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-ihvba0tcos(ωt')eiω0t'dt'=-iVba2h0tej(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,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 “small,” and state precisely what small means in this context, as a constraint on V.

Pab(t)=cb(t)2Vab2hsin2ω0-ωt/2ω0-ω2(9.28)

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


Short Answer

Expert verified

(a)cb(t)=-i2hωrVbaei(ω0-ω)t/2sin(ωrt),=ei(ω-ω0)1/2cos(ωrt)+iω0-ω2ω0sin(ωrt)(b)ca2+cb2=cos2(ωrt)+ω0-ω2ω02sin2(ωrt)+Vab2hωr2sin2(ωrt).(c)Pab(t)=cb(t)2Vab2h2sin2(ω0-ω)t/2(ω0-ω)2(d)ωrt=πt=π/ωr.

Step by step solution

01

(a) Solving the equation 9.13 in wave rotating approximation

ca=-i2hVabeiωte-iω0tcb;cb.=-i2hVbae-iωte-iω0tca

Differentiate the latter, and substitute in the former:

Cb=-iVba2hi(ω0-ω)ei(ω0-ω)tca+ei(ω0-ω)tca˙=i(ω0-ω)-iVba2hei(ω0-ω)tca-iVab2hei(ω0-ω)t-iVab2hei(ω0-ω)tcb=i(ω0-ω)cb˙-Vab22h2cb.d2cbdt2+i(ω-ω0)dcbdt+Vab22h2cb=0.Soluationoftheformcb=eλt:λ2+i(ω0-ω)λ+Vab24h2λ=12-i(ω-ω0)±-(ω-ω0)2-Vab2h2=i-(ω-ω0)2±ωr,withωr.Generalsoluation:cb(t)=Aei(ω-ω0)2+ωrt+Bei(ω-ω0)2+ωrt=e-i(ω-ω0)t/2Aert+Be-rt,or,moreconveniently:cb(t)=e-i(ω-ω0)t/2Ccos(ωrt)+Dsin(ωrt).Butcb(0)=0soC=0cb(t)=Dei(ω0-ω)t/2sin(ωrt).cb=Diω0-ω2ei(ω0-ω)t/2sin˙uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('b21ce702a20675c...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage('" width="0" height="0" role="math">Cb=-iVba2hi(ω0-ω)ei(ω0-ω)tca+ei(ω0-ω)tca˙=i(ω0-ω)-iVba2hei(ω0-ω)tca-iVab2hei(ω0-ω)t-iVab2hei(ω0-ω)tcb=i(ω0-ω)cb˙-Vab22h2cb.d2cbdt2+i(ω-ω0)dcbdt+Vab22h2cb=0.Soluationoftheformcb=eλt:λ2+i(ω0-ω)λ+Vab24h2λ=12-i(ω-ω0)±-(ω-ω0)2-Vab2h2=i-(ω-ω0)2±ωr,withωr.Generalsoluation:cb(t)=Aei(ω-ω0)2+ωrt+Bei(ω-ω0)2+ωrt=e-i(ω-ω0)t/2Aert+Be-rt,or,moreconveniently:cb(t)=e-i(ω-ω0)t/2Ccos(ωrt)+Dsin(ωrt).Butcb(0)=0soC=0cb(t)=Dei(ω0-ω)t/2sin(ωrt).cb=Diω0-ω2ei(ω0-ω)t/2sin(ωrt)+ωrei(ω-ω0)t/2cos(ωrt)˙uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('46fca5c470c333f...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage('" width="0" height="0" role="math">Cb=-iVba2hi(ω0-ω)ei(ω0-ω)tca+ei(ω0-ω)tca˙=i(ω0-ω)-iVba2hei(ω0-ω)tca-iVab2hei(ω0-ω)t-iVab2hei(ω0-ω)tcb=i(ω0-ω)cb˙-Vab22h2cb.d2cbdt2+i(ω-ω0)dcbdt+Vab22h2cb=0.Soluationoftheformcb=eλt:λ2+i(ω0-ω)λ+Vab24h2λ=12-i(ω-ω0)±-(ω-ω0)2-Vab2h2=i-(ω-ω0)2±ωr,withωr.Generalsoluation:cb(t)=Aei(ω-ω0)2+ωrt+Bei(ω-ω0)2+ωrt=e-i(ω-ω0)t/2Aert+Be-rt,or,more

conveniently:cb(t)=e-i(ω-ω0)t/2Ccos(ωrt)+Dsin(ωrt).Butcb(0)=0soC=0cb(t)=Dei(ω0-ω)t/2sin(ωrt).cb=Diω0-ω2ei(ω0-ω)t/2sin(ωrt)+ωrei(ω-ω0)t/2cos(ωrt)˙ca(t)=i2hVbaei(ω0-ω)tcb=˙i2hVbaei(ω0-ω)t/2Diω0-ω2sin(ωrt)+ωrcos(ωrt).Butca1=i2hVbar,orD=-iVba2hωrcb(t)=-i2hωrVbaei(ω0-ω)t/2sin(ωrt),ca(t)=ei(ω-ω0)t/2cos(ωrt)+iω0-ω2sin(ωrt).

