量子化學課程習題及標準答案

1、量子化學習題及標準答案Chapter 01A certain one-particle, one-dimensional system has ； =ae-btemx2/ , where a and b are constants andmis theparticle s mass. Find the potential-energy function V for this system. ( Hint : Use the time-dependent Schrodinger equation.)Solution : As (x,t) is known, we can derive the c

2、orresponding derivatives.一(x,t) 二一 ib： 2 -t；(x,t) = ae也emx/ =(x,t) - 2bmi手 4 4b2m2x2 ,ax2h 412According to time-dependent Schroedinger equation,苴 3ni3+v(x,t)*(x,t) i M 2m &xsubstituting into the derivatives, we getV (x,t)= 2b2 mx2At a certain instant of time, a one-particle, one-dimensional system h

3、as=(2/b3)1/2xe4x|/b, where b = 3.000 nm. If ameasurement of x is made at this time in the system, find the probability that the result (a) lies between 0.9000 nmand 0.9001 nm (treat thisintervalasinfinitesimal); (b) lies between 0 and 2 nm (use the table of integrals,ifnecessary). (c) For what value

4、 of x is the probability density a minimum? (There is no need to use calculus to answer this.) (d) Verify that 甲 is normalized.Solution : a) The probability of findingan particle is given byb) P2 b3 * *_92*102 -2 x / bx edx= 3.29* 1060 b3x2e2x/bdx = 0.0753in a space between x and x+dx c) Clearly, th

5、e minimum of probability density is at x=0, where the probabilitydensity vanishes.d)-He-He-He-He口. T 2r-2-NZ2 -2|x|/b_2_ v2 -2|x|/b_4 f 2 -2x/b .P - dx -3 x e dx - 3 x e dx 一 3 x e d二三b A one-particle, one-dimensional system has the state function 2.22,22.1/4 -x2/ c26.1/4 -x2 /c2b3 _.：b3 0at t = 0,

6、estimate the probability that the result will lie between 2.000? and 2.001 ?.Solution : when t=0, the wavefunction is simplified as(當產(chǎn)xe c4: 2.158 * 10 4Chapter 021. Consider an electron in a one-dimensional box of length 2.000 ? with the left end of the box atx = 0.(a) Suppose we have one million o

7、f these systems, each in the n = 1 state, and we measure the x coordinate of the electron in each system. About how manytimes will the electron be found between 0.600 ? and 0.601 ? Consider the interval to be infinitesimal. Hint: Check whether your calculator is set to degrees or radians. (b) Suppos

8、e 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? and 0.701 ? in126 of the measurements. In about how many measurements will the electron be found between 1.000? and 1.001?Solution: a

9、) In a 1D box, the energyand wave-function of a micro-system are given byx)2./ nsin(一 l ltherefore, the probability density of finding the electron between 0.600 and 0.601 ? isP = 2sin 2 (n llJIx)dx = 6.545 * 10 -4 =655b) From the definition of probability, the probability of finding an electron bet

10、ween x and x+dx is given by2 . 2,n二、, P = - sin (x)dx l l TOC o 1-5 h z As the number of measurements of finding the electron between 0.700 and 0.701?is known, the number of system is12612622 N = = 158712= P = 158712* sin2(-P 20 1* 二2P -sin2(0.7)0.0012222. When a particle of mass 9.1*10-28g ina cert

11、ain one-dimensional box goes fromthe n = 5 level to then = 2 level, itemits a photon of frequency 6.0*10 Find the length of the box.Solution.(n2 - n2 )二 2 2(n2 - n2 )h2up lowerup lower )匚=222ml28ml214 s -11.26646*103. An electron in a stationary state of a one-dimensional box of length0.300 nmemits

12、a photon of frequency 5.05*10 15 s-1.Find the initialandfinalquantumnumbers for this transition.Solution: 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.nupper/ 22、一 22(nupnlower )2ml2二3 n2j lowe

13、rJ(nup2、卜2nlower )h8ml2二 hv= n：p2 nlower4. For the particlein a one-dimensionalSolution:The wavefunction for a particle in a one-dimernsional box can be written as TOC o 1-5 h z 2mE2mE 、(x) = ACos(x) BSin(x)hhIf the coordinate origin is defined at the center of the box, the boundary conditions are g

