量子化学课程习题及标准答案

量子化学习题及标准答案Chapter 011. A certain one-particle,one-dimensional system has/2bmx ibt e ae --=ψ,where a and b are constants and m is theparticle ’s mass. Find thepotential-energy function V for thissystem. (Hint : Use the time-dependentSchrodinger equation.)Solution ：As (x,t) is known, we canderive the corresponding derivatives. ⎪⎪⎩⎪⎪⎨⎧ψ+ψ-=∂ψ∂ψ-=∂ψ∂⇒=ψ--222222/42),(),(),(2 x m b bm xt x ib t t x e ae t x bmx ibtAccording to time-dependentSchroedinger equation,substituting into the derivatives, weget222),(mx b t x V =2. At a certain instant of time, aone-particle, one-dimensional systemhas bx xe b /||2/13)/2(-=ψ, where b = 3.000 nm. If ameasurement of x is made at this time inthe system, find the probability thatthe result (a) lies between 0.9000 nm and0.9001 nm (treat this interval asinfinitesimal); (b) lies between 0 and2 nm (use the table of integrals, ifnecessary). (c) For what value of x isthe probability density a minimum?(There is no need to use calculus toanswer this.) (d) Verify that ψ isnormalized. 222(,)(,)2x t x t V i t m x∂ψ∂ψ=-+∂∂Solution ：a) The probability of findingan particle in a space between x and x+dx is given by6/223210*29.32--==ψ=dx e x b dx P b x b) 0753.02910*20/223==⎰--dx e x bP b x c) Clearly, the minimum of probabilitydensity is at x=0, where the probabilitydensity vanishes. d)4220/223/223/2232===ψ=⎰⎰⎰⎰+∞-+∞∞--+∞∞--+∞∞-dx e x b dx e x b dx e x b dx P b x b x b x3. A one-particle, one-dimensionalsystem has the state function2222/4/16/4/12)/32)((cos )/2)((sin c x c x xe c at e c at --+=ψππwhere a is a constant and c = 2.000 Å.If the particle ’s position is measuredat t = 0, estimate the probability thatthe result will lie between 2.000 Å and2.001 Å.Solution ：when t=0, the wavefunction is simplified as441610*158.2)32(),(22--==ψc x xec t x πChapter 021. Consider an electron in a one-dimensional box of length2.000Åwith the left end of the box at x = 0.(a) Suppose we have one million of these systems, each in the n= 1 state, and we measure the x coordinate of the electron in each system. About how many times will the electron be found between 0.600 Åand 0.601 Å? Consider the interval to be infinitesimal. Hint: Check whether your calculator is set to degrees or radians.(b) Suppose we have a large number of these systems, each in the n =1 state, and we measure the x coordinate of the electron in each system and find the electron between 0.700 Å and 0.701 Å in 126 of the measurements. In about how many measurements will the electron be found between 1.000 Å and 1.001 Å? Solution: a) In a 1D box, the energyand wave-function of a micro-system are given by)sin(2,22222x ln l ml n E πψπ== therefore, the probability density offinding the electron between 0.600 and0.601 Å is65510*545.6)(sin 242⇒==-dx x ln l P πb) From the definition of probability,the probability of finding an electron between x and x+dx is given bydx x l n l P )(sin 22π= As the number of measurements of findingthe electron between 0.700 and 0.701 Å is known, the number of system is1(sin 22*158712158712001.0)7.02*1(sin 2212612622=⇒===πP P N2. When a particle of mass 9.1*10-28 g ina certain one-dimensional box goes from the n = 5 level to the