Legendre Polynomials and Angular Momentum
|
|
- Juliana Stanley
- 6 years ago
- Views:
Transcription
1 University of Connecticut Chemistry Education Materials Department of Chemistry August 006 Legendre Polynomials and Angular Momentum Carl W. David University of Connecticut, Follow this and additional works at: Recommended Citation David, Carl W., "Legendre Polynomials and Angular Momentum" 006). Chemistry Education Materials. 6.
2 Legendre Polynomials and Angular Momentum C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut Dated: August, 006) I. INTRODUCTION USING CARTESIAN COÖRDINATES AND GUESSWORK There are so many different ways to introduce Legendre Polynomials that one searchs for a path into this subject most suitable for chemists. Consider Laplace s Equation: χ = χ x + χ y + χ z = 0.) which is a partial differential equation for χx, y, z). For our purposes, the solutions can be obtained by guessing, starting with χ =, and proceeding to χ = x, χ = y and χ = z. It takes only a little more imagination to obtain χ = xy and its associates χ = yz and χ = xz. It takes just a little more imagination to guess χ = x y, but once done, one immediately guesses its two companions χ = x z and χ = y z. Reviewing, there was one simple order 0) solution, three not so simple, but not particularly difficult solutions of order ) and six slightly more complicated solutions of order. Note that the first third-order solution one might guess would be χ = xyz! If one looks at these solutions, knowing that they are quantum mechanically important, the, x, y, and z solutions, accompanied by the xy, xz, yz, and x y solutions suggest something having to do with wave functions, where the first function ) is associated in some way with s-orbitals, the next three x, y, and z) are associated with p-orbitals, and the next of order ) are associated with d-orbitals. If all this is true, the perceptive student will wonder where is the d z orbital, the fifth one, and how come there are six functions of order two listed, when, if memory serves correctly, there are 5 d-orbitals! Ah. can be rewritten as which is or as y z ) + x z ) y z ) + x z ) + z z ) x + y + z z r z which combines the last two linearly dependent) solutions into one, the one of choice, which has been employed for more than 50 years as the d z orbital. II. CONVERTING TO SPHERICAL POLAR COÖRDINATES In Spherical Polar Coördinates, these solutions become x r sin cos ϕ y r sin sin ϕ z r cos xy r sin cos ϕr sin sin ϕ = r sin cos ϕ sin ϕ xz r sin cos ϕr cos = r sin cos cos ϕ yz r sin sin ϕr cos = r sin cos sin ϕ x y r sin cos ϕ r sin sin = r sin cos ϕ sin ϕ ) r z r r cos = r cos )..) Now, Laplace s equation in spherical polar coördinates is χ = χ χ sin r + r sin sin + χ ϕ = 0.) Next, we notice that our earlier solutions were all of the form χ = r l Y l,ml r,, ϕ).)
3 where l specifies the order, 0,,, etc., and the function of angles is dependent on this order, and on another quantum number as yet unspecified. Memory suggests that this last quantum number, m l is the famous substate quantum number in elementary atomic structure discussions. Substituting this form into the spherical polar form of Laplace s equation results in a new equation for the angular parts alone, which we know from our previous results. We substitute Equation. into Equation. to see what happens, obtaining r l Y l,ml sin l Y l,ml + sin r sin = l Y l,ml r + r l Y l,ml ϕ = 0.4) which, by the nature of partial differentiation, becomes +r l r sin r l Y l,ml = Y l,ml r sin sin l,ml and the first term in Equation.5 is l r + Y l,ml ϕ l = 0.5) = r ll + )rl.6) so, substituting into Equation.5 we obtain r l Y l,ml = ll + )r l Y l,ml + r l sin sin sin l,ml + Y l,ml ϕ = 0.7) which, of course, upon cancelling common terms, gives ll)y l,ml + sin sin sin l,ml + Y l,ml ϕ = 0.8) Laplace s Equation in Spherical Polar Coördinates on the unit sphere, i.e. Legendre s Equation! l = ; cos l = ; sin cos ϕ sin ϕ l = ; sin cos cos ϕ l = ; sin cos sin ϕ l = ; sin cos ϕ sin ϕ ) l = ; cos )..) III. INTERPRETING ATOMIC ORBITAL DESIGNATORS What then are these angular solutions we found? l = 0; l = ; sin cos ϕ l = ; sin sin ϕ and if you check back concerning their origin, you will see where the Hydrogenic Orbitals naming pattern comes from for s, p and d orbitals. These functions have been written in real form, and they have complex, not complicated) forms which allow a different simplification: s l = 0; p x l = ; sin eıϕ + e ıϕ e ıϕ e ıϕ ) p y l = ; sin ı p z l = ; cos e d xy l = ; sin ıϕ + e ıϕ ) e ıϕ e ıϕ ) ı d xz l = ; sin cos eıϕ + e ıϕ
