Develop time-dependent perturbation theory for a multi-level system, starting with the generalization of Equations 9.1 and 9.2:

H^0ψn=Enψn,   ψnψm=δnm (9.79)

At time t = 0 we turn on a perturbation H'(t)so that the total Hamiltonian is

H^=H^0+H^'(t)(9.80).

(a) Generalize Equation 9.6 to read

Ψ(t)=ca(t)ψaeiEat/+cb(t)ψbeiEbt/(9.81).

and show that

c˙m=incnHmn'ei(EmEn)t/ (9.82).

Where

Hmn'ψm|H^'|ψn (9.83).

(b) If the system starts out in the state ψN, show that (in first-order perturbation theory)

cN(t)1i0tHNN'(t')dt'(9.84).

and

cm(t)i0tHmN'(t')ei(EmEN)t'/dt',   (mN)(9.85).

(c) For example, supposeH^'is constant (except that it was turned on at t = 0 , and switched off again at some later time . Find the probability of transition from state N to state M (MN),as a function of T. Answer:

4|HMN'|2sin2[(ENEM)T/2](ENEM)2 (9.86).

(d) Now supposeH^'is a sinusoidal function of timeH^'=Vcos(ωt): Making the usual assumptions, show that transitions occur only to states with energy EM=EN±, and the transition probability is

PNM=|VMN|2sin2[(ENEM±ω)T/2](ENEM±ω)2 (9.87).

(e) Suppose a multi-level system is immersed in incoherent electromagnetic radiation. Using Section 9.2.3 as a guide, show that the transition rate for stimulated emission is given by the same formula (Equation 9.47) as for a two-level system.

Rba=π3ϵ02||2ρ(ω0)Rb (9.47).

Short Answer

Expert verified

(a)cneiEnt/Hmn'=ic˙meiEmt/,orc˙m=incnHmn'ei(EmEn)t/

(b)c˙m=iHmN'ei(EmEN)t/,   or   cm(t)=i0tHmN'(t')ei(EmEN)t'/dt'

(c)PNM=|cM|2=4|HMN'|2(EMEN)2sin2(EMEN2t)

(d)PNM=|VMN|2(EMEN±ω)2sin2(EMEN±ω2t)

(e)RNM=π3ϵ02||2ρ(ω),withω=±(EMEN)

Step by step solution

01

(a) Generalize Equation 9.6

Ψ(t)=cn(t)eiEnt/ψn  HΨ=iΨt;   H=H0+H'(t);   H0ψn=Enψn

so

cneiEnt/Enψn+cneiEnt/H'ψn=ic˙neiEnt/ψn+i(i)cnEneiEnt/ψn

The first and last terms cancel, so

cneiEnt/H'ψn=ic˙neiEnt/ψncneiEnt/ψm|H'|ψn=ic˙neiEnt/ψmψn

Assume orthonormality of the unperturbed states,

ψmψn=δmn,

and define

Hmn'ψm|H'|ψn

cneiEnt/Hmn'=ic˙meiEmt/,orc˙m=incnHmn'ei(EmEn)t/.

02

(b)showing in first-order perturbation theory

Zero order:cN(t)=1,   cm(t)=0   formNcN(t)=1,

Then in first order:

c˙N=iHNN',   or   cN(t)=1i0tHNN'(t')dt',whereasformN:c˙m=iHmN'ei(EmEN)t/,   or   cm(t)=i0tHmN'(t')ei(EmEN)t'/dt'

03

(c) Finding the probability of transition from state N to state M (M≠N),as a function of T

cM(t)=iHMN'0tei(EMEN)t'/dt'=iHMN'[ei(EMEN)t'/i(EMEN)/]|0t=HMN'[ei(EMEN)t/1EMEN]

=HMN'(EMEN)ei(EMEN)t/22isin(EMEN2t)

PNM=|cM|2=4|HMN'|2(EMEN)2sin2(EMEN2t)

04

(d)showing that transitions occur only to states with energyEM=EN±ℏ 

cM(t)=iVMN120t(eiωt'+eiωt')ei(EMEN)t'/dt'=iVMN2[ei(ω+EMEN)t'/i(ω+EMEN)/+ei(ω+EMEN)t'/i(ω+EMEN)/]|0t