n = 2 level, itemits a photon of frequency 6.0*1014 s -1.Find the length of the box. Solution.lml h n n ml n n E lower up lower up 36222222222110*26646.18)(2)(-=-=-=∆ π3. An electron in a stationary state of a one-dimensional box of length 0.300 nmemits a photon of frequency 5.05*1015 s -1.Find the initial and final quantum numbers for this transition. Solution:2,388)(2)(22222222222===-⇒=-=-=∆lower upper lower up lower up lower up n n n n hv ml h n n ml n n E π4. For the particle in a one-dimensional box of length l , we could have put the coordinate origin at the center of the box. Find the wave functions and energylevels for this choice of origin.Solution: The wavefunction for a particle in a one-dimernsional box can be written as)2()2()(x mE BSin x mE ACos x+=ψ If the coordinate origin is defined at the center of the box, the boundary conditions are given as2()22(0)(2()22(0)(22mE BSin l mE ACos x mE BSin l mE ACos x l x lx +⇒=-⇒==-= ψψ Combining Eq1 with Eq2, we get)4(,0)22()3(,0)22(Eq l mE BSin Eq l mE ACos ==Eq3 leads to A=0, or )22(l mE Cos =0. We willdiscuss both situations in the following section.If A=0, B must be non-zero number otherwise the wavefunction vanishes.2220)22(02mlh n E n l mE l mE Sin B π=⇒=⇒=⇒≠If A ≠08)12()21(220)22(00)22(0)22(0222mlh n E n l mE l mE Cos B l mE Sin l mE Cos A ψπ⇒+=⇒+=⇒==⇒≠⇒=⇒≠5. For an electron in a certain rectangular well with a depth of 20.0 eV, the lowest energy lies 3.00 eV above the bottom of the well. Find the width of this well. Hint : Use tanθ = sin θ/cos θ Solution : For the particle in a certainrectangular well, the E fulfill with)2sin()2()2cos()(21010l mE V E l mE E E V ---=- Substituting into the V and E, we get1010011110*64.22)7954.0(7954.022)(2)2()2()2(-----=⇒+-=⇒+-=⇒-=--==lowest l mEn l n l mE V E E E V l mE Tan l mE Cos l mE Sin ππChapter 03 1. If A ˆf (x ) = 3x 2f (x ) + 2xd f /dx , givean expression for Aˆ. Solution ：Extracting f(x) from the known equationleads to the expression of Adx d x x A 23ˆ2+=2. (a) Show that (Aˆ+Bˆ)2= (Bˆ+Aˆ)2for any two operators. (b) Under what conditionsis (A ˆ+B ˆ)2 equal to A ˆ2+2A ˆB ˆ+B ˆ2?Solution:a)2222ˆ()ˆˆ)(ˆˆ(ˆˆˆˆˆˆˆˆˆˆˆ)ˆˆ)(ˆˆ()ˆˆ(B A B A BA B A A B B A B B A A B A B A B A =++=+++=+++=++=+ b)B AA B A B B A A B A ˆˆ2ˆˆˆˆˆˆˆ)ˆˆ(2222++=+++=+If and only if A and B commute, (Aˆ+Bˆ)2equals to A ˆ2+2A ˆB ˆ+B ˆ2 3. If A ˆ = d 2/dx 2 and B ˆ = x 2, find (a) A ˆB ˆx 3; (b) B ˆA ˆx 3; (c) AˆB ˆf (x ); (d)B ˆA ˆf (x )Solution:a)3522320ˆˆx x dxd x B A == b)3322236ˆˆx x dxd x x A B == c))()(4)(2)]()(2[)]([)(ˆˆ2222222x f dxd x x f dx d x x f x f dxd x x xf dx d x f x dx d x f B A ++=+==d))()()(ˆˆ222222x f dxd x x f dx d x x f A B ==4. Classify these operators as linear ornonlinear: (a) 3x 2d 2/dx 2; (b) ( )2; (c) ∫dx ; (d) exp; (e) ∑=n x 1.Solution:Linear operator is subject to the following condition.f A c cf Ag A f A g f A ˆ)(ˆˆˆ)(ˆ=+=+ a) Linearb) Nonlinear c) Linear d) Nonlinear e) Linear5. The Laplace transform operator Lˆ is defined by⎰∞-=0)()(ˆdx x f e x f L px(a) Is L ˆ linear? (b) Evaluate L ˆ(1). (c)Evaluate L ˆe ax , assuming that p >a .Solution:a) L is a linear operatorb),1)1(ˆ0>==⎰∞-p p dx e L px c)pdx e dx e e e L x a p ax px ax ===⎰⎰∞--∞-)(ˆ0)(06. We define the translation operator hTˆ by hT ˆf (x ) = f (x + h ). (a) Is hT ˆ a linear operator? (b) Evaluate(2ˆ3ˆ121+-T T )x 2.Solution:a) The translation operator is linear operator2212212121(2ˆ3ˆ)2ˆ3ˆ(x x x T x T x T T +=+-=+-7. Evaluate the commutators (a) [xˆ,xp ˆ]; (b) [x ˆ,2ˆxp ]; (c) [x ˆ,y p ˆ]; (d) [x ˆ, ),,(ˆz y x V ]; (e)[xˆ,H ˆ]; (f) [z y x ˆˆˆ, 2ˆxp ]. Solution:a)i xx x x i x x i p xx =∂∂--∂∂-=∂∂-=)ˆ1ˆ(],ˆ[]ˆ,ˆ[b)xp i p p x p x p p xx x x x x x∂∂==+=222ˆ2)ˆ]ˆ,ˆ[]ˆ,ˆ[ˆ(]ˆ,ˆ[c)0)ˆˆ(],ˆ[]ˆ,ˆ[=∂∂-∂∂-=∂∂-=yx y x i y x i p xy0ˆ),,(ˆ),,(ˆˆ)],,(ˆ,ˆ[=-=x z y x V z y x V x z y x V xe)mxT x V T x H x ==+=21,ˆ[]ˆ,ˆ[]ˆˆ,ˆ[]ˆ,ˆ[f)xz y p z y xx∂∂=ˆˆ2]ˆ,ˆˆˆ[22Chapter 041：The one-dimensional harmonic-oscillator is at its first excited state and its wavefunction is given as)21exp()()(2)(24/14/31x x x βπβψ-=please evaluate the expectation values(average values) of kinetic energy (T), potential energy (V) and the total energy.Answer: 1) First of all, check the normalization property of the wavefunction.2) Evaluate the expectationvalue of kinetic energy.3) Evaluate the expectation valueof potential energy4) Total Energy = T + V 2. The one-dimensional harmonic-oscillator Hamiltonian is2222ˆ22ˆˆxm v mp H xπ+= The raising and lowering operators forthis problem are defined as]ˆ2ˆ[)2(1ˆ2/1x ivm p m A x π+=+, ]ˆ2ˆ[)2(1ˆ2/1x ivm p m A x π-=-Show thathv H A A 21ˆˆˆ-=-+, hv HA A21ˆˆˆ+=+-, hv A A-=-+]ˆ,ˆ[ ++=A hv A H ˆ]ˆ,ˆ[, ---=A hv A Hˆ]ˆ,ˆ[ Show that +Aˆ and -A ˆ are indeed ladder operators and that the eigenvalues are spaced at intervals of hv . Since both the kinetic energy and the potential energy are nonnegative, we expect the energy eigenvalues to be nonnegative. Hence there must be a state of minimum energy. Operate on the wave function for thisstate first with -A ˆ and then with +Aˆ and show that the lowest energy eigenvalueis hv 21. Finally, conclude that hvn E )21(+=, n = 0, 1, 2, …Answer:1） Write down the definition of operatordx di px -=ˆ2) Expand the operators in full form.]2[21]2[2122ˆ222222vmxi dx di mA vmxi dx d i mA mx v dx d m H πππ--=+-=+-=-+ 3) Evaluate the correspondingcombination of operators]22][2[21ˆ4222[21]2[2]2[2[21]2[21]22[ˆ21ˆ2122]4222[21]]2[2]2[[21]2[21]2[2121ˆ2122]42[21]4222[21]]2[2]2[[21]2[21]2[21222222332222222333222222222222222222222222222222222222222222222mx v dx d m vmxi dx d i mH A v dx d i mx v dx d ix v dx d i v dxd m i m vmxi dx d i mx v vmxi dx d i dxd m m vmxi dx d i mmx v dx d m A H hv H hv mx v dx d m x m v dxd vmx dx d x vm vm dx d m vmxi dxd i vmxi vmxi dx d i dx d i m vmxi dx d i mvmxi dx d i m A A hv H hv mx v dx d m x m v vm dxd m x m v dxd vmx dx d x vm vm dx d m vmxi dxd i vmxi vmxi dx d i dx d i m vmxi dx d i mvmxi dx d i m A A πππππππππππππππππππππππππππππππππ+-+-=+---=+-++--=+-+-=+=++-=+-++-=+--+--=+---=-=-+-=+--=++---=--+---=--+-=+++--++++=+-=+-=+-=-hvA vmxi m dx d i m hv vmxi m hv dxd i m hv mxi v dx d i v m H A A H ]22121[]221[]21[]42[21ˆˆ222ππππIn the same manner, we can get---=-hvA H A A H ˆˆ 4) Substituting the above communicatorsinto the Schroeidnger equation, we getψψψψψψψψψψψψ------++++++-=-=-=+=+=+==A hv E hvA E A hvA H A A H A hv E hvA E A hvA H A A H E H)(]ˆ[ˆ)(]ˆ[ˆˆThis shows that +Aˆ and -A ˆ are indeed ladder operators and that theeigenvalues are spaced at intervals ofhv . 