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 the particle ’s mass. Find the potential-energy function V for this system. (Hint : Use the time-dependent Schrodinger equation.)Solution ：As ψ(x,t) is known, we can derive 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-dependent Schroedinger equation,substituting into the derivatives, we get222),(mx b t x V =2. At a certain instant of time, a one-particle, one-dimensional system has bx xe b /||2/13)/2(-=ψ, where b = 3.000 nm. If a measurement of x is made at this time in the system, find the probability that the result (a) lies between 0.9000 nm and 0.9001 nm (treat this interval as infinitesimal); (b) lies between 0 and 2 nm (use the table of integrals, if necessary). (c) For what value of x is the probability density a minimum? (There is no need to use calculus to answer this.) (d) Verify that ψ is normalized.Solution ：a) The probability of finding an 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 probability density is at x=0, where the probability densityvanishes. d)4220/223/223/2232===ψ=⎰⎰⎰⎰+∞-+∞∞--+∞∞--+∞∞-d e x b dx e x b dx e x b dx P b x b x b x3. A one-particle, one-dimensional system has the state function2222/4/16/4/12)/32)((cos )/2)((sin c x c x xe c at e c at --+=ψππ where a is a constant and c = 2.000 Å. If the particle ’s position is measured at t = 0, estimate the probability that the result will lie between 2.000 Å and 2.001 Å.Solution ：when t=0, the wavefunction is simplified as441610*158.2)32(),(22--==ψc x xe c t x πChapter 021. Consider an electron in a one-dimensional box of length2.000Å with 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 Å and 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 energy and wave-function of a micro-system are given by)sin(2,22222x ln l ml n E πψπ== therefore, the probability density of finding the electron between 0.600 and 0.601 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 finding the 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 in acertain one-dimensional box goes from the n = 5 level to the n = 2 level, it emits a photon offrequency 6.0*1014 s -1. Find the length of thebox.Solution.lml h n n ml n n E lowerup lowerup 36222222222110*26646.18)(2)(-=-=-=∆ π3. An electron in a stationary state of a one-dimensional box of length 0.300 nm emitsa photon of frequency 5.05*1015 s -1. Find theinitial 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 energy levels 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 certain rectangular 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 mE n 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 2 f (x ) + 2xd f /dx , give an expression for Aˆ. Solution ：Extracting f(x) from the known equation leads to the expression of Adx d x x A 23ˆ2+=2. (a) Show that (Aˆ+Bˆ)2 = (Bˆ+Aˆ)2for any twooperators. (b) Under what conditions is (Aˆ+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 AA B A B A B A =++=+++=+++=++=+ b) BA AB A B B A A B A ˆˆ2ˆˆˆˆˆˆˆ)ˆˆ(2222++=+++=+If and only if A and B commute, (Aˆ+B ˆ)2 equals to A ˆ2+2A ˆB ˆ+B ˆ23. If A ˆ = d 2/dx 2 and B ˆ = x 2, find (a) A ˆB ˆx 3; (b)B ˆA ˆx 3; (c) A ˆB ˆf (x ); (d)B ˆA ˆf (x )Solution: a)3522320ˆˆxx 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.fA c cf A g A f A g f A ˆ)(ˆˆˆ)(ˆ=+=+ a) Linearb) Nonlinear c) Linear d) Nonlinear e) Linear5. The Laplace transform operator Lˆ isdefined by⎰∞-=0)()(ˆdx x f e x f L px(a) Is L ˆ linear? (b) EvaluateLˆ(1). (c)Evaluate L ˆe ax , assuming that p >a .Solution:a) L is a linear operator b)0,1)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 hT ˆ 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 operator b)2212212121(2ˆ3ˆ)2ˆ3ˆ(x x x T x T x T T +=+-=+-7. Evaluate the commutators (a) [x ˆ,xpˆ]; (b) [x ˆ,2ˆxp]; (c) [x ˆ,yp ˆ]; (d) [x ˆ, ),,(ˆz y x V ]; (e) [x ˆ,H ˆ]; (f) [z y xˆˆˆ, 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 xyd)0ˆ),,(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 expectation value of kinetic energy.3) Evaluate the expectation value ofpotential energy4) Total Energy = T + V2. The one-dimensional harmonic-oscillator Hamiltonian is2222ˆ22ˆˆxm v mp H xπ+= The raising and lowering operators for thisproblem 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 H A A 21ˆˆˆ+=+-, 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 wavefunction for this state first with -Aˆ and then with +Aˆ and show that the lowest energy eigenvalue ishv 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 corresponding combination ofoperators4242[21]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[21322222222333222222332222222333222222222222222222222222222222222222222222222v dx d ix v dx d i mx v mxi v dxd m i m mx 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 πππππππππππππππππππππππππππππππππππππ+---=+-+-=+---=+-++--=+-+-=+=++-=+-++-=+--+--=+---=-=-+-=+--=++---=--+---=--+-=+++--++++=+-=+-=+-=-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 communicators into 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 the eigenvalues are spaced at intervals of hv .5) Suppose that ψ is the eigenfunction with the lowest eigenvalue. ψψlowest E H =ˆAccording to the definition of A_ operator, we haveψψ---=A hv E A H )(ˆAs ψ is the eigenfunction with the lowest eigenvalue, the above equation is fulfilled if and only if 0=-ψAOperating on the wave function for this statefirst with -A ˆ 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 the one-dimensional harmonic oscillator, compute the standard deviations ∆x and ∆p x and check that the uncertainty principle is obeyed.Answer:1) The ground state wavefunction of the one-dimensional harmonic oscillator is given by2214141)(x e ααπψ--=2) The standard deviations ∆x and ∆p x are defined as222x x x -=∆22∆-∆=p(p∆)(pThe product of ∆x and ∆p is given by2422122 ==∙=∆∆ααp x It shows that the uncertainty principle is obeyed.2. (a) Show that the three commutation relations [x L ˆ,y L ˆ] = z L i ˆ , [y L ˆ,z L ˆ] = x L i ˆ , [z L ˆ,x L ˆ] = y L i ˆare equivalent to the single relation L L Lˆˆˆ i =⨯ (b) Find [2ˆx L,y L ˆ] Answer:1): z y x y x z xz 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 between L and the z axis for l = 2.Answer:The possible angles between L and the z axis are equivalent the angles between L and L z . Hence, the angles are given by:Lm Cos L z ==+=θ6)12(2 ︒︒︒︒=7.144,10.114,00.90,91.65,26.35θ4. 