14、iven as7 2mE BSin(2mEBSin(一72mE l(x) x2(x) x2ACos(-)2mE l、ACos( )2Combining Eq1 with Eq2, we getv2mE lACos( 廠 0,(Eq3)2,2mE lBSin(二廠 0,(Eq4)2Eq3 leads to A=0, orcos(-mE2)=0. We willdiscuss both situations in the following section.If A=Q B must be non-zero number 2mE ln =h On2h2mlotherwise the wavefun

15、ction vanishes.2mE lB=0= Sin()=0 =2IfAw 0J2mE lSin(J2mE lB= 0Cos(J2mE l,1、(n 一)二222(2n 1)2h2 _ r8ml25. 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 tan 6 = sin 8/cos 6 Soluti

16、on : For the particle in a certainrectangular well, the E fulfill with2(V0 - E)Ecos(、2mE,l) = (2E -V0)sin(/2mE -1l)Substituting into the V and E, we getSin(A2mE 1)Cos(2mE 1l)=Tan(2mE 1l)= 2mE 1l )-0.7954 n = l =2y(V0- E)E2E -V。( 0.7954 n )V2mE=llowes2.64*10 10Chapter 032IfAf(x)=3xf(x)+2xdf / dx,give

17、an expression for A.Solution :Extracting f(x) from the known equation leads to the expression of AA = 3x22xdx(a) Show that (及+)2 = ( +a)2 for anytwo operators. (b) Under what conditions is ( a?+b) 2 equal toa2+2a? ?+/？Solution:(/?E?)2 = (Re?)(A 自廠 A2 A? BA二 e?2 BA A? R2 =自用(g R)= (B(A e?)2 二 A2 A? B

18、A B2 二 A 2A?If and only if A and B commute, (a?+e?)22equals to a +2a+bIf a = d2/ dx2 and = x2, find (a) Ax3;3(b)國 Ax； (c)AE?f(X)； (d) E?Af (x)Solution:a)RBx3 =b)BAx3 =c)A?f(x)=會 dxd 532 x5 - 20 x3dx22 d 3 八 3x -2 x = 6xdx TOC o 1-5 h z 2d2 d:f(x)2xf(x) x f (x) HYPERLINK l bookmark146 o Current Docum

19、ent dxdx=2f (x)d2 d24xf (x) x 2 f (x)dxdx2d)BAf (x)=x2d2f(x)= x2d2dx2f(x)Classify these operators as linear or nonlinear: (a) 3x2d2/ dx2; (b) ( )2; (c) / dx; (d) exp; (e) 士 . x4Solution:Linear operator is subject to the following condition.-f + g 廠 Af + Ag區(qū)(cf)cAfLinearNonlinearLinearNonlinearLinear

20、The Laplace transform operator L? is defined byLf (x) = e-pxf (x)dx(a) Is l? linear? (b) Evaluate 式1). (c) Evaluate L?eax, assuming thatpa.Solution:L is a linear operator b)QOL?(1) = e pxdx = , p 0 TOC o 1-5 h z 0pc)QOOOI?/ axpx ax(p a)x?(e ) = e e dx= e dx=00PWe define the translation operatorTh by

21、 Thf (x) = f ( x + h). (a) IsTh alinear operator? (b) Evaluate (砰-3T? +2)x2.Solution:a) The translation operator is linear operatorb)(T?2 3T? 2)x2 = T?2x23#x2 2x2 = (xEvaluate the commutators (a) 凡？x;(b)及，叼；(c)及，？y; (d)夕，V?(x,y,z); (e)?,町(f) 卓,町Solution:X，?x 二i ?,一 二C zi (?一C z1次一）二iC z?,82:(僅區(qū)伙x,?x

22、?x)= 2i ?x= 2 2-死？y = i 死一=i (5?)=0yy yd)?,V?(x, y,z)二姆(x, y,z) V(x, y,z)?= 0e)1尺田%P V出國x,廣2mf)槌優(yōu)2 2超一xChapter 041:The one-dimensionalharmonic-oscillator is at its first excited state and its wavefunction is given as，/、, 2( ：)3/4 / 1 ： 2、 - 1 (x) =11；4 xexp(-二 x )(二)2please evaluate the expectation

23、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.Exp 1f Asstnptln59:=phi Sqrt 2 A 3 4 Sqrt Sqrt Pi x IntegrateInfinityz InfinityOut59-Out60= 12)Evaluate the expectationvalue of kinet

24、ic energy.In63:= T1 D D phiz x , xIntegrate 1 2 A2 mAssunptions 03 2 多 x 7 42CUt63=-32Out64=4mT1 phi, x. Infinity, Inf ini勺,3 11 4flEvaluate the expectation valueof potential energyln67:= V人2 2m A2 xA2Integrate V phi人2 r xz Infinityf Infinity f Assunptiony2 2 2Out 67= 2m32Out68= 4 mTotal Energy = T

