基础量子化学练习定稿版

量子化学考试试题一、选择题（每题 5 分，共 30 分）1、量子化学中，描述微观粒子运动状态的函数被称为（）A 波函数B 概率密度C 哈密顿量D 薛定谔方程2、下列哪个量子数决定了原子轨道的形状（）A 主量子数B 角量子数C 磁量子数D 自旋量子数3、对于氢原子的 1s 轨道，其电子出现概率最大的位置是（）A 原子核处B 离核无穷远处C 离核一定距离处D 无法确定4、量子化学中，计算分子能量常用的方法是（）A 半经验方法B 从头算方法C 密度泛函理论D 以上都是5、下列哪种化学键具有明显的量子力学特征（）A 离子键B 共价键C 金属键D 氢键6、在量子化学中，分子轨道是由原子轨道线性组合而成，这一原理被称为（）A 杂化轨道理论B 价键理论C 分子轨道理论D 晶体场理论二、填空题（每题 5 分，共 30 分）1、量子力学的基本假设包括波函数假设、算符假设、测量假设、全同性原理和＿_________________ 。

2、氢原子的薛定谔方程在球坐标下的解中，径向波函数 R(r) 与＿_________________ 有关。

3、多电子原子的电子排布遵循的原则有能量最低原理、泡利不相容原理和＿_________________ 。

4、分子的偶极矩是衡量分子＿_________________ 的物理量。

5、密度泛函理论的核心思想是将体系的能量表示为＿_________________ 的泛函。

6、量子化学计算中，常用的基组有 STO-3G、6-31G 等，其中 6-31G 表示的是＿_________________ 。

三、简答题（每题 10 分，共 20 分）1、简述量子化学中 HartreeFock 方法的基本思想。

2、解释为什么分子的振动光谱通常具有一系列的吸收峰，而不是单一的吸收峰。

四、计算题（共 20 分）已知氢原子处于某一激发态的波函数为：ψ ＝1/√8π a₀³（r/a₀) exp(r/2a₀) ，其中 a₀为玻尔半径。

结构化学练习之量子力学基础习题附参考答案量子力学基础习题一、填空（用正确答案填空）1101、光波粒二象性的关系式为_______________________________________。

1102、德布罗意关系式为____________________；宏观物体的λ值比微观物体的λ值_______________。

1103. 在电子衍射实验中，│? 对于电子，它代表___________________。

1104.它解释了不确定性关系为。

1105.一组正交和归一化的波函数？1.2.3，….正交性的数学表达式是，归一性的表达式为。

1106、│?(x1，y1，z1，x2，y2，z2)│2代表______________________。

1107、物理量xpy-ypx的量子力学算符在直角坐标系中的表达式是_____。

1108、质量为m的一个粒子在长为l的一维势箱中运动，(1)体系哈密顿算符的本征函数集为_______________________________；(2)体系的本征值谱为____________________，最低能量为____________；(3)体系处于基态时，粒子出现在0─l/2间的概率为_______________；(4)势箱越长，其电子从基态向激发态跃迁时吸收光谱波长__________；(5)若该粒子在长l、宽为2l的长方形势箱中运动，则其本征函数集为____________，本征值谱为_______________________________。

1109.质量为m的粒子被限制在边长为a的立方体中。

波函数？______;；当粒子处于某种状态时？二百一十一211(x，y，z)=当，最大概率密度处的坐标为7h2_______________________；若体系的能量为，其简并度是_______________。

24ma3h21110、在边长为a的正方体箱中运动的粒子，其能级e=的简并度是_____，24ma27h2e'=的简并度是______________。

量子化学试题及答案一、选择题（每题2分，共10分）1. 量子化学中，描述电子运动状态的基本方程是：A. 薛定谔方程B. 牛顿运动方程C. 麦克斯韦方程D. 热力学第一定律答案：A2. 以下哪个不是量子化学中的量子数？A. 主量子数B. 角量子数C. 磁量子数D. 动量量子数答案：D3. 根据泡利不相容原理，一个原子轨道最多可以容纳：A. 1个电子B. 2个电子C. 3个电子D. 4个电子答案：B4. 量子化学中，分子轨道理论认为分子轨道是由：A. 原子轨道的叠加B. 原子轨道的差分C. 原子轨道的线性组合D. 原子轨道的非线性组合答案：C5. 以下哪个不是量子化学中的波函数？A. 波恩-奥本海默近似B. 哈密顿算符C. 原子轨道D. 分子轨道答案：B二、填空题（每题2分，共10分）1. 量子化学中，电子的波动性可以通过______方程来描述。

答案：薛定谔2. 根据量子化学理论，原子轨道的能级是由______量子数决定的。

答案：主3. 量子化学中，分子的电子云分布可以通过______轨道理论来分析。

答案：分子轨道4. 量子化学中，电子的自旋量子数可以取值为______。

答案：±1/25. 量子化学中，原子轨道的径向分布函数通常用______来表示。

答案：R(r)三、简答题（每题5分，共20分）1. 简述量子化学中波函数的物理意义。

答案：波函数是量子化学中描述电子状态的数学函数，它包含了电子在空间中出现的概率分布信息。

2. 解释量子化学中量子数的作用。

答案：量子数是量子化学中用来描述电子在原子轨道中运动状态的一组整数或半整数，包括主量子数、角量子数、磁量子数和自旋量子数，它们决定了电子的能量、角动量和自旋状态。

