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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.

Short Answer

Expert verified

ca2=1-ih∫0tHaat'dt'1-ih∫0tHaat'dt'=11-ih∫0tHaat'dt'2=1ca2=1-ih∫0tHaat'dt'1-ih∫0tHaat'dt'=0Soca2+cb2=1tofirstorder

(b) role="math" localid="1658488366910" da=-ihein∫0tHtdtcbHabeiӬ0tdb=ein∫0tHtdtihHbbcb+eih∫0tHtdtcb

(c) db=ein∫0tHtdtihHbbcb+eih∫0tHtdtcb

Step by step solution

01

(a) Finding  ca(t) and cb(t)

ca=-ihcaHaa+cbHabeiӬ0t⇒cb=-ihcaHaa+cbHabeiӬ0t

cb=-ihcaHaa+cbHabe-iEb-Ebt/h

cb=-ihcaHaa+cbHabe-iEb-Ebt/h

Initial conditions:cat=1,cb0=0,

Zero order:.

First order:

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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 frequencyÓ¬0=γµþ0 (Example 4.3). Now we turn on a small transverse radiofrequency (rf) field,Brf[cos(Ó¬t)ı^−sin(Ó¬t)j^]$, so that the total field is

role="math" localid="1659004119542" B=Brfcos(Ӭt)ı^−Brfsin(Ӭt)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

aË™=i2(Ω±ðiÓ¬tb+Ó¬0a): â¶Ä‰â¶Ä‰bË™=i2(Ω±ð−iÓ¬ta−Ӭ0b)

where Ω≡γµþrfis 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(Ӭ't/2)+iӬ'[a0(Ӭ0−Ӭ)+b0Ω]sin(Ӭ't/2)}eiӬt/2b(t)={b0cos(Ӭ't/2)+iӬ'[b0(Ӭ−Ӭ0)+a0Ω]sin(Ӭ't/2)}e−iӬt/2

Where

Ӭ'≡(Ӭ−Ӭ0)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/[(Ӭ−Ӭ0)2+Ω2]}sin2(Ӭ't/2)

(e) Sketch the resonance curve,

role="math" localid="1659004767993" P(Ӭ)=Ω2(Ӭ−Ӭ0)2+Ω2,

as a function of the driving frequencyӬ (for fixed Ӭ0andΩ ). Note that the maximum occurs atӬ=Ӭ0 Find the "full width at half maximum,"ΔӬ

(f) Since Ó¬0=γµþ0we 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.)

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.

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?

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(Ӭ)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(Ó¬) (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.

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)

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