5) Suppose that is the eigenfunctionwith the lowest eigenvalue. ψψlowest E H =ˆAccording to the definition of A_operator, we haveψψ---=A hv E A H )(ˆ As is the eigenfunction with thelowest eigenvalue, the above equation isfulfilled if and only if 0=-ψAOperating on the wave function for thisstate first with -Aˆ and then with +A ˆ leads to ψψψψhv H hv H A A 21ˆ]21ˆ[0=⇒-==-+Therefore, the lowest energy is 1/2 hv.3,2,1,0,)(21ˆ=+=n hv n H ψψChapter 051. For the ground state of theone-dimensional harmonic oscillator,compute the standard deviations x andp x and check that the uncertaintyprinciple is obeyed.Answer:1) The ground state wavefunction of theone-dimensional harmonic oscillator isgiven by2214141)(x e ααπψ--=2) The standard deviations x and p xare defined as222x x x -=∆222)∆∆-=p(p∆)(pThe product of x and p is given by2422122 ==∙=∆∆ααp xIt shows that the uncertainty principleis obeyed.2. (a) Show that the three commutationrelations [x L ˆ,y L ˆ] = z L i ˆ , [y L ˆ,z L ˆ] = x L i ˆ , [z L ˆ,x Lˆ] = y Li ˆ are equivalent to the single relation L L Lˆˆˆ i =⨯ (b) Find [2ˆx L ,y L ˆ] Answer:1): zy x y x z x z y z y x y x x z z y x y y x z x x z y z z y y z x z z y x y z x y x z y x z y x z y x L i L L L i L L L i L L k L j L i L i k L L j L L i L L k L L L L j L L L L i L L L L i L L j L L i L L k L L j L L k L L k L j L i L k L j L i L L L k L j L i L L ===⇒++=++=-+-+-=-++--=++⨯++=⨯++=],[],[],[)(],[],[],[)()()()()(ˆˆˆ2):)()()(],[],[],[2x z z x xz z x x y x y x x y x L L L L i L L i L i L L L L L L L L L +=+=+=3. Calculate the possible angles betweenL and the z axis for l = 2.Answer:The possible angles between L and thez axis are equivalent the angles betweenL and L z . Hence, the angles are given by:Lm Cos L z ==+=θ6)12(2 ︒︒︒︒=7.144,10.114,00.90,91.65,26.35θ 4. Complete this equation:m l m l z Y m Y L 333ˆChapter 061. Explain why each of the following integrals must be zero, where the functions are hydrogenlike wave functions: (a) <2p 1|z L ˆ|3p -1>; (b) <3p 0|z L ˆ|3p 0>Answer:Both 3p -1 and 3p 0 are eigenfunctions of L z , with eigenvalues of -1 and 0, respectively. Therefore, the above integrals can be simplified asa) due to orthogonalization properties of eigenfunctions03|213ˆ21111=-=--p p p L p z b) 02. Use parity to find which of the following integrals must be zero: (a)<2s |x |2p x >; (b) <2s |x 2|2p x >; (c)<2p y |x |2p x >. The functions in these integrals are hydrogenlike wave functions.Answer:1) b) and c) must be zero.3. For a hydrogen atom in a p state, the possible outcomes of a measurement of L z are – ħ, 0, and ħ. For each of the following wave functions, give the probabilities of each of these three results: (a)z p 2ψ; (b) y p 2ψ; (c) 12p ψ. Then find <L z > for each of these three wave functions.Answer:a) 022p p z ψψ=, therefore, the probabilities are: 0%, 100%, 0% )(2111222-+=p p p x ψψψ, the probabilities are 50%, 0%, 50%.12p ψ,the probabilities are 100%, 0%, 0% b) 0，0，14. A measurement yields 21/2ħ for themagnitude of a particle’s orbital angular momentum. If L x is now measured,what are the possible