25、+ V2.Theone-dimensionalharmonic-oscillator Hamiltonian isH?= -p2 2v2mX22mThe raising and lowering operators forthis problem are defined as1kW2iv峭Show that1hvA_A.=H? 1hv A?.,/?= hv2 ,-2 , 一H?, A+ = hv/+，H?, A=-hvA_Show that a and a_ are indeed ladder operators and that the eigenvalues are spaced at i

26、ntervals 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 this state first with A and then with 凡 and show that the lowest energy eigenval

27、ue is ihv. Finally, conclude thatE =（n ；）hv, n = 0, 1,2,ofAnswer:Write down the definitionoperatordx?xExpand the operators in full form.2 d22m dx22 2 2v mxV2mdx2 vmxi3) Evaluatedx2 vmxithe correspondingcombination of operators力+鄴X3 ，+ P ,乙耳1xp-pxwj乙憶一/xwj乙力一:xp山乙山乙r l =XIJU A1乙乙乙xpw乙Gxpw乙廠-lXLUA u乙

28、+ - m ! - =4+/PVxp:2+ 一產(chǎn)八xp紅八u乙-Pxp山乙山3廣Ft 二V：JXIJUA 記 +”小XWAP, 乙 乙乙xp-pxpUJ 乙p-FIXIJUA 化xpPIXIJUAxp=Wxp xp UJS山乙六u 乙 +!XUJAl乙 +!仆-= =Pxp 山乙4xpW乙Qxuj八化 *xw 八1 +=+如P L乙乙乙 7P 7邛G G乙乙xp UJSAM - +zH = Aq + XUJ a +-LL乙乙乙 ,P *G Gxpxpxp UJ 乙X UJ A力+XUJ八化一X4UJ八化一刈八1+ J =ccccpPP 6 l，乙xpIXIUA 遼 + vl - |XIJUA

29、工乙-IXIJUA L乙P乙乙xp LUSAU - UI = AU - XIJU A 化 + 上=LL乙乙乙？P -G G乙xp嚴乙X UJ A巾+9底乙一三廠一 = TOC o 1-5 h z G G G GF j GI乙kX LU A 巾1乙乙乙乙xpxpxpUJ 乙JW/U乙 +X VIUA kS -耳WAu乙乙學一=PPP I*xp 乙F pxp乙 Pxpiu!XUJ 八憶 +!一二VPIXIJUA 化-v! -|XIJUA 化 + IXIJUA ir PXp LUQxujA 憶-1/1 -=p TOC o 1-5 h z Ha - A H? = - 2： v 2i 4二 2V2mxi

30、 = hv-i ,2mdx2m dx2 二 vmxi = hvA11 dhv2 二 vmxi = hv-i 2m 2m dx x 2mIn the same manner, we can getHA_ - A_H? = hvA_Substituting the above communicatorsinto the Schroeidnger equation, we getH? = EHA=A H? hvA =A E hvA= (E hv)AHA=AN - hvA=A_E- hvA= (E - hv)A_This shows that 及+ andA_ are indeedladder ope

31、rators and that the eigenvalues are spaced at intervals of hv.Suppose that is the eigenfunction with the lowest eigenvalue.HEof AHElowestAccording to the definitionoperator, we haveHa (e hv)AAs is the eigenfunction with the lowest eigenvalue, the above equation is fulfilled if and only ifA 0Operatin

32、g on the wave function for this state first with 及一 and then with a leads toA A 0 M 1 hv = H? = 1 hv 22Therefore, the lowest energy is 1/2 hv.H? = 1(n hv) , n = 0,1,2,3Chapter 05For the ground state of the one-dimensional harmonic oscillator, compute the standard deviations x and px and check that t

33、he uncertainty principle is obeyed.Answer:1) The ground state wavefunction of the one-dimensional harmonic oscillator is given by()4 4e-：x2 *2Px2x2In33:= psi Pi A 1 4 A 1 41 2 xA2 ;Integrate psi八2, x, Infinity, Infinity , Assunptiai &qp2 Integrate psiA2 xA2, x, Infinity, Infinity &qpxs Integrate psi

