acts on $A$, while $\overrightarrow{\partial_p}$ acts on $B$.

It is convenient to define the bidifferential operator

$$A\Lambda B = \frac{\partial A}{\partial x} \frac{\partial B}{\partial p} - \frac{\partial A}{\partial p} \frac{\partial B}{\partial x}.$$

Then

$$A\star B = A\exp\left( \frac{i\hbar}{2}\Lambda \right)B.$$

The first term in $\Lambda$ is the classical Poisson bracket structure. The exponential contains the quantum corrections required by noncommutativity.

Expansion in Powers of hbar

Expanding the exponential gives

$$A\star B = AB + \frac{i\hbar}{2}\{A,B\}_{\rm PB} + O(\hbar^2),$$

where

$$\{A,B\}_{\rm PB} = \frac{\partial A}{\partial x} \frac{\partial B}{\partial p} - \frac{\partial A}{\partial p} \frac{\partial B}{\partial x}.$$

Keeping terms through second order,

$$\begin{aligned} A\star B &= AB + \frac{i\hbar}{2} \left( A_xB_p-A_pB_x \right) \\ &\quad - \frac{\hbar^2}{8} \left( A_{xx}B_{pp} -2A_{xp}B_{xp} +A_{pp}B_{xx} \right) + O(\hbar^3). \end{aligned}$$

Here $A_x=\partial A/\partial x$, $A_{xp}=\partial^2A/\partial x\,\partial p$, and similarly for $B$. The expansion is formal unless the functions are smooth enough and the series is controlled in the regime of interest.

A compact all-orders expression in one dimension is

$$\begin{aligned} A\star B &= \sum_{n=0}^{\infty} \frac{1}{n!} \left( \frac{i\hbar}{2} \right)^n \sum_{r=0}^{n} (-1)^r \binom{n}{r} \\ &\quad\times \left( \partial_x^{\,n-r}\partial_p^{\,r}A \right) \left( \partial_x^{\,r}\partial_p^{\,n-r}B \right). \end{aligned}$$

The compact exponential notation is usually less error-prone than writing this full sum.

Basic Examples

For the canonical coordinates,

$$x\star p = xp+\frac{i\hbar}{2}, \qquad p\star x = xp-\frac{i\hbar}{2}.$$

Therefore

$$x\star p-p\star x=i\hbar.$$

If both functions depend only on $x$, then

$$f(x)\star g(x)=f(x)g(x),$$

because all $p$ derivatives vanish. Similarly,

$$f(p)\star g(p)=f(p)g(p).$$

Noncommutativity enters when $x$ and $p$ dependence mix across the two factors.

For a general function $B(x,p)$,

$$x\star B = \left( x+\frac{i\hbar}{2}\partial_p \right)B,$$

and

$$p\star B = \left( p-\frac{i\hbar}{2}\partial_x \right)B.$$

These are examples of Bopp shifts. They are often the fastest way to compute simple star products.

Operator Products and Associativity

Because ordinary operator multiplication is associative,

$$(AB)C=A(BC),$$

the star product is associative:

$$(A_W\star B_W)\star C_W = A_W\star(B_W\star C_W).$$

It is generally not commutative:

$$A\star B\neq B\star A.$$

Associativity is essential. It means the phase-space representation preserves the operator algebra, even though it represents the algebra using functions and a deformed product.

The identity operator maps to the constant function one:

$$I_W=1.$$

Accordingly,

$$1\star A=A\star1=A.$$

Commutators and Anticommutators

The star commutator represents the operator commutator:

$$([A,B])_W = A_W\star B_W-B_W\star A_W.$$

The Moyal bracket is the normalized version:

$$\{A_W,B_W\}_M = \frac{1}{i\hbar} \left( A_W\star B_W-B_W\star A_W \right).$$

Its leading term is the Poisson bracket:

$$\{A,B\}_M = \{A,B\}_{\rm PB} + O(\hbar^2).$$

The symmetric part represents the operator anticommutator:

$$(\{A,B\}_+)_W = A_W\star B_W+B_W\star A_W,$$

where $\{A,B\}_+=AB+BA$. This Jordan-product side is important for variances, covariances, and symmetrically ordered observables.

Trace and Expectation Values

For suitable operators,

$$\operatorname{Tr}(AB) = \int\frac{dx\,dp}{2\pi\hbar}\, A_W(x,p)B_W(x,p).$$

One can also write

$$\operatorname{Tr}(AB) = \int\frac{dx\,dp}{2\pi\hbar}\, (A_W\star B_W)(x,p).$$

Under the trace integral, the star product may be replaced by ordinary multiplication when boundary terms vanish:

$$\int dx\,dp\,A\star B = \int dx\,dp\,AB.$$

This cyclic property is why expectation values with a Wigner function can look classical:

$$\operatorname{Tr}(\rho A) = \int dx\,dp\, W_\rho(x,p)A_W(x,p).$$

The simplicity of this formula does not mean operator products are classical. For products such as $\operatorname{Tr}(\rho AB)$, one must use

$$\operatorname{Tr}(\rho AB) = \int dx\,dp\, W_\rho(x,p)(A_W\star B_W)(x,p).$$

Classical Limit

The star product is a deformation of ordinary multiplication:

$$\lim_{\hbar\to0}A\star B=AB,$$

when the functions are smooth on the relevant phase-space scales. The star commutator has the corresponding limit

$$\lim_{\hbar\to0} \frac{1}{i\hbar} \left( A\star B-B\star A \right) = \{A,B\}_{\rm PB}.$$

This is the algebraic core of the phase-space classical limit: operator multiplication becomes commutative multiplication, and commutators become Poisson brackets. Physically, the limit also requires states and observables that do not probe $\hbar$-scale oscillatory structure.

