In Quantum-Theoretical Re-Interpretation (1925), Heisenberg gives the following:

In order to characterize this radiation [of an electron] we first need the frequencies which appear as functions of two variables. In quantum theory these functions are of the form $$ \nu(n,\alpha) = \frac{1}{h}\left[W(n)-W(n-\alpha)\right], $$ and in classical theory of the form $$ \nu(n,\alpha) = \alpha\nu(n) = \alpha \frac{1}{h} \frac{dW}{dn}. $$

Here, $\nu$ is frequency, $n$ in the first equation is the quantum number of the first state, $\alpha$ is an integer such that the second state's quantum number is $n-\alpha$, and $W$ is the energy. Where does the classical equation come from, and how is $n$ to be interpreted? (It might be more specific to ask, how do we derive the classical equation from the correspondence principle?)


1 Answer 1


It took me a bit to track down this reference, it doesn't seem to be mentioned in many of my usual references on the old quantum theory, but fortunately the answer can be found in From c-Numbers to q-Numbers.

This equation, closely connected with the correspondence principle, appears to come from:

  • Bohr, N. (1914). On the effect of electric and magnetic fields on spectral lines. The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 27(159), 506–524. https://doi.org/10.1080/14786440308635119

There, Bohr treats the problem of emission spectra by assuming only certain stationary states are allowed, each state being characterized by the quantum number $n$. The formula Heisenberg gives is not actually the classical formula, but the correspondence limit which (by the principle) should agree with classical physics. In particular, from Bohr in the reference cited:

On the present theory the frequency of the radiation emitted by the transition from the $(n+1)$th to the $n$th stationary state is equal to $\frac{1}{h}(A_{n+1}-A_n)$. When $n$ is large, this approaches to $\frac{1}{h} \frac{dA_n}{dn}$. On the ordinary electrodynamics we should expect the frequency of the radiation to be equal to the frequency of revolution, and consequently it is to be anticipated that for large values of $n$ $$ \frac{dA_n}{dn} = h\omega_n.\quad .\ .\ .\ .\ .\ (8) $$

Now, by taking $\alpha$ to be the classical harmonic in the Fourier series representation of the (periodic) motion of the electron, we get $$ \alpha \omega_n = \alpha \frac{1}{h}\frac{dA_n}{dn} $$ where $A_n$ denotes the energy in the $n$th state. .


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.