# The Correspondence Principle

Thus far, we have been focusing our attention on experiments that defy explanation in terms of classical mechanics and, at the same time, isolate certain aspects of the laws of quantum mechanics. We must not lose sight, however, of the fact that there exists an enormous domain, the domain of the macroscopic physics, for which classical physics works extremely well. There is thus an obvious requirement that quantum mechanics must satisfy - in the appropriate limit, namely the classical limit, it must lead to the same predictions as classical mechanics. Mathematically, this limit is that in which $\hbar$ may be regarded as small. For the electromagnetic field, for example, this means that the number of quanta in the field must be very large. For particles, it means that the de Broglie wavelengths must be very small compared to all relevant length scales. Of course, the statements of quantum mechanics are, as stated earlier, probabilistic in nature, while those of classical mechanics are completely deterministic. Thus, in the classical limit, quantum mechanical probabilities must become practical certainties; fluctuations must become negligible. This principle, that, in the classical limit, the predictions of the laws of quantum mechanics must be in one-to-one correspondence with the predictions of classical mechanics, is called the correspondence principle.

For example, in classical mechanics, physical quantities are functions $A(\mathbf{r},\mathbf{p})\!$ of the position and momentum variables. The correspondence principle leads us to assume that, in quantum mechanics, these quantities are replaced with operators whose functional form in terms of position and momentum, also now operators, is the same. In other words, to a given classical quantity $A\left (\mathbf{r},\mathbf{p}\right ),$ there corresponds an operator, $A\left (\hat{\mathbf{r}} ,\hat{\mathbf{p}} \right ).$ For instance, the Hamiltonian, $H=\frac{p^2}{2m}+V(\mathbf{r}),$ becomes

$\hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+V(\hat{\mathbf{r}})=-\frac{\hbar^2}{2m}\nabla^2+V(\hat{\mathbf{r}}),$

and the angular momentum, $\mathbf{L}=\mathbf{r}\times\mathbf{p},$ becomes

$\hat{L}=\hat{\mathbf{r}}\times\hat{\mathbf{p}}=\frac{\hbar}{i} \left( \vec{r}\times\vec{\nabla} \right).$