4 e ıϕ e ıϕ ) d yz l = ; sin cos ı e d x y sin l = ; ıϕ + e ıϕ ) e ıϕ e ıϕ ) ) ı d z l = ; cos )..) Once they have been written in imaginary form, we can create intelligent linear combinations of them which illuminate their underlying structure. Consider the linear combination of p x and p y, i.e., p x + ıp y which becomes sin e ıϕ aside from irrelevant constants. Next consider p x ıp y which becomes We have a trio of functions sin e ıϕ p ml = = sin e ıϕ p ml =0 = cos p ml = = sin e +ıϕ.).4) which correspond to the m l values expected of p-orbitals. You will see that the same trick can be applied to d xz and d yz. Finally, the same trick, almost, can be applied to d xy and d x y with the m l = 0 value reserved for d z, all by itself. IV. CONNECTION TO THE H-ATOM Perhaps the next best place to introduce Spherical Harmonics and Legendre Polynomials is the Hydrogen Atom, since its eigenfunctions have angular components which are known to be Legendre Polynomials. The Schrödinger Equation for the H-atom s electron is h n, l, m l > Ze m e r n, l, m l >= E n n, l, m l > 4.) We know that this equation is variable separable, in the sense that n, l, m l >= R n,l r)y l,ml, ϕ) Here, R n,l is the radial wave function, and Y l,ml is the angular wave function a function of two variables), Since = r r + r sin sin sin + ϕ n, l, m l >= R n,l r)y l,ml, ϕ) = Y r R which means that the Schrödinger Equation has the form { h Y R R sin m e r + r sin sin Dividing through by R n,l Y l,ml abreviated as RY) we have { h R sin m e Rr + r Y sin sin + R sin r sin sin } + Y ϕ Ze r RY = E nry } + Y ϕ Ze = E n r + Y ϕ Multiplying through by r shows that the expected variable separation has resulted in a proper segregation of the angles from the radius. R sin + R Y sin sin + Y ϕ + mze r h = m h E nr }{{}
5 4 We can see this by noting that the underbracketed part of this last equation is a pure function of angles, with no radius explicitly evident. It is now standard to obtain recover?) the equation.8) Y sin sin sin + Y ϕ = ll + ) 4.) where the minus sign which is essentially arbitrary) is demanded by convention. Cross multipilying by Y, one has sin sin sin + Y ϕ = ll + )Y 4.) and so = sin = cos = µ = ) which is written in traditional eigenfunction/eigenvalue form. which is, substituting into Equation 4. V. CHANGING THE ϑ VARIABLE A change of variables can through this equation into a special form, which is sometimes illuminating. We write = ) µ µ = cos µ ) µ µ µ µ + Y ϕ = ll + )Y 5.) where the ll) form is a different version of a constant, looking ahead to future results! where ) µ ) + Y ϕ = ll + )Y 5.) Y l,ml, ϕ) = e ±ım lϕ S l,ml ) For m l = 0 we have ) µ ) S l,0 = ll + )S l,0 5.) which is Legendre s equation. VI. SCHMIDT ORTHOGONALIZATION We assume polynomials in µ n and seek orthonormal combinations which can be generated starting with a constant n=0) term. Then the normalization integral implies that < 0 0 >= 0 >= ψ 0dµ = We now seek a function > orthogonal to 0 > which itself is normalizeable. We have < 0 >= ψ dµ = 0 which has solution ψ = N x. Normalizing, we have which yields < >= ψ dµ = = = N N µ dµ
6 5 i.e., >= x To proceed, we need to generate a function > which is not only normalizeable but orthogonal to 0 > and >. < 0 >= ψ dµ = 0 and the other integral vanishes automatically. i.e., i.e., a = b so a + b = 0 >= bµ + b = b µ + ) in unnormalized form. Normalizing gives and < >= ψ µdµ = 0 where > is a polynomial of order in µ, i.e., i.e., < >= bµ + b) dµ = Substituting, we have and < 0 >= < >= ψ = >= aµ + b aµ + b) dµ = 0 aµ + b) µdµ = 0 We obtain two equations in two unknowns, a and b, ) a + b = 0 which gives < >= b µ + ) dµ = b = µ + ) dµ where, of course, we recognize the sign ambiguity in this result. One can continue forever with this Schmidt Orthogonalization, but the idea is clear, and there are better ways, so why continue? The limits correspond to = 0 to = π
ONE AND MANY ELECTRON ATOMS Chapter 15
See Week 8 lecture notes. This is exactly the same as the Hamiltonian for nonrigid rotation. In Week 8 lecture notes it was shown that this is the operator for Lˆ 2, the square of the angular momentum.