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+L=LL(=⇒21)1,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 Hermitian if⎰⎰=dx A dx A **)ˆ(ˆψψψψa)⎰⎰⎰⎰-=-=-=∞∞-dxdxd dxdxd dx dx d dx dx d *****)(ψψψψψψψψψψb)⎰⎰⎰⎰=-=-=∞∞-dxdxd i dd i dx dx d i i dx dx d i ****)(ψψψψψψψψψψc)⎰⎰⎰⎰⎰=+-=-=-=∞∞-∞∞-dxdxddddxddxdxddxddxddxddxddxdxdψψψψψψψψψψψψψ2*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 Aˆand Bˆare Hermitian operators, prove that their product AˆBˆis Hermitian if and only if Aˆand Bˆcommute. (b) If Aˆand Bˆare Hermitian, prove that 1/2(AˆBˆ+BˆAˆ) is Hermitian.(c) Is x p xˆˆHermitian? (d) Is 1/2(x p xˆˆ+x p xˆˆ)Hermitian? Answer: 1)If operator A and B commute , we have⎰=-⇒=-⇒=0])ˆˆˆˆ[(0ˆˆˆˆˆˆˆˆˆ*τψψd A B B A A 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, AB is 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 A Operator 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 the operator 1/2[AB+BA] is Hermitian.c) xp x is not Hermitian since both x and px are Hermitian and do not commute. d) YesChapter 081. Apply the variation function cr e -=φto the hydrogen atom; choose the parameter c to minimize the variational integral, and calculate the percent error in the ground-state energy. Solution ：1） The requirement of the variation function being a well-behaved function requires that c must be a positive number.2） check the normalization of the variation function.322*)(c d d Sin dr r ed 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 r r r e c d r cc d r cd d H w cr cr τφφπφπτφφπτφφτφφ4) The minimum of the variation integral is21101-=⇒=⇒=-=∂∂w c c c w5) The percent error in the ground state is 0%2. If the normalized variation function x l 2/13)/3(=φ for 0 ≤ x ≤ l is applied to the particle-in-a-one-dimensional-box problem, one finds that the variation integral equals zero, which is less than the 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 basedon this incorrect variation function does not make any sense.3. Application of the variation function 2cx e-=φ(where c is a variation parameter) toa problem with V = af (x), where a is a positive constant and f (x ) is a certain function of x , gives the variation integral as W = c ħ2/2m+15a /64c 3. Find the minimum value of W for this variation function. Solution:23434123434141min4141413272598.03)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 variation functionN exp(-br 2a 02-cr /a 0) to the hydrogen 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 variation function N exp(-br 2a 02-cr /a 0) when b equals 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, if φ is any normalized, well-behaved function that is not equal to the true ground-state wavefunction. (E 0 is the lowest-energy eigenvalueof Hˆ) Solution:As the eigenfunctions of the Hermitian operator H form a complete set, any well-behaved function which is subject to 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 ψis are eigenfunctions of Hermitian operator H, c i s are constant.The expectation value of φ with respect to the Hermitian operator is20102020102002*00**00*0*0*ˆˆ)(ˆE c E E c E c c E c E cE c c E c cH c c d c H c d H i i i i i ii i ii iii ijj j ji ii 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 xeαπαψ-=Hence, the first order correct to energy of the first excited state is given by6213422134134324131'*1415)4()4()4)(()4(ˆ222απαπαπαπαψψαααd dx ex d dx x d ex xx d x c xedx H x xx ==∙=∙+∙=---⎰⎰⎰⎰2. Consider the one-particle, one-dimensional system with potential-energy V = V 0 forl x l 4341<<, V = 0 forl x 410≤≤ andl 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 nπψ=Take this as unperturbed wavefunction, andthe perturbation H ’ is given by V .a) The first-order energy correction for ψn 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 lln n-+=-+====⎰⎰⎰b) The first correction to the wavefunction is given by)0()0()0()0()0()1(2)0()0(222)0('(818mnm mnnm nm n nEEH n E E ml h n Eψψψψ∑∞≠-==-⇒=3. For an anharmonic oscillator with3222212ˆcx kx dx d m H ++-= , take 'ˆHas cx 3. (a) Find E (1) for the state with quantum number v . (b) FindE (2)for the state with quantum number 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 αδαSolution: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, the first 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 hv n n n hv n hv n hv n n n c E E nn n n n n n c E E cx EEH H En m mn n n n n n n nm mn mnnm mnmn n mnαααααδαδαδαψψψψψψ4. 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 θ 5. (a) Show that12ˆP and23ˆP 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 have ψψψψψψψ2312121223231223ˆˆ)(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:2G, 2F, 2D, 2P, 2S4F, 2F, 4D, 2D, 4P, 2P。
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