outcomes?Answer:1): Since the wavefunction is the eigenfunction of L2, a measurement of the magnitude of the orbital angular momentum should be+LLL=)1⇒21 (=,The possible outcomes when measure L x are-1, 0, 1Chapter 071. Which of the following operators areHermitian: d /dx , i (d /dx ), 4d 2/dx 2,i (d 2/d x 2)?Answer ：An operator in one-D space is Hermitianif⎰⎰=dx A dx A **)ˆ(ˆψψψψ a)⎰⎰⎰⎰-=-=-=∞∞-dx dx d dx dx d dx dx d dx dxd *****)(ψψψψψψψψψψ b)⎰⎰⎰⎰=-=-=∞∞-dx dx d i dx d i dx dx d i i dx dx d i ****)(ψψψψψψψψψψc)⎰⎰⎰⎰⎰=+-=-=-=∞∞-∞∞-dxdxddxddxddxdxddxddxddxddxddxdxdψψψψψψψψψψψψψ2*22****22*44 444 44This operator can be written as a product of 1D kinetic operator and a constant. Hence, it’s Hermitian.d) As the third operator is Hermitian, this operator is not Hermitian.2. If Aˆ and Bˆ are Hermitian operators, prove that their product AˆBˆ is Hermitian if and only if Aˆ and Bˆcommute. (b) If Aˆ and Bˆ are Hermitian,prove that 1/2(A ˆB ˆ+B ˆA ˆ) is Hermitian. (c)Is x px ˆˆHermitian? (d) Is 1/2(x p x ˆˆ+x p xˆˆ) Hermitian?Answer:1)If operator A and B commute , we have⎰=-⇒=-⇒=0])ˆˆˆˆ[(0ˆˆˆˆˆˆˆˆˆ*τψψd A B B AA B B A A B B A ⎰⎰⎰=⇒=-⇒τψψτψψτψψd A B d B A d A B B A ***]ˆˆ[]ˆˆ[0])ˆˆˆˆ[( Operator A and B are Hermitian, we have⎰⎰⎰==⇒τψψτψψτψψd B A d B A d A B ˆˆ)ˆ()ˆ(]ˆˆ[***Therefore, when A and B commute, thefollowing equation fulfills. Namely, ABis also Hermitian.⎰⎰=τψψτψψd B A d B A ˆˆ]ˆˆ[** 2)]ˆˆˆˆ[21)]ˆˆˆˆ(21[***⎰⎰⎰+=+τψψτψψτψψd A B d B A d A B B AOperator A and B are Hermitian, we get⎰⎰⎰⎰⎰⎰⎰+=+⇒+=+=+τψψτψψτψψτψψτψψτψψτψψd A B B A d A B B A d B A A B d B A d A B d A B d B A *******])ˆˆˆˆ(21[)ˆˆˆˆ(21])ˆˆˆˆ(21[)ˆˆ()ˆˆ([21]ˆˆˆˆ[21The above equation shows that theoperator 1/2[AB+BA] is Hermitian.c) xp x is not Hermitian since both x andpx are Hermitian and do not commute.d) YesChapter 081. Apply the variation function cr e -=φtothe hydrogen atom; choose the parameterc to minimize the variational integral,and calculate the percent error in theground-state energy.Solution ：1） The requirement of the variationfunction being a well-behaved functionrequires that c must be a positivenumber.2） check the normalization of the variation function.322*)(c d d Sin dr r e d cr πϕθθτφφ==⎰⎰⎰⎰- 3) The variation integral equals to)2(214])2[(2)1(21()121(ˆ32223*32*32*3**-=-∂∂+∂∂-=-∇-=-∇-==⎰⎰⎰⎰⎰⎰--c c c dr r e rr r e c d r c c d r c d d H w cr cr τφφπφπτφφπτφφτφφ4) The minimum of the variation integral is 21101-=⇒=⇒=-=∂∂w c c c w5) The percent error in the ground stateis 0%2. If the normalized variation functionx l 2/13)/3(=φ for 0 ≤ x ≤ l is applied tothe particle-in-a-one-dimensional-boxproblem, one finds that the variationintegral equals zero, which is less thanthe true ground-state energy. What is wrong?Solution:The correct trail variation function must be subject to the same boundary condition of the given problem. For the particle in a 1D box problem, the correct wavefunction must equal to zero at x=0 and x=l. However, the trial variation function x l 2/13)/3(=φ does not fulfill these requirement. The variation integral based on this incorrect variation function does not make any sense.3. Application of the variation function 2cx e -=φ(where c is a variationparameter) to