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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.

Short Answer

Expert verified

ca2=1-ih0tHaat'dt1-ih0tHaat'dt=1+ih0tHaat'dt2=1cb2=-ih0tHbat'ei0tdt'ih0tHabt'ei0tdt'=0Soca2+cb2=1(tofirstorder)

da=-iheih0tHaat'dt'cbH'abei0t.db=-iheih0tHaat'dt'ihH'bbcb+eih0tHbbt'dt'cb

db=-ihe-iH'baei0tdb=-ih0te-itH'bat'ei0tdt'

Step by step solution

01

(a) Finding  ca(t) and cb(t)

ca=-ihcaH'aa+caH'abei0tca=-ihcbH'bb+caH'baei0t

role="math" localid="1658555553364" ca=-ihcaH'aa+cbH'abe-iEb-Eat/h

ca=-ihH'abei0tcb,cb=-ihH'baei0tca

role="math" localid="1658555600978" cb=-ihcbH'bb+caH'bae-iEb-Eat/h

Initial conditions: role="math" localid="1658555900086" ca0=1,cb0=0,

Zero order: ca0=1,cb0=0,

First order:ca=-ihH'aacat=1-ih0tH'aat'dt'cb=-ihH'baei0tcbt=1-ih0tH'bat'ei0tdt'

ca2=1-ih0tHaat'dt1-ih0tHaat'dt=1+ih0tHaat'dt2=1tofirstorderinH'cb2=-ih0tHbat'ei0tdt'ih0tHabt'ei0tdt'=0tofirstorderinH'Soca2+cb2=1(tofirstorder)

02

(b) Showing  da=-iheiϕH'abeiω0tdb;da=-iheiϕH'baeiω0tda

db=-iheih0tHaat'dt'ihH'bbcb+eih0tHbbt'dt'cbButca=-ihcaH'aa+cbH'abe-i0t

Two terms cancel, leaving

db=-iheih0tHaat'dt'cbHabe-i0t.Butcb=e-ih0tH'bbt'dt'db=-iheih0tH'aat'-H'bbtdtHabei0tdb,orda=-iheiH'abei0tdb

Similarly,db=-iheih0tHaat'dt'ihH'bbcb=e-ih0tH'bbt'dt'db.Butcb=-ihcbHbb+caH'baei0t=-iheih0tHaat'dt'caH'baei0t.Butca=e-ih0tHaat'dt'da=-iheih0tH'bbt'-H'aa(t')dt'H'baei0tda=-iheiH'baei0tda

03

(c) Using the method in part (b) to obtain  ca(t) and cb(t)

Initial conditions:ca0=1da(0)=1;cb(0)=0db(0)=0.

Zero order:da(t)=1,dbt=0.

First order:da=0dat=1cat=e-ih0tH'aa(t')dt'

db=-ihe-iH'baei0tdb=-ih0te-i(t')H'ba(t')ei0t'dt'

These don鈥檛 look much like the results in (a), but remember, we鈥檙e only working to first order in H'Hsoca(t)1-ih0tH'aa(t')dt'(tothisorder),whileforcb,thefactorH'bain the integral means it is already first order and hence both the exponential factor in front and e-ishould be replaced by 1. Then cbt1-0tihH'aat'ei0t'dt', and we recover the results in (a).

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Most popular questions from this chapter

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(kr蝇t).

If the atom is centered at the origin, thenkr1 over the relevant volume,|k|=2/ sokr~r/1) and that's why we could afford to drop this term. Suppose we keep the first-order correction:

E(r,t)=E0[cos(t)+(kr)sin(t)].

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" Rba=q250c5|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=q23n(n1)15蟺系0m2c5.

(Note: Here 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 the2S1S 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

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).

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 鈥渁ccessible鈥 from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration鈥攁ssuming transitions to higher excited states can be ignored.

Calculate the lifetime (in seconds) for each of the four n = 2 states of hydrogen. Hint: You鈥檒l need to evaluate matrix elements of the form <100x200>,<100y211>, and so on. Remember that role="math" localid="1658303993600" x=r蝉颈苍胃肠辞蝉,y=r蝉颈苍胃蝉颈苍andz=r肠辞蝉胃. Most of these integrals are zero, so inspect them closely before you start calculating. Answer: 1.6010-9seconds for all except role="math" localid="1658304185040" 200, which is infinite.

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

(9.13).

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