Simple Nature - Light and Matter

Photons from a lightbulb example 7

.Roughly how many photons are emitted by a 100 watt lightbulb

in 1 second?

.People tend to remember wavelengths rather than frequencies

for visible light. The bulb emits photons with a range of frequen-

cies and wavelengths, but let’s take 600 nm as a typical wave-

length for purposes of estimation. The energy of a single photon

is

Ephot on=hf

=hc/λ

A power of 100 W means 100 joules per second, so the number

of photons is

(100 J)/Ephot on= (100 J)/(hc/λ)

≈ 3 × 1020

This hugeness of this number is consistent with the correspon-

dence principle. The experiments that established the classical

theory of optics weren’t wrong. They were right, within their do-

main of applicability, in which the number of photons was so large

as to be indistinguishable from a continuous beam.

Measuring the wave example 8

When surfers are out on the water waiting for their chance to

catch a wave, they’re interested in both the height of the waves

and when the waves are going to arrive. In other words, they ob-

serve both the amplitude and phase of the waves, and it doesn’t

matter to them that the water is granular at the molecular level.

The correspondence principle requires that we be able to do the

same thing for electromagnetic waves, since the classical theory

of electricity and magnetism was all stated and verified experi-

mentally in terms of the fieldsEandB, which are the amplitude

of an electromagnetic wave. The phase is also necessary, since

the induction effects predicted by Maxwell’s equation would flip

their signs depending on whether an oscillating field is on its way

up or on its way back down.

This is a more demanding application of the correspondence prin-

ciple than the one in example 7, since amplitudes and phases

constitute more detailed information than the over-all intensity of

a beam of light. Eyeball measurements can’t detect this type of in-

formation, since the eye is much bigger than a wavelength, but for

example an AM radio receiver can do it with radio waves, since

the wavelength for a station at 1000 kHz is about 300 meters,

which is much larger than the antenna. The correspondence prin-

ciple demands that we be able to explain this in terms of the pho-

ton theory, and this requires not just that we have a large number

876 Chapter 13 Quantum Physics