量子力学-斯坦福大学-Chapter09-08

1、Non-degenerate Perturbation TheoryProblem :nnnHE cant solve exactly.But02HHHH0000Lim 0nnnHEwithUnperturbed eigenvalue problem.Can solve exactly.0nE0n 2HHTherefore, know and .called perturbationsCopyright Michael D. Fayer, 2007Solutions of 0n 000012,00nmmn 0000nnnHE complete, orthonormal set of state

2、swith eigenvaluesand000012,EEEKronecker delta10nmnmnm Copyright Michael D. Fayer, 2007Expand wavefunction02nnnn 02nnnnEEEEandHn nEHave series for Substitute these series into the original eigenvalue equationnnnHE Copyright Michael D. Fayer, 200702HHHHalso have 000000000nnnnnnnnnHEHHEE 00020nnnnnnnnn

3、HHHEEE 0Sum of infinite number of terms for all powers of equals 0.Coefficients of the individual powers of must equal 0.00000nnnHE00000nnnnnnnnnHHHEEE00000nnnnnnHHEEzerothorder - 0firstorder - 1secondorder - 2Copyright Michael D. Fayer, 2007000nnnnnHEEH nE n First order correctionWant to find and .

4、 n 0iiinc Expand 00000niiiiiiiHc Hc EThenSubstituting this result.also substitutingAfter substitution0000iininnicEEEHCopyright Michael D. Fayer, 2007After substitution0000iininnicEEEH0n 000000niininnnicEEEHLeft multiply by000000iinninnnicEEEH 000ni 000nnEEunless n = i,but thenTherefore, the left sid

5、e is 0.Copyright Michael D. Fayer, 2007 000nnnEH We have00000nnnnnEH 00nnnEH The first order correction to the energy.Then0nnnEEE 0nnnEEE 00nnnnnEHHAbsorbing into and nnnnHEThe first order correction to the energy is the expectation value of . HCopyright Michael D. Fayer, 20070000iininnicEEEHFirst o

6、rder correction to the wavefunctionAgain using the equation obtained after substituting series expansions 000000jiinijnnicEEEH 0000jjnjnncEEEH 0000jjnjncEEH 0000jnjnjHcjnEE 0j Left multiply byEquals zero unless i = j.Coefficients in expansion of ket in terms of the zeroth order kets.Copyright Michae

7、l D. Fayer, 2007 0000jnjnjHcjnEE 00jnjnjHcEE jnHH0j 0n is the bracket of with and . 0000()jnnnjjnjHEE ThereforeThe prime on the sum meanj n.zeroth order ket correction to zeroth order ketenergy denominatorCopyright Michael D. Fayer, 2007First order corrections000000()jnnnjjnjnjnjHHHEE 000nnnnnnnnEEH

8、HH Copyright Michael D. Fayer, 2007Second Order CorrectionsUsing 2 coefficientExpandingn n 00niinnnniniH HEHEE Substituting and following same type of procedures yields 2 coefficients have been absorbed.Second order correction dueto first order piece of H.Second order correction due to anadditional

9、second order piece of H. 02000000kmmnnnknnkkmnknmnkHHHHEEEEEE 000()knkknkHEE Second order correction dueto first order piece of H.Second order correction due to anadditional second order piece of H.0000niinniinH HHH Copyright Michael D. Fayer, 2007 00niinnniniH HHEE 0nnEEH0000()jnnnjjnjHEE 02000000k

10、mmnnnknkkmnknmnkHHHHEEEEEE 000()knkknkHEE Energy and Ket Corrected to First and Second OrderCopyright Michael D. Fayer, 2007xE Example: x3 and x4 perturbation of the Harmonic OscillatorVibrational potential of molecules not harmonic.Approximately harmonic near potential minimum.Expand potential in p

11、ower series.First additional terms in potential after x2 term are x3 and x4. Copyright Michael D. Fayer, 20072234122pHkxcxqxmcubic “force constant”quartic “force constant”202122pHkxm 012Haaa a 0012En harmonic oscillator know solutionsnzeroth order eigenvalueszeroth order eigenkets34Hcxqxperturbation

12、c and q are expansion coefficients like .0When and 0,cqHHCopyright Michael D. Fayer, 2007nnHn H n 34n cxqx n34c n xnq n x n 1202xaak In Dirac representation 33xaa First consider cubic term.323,.aa aaa aa30n xn Multiply out. Many terms.None of the terms have the same number of raising and lowering op

