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Chapter 9: Q8P (page 356)

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?

Short Answer

Expert verified

The spontaneous emission dominates.

Step by step solution

01

Formula for the spontaneous emission rate

The spontaneous emission rate is given by:

A=ω3℘23πo0hc3

The thermally stimulated emission rate is given by:

role="math" localid="1658378927515" R=π3o0h2℘2ρω

where:

ρω=hπ2c3ω3ehω/kBT-1

The ratio of the spontaneous emission rate to the thermally stimulated emission rate is,

AR=ω3℘23πo0hc3.3πo0h2π℘2.π2c3ehω/KBT-1hω3=ehω/KBT-1

02

Find out the result

Seek the point at which the ratio is, that is:

1=ehω/KBT-1ehω/KBT-2hωkBT=In2ω=kBThIn2

The frequency is,

v=ω2π=kBThIn2

At room temperature, we get:

v=1.38×10-23J/k300k6.63×10-34J.s=4.33×1012Hz

For a frequency higher than this frequency the spontaneous emission dominates, note that higher frequencies than this one includes the visible light.

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Most popular questions from this chapter

The first term in Equation 9.25 comes from the eiωt/2, and the second from e-iωt/2.. Thus dropping the first term is formally equivalent to writing H^=(V/2)e-iωt, which is to say,

cbl-ihvba∫0tcos(ωt')eiω0t'dt'=-iVba2h∫0tej(ω0+ω)t'+ej(ω0-ω)t'dt'=--iVba2hej(ω0+ω)t'-1ω0+ω+ej(ω0-ω)t'-1ω0-ω(9.25).Hba'=Vba2e-iωt,Hab'=Vab2eiωt(9.29).

(The latter is required to make the Hamiltonian matrix hermitian—or, if you prefer, to pick out the dominant term in the formula analogous to Equation 9.25 forca(t). ) Rabi noticed that if you make this so-called rotating wave approximation at the beginning of the calculation, Equation 9.13 can be solved exactly, with no need for perturbation theory, and no assumption about the strength of the field.

c.a=-ihHab'e-iω0tcb,c.b=-ihHba'e-iω0tca,

(a) Solve Equation 9.13 in the rotating wave approximation (Equation 9.29), for the usual initial conditions: ca(0)=1,cb(0)=0. Express your results (ca(t)andcb(t))in terms of the Rabi flopping frequency,

ωr=12(ω-ω0)2+(Vab/h)2 (9.30).

(b) Determine the transition probability,Pa→b(t), and show that it never exceeds 1. Confirm that.

ca(t)2+cb(t)2=1.

(c) Check that Pa→b(t)reduces to the perturbation theory result (Equation 9.28) when the perturbation is “small,” and state precisely what small means in this context, as a constraint on V.

Pa→b(t)=cb(t)2≈Vab2hsin2ω0-ωt/2ω0-ω2(9.28)

(d) At what time does the system first return to its initial state?

Suppose the perturbation takes the form of a delta function (in time):

H^'=U^δ(t);

Assume thatUaa=Ubb=0,andletUab=Uba+=αif ca(-∞)=1and cb(-∞)=0,

find ca(t)andcb(t),and check that lc(t)l2+lcb(t)l2=1. What is the net probability(Pa→bfort→∞) that a transition occurs? Hint: You might want to treat the delta function as the limit of a sequence of rectangles.

Answer:Pa→b=sin2(|α|lh)

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.

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.

In Equation 9.31 assumed that the atom is so small (in comparison to the wavelength of light) that spatial variations in the field can be ignored. The true electric field would be E(r,t)=E0cos(k⋅r−ωt).

If the atom is centered at the origin, thenk⋅r≪1 over the relevant volume,|k|=2π/λ sok⋅r~r/λ≪1) and that's why we could afford to drop this term. Suppose we keep the first-order correction:

E(r,t)=E0[cos(ωt)+(k⋅r)sin(ωt)].

The first term gives rise to the allowed (electric dipole) transitions we considered in the text; the second leads to so-called forbidden (magnetic dipole and electric quadrupole) transitions (higher powers of k.rlead to even more "forbidden" transitions, associated with higher multipole moments).

(a) Obtain the spontaneous emission rate for forbidden transitions (don't bother to average over polarization and propagation directions, though this should really be done to complete the calculation). Answer:role="math" localid="1659008133999" Rb→a=q2ω5πϵ0ℏc5|⟨a|(n^⋅r)(k^⋅r)|b⟩|2.

(b) Show that for a one-dimensional oscillator the forbidden transitions go from leveln to levelrole="math" localid="1659008239387" n-2 and the transition rate (suitably averaged over n^andk^) isR=ℏq2ω3n(n−1)15πϵ0m2c5.

(Note: Hereω is the frequency of the photon, not the oscillator.) Find the ratio of the "forbidden" rate to the "allowed" rate, and comment on the terminology.

(c) Show that the2S→1S transition in hydrogen is not possible even by a "forbidden" transition. (As it turns out, this is true for all the higher multipoles as well; the dominant decay is in fact by two-photon emission, and the lifetime it is about a tenth of a second
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