IfEM>EN, the second term dominates, and transitions occur only forω(EMEN)/:

cM(t)iVMN21(i/)(EMENω)ei(EMENω)t/22isin(EMENω2t)

so

PNM=|cM|2=|VMN|2(EMENω)2sin2(EMENω2t)

IfEM<EN the first term dominates, and transitions occur only forω(ENEM)/

cM(t)iVMN21(i/)(EMEN+ω)ei(EMEN+ω)t/22isin(EMEN+ω2t)

and hence

PNM=|VMN|2(EMEN+ω)2sin2(EMEN+ω2t)

Combining the two results, we conclude that transitions occur to states with energy EMEN±and

PNM=|VMN|2(EMEN±ω)2sin2(EMEN±ω2t)

05

(e)show that the transition rate for stimulated emission is given by the same formula (Equation 9.47) as for a two-level system.

For lightVba=E0, (Eq. 9.34). The rest is as before (Section 11.2.3), leading to Eq. 9.47:

Vba=E0 (9.34).

Rba=π3ϵ02||2ρ(ω0) (9.47).

RNM=π3ϵ02||2ρ(ω),withω=±(EMEN)

(+signabsorption,signstimulatedemission)

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Most popular questions from this chapter

Magnetic resonance. A spin-1/2 particle with gyromagnetic ratio γ at rest in a static magnetic fieldB0k^ precesses at the Larmor frequencyω0=γB0 (Example 4.3). Now we turn on a small transverse radiofrequency (rf) field,Brf[cos(ωt)ı^sin(ωt)j^]$, so that the total field is

role="math" localid="1659004119542" B=Brfcos(ωt)ı^Brfsin(ωt)j^+B0k^

(a) Construct the 2×2Hamiltonian matrix (Equation 4.158) for this system.

(b) If χ(t)=(a(t)b(t))is the spin state at time t, show that

a˙=i2(Ωeiωtb+ω0a):   b˙=i2(Ωeiωtaω0b)

where ΩγBrfis related to the strength of the rf field.

(c) Check that the general solution fora(t) andb(t) in terms of their initial valuesa0 andb0 is

role="math" localid="1659004637631" a(t)={a0cos(ω't/2)+iω'[a0(ω0ω)+b0Ω]sin(ω't/2)}eiωt/2b(t)={b0cos(ω't/2)+iω'[b0(ωω0)+a0Ω]sin(ω't/2)}eiωt/2

Where

ω'(ωω0)2+Ω2

(d) If the particle starts out with spin up (i.e. a0=1,b0=0,), find the probability of a transition to spin down, as a function of time. Answer:P(t)={Ω2/[(ωω0)2+Ω2]}sin2(ω't/2)

(e) Sketch the resonance curve,

role="math" localid="1659004767993" P(ω)=Ω2(ωω0)2+Ω2,

as a function of the driving frequencyω (for fixed ω0andΩ ). Note that the maximum occurs atω=ω0 Find the "full width at half maximum,"Δω

(f) Since ω0=γB0we can use the experimentally observed resonance to determine the magnetic dipole moment of the particle. In a nuclear magnetic resonance (nmr) experiment the factor of the proton is to be measured, using a static field of 10,000 gauss and an rf field of amplitude gauss. What will the resonant frequency be? (See Section for the magnetic moment of the proton.) Find the width of the resonance curve. (Give your answers in Hz.)

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).

An electron in the n=3,l=0,m=0state of hydrogen decays by a sequence of (electric dipole) transitions to the ground state.

(a) What decay routes are open to it? Specify them in the following way:

|300|nlm|n'l'm'|100.

(b) If you had a bottle full of atoms in this state, what fraction of them would decay via each route?

(c) What is the lifetime of this state? Hint: Once it’s made the first transition, it’s no longer in the state |300\rangle∣300⟩, so only the first step in each sequence is relevant in computing the lifetime.

Calculate ca(t)andcb(t), to second order, for a time-independent perturbation in Problem 9.2. Compare your answer with the exact result.

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(Pabfort) that a transition occurs? Hint: You might want to treat the delta function as the limit of a sequence of rectangles.

Answer:Pab=sin2(|α|lh)

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