3. 描述量子化学中分子轨道理论的基本原理。

答案：分子轨道理论是基于量子力学的基本原理，认为分子轨道是由原子轨道的线性组合形成的，分子轨道的形成可以解释分子的化学性质和稳定性。

4. 量子化学中，如何理解电子的波粒二象性？答案：电子的波粒二象性是指电子既可以表现出粒子的性质，如在原子中占据特定的轨道，也可以表现出波动的性质，如干涉和衍射现象。

1 能量量子化根底稳固1.关于对黑体的认识,以下说法正确的选项是()A.黑体只吸收电磁波,不反射电磁波,看上去是黑的B.黑体辐射电磁波的强度按波长的分布除与温度有关外,还与材料的种类及外表状况有关C.黑体辐射电磁波的强度按波长的分布只与温度有关,与材料的种类及外表状况无关D.如果在一个空腔壁上开一个很小的孔,射入小孔的电磁波在空腔内外表经屡次反射和吸收,最终不能从小孔射出,这个空腔就成了一个黑体解析:黑体自身辐射电磁波,不一定是黑的,选项A错误。

黑体辐射电磁波的强度按波长的分布只与黑体的温度有关,选项B错误,选项C正确。

小孔只吸收电磁波,不反射电磁波,因此小孔成了一个黑体,而不是空腔,选项D错误。

答案:C2.能正确解释黑体辐射实验规律的是()A.能量的连续经典理论B.普朗克提出的能量量子化理论C.以上两种理论体系任何一种都能解释D.牛顿提出的能量微粒说解析:根据黑体辐射的实验规律:随着温度的升高,一方面各种波长的辐射强度都增加;另一方面辐射强度的极大值向波长较短的方向移动。

只能用普朗克提出的能量量子化理论才能得到较满意的解释,应选项B正确。

答案:B3.红、橙、黄、绿四种单色光中,光子能量最小的是()A.红光B.橙光C.黄光D.绿光解析:在红、橙、黄、绿四种颜色的光中,红光的波长最长而频率最小,由光子的能量ε=hν可知红光光子能量最小。

答案:A4.某种单色光的波长为λ,在真空中光速为c,普朗克常量为h,那么电磁波辐射的能量子ε的值为()A.ℎℎℎB.ℎℎC.ℎℎℎD.以上均不正确解析:由波速公式v=λν可得ν=ℎℎ由光的能量子公式ε=hν=ℎℎℎ应选项A正确。

答案:A5.以下描绘两种温度下黑体辐射强度与波长关系的图中,符合黑体辐射实验规律的是()解析:随着温度的升高,辐射强度增加,辐射强度的极大值向着波长较短的方向移动,选项A正确。

答案:A6.(2021·上海卷)光子的能量与其()A.频率成正比B.波长成正比C.速度成正比D.速度的二次方成正比解析:由E=hν、c=λν得E=hν=ℎℎℎ,可见光子的能量与其频率成正比,与其波长成反比,A正确,B 错误;任意能量的光子在真空中传播的速度都是一样的,故C、D错误。

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。

Chapter1Chapter3/4求矩阵的本征值及其近似值例2.在F 表象中Q 算符为 ，求Q 的本征值和正交归一的本征函数。

F*ψF⎦⎤⎢⎣⎡-=33i i Q解：由 得即又C1,C2均不为0，则有 解得入1=4，入2=2当入1=4时，解得此时当入2=2时，解得 此时例3. 一个正交基构成了如下的哈密顿算符矩阵⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-=c c c H 000000200030001 c 为常数1） 求H 2） 3） 比较（1）和（2解：（1）因为 023120030=----=---λλλλλc cc c c)3)(1[(---λλ212c +±(2)修正到二级近似<+=||''W E E E 21211c E -= 22213c E += 23-=c E （3）)211(21222c c +±=+±2213c +=和)211(2c - 比较结果即可知。