34、A2 x, x7 Infinity, Infinity r xdev 日c2Out34= 11OLrt35=Out閔二 01Out37|= 2(Ap)Out 47= Cut網(wǎng)二 0cut4sg= =:(Ap)2) -(Ap)2In44-psi Pi A 1 4 A 1 4 由斗 1 2x 八2 ;由si D psi, x si2 A2 D D psi, x , x expz2 Integrate psi csi2, x, Infinityf Infinity , Assurptions ( expxs Integrate psix, Infinity, infinity , Assunptio

35、ns 0pdev expx2 expxs A2Cut45=Out46-1 41 4The product ofp is given byx andAIt shows that the uncertainty principle is obeyed.(a) Show that the three commutation relations , L?y = i J , & = i& ,旦 =i L?y are equivalent to the single relation 。l? l? (b) Find ?x2, LyAnswer: 1):l?= LxiLyj Lzk一 一 . 一 一 L?

36、I?= (LxiLyj Lzk) (LxiLyj Lzk)二 LxLyk - LxLzj - LyLxkLyLziLzLxj - LzLyi二(LyLz - LzLy)i (LzLx - LxLz)j (LxLy - LyLx)k=Ly,Lzi Lz,Lxj Lx,Lyk = i (LxiLyj Lzk)二Ly,Lz二 i Lx Lz,Lx= i Ly Lx,Ly= i Lz2)：Lj,Ly = LxLx,Ly Lx,LyLx = Lx(i Lz) (i LZ)LX =i (LxLz LzLx)Calculate the possible angles betweenL and the z a

37、xis for l = 2.Answer:The possible angles between L and the z axis are equivalent the angles between L and Lz. Hence, the angles are given by:L112(2+ 1y =而Cosmz二 35.26 ,65.91 ,90.00 ,114.10 ,144/4.Completethisequation:L?Ym313YmChapter 06Explain why each of the following integralsmust be zero, where t

38、hefunctions are hydrogenlike wave functions:(a);(b) Answer:Both 3p-1 and 3P0 are eigenfunctions of Lz, with eigenvalues of -1 and 0, respectively. Therefore, the above integrals can be simplified as a) due to orthogonalization properties of eigenfunctions (2p1 &3pJ = 1(25 |3pj=0 b) 0Use parity to fi

39、nd which of the following integrals must be zero: (a);(b);(c).The functions in theseintegrals are hydrogenlike wave functions.Answer:1) b) and c) must be zero.For a hydrogen atom in a p state, the possible outcomes of a measurement of Lz are - ?, 0, and ?. For each of the followingwave functions,giv

40、e theprobabilities of each of these three results: (a) 2Pz; (b) z,; (c)1一 Thenfind for each of these three wave functions.Answer:2 Pz = 2po ,therefore, theprobabilities are: 0%, 100%, 0%.1 ,.、2px - 2( 2pi2p4) , the probabilitiesare 50%, 0%, 50%.25 ,the probabilities are 100%, 0%, 0%0 , 0, 1A measure

41、ment yields21/2? for the magnitude of a particle s orbital angular momentum.If Lx 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 orbitalangularmomentum should beThe possible outcomes when measure Lx

42、are -1,0, 1Chapter 07Which of the following operators are Hermitian:d/dx,i(d/dx),4d2/dx2,i (d2/d x2)?Answer:An operator in one-D space is Hermitian if,w A dx(父)dxa)dxb)dx*I.dx 二一dx, *ddx dx)dxdx 二dx(iddxdxdx 二d_ ddxc)dx 二dxdx .dx dxdxdxdx dxdxdxThis operator can be written as a product of 1D kinetic

43、 operator and a constant.Hence, it s Hermitian.d) As the third operator is Hermitian, this operator is not Hermitian.If a and i? are Hermitian operators, prove that their product a i? is Hermitian if and only if A? andi?commute. (b) If A? and 目 are Hermitian, prove that 1/2( Ae?+ma) is Hermitian. (c

44、)Is 祝 Hermitian? (d) Is 1/2(邠x + 成父) Hermitian?Answer:If operator A and B commute , we have A的欲=A?-B？ 二0(Ag- BAy *d=0=(Ag- BA) *d =0= 1 4? *d=1 蹶*dOperator A and B are Hermitian, we have =u BA， d7=mA，)(由心=產(chǎn) A臥d7Therefore, when A and B commute, the following equation fulfills. Namely, AB is also Herm

