In the Heisenberg picture of Quantum Mechanics, for an observable $\hat{A}$, we have the famous Heisenberg equation giving the time evolution of the operator: ($\hat{H}$ is the Hamiltonian operator) $$ i\hbar \frac{\partial \hat{A}}{\partial t} = -[\hat{H},\hat{A}] $$ Which one can rewrite by defining the Lioville operator as: $$ \hat{L}=\frac{1}{\hbar}[H,.] $$ Thus \begin{align} \hat{A}(t) = e^{i\hat{L}t}\hat{A}(0) = e^{i\hat{H}t/\hbar}\hat{A}(0)e^{-i\hat{H}t/\hbar} \tag{1} \end{align}

Similarly in classical statistical mechanics, for some classical differentiable variable $A$ we have the Poisson equation: ($H$ being the classical Hamiltonian here) $$ \frac{\partial A}{\partial t} = \{A,H\} $$ Now defining the classical version of the Liouville operator: \begin{align} \mathcal{L}:=i\{H,.\}=\sum_{i=1}^n [\frac{\partial H}{\partial p_i}\frac{\partial}{\partial q_i}-\frac{\partial H}{\partial q_i}\frac{\partial}{\partial p_i}] \tag{*} \end{align} Thus again defining the time evolution of $A$ with the corresponding propagator: \begin{align} A(t) = e^{i\mathcal{L}t}A(0) \tag{2} \end{align}


  • In $(1)$ we were able to further expand the unitary operator $e^{i\hat{L}t}$ into the two unitary time translations generated by $\hat{H}$ (sandwiching the value of the operator at time $t=0$), but looking at $(2)$, is there a way to further expand the propagator $e^{i\mathcal{L}t}$, using $(*)$, into a type of expression as was obtained in the QM version, namely equation $(1)$?
  • $\begingroup$ The expression involving the Liouville operator is a formal one, i.e. a way to summarise a series of operations to perform. Observe that in the classical setting one deals with commutative structures, so that conjugation by unitaries is ineffective. $\endgroup$
    – Phoenix87
    Commented Sep 14, 2015 at 22:08

1 Answer 1


OP's question (v1) is essentially asking

Does the operator identity $$ e^{\frac{it}{\hbar}[\hat{H},~\cdot~]}\hat{A}~ =~ e^{i\hat{H}t/\hbar}\hat{A}e^{-i\hat{H}t/\hbar} \tag{1} $$ have an analog using functions/symbols $H$ and $A$ rather than operators $\hat{H}$ and $\hat{A}$, respectively?

The answer is: Yes, in terms of the Groenewold-Moyal star product $\star$. If the Poisson bracket is written as

$$ \{A,B\}_{PB}~:=~A \stackrel{\leftarrow}{\partial_I} \omega^{IJ} \stackrel{\rightarrow}{\partial_J} B, \qquad \partial_I~\equiv~\frac{\partial}{\partial z^I}, \qquad \{z^I,z^J\}_{PB}~=~\omega^{IJ}, \tag{2} $$

where $z^1, \ldots, z^{2n}$ are canonical coordinates, and $A,B$ are functions/symbols (as opposed to operators), then the star product reads

$$ A\star B~:=~A \exp\left(\stackrel{\leftarrow}{\partial_I}\frac{i\hbar}{2} \omega^{IJ} \stackrel{\rightarrow}{\partial_J}\right) B~=~AB+\frac{i\hbar}{2}\{A,B\}_{PB} +{\cal O}(\hbar^2). \tag{3} $$

And then an analog of eq. (1) is

$$ e^{\frac{it}{\hbar}[~H~\stackrel{\star}{,}~\cdot~]}A~=~ e_{\star}^{iHt/\hbar}\star A\star e_{\star}^{-iHt/\hbar}, \tag{4} $$


$$[A\stackrel{\star}{,}B]~:= A\star B-B\star A~=~i\hbar\{A,B\}_{PB} +{\cal O}(\hbar^3)\tag{5} $$

is the star commutator, and

$$e_{\star}^B~:=~1+B\sum_{n=1}^{\infty}(\star B)^{n-1}/n!\tag{6} $$

is the star exponential.

  • $\begingroup$ Thanks a lot, I was indeed looking for something along these lines, but I really have trouble understanding what the Moyal product means here. Would you be so kind as to give a brief basic explanation for it? (the given link is extremely terse). $\endgroup$
    – user098876
    Commented Sep 14, 2015 at 23:12
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Sep 14, 2015 at 23:32
  • 1
    $\begingroup$ The identity (4) is a semiclassical power series in $\hbar$. The zeroth-order term in $\hbar$ denotes the classical contribution. $\endgroup$
    – Qmechanic
    Commented Sep 16, 2015 at 11:50

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