量子力学课后习题

十万种考研考证电子书、题库视频学习平台
，即得
最后，计算（r－3）． 对于 S 态（l＝0），r→0 处φ→C（常数），所以
当 l≠0，利用题（5．7）式（7b），即得
因此
当 l→0，上式右端→∞，所以上式实际上适用于一切 l 值．
讨论：由于总能量算符及径向方程均与磁量子数优无关，所以 与 m 无关．但
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圣才电子书 十万种考研考证电子书、题库视频学习平台

（c）μ子偶素（muonium，指μ＋－μ－束缚体系）． 解：（a）由于正负电子的质量均为 me，电子偶素的约化质量为
此体系的能谱为
（b）μ原子中μ子质量为ｍμ≈207me，原子核的质量为 M，而约化质量为：
a 为 Bohr 半径，上式右边第 2 项为屏蔽 Coulomb 势，求价电子的能级．
（c）r2 的平均值也已在题 5．9 中算出．对于本题，
因此，r 的涨落为
可见 n 越大，
越小，量子力学的结果和 Bohr 量子化轨道的图像越加接近．
5.7 按（5.1）节，式（8），中心力场Ｖ（r）中的粒子的径向方程可以写成
利用 Feynman－Hellmann 定理（见 4.7 题），证明对于处在能量本征态下的三维各向同性 谐振子，有
体系的能谱为
（c）设μ子质量为 mμ，则μ子偶素的约化质量为
，体系的能谱为
概括起来，如采用自然单位（能量自然单位是
，则这几个体系的能级公式都与
氢原子相同，即 μ的大小，其顺序如下
但每个体系的约化质量μ不同．按能量自然单位或按约化质量
电子偶素 氢原子
μ子偶素
μ原子
5.4 对于氢原子基态，计算△x△p．
解：氢原子基态波函数为

练习28.1 证明: ()[]()t G t G -=-++00证明: 根据公式(28.4)()()()00H t t ie t t it t G '--±'±='-θ可知()()00tH ie t it G-+-=θ()()()00H t i e t i t G ---+=-θ则()[]()()000tH i tH i e t ie t i t G θθ=⎥⎦⎤⎢⎣⎡-=+-++()()()t G e t i H t i-==---00θ #28.2证明下列二式成立：()()()()⎰∞∞-±±±±--+-=-''dt 't t VG ''t t G 't t G 't t G 00()()()()⎰∞∞-±±±±--+-=-''dt 't ''t VG ''t t G 't t G 't t G 00证明：因为：()()()⎰∞+∞---±±π=-dE e E G 21't t G 't t E i()()()⎰∞+∞---±±π=-dE e E G 21't t G 't t E i00又因为：()()()()E VG E G E G E G 00±±±±+=即有()()()()()()[]()()()()()()()()()()()()()''dt t ''t VG ''t t G 't t G dE e E VG E G 21't t G dE e E VG E G 21dE e E G 21dE e E VG E G E G 21dE e E G 21't t G '00't t E i00't t E i 0't t E i 0't t E i00't t E i00--+-=π+-=π+π=+π=π=-±∞+∞-±±∞+∞---±±±∞+∞---±±∞+∞---±∞+∞---±±±∞+∞---±±⎰⎰⎰⎰⎰⎰又因为()()()()()()()E VG E G E G E VG E G E G E G 0000±±±±±±±+=+=同理可证得()()()()''dt t ''t VG ''t t G 't t G 't t G '00--+-=-±+∞∞-±±±⎰综上所述()()()()()()()()''dt t ''t VG ''t t G 't t G 't t G ''dt t ''t VG ''t t G 't t G 't t G '0'00--+-=---+-=-±∞+∞-±±±±+∞∞-±±±⎰⎰两式成立。

