I have studied the quantum harmonic oscillator and solved the Schrodinger equation to find the eigen-energies given by
$$ E_n = \left(n+\frac{1}{2}\right)\hbar \omega. $$
Which means the energy levels are separated by
$$ \Delta E = \hbar \omega = hf $$
I have also studied blackbody radiation and one of the assumptions Planck made was that the energy levels accessible to the cavity oscillators are separated by this same amount $\Delta E$. This makes sense to me as the source of blackbody radiation is just an oscillator in the wall so I can see the connection between a quantum harmonic oscillator and a cavity oscillator. However a photon that is emitted when an electron transitions between energy levels in an atom is not due to any oscillators, is it just a coincidence that the energy of a non-blackbody photon is given by $ E = \hbar \omega $ or is there some connection between photons and oscillators?
Best Answer
It's not a coincidence! You can see the reason even in classical mechanics: if you take a charge and shake it sinusoidally at frequency $\omega_q$, it makes light with equal frequency $\omega_{\gamma} = \omega_q$.
If you quantize light wave emission into individual photons, so that $E = \hbar \omega_{\gamma}$, the spacing between harmonic oscillator energy levels must be $\hbar \omega_{\gamma}$. But since $\omega_\gamma = \omega_q$, this is equal to $\hbar \omega_q$, so $$E_n = n \hbar \omega_q + \text{const.}$$ as you observed.