Chapter4 会写slater 行列式λψψ=Q 211=C i C 212=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=ψ221)(i x 211=C 22iC -=⎥⎥⎤⎢⎢⎡=ψ21)(x ⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-2100212133C C C C C C i i λλλ033=---λλi iC H λ=ˆHe 基态原子的组态1S 2，其波函数为会算简单的行列式（二阶He 、三阶Li ） He 1S 2Li 1S 22S 1 和什么是变分原理？答:给定一个体系的哈密顿算符 ，如果 是任意一个合格条件下的函数，则有 E 0 （E 0为基态能量， 未归一化）E 0 （E 0为基态能量， 归一化） Chapter5())2(1)1(1)2(1)1(121ββααψs s s s =Hφ≥**=⎰⎰τφφτφφd d H Eφ≥*=⎰τφφd H Eφ())2(1)1(1)2(1)1(12121s s s s s s ϕϕϕϕψ=)3(1)2(2)1(2)3(1)2(1)1(1)3(1)2(1)1(161)321(αααβββαααψs s s s s s s s s =)3(1)2(2)1(2)3(1)2(1)1(1)3(1)2(1)1(161)321(ββββββαααψs s s s s s s s s =1.3、C 2h 群顺式丁二烯为例：基态：2221ψψ到第一激发态：131221ψψψ跃迁是否会产生紫外光谱？（分子平面为YZ 面）解：根据HMO 方法，得到其分子轨道432113717.06015.06015.03717.0φφφφψ+++= b 1 432126015.03717.03717.06015.0φφφφψ--+= a 2 432136015.03717.03717.06015.0φφφφψ+--= b 1432143717.06015.06015.03717.0φφφφψ-+-= a 2基态：2221ψψ 所属的不可约表示为：12211A a a b b =⊗⊗⊗ 第一激发态：131221ψψψ 所属的不可约表示为：21211B b a b b =⊗⊗⊗跃迁矩阵元的不可约表示*xx x dx μψμψ<>=⎰基态第一激发态 1122A B B A ∈⊗⊗= *yy y dy μψμψ<>=⎰基态第一激发态 1221A B B A ∈⊗⊗= *zz z dz μψμψ<>=⎰基态第一激发态 1122A A B B ∈⊗⊗=所以，上述跃迁会产生紫外光谱。

量子力学基础习题一、填空题（在题中的空格处填上正确答案）1101、光波粒二象性的关系式为_______________________________________。

1102、德布罗意关系式为____________________；宏观物体的λ值比微观物体的λ值_______________。

1103、在电子衍射实验中，│ψ│2对一个电子来说，代表___________________。

1104、测不准关系是_____________________，它说明了_____________________。

1105、一组正交、归一的波函数ψ1， ψ2， ψ3，…。

正交性的数学表达式为 ，归一性的表达式为 。

1106、│ψ (x 1， y 1， z 1， x 2， y 2， z 2)│2代表______________________。

1107、物理量xp y - yp x 的量子力学算符在直角坐标系中的表达式是_____。

1108、质量为 m 的一个粒子在长为l 的一维势箱中运动，(1)体系哈密顿算符的本征函数集为_______________________________ ；(2)体系的本征值谱为____________________， 最低能量为____________ ；(3)体系处于基态时， 粒子出现在0 ─ l /2间的概率为_______________ ；(4)势箱越长， 其电子从基态向激发态跃迁时吸收光谱波长__________ ；(5)若该粒子在长l 、宽为2l 的长方形势箱中运动， 则其本征函数集为____________，本征值谱为 _______________________________。

1109、质量为m 的粒子被局限在边长为a 的立方箱中运动。

波函数ψ211(x ，y ，z )= _________________________；当粒子处于状态ψ211时，概率密度最大处坐标是_______________________；若体系的能量为2247ma h ，其简并度是_______________。

基础量子化学练习 TTA standardization office【TTA 5AB- TTAK 08- TTA 2C】2010基础量子化学练习（1）一、 判断正误（ ）1、一个态函数总是等于时间的函数乘以坐标的函数。

（ ）2、 态函数总是Hamiltonian 算符的本征函数。

（ ）3、 Hamiltonian 算符的本征函数的任意线性组合是Hamiltonian 算符的本征函数。

（ ）4、 如果态函数不是算符ˆA的本征函数，则性质A 的一次测量可给出一个不是ˆA的本征值的值。

（ ）5、几率密度与时间无关。

（ ）6、如果两个算符具有共同的本征函数，那么这两个算符可对易。

（ ）7、算符ˆx 与d i dx -可对易。

（ ）8、氢原子Hamiltonian 算符的束缚态的本征函数构成完备集。

（ ）9、厄米算符的本征函数是正交的。

（ ）10、 描述电子轨道运动的波函数必须是奇函数。

二、已知：2ˆˆˆ,Ad dx B x ==，计算2ˆˆˆˆ,()A B A B ⎡⎤+⎣⎦及 三、已知：11223344ˆˆˆˆ,,,,A a A b A a A d ϕϕϕϕϕϕϕϕ====如果任意状态可以表示为12343253,ψϕϕϕϕ=+++那么当我们对该状态进行测量时，获得a 和d 的几率各是多少？求任意状态 的性质A 的平均值。

2010基础量子化学练习（2）一、 判断正误（ ）11、 算符ˆˆˆ,,A B C 满足ˆˆˆˆ,0,,0A B A C ⎡⎤⎡⎤==⎣⎦⎣⎦，则三个算符存在共同的本征函数集。

（ ）12、 不能对易的算符不可能具有共同的本征函数。

（ ）13、 当对本征态的性质A 进行测量时，能够得到的唯一仅有的值是算符ˆA的本征值。

（ ）14、 如果一个算符的平方等于单位算符，那么这个算符的本征值等于＋1或者－1。

（ ）15、 所有品优的奇函数和偶函数都是宇称算符的本征函数。

（ ）16、 满足[]1212ˆˆˆ()()()()A c f x c g x c Af x c Ag x +=+的算符称为线性算符。