More Degrees of Freedom

For canonical coordinates $q^a,p_a$, define

$$\Lambda = \sum_a \left( \overleftarrow{\partial_{q^a}} \overrightarrow{\partial_{p_a}} - \overleftarrow{\partial_{p_a}} \overrightarrow{\partial_{q^a}} \right).$$

Then

$$A\star B = A\exp\left( \frac{i\hbar}{2}\Lambda \right)B.$$

Equivalently, using phase-space coordinates $z^i$ and a constant symplectic matrix $\Omega^{ij}$,

$$A\star B = A\exp\left[ \frac{i\hbar}{2} \overleftarrow{\partial_i} \Omega^{ij} \overrightarrow{\partial_j} \right]B.$$

This form makes the symplectic structure explicit and is the natural notation for many degrees of freedom.

Common Mistakes

• Replacing operator multiplication by ordinary multiplication of Weyl symbols.
• Forgetting that $A\star B$ and $B\star A$ usually differ.
• Using the Poisson bracket when higher Moyal corrections are not negligible.
• Dropping boundary terms under phase-space integrals without checking decay or periodicity.
• Mixing Weyl symbols with normally ordered or antinormally ordered symbols.
• Treating the formal $\hbar$ expansion as convergent without checking the functions and regime.

Cross-Links

• Weyl Transform
• Wigner Function
• Phase-Space Conventions
• Moyal Bracket
• Poisson Brackets
• Commutators
• Formula Sheet

References

• H. Weyl, The Theory of Groups and Quantum Mechanics, Dover, 1950.
• J. E. Moyal, “Quantum mechanics as a statistical theory,” Proceedings of the Cambridge Philosophical Society 45, 99, 1949.
• F. Bayen, M. Flato, C. Fronsdal, A. Lichnerowicz, and D. Sternheimer, “Deformation theory and quantization,” Annals of Physics 111, 61, 1978.
• C. K. Zachos, D. B. Fairlie, and T. L. Curtright, eds., Quantum Mechanics in Phase Space, World Scientific, 2005.
• T. Curtright, D. Fairlie, and C. Zachos, “Features of time-independent Wigner functions,” Physical Review D 58, 025002, 1998.
• M. de Gosson, Symplectic Methods in Harmonic Analysis and in Mathematical Physics, Birkhäuser, 2011.

Exercises

1. Compute $x\star p$ and $p\star x$ from the definition.

Solution

For $x\star p$, only the first derivative term contributes:

$$x\star p = xp + \frac{i\hbar}{2} \left( \frac{\partial x}{\partial x} \frac{\partial p}{\partial p} - \frac{\partial x}{\partial p} \frac{\partial p}{\partial x} \right).$$

Thus

$$x\star p = xp+\frac{i\hbar}{2}.$$

Similarly,

$$p\star x = px + \frac{i\hbar}{2} \left( \frac{\partial p}{\partial x} \frac{\partial x}{\partial p} - \frac{\partial p}{\partial p} \frac{\partial x}{\partial x} \right) = xp-\frac{i\hbar}{2}.$$

Therefore

$$x\star p-p\star x=i\hbar.$$

2. Show that $f(x)\star g(x)=f(x)g(x)$.

Solution

If both functions depend only on $x$, then

$$\partial_p f=0, \qquad \partial_p g=0.$$

Every nonzero term in the star-product correction contains at least one $p$ derivative acting on one of the two factors. Therefore all correction terms vanish and

$$f(x)\star g(x)=f(x)g(x).$$

3. Use the Bopp shift to compute $x\star B$.

Solution

Since $A=x$ has only one nonzero derivative, $\partial_xA=1$, the star product truncates after first order:

$$x\star B = xB + \frac{i\hbar}{2} \frac{\partial B}{\partial p}.$$

Thus

$$x\star B = \left( x+\frac{i\hbar}{2}\partial_p \right)B.$$

4. Verify that the Moyal bracket of $x$ and $p$ is one.

Solution

Using the first exercise,

$$x\star p-p\star x=i\hbar.$$

Therefore

$$\{x,p\}_M = \frac{1}{i\hbar} \left( x\star p-p\star x \right) = 1.$$

This matches the operator identity $[\hat x,\hat p]/(i\hbar)=1$.

5. Explain why $\operatorname{Tr}(\rho AB)$ uses $A_W\star B_W$ rather than $A_WB_W$.

Solution

The Weyl symbol of the product $AB$ is

$$(AB)_W=A_W\star B_W.$$

Therefore

$$\operatorname{Tr}(\rho AB) = \int dx\,dp\, W_\rho(x,p)(AB)_W(x,p) = \int dx\,dp\, W_\rho(x,p)(A_W\star B_W)(x,p).$$

Using $A_WB_W$ would replace the noncommutative operator product by a commutative classical product and would miss ordering corrections.