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).

Short Answer

Expert verified

The second order perturbation theory for the equation is

cα(2)(t)=α-ibh0tHabc(tc)e-iω0tcdtc-αh20tHabc(tc)e-iω0tc0tcHbac(tcc)eiω0tccdtccdtccb(2)(t)=b-iah0tHabc(t¢)eiω0tcdtc-bh20tHabc(t¢)eiω0tcdtc0tcHbac(tcc)e-iω0tccdtccdtc

Step by step solution

01

Concept.

Consider a time dependent potential, we can solve the Schrodinger equation for this potential in a two state system if we split the Hamiltonian into a time independent part H°and time dependent part H', that is:

H=H°+H1

This is done in the section 9.1 to get the solution.

02

Solving the given equation to second order perturbation theory

Wehave,Zeroorder:cα(0)(t)=α,cb(0)(t)=bThecoefficientsaresolutionsofthecouplesODEs,thatis:Theequation9.13isc.a=-iHab'e-iω0tcb,c.b=-iHab'e-iω0tcαFirstorder;c.a=-iHab'e-0tbcα(1)(t)=α-ib0tHab'(t)e-0tdt'c.a=-iHab'e-0tαcb(1)(t)=b-0tHab'(t)e-0tdt'Secondorder:c.a=-iHab'e-0tb-0tHab'(t)e-0tdt'cα(2)(t)=α-ibHab'(t)e-0tdt'-α20tHab'(t)e-0tdt'0tHba'(t)e-0tdt'dt'Togetcbjustswitchab(whichentailsalsochangingthesignofω0)cb(2)(t)=b-iaHba'(t')e0tdt'-b20tHab'(t')e0tdt'0tHba'(t')e-0tdt'dt'

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Most popular questions from this chapter

Show that the spontaneous emission rate (Equation 9.56) for a transition from n,lton',l' in hydrogen is

e2ω2l23πo0hc3˙×{l+12l+1ifl'=l+1l2l-1ifl'=l-1

where

l=0r3Rnl(r)Rn'J'(r)dr

(The atom starts out with a specific value of m, and it goes toamyof the state’s mconsistent with the selection rules:m'=m+1,m or m -1 . Notice that the answer is independent of m .) Hint: First calculate all the nonzero matrix elements of x,y,and z between role="math" localid="1658313179553" |n|m>andn'l'm'>for the case . From these, determine the quantity

|n'.1+1.m+1rn|m|2+|n'.1+1,mr|nm|2+|n'.1+1,m-1r|nm|2

Then do the same forl'=l-1.

Calculate the lifetime (in seconds) for each of the four n = 2 states of hydrogen. Hint: You’ll need to evaluate matrix elements of the form <ψ100xψ200>,<ψ100yψ211>, and so on. Remember that role="math" localid="1658303993600" x=rsinθcosϕ,y=rsinθsinϕandz=rcosθ. Most of these integrals are zero, so inspect them closely before you start calculating. Answer: 1.60×10-9seconds for all except role="math" localid="1658304185040" ψ200, which is infinite.

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?

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).

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 “accessible” from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration—assuming transitions to higher excited states can be ignored.

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