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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 of mass m is initially in the ground state of the (one-dimensional) infinite square well. At time t = 0 a “brick” is dropped into the well, so that the potential becomes

V(x)={V0,0xa/20,a/2<xa;,otherwise

where V0E1After a time T, the brick is removed, and the energy of the particle is measured. Find the probability (in first-order perturbation theory) that the result is nowE2 .

We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption?

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.

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.)

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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