45、itian.A學*d=1 *A/ d?1 *1(ag+ B?)r d。= 1r *a/ d，+ 產(chǎn) *b? d。22Operator A and B are Hermitian, we get1 *cc*cc1cc*-A? d BA d (B? ) d22(A? )*d =(w ；(ft?+ 險*dT1 *1(A?+ BA)w d。=產(chǎn)1(Af?+ BAy *d。The above equation shows that the operator 1/2AB+BA is Hermitian.xp x is not Hermitian since both x and px are Hermit

46、ian and do not commute.YesChapter 081. Apply the variation function = ecr to the hydrogen atom; choose the parameter c to minimize the variational integral, and calculate the percent error in the ground-state energy.Solution :The requirement of the variation function being a well-behaved function re

47、quires that c must be a positive number.check the normalization of the variation function.31 *-22_d = e r dr Sin. )dd 二二cThe variation integral equals toL 2 一 1)d ;*(-1 22 r23c *1(-)dr=工 c(c - 2)2一 23-crr2c e (-r2cr 23)e r dr - 4c r r*H?dc3*w =;二(dThe minimum of the variation integral isc 1 = 0 = c

48、= 1 = w =The percent error is 0%in the ground statethe 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 a

49、nd x=l. However, the trial variation function=(3/i3)1/2x does not fulfill theserequirement.The variationintegralbased on this incorrect variation function does not make any sense.Applicationof the variationfunction二- cx2一 e (where c is a variation parameter) to a problem with V = af (x), where a is

50、a positive constant andf (x)is a certain function ofx, gives the2 、 variation integral as W =c? /2 m+15a/64 c3. Find the minimum value of Wfor this variation function.Solution:d(2m1564c3dc2m4%/113 35(一)4a4m4 213 3= wmin = 2一， 0.72598a4m4 2In 1971 a paper was published that applied the normalized var

51、iation functionNfexp(- br2a02-cr/a0)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 wro

52、ng. 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 functionNfexp(- br2a02-cr/a0)when bequals to zero. Therefore, adopting the normalized variatio

53、n function as a trial variation function should also have no error in the ground state energy for hydrogen atom.Prove that, for a system with a nondegenerate ground state, *H?d e。，if is any normalized,well-behavedfunction that is not equal to the true ground-state wave function. ( E)is the lowest-en

54、ergy eigenvalue of h?) 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,ci，where is are eigenfunction

55、sof Hermitian operator H,csareconstant.The expectation value ofwithrespect to the Hermitian operator isQOQOQOOOH? d = C G J H? qd 八cjcj i H?i-0j=0i=0j=0oOoOoO_*_一 K_*一:cicjEj, ij 一 GGEii=0 j=0i=0oOQO工c|2Ei =|co|2Eo+ l Igi=0i=1QOQOCo2Eo + z |c/E0= Ez |q|2= Eoi =1i = 0Chapter 09, 101. For the anharmon

56、ic oscillator with2 d 1Hamiltonian 一二二42 cx3 dx4, evaluate 2m dx22E(1) for the first excited state, taking the unperturbed system as the harmonic oscillator.Solution:The wavefunction of the first excitedstate of the harmonic oscillator is4 3 11 =()4 xe二區(qū)2Hence, the first order correct to energy of t

57、he first excited state is given by24 31 /34 4 3 -()4xe 2 (c , x d , x )()4x冗n15(43 12.2 2 - x4-)2 x e d , x dx ;3 14 X2 6 - x2 .d()2 x e dx ;ji2.Considertheone-particle,one-dimensionalsystemwithpotential-energyV = V)for 1i x 3, V = 0 for ox1i and 44 143 -l x l 4and V = 0 elsewhere, whereV)= 先2/mi2.T

58、reat 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 bo

59、x is given by, (o)2n 二、：=，Sin( x) ，l lTake this as unperturbed wavefunction, and the perturbation H is given by V. a) The first-order energy correction for n isE(1)叫0件n0)dx2 Sin( ln 二 一-x)VSin(x)dx3l=2 Sin(n lll4二 V0乂2 2nJIx)Sin(nlJIx)Vodx =2V0l7 4nSiny 74n(Siny- Sin32)b) The first correction wavefu

60、nction is given byto theE(0)En8ml2E(0) .EnE(0)Em(n(0) mn0)匚(0) - 匚(0) m： n e- Em(0) mln20|：= 2 1 Integrate Sin n Pi 1 x Sin m Pi 1 x fSin 1 zn n48m 1 nA2 in A2CDut20= m n Sin m81m4,1 m n Sin m n 43.For an anharmonic oscillatorwithh? = ./g gkx2 cx3, take 田 as cx3. (a) FindE(1)for the state with quant