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The half-life of (t1/2)an excited state is the time it would take for half the atoms in a large sample to make a transition. Find the relation betweenrole="math" localid="1658300900358" t1/2andT(the 鈥渓ife time鈥 of the state).

Short Answer

Expert verified

The relation between t1/2andTis thet1/2=蟿滨苍2

Step by step solution

01

Half-Life

The half-life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach 50% of its initial concentration (I.e., the time taken for the reactant to reach half of its initial value). It is denoted by the symbol t1/2and is usually expressed in seconds.

02

Finding the relation between t1/2 and T (the “life time” of the state)

By definition, the half-life of an excited state is the time it would take for half the atoms to make a transition to a lower state. We use the formula for decay:

Nt=e-tlN0 (Equation-9.58 and 9.59).

Nbt=Nb0e-At (Equation-9.58).

=1A (Equation-9.59).

The half-life of an excited state is the time it would take for half the atoms in a large sample to make the transition.

After one half-life,

Nt=12N0,so12=e-t/,or2=et/,sot/=In2,ort1/2=蟿滨苍2

Thus, the relation betweenn role="math" localid="1658302506697" t1/2&Tist1/2=蟿滨苍2

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Most popular questions from this chapter

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

(9.13).

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

cbl-ihvba0tcos(t')ei0t'dt'=-iVba2h0tej(0+)t'+ej(0-)t'dt'=--iVba2hej(0+)t'-10++ej(0-)t'-10-(9.25).Hba'=Vba2e-it,Hab'=Vab2eit(9.29).

(The latter is required to make the Hamiltonian matrix hermitian鈥攐r, 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-i0tcb,c.b=-ihHba'e-i0tca,

(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,

r=12(-0)2+(Vab/h)2 (9.30).

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

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

(c) Check that Pab(t)reduces to the perturbation theory result (Equation 9.28) when the perturbation is 鈥渟mall,鈥 and state precisely what small means in this context, as a constraint on V.

Pab(t)=cb(t)2Vab2hsin20-t/20-2(9.28)

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


Suppose you don鈥檛 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

daeih0tHaac(tc)dtc0ca,dbei20tHbbc(tc)dtc0cb.

Show that

role="math" localid="1658561855290" d-a=-iheiH'abe-i0tdb;db-=-iheiH'bae-i0tda

where

(t)1h0l[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" eitackedontoH^')

ca-=-ihH'abe-i0tcb,cb-=-ihH'baei0tca

(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.

Solve Equation 9.13 for the case of a time-independent perturbation, assumingthatandcheck that

. Comment: Ostensibly, this system oscillates between 鈥溾 Doesn鈥檛 this contradict my general assertion that no transitions occur for time-independent perturbations? No, but the reason is rather subtle: In this are not, and never were, Eigen states of the Hamiltonian鈥攁 measurement of the energy never yields. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end,are Eigen states of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time T 鈥攖his doesn鈥檛 affect the calculations, but it allows for a more sensible interpretation of the result.

ca=-ihHabeigtcb,cb=-ihHbaeigtca 鈥(9.13).

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.(rp)=0, to demonstrate that

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

The generalization from z to r is trivial.

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