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1、word量子化学习题与标准答案Chapter 011. A certain one-particle, one-dimensional system has , where a and b are constants and m is the particles mass. Find the potential-energy function V for this system. (Hint: Use the time-dependent Schrodinger equation.)Solution：As Y(x,t) is known, we can derive the correspon
2、ding derivatives. According to time-dependent Schroedinger equation, substituting into the derivatives, we get 2. At a certain instant of time, a one-particle, one-dimensional system has , where b = 3.000 nm. If a measurement of xis made at this time in the system, find the probability that the resu
3、lt(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 is normalized.So
4、lution：a) The probability of finding an particle in a space between x and x+dx is given by b)c) Clearly, the minimum of probability density is at x=0, where the probability density vanishes. d) 3. A one-particle, one-dimensional system has the state functionwhere a is a constant and c = 2.000 . If t
5、he particles position is measured at t and 2.001 .Solution：when t=0, the wavefunction is simplified as Chapter 021 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 h
6、ow many times will the electron be found between 0.600 ? 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 ?Solution: a) In a 1D box, the energ
7、y and wave-function of a micro-system are given by therefore, the probability density of finding the electron between 0.600 and 0.601 is b) From the definition of probability,the probability of finding an electron between x and x+dx is given byAs the number of measurements of finding the electron be
8、tween 0.700 and 0.701 is known, the number of system is 2. When a particle of mass 9.1*10-28 g in a certain one-dimensional box goes from the n = 5 level to the n = 2 level, it emits a photon of frequency 6.0*1014 s-1. Find the length of the box. Solution. 3. An electron in a stationary state of a o
9、ne-dimensional box of length 0.300 nm emits a photon of frequency 5.05*1015 s-1. Find the initial and final quantum numbers for this transition.Solution: 4. 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 function
10、s and energy levels for this choice of origin.Solution: The wavefunction for a particle in a one-dimernsional box can be written asIf the coordinate origin is defined at the center of the box, the boundary conditions are given asbining Eq1 with Eq2, we getEq3 leads to A=0, or =0. We will discuss bot
11、h situations in the following section. If A=0, B must be non-zero number otherwise the wavefunction vanishes. If A05. 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 = sin/c
12、osSolution: For the particle in a certain rectangular well, the E fulfill with Substituting into the V and E, we getChapter 031. If f (x) = 3x2 f(x) + 2xdf /dx, give an expression for .Solution：Extracting f(x) from the known equation leads to the expression of A2. (a) Show that (+)2 = (+)2 for any t
13、wo operators. (b) Under what conditions is (+)2 equal to 2+2+2?Solution: a) b) If and only if A and B mute, (+)2 equals to 2+2+23. If = d2/dx2 and = x2, find (a) x3; (b) x3; (c) f(x); (d)f(x)Solution: a) b) c) d)4. Classify these operators as linear or nonlinear: (a) 3x2d2/dx2; (b) ( )2; (c) dx; (d)
14、 exp; (e) .Solution: Linear operator is subject to the following condition. a) Linearb) Nonlinearc) Linear d) Nonlineare) Linear5. The Laplace transform operator is defined by(a) Is linear? (b) Evaluate (1). (c) Evaluate eax, assuming that pa.Solution:a) L is a linear operator b) c)6. We define the
15、translation operator by f (x) = f (x + h). (a) Is a linear operator? (b) Evaluate ()x2. Solution: a) The translation operator is linear operatorb) 7. Evaluate the mutators (a) ,; (b) ,; (c) ,; (d) , ; (e) ,; (f) , .Solution: a)b)c)d)e)f)Chapter 041：The one-dimensional harmonic-oscillator is at its f
16、irst excited state and its wavefunction is given asplease 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 ene
17、rgy. 3) Evaluate the expectation value of potential energy 4) Total Energy = T + V2. The one-dimensional harmonic-oscillator Hamiltonian isThe raising and lowering operators for this problem are defined as, Show that , , , Show that and are indeed ladder operators and that the eigenvalues are spaced
18、 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 this state first with and then with and show that the lowest energy eigenva
19、lue is . Finally, conclude that, n = 0, 1, 2, Answer: 1） Write down the definition of operator2) Expand the operators in full form. 3) Evaluate the corresponding bination of operators In the same manner, we can get 4) Substituting the above municators into the Schroeidnger equation, we get This show
20、s that and are indeed ladder operators and that the eigenvalues are spaced at intervals of hv. 5) Suppose that Y is the eigenfunction with the lowest eigenvalue. According to the definition of A_ operator, we haveAs Y is the eigenfunction with the lowest eigenvalue, the above equation is fulfilled i
21、f and only if Operating on the wave function for this state first with and then with leads to Therefore, the lowest energy is 1/2 hv. Chapter 051. For the ground state of the one-dimensional harmonic oscillator, pute the standard deviations Dx and Dpx and check that the uncertainty principle is obey
22、ed. Answer: 1) The ground state wavefunction of the one-dimensional harmonic oscillator is given by2) The standard deviations Dx and Dpx are defined as The product of Dx and Dp is given by It shows that the uncertainty principle is obeyed. 2. (a) Show that the three mutation relations , = , , = , ,