13、erators.Copyright Michael D. Fayer, 2007 2244024n x nn aank 4aa 40n x n has terms with same number of raising and lowering operators.Therefore, 12n aaa annn 1n a a aa nn n 21n aa aann2n a aa a nn 1n aa a a nn n 1n a aaannn Using1/21/21 and (1)1a nnnannn Only terms with the same number ofraising and

14、lowering operators arenon-zero.There are six terms.Copyright Michael D. Fayer, 2007 426(1 2)n aannn 222023122nnqHnnk Sum of the six termsTherefore0k m 2420km 220220131222nEnqnnm WithEnergy levels not equally spaced.Real molecules, levels get closer together q is negative.Correction grows with n fast

15、er than zeroth order termdecrease in level spacing.Copyright Michael D. Fayer, 200711HE Perturbation Theory for Degenerate States22HE 1 2 andnormalize and orthogonal1 2 andDegenerate, same eigenvalue, E.11221 1221If with ccc cc cHE Any superposition of degenerate eigenstates is also an eigenstatewit

16、h the same eigenvalue.Copyright Michael D. Fayer, 2007n linearly independent states with same eigenvaluesystem n-fold degenerateCan form an infinite number of sets of .Nothing unique about any one set of n degenerate eigenkets.i Can form n orthonormal from the degerate .innCopyright Michael D. Fayer

17、, 2007 0jjjHHE0000jjjHE Want approximate solution tozeroth orderHamiltonianperturbation0 zeroth ordereigenketzeroth orderenergy00012,m0iEBut is m-fold degenerate.Call these m eigenkets belonging to the m-fold degenerate E00000121mEEEEorthonormalWithCopyright Michael D. Fayer, 20070iiHere is the diff

18、iculty0 perturbed ketzeroth order ket having eigenvalue, 01E0i 00001122immccc0i 0.i But, is a linear combination of the We dont know which particular linear combination it is.is the correct zeroth order ket, but we dont know the ci.Copyright Michael D. Fayer, 2007To solve problemi 01EEE 01mijjijc Ex

19、pand E and Some superposition, but we dont know the cj.Dont know correct zeroth order function.0000111mmjjjjjjHcEc 00011mijjjHEcEH Substituting the expansions for E and into 0iiiHHEi and obtaining the coefficients of powers of , giveszerothorderfirstorderwant theseCopyright Michael D. Fayer, 20070ik

20、kkA 00011mijjjHEcEH To solvesubstitute0jH 0000jkkjkHH 00kk0jH 0k NeedUse projection operatorThe projection operator gives the piece of that is .0i Then the sum over all k gives the expansion of in terms of the .0jH 00kjkjHH 00jkjkkHH DefiningKnown know perturbation piece of theHamiltonian and the ze

21、roth order kets.Copyright Michael D. Fayer, 2007 00011mijjjHEcEH 0011mmjjjkjkjjkc Hc H this piece becomes0i 00000000111mmkkikjijjkjikkjkjEEAE cc H Left multiplying bySubstituting this and gives 0ikkkA 00000111mmkkkjjjkjkkjkjEEAE cc H Result of operating H0 on the zeroth order kets.Copyright Michael

22、D. Fayer, 2007 00000000111mmkkikjijjkjikkjkjEEAE cc H Correction to the EnergiesTwo cases: i m (the degenerate states) and i m.i mLeft hand side sum over k equals zero unless k = i.But with i m, 0000110 Therefore, iiEEEEThe left hand side of the equation = 0.Right hand side, first term non-zero when

23、 j = i. Bracket = 1, normalization.Second term non-zero when k = i. Bracket = 1, normalization.The result is10mijjijH cE c We dont know the cs and the . E s Copyright Michael D. Fayer, 200710mijjijH cE c is a system of m of equations for the cjs. 11112210mmHEcH cHc 21 122220mmH cHEcHc 1 1220mmmmmHcH

24、cHEcOne equation for each index i of ci.120mcccBesides trivial solution ofonly get solution if the determinant of the coefficients vanish. 11121222120mmmmmmHEHHHEHHHHE 00jkjkHH We know theHave mth degreeequation for the .E s Copyright Michael D. Fayer, 2007Solve mth degree equation get the . Now hav

25、e the corrections to energies. 11112210mmHEcH cHc 21 122220mmH cHEcHc 1 1220mmmmmHcHcHEciE s iE To find the correct zeroth order eigenvectors, one for each , substitute (one at a time) into system of equations.iE Get system of equations for the coefficients, cjs.11221*,mmc cc cc cThere are only m 1 conditions because can multiply everything by constant.Use normalization for mth condition.Now we have the correct zeroth order functions.ijH Know the . Copyright Michael D. Fayer, 20071,2,mE EE011iiEEEim The solutions to the mth degree equation (ex