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Chapter 9: Q19P (page 365)

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

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.

Prove the commutation relation in Equation 9.74. Hint: First show that

[L2,z]=2ih(xLy-yLx-ihz)

Use this, and the fact that localid="1657963185161" r.L=r.(r×p)=0, to demonstrate that

[L2,[L2,z]]=2h2(zL2+L2z)

The generalization from z to r is trivial.

A hydrogen atom is placed in a (time-dependent) electric fieldE=E(t)k.⏜calculateallfourmatrixelementsHij,oftheperturbationH,⏜=eEzbetween the ground state (n = 1 ) the (quadruply degenerate) first excited states (n = 2 ) . Also showthatHii,=0 for all five states. Note: There is only one integral to be done here, if you exploit oddness with respect to z; only one of the n = 2 states is “accessible” from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration—assuming transitions to higher excited states can be ignored.

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 assumeH'aa=H'aa=0

(a) Findca(t)andcb(t)in first-order perturbation theory, for the case

ce(0)=1,cb(0)=0.show that |ca(1)(t)|2+|cb(1)(t)|2=1, to first order inH^' .

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

daeih∫0tHaac(tc)dtc0ca,dbei2∫0tHbbc(tc)dtc0cb.

Show that

role="math" localid="1658561855290" d-a=-iheiϕH'abe-iω0tdb;db-=-iheiϕH'bae-iω0tda

where

ϕ(t)≡1h∫0l[H'aa(t')-H'bb(t')]dt'.

So the equations fordaanddbare identical in structure to Equation 11.17 (with an extra role="math" localid="1658562498216" eiϕtackedontoH^')

ca-=-ihH'abe-iω0tcb,cb-=-ihH'baeiω0tca

(c) Use the method in part (b) to obtainrole="math" localid="1658562468835" ca(t)andcb(t)in first-order
perturbation theory, and compare your answer to (a). Comment on any discrepancies.
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