02

(b) Determining the transition probability

Cb=-iVba2hi(ω0-ω)ei(ω0-ω)tca+ei(ω0-ω)tca˙=i(ω0-ω)-iVba2hei(ω0-ω)tca-iVab2hei(ω0-ω)t-iVab2hei(ω0-ω)tcb=i(ω0-ω)cb˙-Vab22h2cb.d2cbdt2+i(ω-ω0)dcbdt+Vab22h2cb=0.Soluationoftheformcb=eλt:λ2+i(ω0-ω)λ+Vab24h2λ=12-i(ω-ω0)±-(ω-ω0)2-Vab2h2=i-(ω-ω0)2±ωr,withωr.Generalsoluation:cb(t)=Aei(ω-ω0)2+ωrt+Bei(ω-ω0)2+ωrt=e-i(ω-ω0)t/2Aert+Be-rt,or,moreconveniently:cb(t)=e-i(ω-ω0)t/2Ccos(ωrt)+Dsin(ωrt).Butcb(0)=0soC=0cb(t)=Dei(ω0-ω)t/2sin(ωrt).cb=Diω0-ω2ei(ω0-ω)t/2sin(ωrt)+ωrei(ω-ω0)t/2cos(ωrt)˙uncaught exception: Invalid chunk

in file: /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 68
#0 /var/www/html/integration/lib/php/Boot.class.php(769): com_wiris_plugin_impl_HttpImpl_1(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Invalid chunk') #1 /var/www/html/integration/lib/haxe/Http.class.php(532): _hx_lambda->execute('Invalid chunk') #2 /var/www/html/integration/lib/php/Boot.class.php(769): haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Invalid chunk') #3 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(30): _hx_lambda->execute('Invalid chunk') #4 /var/www/html/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Invalid chunk') #5 /var/www/html/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), Object(sys_net_Socket), NULL) #6 /var/www/html/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(43): haxe_Http->request(true) #7 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(268): com_wiris_plugin_impl_HttpImpl->request(true) #8 /var/www/html/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(307): com_wiris_plugin_impl_RenderImpl->showImage('46fca5c470c333f...', NULL, Object(PhpParamsProvider)) #9 /var/www/html/integration/createimage.php(17): com_wiris_plugin_impl_RenderImpl->createImage('" width="0" height="0" role="math">Pab(t)=cb(t)2=Vab2hωr2sin2(ωrt).Thelargestthisgets(whwnsin2=1whensin2=1)isPab(t)=cb(t)2=Vab2hωr2sin2(ωrt).Thelargestthisgets(whwnsin2=1whensin2=1)isVab2/h24ωr2Andthedenominatorexceedsthenumerator,soP>1(and1onlyifω=ω0)ca2+cb2=cos2(ωrt)+ω0-ω2ωr2sin2(ωrt)+Vab2hωr2sin2(ωrt).=cos2(ωrt)+(ω=ω0)2+(Vab/h)24ωr2sin2(ωrt)=cos2(ωrt)+sin2(ωrt)=1

03

:(c) Checking Pa→b(t)reduces to perturbation theory

If

Vab2×h2(ω-ω0)2,thenωr12ω-ω0,andPabVab2h2sin2ω-ω02t(ω-ω0)2Pab(t)=cb(t)2Vab2h2sin2ω-ω0)t/2(ω-ω0)2

04

(d) At time the system first returns to its initial stage

ωrt=πt=π/ωr.

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Most popular questions from this chapter

A particle starts out (at time t=0 ) in the Nth state of the infinite square well. Now the “floor” of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent:V0(t),withV0(0)=V0(T)=0.

(a) Solve for the exact cm(t), using Equation 11.116, and show that the wave function changes phase, but no transitions occur. Find the phase change, role="math" localid="1658378247097" ϕ(T), in terms of the function V0(t)

(b) Analyze the same problem in first-order perturbation theory, and compare your answers. Compare your answers.
Comment: The same result holds whenever the perturbation simply adds a constant (constant in x, that is, not in to the potential; it has nothing to do with the infinite square well, as such. Compare Problem 1.8.

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.

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

. Comment: Ostensibly, this system oscillates between “” Doesn’t 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—a 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 —this doesn’t affect the calculations, but it allows for a more sensible interpretation of the result.

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

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(krωt).

If the atom is centered at the origin, thenkr1 over the relevant volume,|k|=2π/λ sokr~r/λ1) and that's why we could afford to drop this term. Suppose we keep the first-order correction:

E(r,t)=E0[cos(ωt)+(kr)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" Rba=q2ω5πϵ0c5|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=q2ω3n(n1)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 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

For the examples inProblem 11.24(c) and (d), calculate cm(t)to first order. Check the normalization condition:

m|cmt|2=1,

and comment on any discrepancy. Suppose you wanted to calculate the probability of remaining in the original state ψN ; would you do better to use |cNt|2,or1-mN|cmt|2 ?

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