23、= are equivalent to the single relation (b) Find ,Answer: 1): 2):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 Lz. Hence, the angles are given by: 4. plete this equation: Chapter 061. E
24、xplain why each of the following integrals must be zero, where the functions 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 prop
25、erties of eigenfunctions b) 02. Use parity to find which of the following integrals must be zero: (a) ; (b) ; (c) . 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 outes of a measurement of Lz are ,
26、0, and . For each of the following wave functions, give the probabilities of each of these three results: (a); (b); (c). Then find for each of these three wave functions.Answer:a) , therefore, the probabilities are: 0%, 100%, 0% , the probabilities are 50%, 0%, 50%. ,the probabilities are 100%, 0%,
27、0% b) 0，0，14. A measurement yields 21/2 for the magnitude of a particles orbital angular momentum. If Lx is now measured, what are the possible outes?Answer: 1): Since the wavefunction is the eigenfunction of L2, a measurement of the magnitude of the orbital angular momentum should be , The possible
28、 outes when measure Lx are -1, 0, 1Chapter 071. Which of the following operators are Hermitian: d/dx, i(d/dx), 4d2/dx2, i(d2/dx2)?Answer：An operator in one-D space is Hermitian if a) b)c) This operator can be written as a product of 1D kinetic operator and a constant. Hence, its Hermitian. d) As the
29、 third operator is Hermitian, this operator is not Hermitian. 2. If and are Hermitian operators, prove that their product is Hermitian if and only if and mute. (b) If and are Hermitian, prove that 1/2(+) is Hermitian. (c) IsHermitian? (d) Is 1/2(+) Hermitian?Answer: 1) If operator A and B mute, we h
30、ave OperatorA and B are Hermitian, we have Therefore, when A and B mute, the following equation fulfills. Namely, AB is also Hermitian.2) Operator A and B are Hermitian, we getThe above equation shows that the operator 1/2AB+BA is Hermitian. c) xpx is not Hermitian since both x and px are Hermitian
31、and do not mute. d) YesChapter 081. Apply the variation functionto 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
32、that c must be a positive number.2） check the normalization of the variation function.3) The variation integral equals to4) The minimum of the variation integral is5) The percent error in the ground state is 0%2. If the normalized variation function for 0 xl is applied to the particle-in-a-one-dimen
33、sional-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
34、wavefunction must equal to zero at x=0 and x=l. However, the trial variation function 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 (where c is a variation parameter) to a proble
35、m 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 = c2/2m +15a/64c3. Find the minimum value of W for this variation function.Solution: 4. In 1971 a paper was published that applied the normalized variation function Nexp(-br2a02-cr
36、/a0) 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 evaluati
37、on 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 Nexp(-br2a02-cr/a0) when b equals to zero. Therefore, adopting the normalized variation function as a trial variati
38、on function should also have no error in the ground state energy for hydrogen atom. 5. Prove that, for a system with a nondegenerate ground state, , if is any normalized, well-behaved function that is not equal to the true ground-state wave function. (E0 is the lowest-energy eigenvalue of )Solution:
39、As the eigenfunctions of the Hermitian operator H form a plete set, any well-behaved function which is subject to the same boundary condition can be expanded as a linear bination of the eigenfunction of the Hermitian operator, namely, , where Yis are eigenfunctions of Hermitian operator H, cis are c
40、onstant. The expectation value of f with respect to the Hermitian operator is Chapter 09, 101. For the anharmonic oscillator with Hamiltonian , 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
41、 the harmonic oscillator is Hence, the first order correct to energy of the first excited state is given by 2. Consider the one-particle, one-dimensional system with potential-energyV = V0 for , V = 0 for and and V = elsewhere, where V0 = . Treat the system as a perturbed particle in a box. (a) Find
42、 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 byTake this as unperturbed wavefunction, and t
43、he perturbation H is given by V. a) The first-order energy correction for Yn is b) The first correction to the wavefunction is given by 3. For an anharmonic oscillator with , take as cx3. (a) Find E(1) for the state with quantum number v. (b) Find E(2) for the state with quantum number v. You will n
44、eed the following integral: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, the first order energy correction is zero.b) The second order energy corr
45、ection is given by4. Calculate the angle that the spin vector S makes with the z axis for an electron with spin function . Solution: For an electron, both S and Sz equal to one half. The magnitude of S is 5. (a) Show that and do not mute with each other. (b) Show that and mute when they are applied
46、to antisymmetric functions.Solution: a) Set the wavefunction to beb) When the function is antisymmetric, we have6. Which of the following functions are (a) symmetric? (b) antisymmetric? (1) ; (2) ; (3) ; (4) ; (5) ; (6) .Solution: (2) is antisymmetric(3), (5) and (6) are symmetricChapter 11, 131. Ho
47、w 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) 2n2,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 moment
