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We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption?

Short Answer

Expert verified

Spontaneous absorption would involve taking energy (a photon) from the ground state of the electromagnetic field

Step by step solution

01

Absorption

The process of one material (absorbate) being retained by another (absorbent); this may be the physical solution of a gas, liquid, or solid in a liquid, attachment of molecules of a gas, vapour, liquid, or dissolved substance to a solid surface by physical forces, etc. In spectrophotometry, absorption of light at characteristic wavelengths or bands of wavelength is used to identify the chemical nature of molecules, atoms, or ions and t measure the concentration of these species.

A phenomenon in which radiation transfers to matter which it traverses some of or all its energy.

02

Step 2: Spontaneous absorption involves taking energy

Spontaneous absorption would involve taking energy (a photon) from the ground state of the electromagnetic field. But you can’t do that, because the ground state already has the lowest allowed energy.

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Most popular questions from this chapter

As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which blackbody radiation is the source). Show that at room temperature (T = 300 K) thermal stimulation dominates for frequencies well below 5×1012Hz , whereas spontaneous emission dominates for frequencies well above . Which mechanism dominates for visible light?

Suppose you don’t assume Haa=Hbb=0

(a) Find ca(t)and cb(t) in first-order perturbation theory, for the case

.show that , to first order in .

(b) There is a nicer way to handle this problem. Let

.

Show that

where

So the equations for are identical in structure to Equation 11.17 (with an extra

(c) Use the method in part (b) to obtain in first-order
perturbation theory, and compare your answer to (a). Comment on any discrepancies.

You could derive the spontaneous emission rate (Equation 11.63) without the detour through Einstein’s A and B coefficients if you knew the ground state energy density of the electromagnetic field P0(ω)for then it would simply be a case of stimulated emission (Equation 11.54). To do this honestly would require quantum electrodynamics, but if you are prepared to believe that the ground state consists of one photon in each classical mode, then the derivation is very simple:

(a) Replace Equation 5.111by localid="1658381580036" N0=dkand deduce P0(ω) (Presumably this formula breaks down at high frequency, else the total "vacuum energy" would be infinite ... but that's a story for a different day.)

(b) Use your result, together with Equation 9.47, to obtain the spontaneous emission rate. Compare Equation 9.56.

Close the “loophole” in Equation 9.78 by showing that ifl'=l=0thenn'l'm'|r|nlm=0

Solve Equation 9.13 to second order in perturbation theory, for the general case ca(0)=a,cb(0)=bca=-ihH'abe-0tcb,cb=-ihH'bae-0tca

(9.13).

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