Introduction to Quantum MechanicsOverviewQuantum Mechanics is a branch of Physics that describes the behavior of matter and energy at a microscopic level. This discipline has had a significant impact on modern science and technology, and its principles have been applied to the development of various fields, such as computing, cryptography and medicine. The study of Quantum Mechanics requires a basic understanding of the principles of Mathematics and Physics. The m of this document is to provide an introduction to Quantum Mechanics and to provide a set of practice exercises with answers that will allow students to test their knowledge and understanding of the subject.Fundamental PrinciplesThe fundamental principles of Quantum Mechanics are based on the concept of a wave-particle duality, which means that particles can behave as both waves and particles simultaneously. The behavior of particles at the microscopic level is probabilistic, and it is described by a wave function. A wave function is a complex function that describes the probability of finding a particle at a givenlocation. The square of the amplitude of the wave function gives the probability density of finding the particle at that point in space. The wave function can be used to calculate various physical quantities, such as the position, momentum and energy of a particle.Operators and ObservablesIn Quantum Mechanics, physical quantities are represented by operators. An operator is a mathematical function that acts on a wave function and generates a new wave function as a result. Operators are used to represent physical observables, such as the position, momentum and energy of a particle. The eigenvalues of an operator correspond to the possible results of a measurement of the corresponding observable. The eigenvectors of an operator correspond to the possible states of a particle. The state of a particle is described by a linear combination of its eigenvectors, which is called a superposition.Schrödinger EquationThe Schrödinger Equation is a mathematical equation that describes the time evolution of a wave function. It is based on the principle of conservation of energy, and it representsthe motion of a quantum system in terms of its wave function. The equation is given by:$$\\hat{H}\\Psi=E\\Psi$$where $\\hat{H}$ is the Hamiltonian operator, $\\Psi$ is the wave function, and E is the energy of the system. The Schrödinger Equation is the foundation of Quantum Mechanics, and it is used to calculate various physical properties of a particle, such as its energy and momentum.Practice Exercises1.Calculate the wave function for a particle that isin a 1D box of length L.–Answer: The wave function for a particle in a 1D box is given by:$$\\Psi(x)=\\sqrt{\\frac{2}{L}}\\sin{\\frac{n\\pi x}{L}}$$where n is a positive integer.2.Derive the time-dependent Schrödinger Equation.–Answer: The time-dependent SchrödingerEquation is given by:$$i\\hbar\\frac{\\partial\\Psi}{\\partialt}=\\hat{H}\\Psi$$3.Calculate the momentum operator for a particle in1D.–Answer: The momentum operator for a particle in 1D is given by:$$\\hat{p_x}=-i\\hbar\\frac{\\partial}{\\partial x}$$4.What is the uncertnty principle?–Answer: The uncertnty principle is afundamental principle of Quantum Mechanics thatstates that the position and momentum of a particlecannot be measured simultaneously with arbitraryprecision. Mathematically, it is given by: $$\\Delta x\\Delta p_x\\geq\\frac{\\hbar}{2}$$5.Calculate the energy of a particle in a 1D box oflength L with quantum number n.–Answer: The energy of a particle in a 1D box is given by:$$E_n=\\frac{n^2\\pi^2\\hbar^2}{2mL^2}$$ConclusionQuantum Mechanics is a fascinating and challenging fieldof study that has provided a deeper understanding of the behavior of matter and energy at the microscopic level. Theprinciples of Quantum Mechanics have been applied to various fields of study, including computing, cryptography and medicine, and they have contributed to significant advances in these fields. The practice exercises provided in this document are intended as a tool for students to test their knowledge and understanding of Quantum Mechanics. By solving these exercises, students will gn a deeper understanding of the fundamental principles of Quantum Mechanics and strengthen their problem-solving skills in this exciting field of study.。

(1) a 0
n 2n
0 a (1)k 1 a 2 k 2 A2 k 3 A2 k 2 0 a 0
即： A
2 n 1
0 a 0 k 1 2 k 2 k 1 2 k 3 (1) a a 0 (1) a 0
(1)k 1 a 2 k 3 0
0 (1)n a 2 n1 n 2 n 1 0 (1) a
（3）由上归纳得，n=0,1,2,……, A
2n
(1)n a 2 n 0
0 (1)n a 2 n1 2 n 1 ，A n 2 n 1 (1) a 0 (1) a
0 2 X 2 5 5 1 5 5
6 65 65 2 X1 65 65 5 65 65
6 65 65 2 1 2 (0, 5, 5) 65 0 65 5 5 5 65 65
n 2 2
n
n
因为 t 时恢复原状态，故 2 ( x, ) 10 ( x) 因此，
ein2 1 （对所有 n，n 为偶数）
m 2k
(l=1,2,3,……)
因此， l / 2 l
4.9 试将 exp
0 a 表示成 2×2 的矩阵，a 是个正的常数。 a 0
0 (1)k a 2 k 1 0 a (1)k 1 a 2 k 2 0 a A2 k 2 A2 k 1 k 2 k 1 0 0 a 0 (1) a a 0
即： A
2n
(1)n a 2 n 0