Lets look at the transition amplitude $U(x_{b},x_{a})$ for a free particle between two points $x_{a}$ and $x_{b}$ in the Feynman path integral formulation

  • $U(x_{b},x_{a}) = \int_{x_{a}}^{x_{b}} \mathcal{D} x e^{\frac{i}{\hbar}S}$

($S$ is the classical action). It is often said that one gets classical mechanics in the limit $\hbar \rightarrow 0$. Then only the classical action is contributing, since the terms with non-classical $S$ cancel each other out because of the heavily oscillating phase. This sounds reasonable.

But when we look at the Heisenberg equation of motion for an operator $A$

  • $\frac{dA}{dt} = \frac{1}{i \hbar} [A,H]$

the limit $\hbar \rightarrow 0$ does not make any sense (in my opinion) and does not reproduce classical mechanics. Basically, the whole procedure of canonical quantization does not make sense:

  • $\{\cdots,\cdots\} \rightarrow \frac{1}{i \hbar} [\cdots,\cdots]$

I don't understand, when $\hbar \rightarrow 0$ gives a reasonable result and when not. The question was hinted at here: Classical limit of quantum mechanics. But the discussion was only dealing with one particular example of this transition. Does anyone has more general knowledge about the limit $\hbar \rightarrow 0$?


4 Answers 4


The theory of deformation quantization provides a framework in which the quantum to classical transition can be carried out and understood.

According to this theory, for (practically any) quantum system, one can find (may be nonuniquely) a Poisson manifold $\mathcal{M}$ (phase space) equipped with an associative product called the "star product" such that the quantum observables are represented by smooth functions on $\mathcal{M}$ and the quantum operator product is given by the star product.

Furthermore, the star product of two functions has a formal power series in $\hbar$

$f\star g = \sum_{k=0}^{\infty} \hbar^k B_k(f,g)$

Such that:

$B_0(f,g) = fg$

$B_1(f,g)-B_1(g, f) = \{f,g\}$, (Poisson bracket)

Thus we obtain:

$f\star g - g\star f = \hbar\{f,g\} + \sum_{k=2}^{\infty} \hbar^k (B_k(f,g)-B_k(g,f))$

Please notice that according to the deformation Philosophy, the quantum observables are just functions on the phase space just as the classical observables and all the quantum noncommutativity is provided by the star product. Thus if we define $\hat{f} = \frac{\hbar}{i} f $, we get the required classical limit.

It should be emphasized that this procedure can be carried out even for quantum systems defined by matrix algebras for example an appropriate phase for spin iis the two-sphere $S^2$, please, see the following article by Moreno and Ortega-Navarro. Morover,

Kontsevich in his seminal work provided a constructive method to construct this star product on every finite dimensional Poisson manifold, Please see the following Wikipedia page.

It is also worthwhile to mention that there are efforts to generalize the deformation construction to field theories and incorporate renormalization into it, please see the following work by Dito.

  • $\begingroup$ Thank you! This is surely an interesting approach. It is hard for me, though, to understand the physical significance of this mathematical structure. Is there any physical justification for the introduction of the $\star$-product or was it invented to make the limit $\hbar \rightarrow \infty$ work? $\endgroup$
    – stankowait
    Commented Mar 8, 2013 at 1:22
  • $\begingroup$ @stankowait Quantum mechanical amplitudes can be represented by expectations of products of operators. Thus, the star product allows to compute physical amplitudes. It is true that for elementary systems, other quantiation methods are more suitable. However, there are some systems such as moduli spaces and certain infinte dimensional phase spaces of physical interest, where deformation quantization is may be the only available quantization method. $\endgroup$ Commented Mar 9, 2013 at 5:37
  • $\begingroup$ @stankowait cont. Please, see the following lecture by Daniel Sternheimer on the current status of deformation quantization: guests.mpim-bonn.mpg.de/deform/dsMPIMaugust08.pdf $\endgroup$ Commented Mar 9, 2013 at 5:39

To the other answers, I might add one important detail: The limit $\hbar \rightarrow 0$ is just a convenient way to calculate. What is really happening is that $S[x] \gg \hbar$ or equivalently $\frac{S[x]}{\hbar} \to \infty$. So the justification why in the path integral one expects to find classical physics in the limit $\hbar \to 0$ is that all paths beyond the classical extreme path oscillate to death. So $\hbar \to 0$ is just a convenient way of calculating, really important is the size of the action. In cases where classical physics applies, non-extremal paths have actions that are much larger, where much larger is measured in units of $\hbar$, i.e. to be exact: $$ \frac{S[x]-S[x_0]}{\hbar} \gg 1 $$ where $x$ is a non-extremal path and $x_0$ the classical one. However the Heisenberg time evolution does not contain the action.


This sounds reasonable.

My rather rough understanding is that, (if there is a classical action for the transition,) in the limit it is only the neighbourhood of the classical action that is contributing. The contribution of the classical action itself maintains to have measure $0$ relative to the contribution of the neighbourhood, even when taking the limit (wherein that neighbourhood goes to "size" or "spread" $0$.)

If there isn't a classical action for the transition, then the whole thing fails anyway.


In the operator language of quantum mechanics one does not (blindly) perform the limit $\hbar \rightarrow 0$ in the equations of motion (the schrödinger equation), but expand the states as powerseries in $\hbar$. In first order: $\psi(x) = a(x) e^{S(x)}$. If you insert this expression in the schrödinger equation you get (in first order) the classical Hamilton-Jacobi equation. See for example: http://en.wikipedia.org/wiki/WKB_approximation#Application_to_Schr.C3.B6dinger_equation or Bates, Weinstein: Lectures on the geometry of quantization for an geometric interpretation


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.