a problem with V = af (x), where a is a positive constant and f (x ) is a certain function of x , gives thevariation integral as W = c ħ2/2m+15a /64c 3. Find the minimum value of Wfor this variation function.Solution:23434123434141min 4141413272598.03)25(23)25(0)64152(m a m a w m a c dc c a m c d c w ==⇒±=⇒=+=∂∂4. In 1971 a paper was published that applied the normalized variationfunction N exp(-br 2a 02-cr /a 0) to thehydrogen atom and stated that minimization of the variation integral with respect to the parameters b and c yielded an energy 0.7% above the true ground-state energy for infinite nuclear mass. Without doing any calculations, state why this result must be wrong. Solution:From the evaluation of exercise 1, we know that the variation function exp(-cr) gives no error in the ground state of hydrogen atom. This function is a special case of the normalized variationfunction N exp(-br 2a 02-cr /a 0) when bequals to zero. Therefore, adopting the normalized variation function as a trial variation function should also have no error in the ground state energy for hydrogen atom.5. Prove that, for a system with anondegenerate ground state, 0*ˆE d H >⎰τφφ, ifφ is any normalized, well-behaved function that is not equal to the true ground-state wave function. (E 0 is thelowest-energy eigenvalue of H ˆ) Solution:As the eigenfunctions of the Hermitian operator H form a complete set, any well-behaved function which is subjectto the same boundary condition can be expanded as a linear combination of the eigenfunction of the Hermitian operator, namely,∑∞==0i i i c ψφ, where i s are eigenfunctions of Hermitian operator H, c i s are constant.The expectation valueof with respect to the Hermitian operator is0201020201020020*00**00*0*0*ˆˆ)(ˆE c E E c E c c E c E c E c c E c cH c c d c H c d H i i i i i i i i i i i i i ij j j j i i i j j i i j j j i i i ==+>+======∑∑∑∑∑∑∑⎰∑∑⎰∑∑⎰∞=∞=∞=∞=∞=∞=∞=∞=∞=∞=∞=δψψτψψτφφChapter 09, 101. For the anharmonic oscillator with Hamiltonian43222212ˆdx cx kx dx d m H +++-= , evaluate E (1) for the first excited state, taking the unperturbed system as the harmonic oscillator.Solution:The wavefunction of the first excited state of the harmonic oscillator is241312)4(x xe απαψ-=Hence, the first order correct to energy of the first excited state is given by6213422134134324131'*1415)4()4()4)(()4(ˆ222απαπαπαπαψψαααd dx e x d dx x d e x xe x d x c xe dx H x x x ==∙=∙+∙=---⎰⎰⎰⎰2. Consider the one-particle, one-dimensional system withpotential-energyV = V 0 for l x l 4341<<, V = 0 for l x 410≤≤ and l x l ≤≤43and V = ∞ elsewhere, where V 0 = 22/ml .Treat the system as a perturbed particle in a box. (a) Find the first-order energy correction for the general stationary state with quantum number n . (b) Find the first-order correction to the wave function of the stationary state with quantum number n .Solution:The wavefunction of a particle in 1D box is given by)(2)0(x ln Sin l n πψ= Take this as unperturbed wavefunction, and the perturbation H ’ is given by V. a) The first-order energy correction forn is])23[]2[(224]2[4[2)()(2)()(2ˆ00004341)0('*)0()1(πππππππππψψn Sin n Sin n V V n S n Sin l l V dx V x l n Sin x l n Sin l dx x ln VSin x l n Sin l dx H E l ln n -+=-+====⎰⎰⎰b) The first correction to the wavefunction is given by)0()0()0()0()0()1(2)0()0(222)0('(818mn m mn nm n mn n E E H n E E ml h n E ψψψψ∑∞≠-==-⇒=3. For an anharmonic oscillator with3222212ˆcx kx dx d m H ++-= , take 'ˆH as cx 3. (a) Find E (1) for the state with quantum number v .