– Atom in its normal (non-excited) state – Atom in excited state Definition of the symbols:

Photon 0 Absorption of the photon by the atom pro- duces: Now, I want you to solve a simple problem: ● The typical lifetime of an electron on an excited orbit is ~ 1 ns; ● The linear speed of an electron on the n = 1 orbit in a H atom 1/137 of the speed of light (it was a homework problem, remember?); ● The radius of the n = 1 orbit in hydrogen (i.e., the “Bohr radius”) is approximately a 0 = 0.05 nm ; I want you to find how many revolutions the electron on the excited n = 2 orbit makes before it returns to the “ground state” (i.e., the n=1 state)? – only the order of magnitude Is of interest to us, don’t do very accurate calculations!

The electron makes nearly 100,000 revolutions, so it is clear that it “does not remember” from which direction came the photon which pushed it to the higher orbit.

The excited atom returns To the ground state, re-emitting the photon which flies away in a completely random direction: Gas of atoms Re-emitted photon

Gas of atoms

OK, why we needed to explain that will become clear little later…. Please remember: this type of photon emission we Considered is called a “spontaneous emission”. Now, lets go back to the year Max Planck first finds an empirical formula describing the spectrum of the blackbody radiation. Shortly afterwards, he derives the same formula theoretically, based on an assumption that electromagnetic radiation can be absorbed and emitted by the atoms comprising the blackbody only in discrete “portions” that he calls quanta. Quantum physics is born at this moment!

A few years later, Albert Einstein explains the photoelec- tric effect, based on the hypothesis that light consists of particle-like photons of energy h. Max Planck does not believe in “light quanta” and strongly opposes the Einstein’s hypothesis. He insists that only atoms has quantum properties, but electromagnetic radiation is continuous. But Einstein strongly believes in his own theory. Therefore, he does not like the way Plank derived his famous formula. He thinks of how it could be done in a way consistent with his “photon theory”

Einstein reaches the conclusion that in addition to the “spontaneous emission” of photons there must be another emission process: Photon 0 An atom that has already been excited by a photon with frequency 0 is hit by another photon of the same frequency The second photon “catalyzes” an emission process. The atom returns to the ground state. Photon 0 Two photons of frequency 0 fly away in the same direction! This process becomes known as a “Stimulated Emission” But at the time Einstein constructs this hypothesis, there is no way of verifying it by an experiment… However, Einstein obtains the proof in an indirect way…

The gray is a “gas” of photons with frequency 0 In 1917, Einstein attacs the “blackbody problem” using a completely different approach than that of Planck. He considers a container of atoms, N A of which are in the non-excited state, and N B are in the excited state. The atoms are “immer- sed” in a “photon gas”.  ( 0 ) is the density of of photons with freq- ency 0 in this gas.

Now, there comes a moment for a very important step in our reasoning. We assume that the system is in EQUILIBRIUM. The proportion of the excited and non- excited atoms remains constant in time. Therefore, the above derivative is zero. From that, we obtain in a very simple way:

Now comes the only moment when you have to accept something I tell you without a proof. Let’s call the energy of the atom in the ground state as E A, and the energy of the excited atom as E B. From thermal physics it is known that if a system is in equilibrium at temperature T, then the proportion of the # of particles with energy E A to the # of particles with energy E B is given by the so-called “Boltzmann Factor”:

Einstein noticed that his formula becomes essentially identical with the Planck’s one, if B=C.

B=C: what does it mean? At the moment Einstein reached that conclusion, he discovered something that is of crucial importance in LASER physics! Let’s recall: B=C means that the transition rate for stimulated emission is the same as for absorption. Translating this into a somewhat more “human” language: the above two expressions determine the way a photon is going TO BE USED.

Photon 0 Let’s consider not a “gas” of photons, but a SINGLE photon incident on the system: If the number of the non-excited (“blue”) atoms is larger than of these excited (“red”), then the photon most likely will engage into an “ordinary absorption act”; It will be scattered “out of the system” Gas of atoms Recall:

Photon 0 Only if the number of excited (“red”) atoms is HIGHER than the number of non-excited (“blue”) atoms, THE PHOTON HAS A GREATER CHANCE “TO BE USED” for triggering a STIMULATED EMISSION

But “chain reaction” is a word from the nuclear weapon vocabulary. And since we are PEACEFUL PEOPLE, we prefer the term LASER ACTION LASER = L ight A mplification through S timulated E mission of R adiation

The most important message to remember: Laser action may occur only if: OR: Such situation is called “INVERSE POPULATION” (or “inverted population”) THE GREATEST CHALLENGE IN LASER PHYSICS IS TO ACHIEVE SUCH A SITUATION!!!!! OR: