Classical wave energy vs Photon energy

Question

The classical theory says that the intensity of light is proportional to the square of the amplitude of an oscillating electric field. Quantum theory gives the intensity of light as proportional to the number of emitted photons.

Are the two concepts equivalent in any way?

When we observe light, is its brightness a measure of separation between the two extremes of an oscillating field (say +10 to -10 V/m) or a measure of the number of relaxation events where an electron drops to a lower energy state and emits a photon? — Ladd Webster, Jul 18 '19 at 23:38

garyp · Answer 1 · 2019-07-19T10:41:28.720

3

More precisely, quantum mechanics gives intensity as number of emitted photons per second, per unit area, $\frac{d^2n}{dtdA}$. The correspondence is (in SI units) $$\frac{d^2n}{dtdA}h\nu = \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}}|E|^2$$

edited Jul 19 '19 at 10:41

answered Jul 19 '19 at 00:34

garyp

22,210

1

Check the units. It appears the left side is in W and right side in W/$m^2$. – amateurAstro Jul 19 '19 at 01:58
@amateurAstro Whoops. Corrected. Thanks. – garyp Jul 19 '19 at 10:42

Classical wave energy vs Photon energy

1 Answers1