(b) Find E (2) for the state with quantumnumber v . You will need the following integral:3,'2/13)0(3)0('2/)1[(3]8/)3)(2)(1[(||++++++>=<v v v v v v v v x αδαψψ1,'2/38/)2)(1([)2/(3---++v v v v v v αδα Solution: a) As the potential of the unperturbed is a even function, the eigenfunctions of the unperturbed system are either even or odd. The perturbation is an odd function with respect to x. Hence, thefirst order energy correction is zero.b) The second order energy correction is given by)51212(8]38)2)(1(838)1(338)3)(2)(1([()2(3)21(38)3)(2)(1((''2323333332)0()0(1,2/31,2/33,32)0()0(2)0(3)0()0()0()0()0()0()0()2(++-=--++-++-+++=-+++++++=-=-=∑∑∑∞≠-++∞≠∞≠n n hvc hvn n n hv n hv n hv n n n c E E n n n n n n n c E E cx E E H H E n m mn n n n n n n n m mn m n n m m n m n nm n αααααδαδαδαψψψψψψ4. Calculate the angle that the spin vector S makes with the z axis for an electron with spin function α.Solution:For an electron, both S and S z equal to one half. The magnitude of S is74.54]31[23)1(===+=ArcCos S S S θ5. (a) Show that 12ˆP and 23ˆP do not commutewith each other. (b) Show that 12ˆPand 23ˆP commute when they are applied to antisymmetric functions. Solution:a) Set the wavefunction to be)1()3()2()2()1()3(321321φφφφφφ≠b) When the function is antisymmetric, we haveψψψψψψψ2312121223231223ˆˆ)(ˆˆˆ)(ˆˆˆP P P P P P P P =-=-==-=-=6. Which of the following functions are (a) symmetric? (b) antisymmetric?(1) )2()1()2()1(ααg f ; (2) )]2()1()2()1()[2()1(αββα-f f ;(3) )3()2()1()3()2()1(βββf f f ; (4) )(21r r a e --;(5) )]1()2()2()1()][2()1()2()1([βαβα--f g g f ; (6) )(21221r r a e r +-. Solution:(2) is antisymmetric(3), (5) and (6) are symmetricChapter 11, 131. How many electrons can be put in each of the following: (a) a shell with principal quantum number n ; (b) a subshell with quantum numbers n and l ; (c) an orbital; (d) a spin-orbital? Solution:a) 2n 2, b) 2*(2l+1), c) 2, d) 12. Give the possible values of the total-angular-momentum quantum number J that result from the addition of angular momentum with quantum numbers (a) 3/2 and 4; (b) 2, 3, and 1/2 Solution:Coupling between two angular momentums with quantum number j 1 and j 2 gives the possible quantum number J of the total angular momentum as:2121j j J j j +<<-a) The possible values are 11/2, 9/2, 7/2,5/2b) The possible values are:11/2, 9/2, 9/2, 7/2, 7/2, 5/2, 5/2, 3/2, 3/2, 1/23. Find the terms that arise from each of the following electron configurations: (a) 1s22s22p63s23p5g; (b) 1s22s22p3p3d (c) 1s22s22p24dSolution:As fully-filled sub-shells do not contribution the total orbital and spin angular momentum, we can ignore the electrons in these sub-shells while considering the atomic terms. Hence, a) The atomic terms can be:3H, 1H, 3G, 1G, 3F, 1Fb) The atomic terms can be:4G, 2G, 4F, 2F, 4D, 2D, 4P, 2P, 4S, 2S4F 2F, 4D, 2D, 4P, 2P4D, 2D, 4P, 2P, 4S, 2Sc) The atomic terms can be:。