More informationOne-electron Atom. (in spherical coordinates), where Y lm. are spherical harmonics, we arrive at the following Schrödinger equation:
One-electron Atom The atomic orbitals of hydrogen-like atoms are solutions to the Schrödinger equation in a spherically symmetric potential. In this case, the potential term is the potential given by Coulomb's
More informationFourier Series by Least Square Minimization
University of Connecticut DigitalCommons@UConn Chemistry Education Materials Department of Chemistry October 6 Fourier Series by east Square Minimization Carl W. David University of Connecticut, Carl.David@uconn.edu
More informationPhysics 342 Lecture 23. Radial Separation. Lecture 23. Physics 342 Quantum Mechanics I
Physics 342 Lecture 23 Radial Separation Lecture 23 Physics 342 Quantum Mechanics I Friday, March 26th, 2010 We begin our spherical solutions with the simplest possible case zero potential. Aside from
More information8.1 The hydrogen atom solutions
8.1 The hydrogen atom solutions Slides: Video 8.1.1 Separating for the radial equation Text reference: Quantum Mechanics for Scientists and Engineers Section 10.4 (up to Solution of the hydrogen radial
More informationAngular momentum. Quantum mechanics. Orbital angular momentum
Angular momentum 1 Orbital angular momentum Consider a particle described by the Cartesian coordinates (x, y, z r and their conjugate momenta (p x, p y, p z p. The classical definition of the orbital angular
More informationLecture #1. Review. Postulates of quantum mechanics (1-3) Postulate 1
L1.P1 Lecture #1 Review Postulates of quantum mechanics (1-3) Postulate 1 The state of a system at any instant of time may be represented by a wave function which is continuous and differentiable. Specifically,
More informationQuantum Mechanics in 3-Dimensions
Quantum Mechanics in 3-Dimensions Pavithran S Iyer, 2nd yr BSc Physics, Chennai Mathematical Institute Email: pavithra@cmi.ac.in August 28 th, 2009 1 Schrodinger equation in Spherical Coordinates 1.1 Transforming
More informationQuantum Theory of Angular Momentum and Atomic Structure
Quantum Theory of Angular Momentum and Atomic Structure VBS/MRC Angular Momentum 0 Motivation...the questions Whence the periodic table? Concepts in Materials Science I VBS/MRC Angular Momentum 1 Motivation...the
More informationH atom solution. 1 Introduction 2. 2 Coordinate system 2. 3 Variable separation 4
H atom solution Contents 1 Introduction 2 2 Coordinate system 2 3 Variable separation 4 4 Wavefunction solutions 6 4.1 Solution for Φ........................... 6 4.2 Solution for Θ...........................
More informationA Quantum Mechanical Model for the Vibration and Rotation of Molecules. Rigid Rotor
A Quantum Mechanical Model for the Vibration and Rotation of Molecules Harmonic Oscillator Rigid Rotor Degrees of Freedom Translation: quantum mechanical model is particle in box or free particle. A molecule
More informationd 1 µ 2 Θ = 0. (4.1) consider first the case of m = 0 where there is no azimuthal dependence on the angle φ.
4 Legendre Functions In order to investigate the solutions of Legendre s differential equation d ( µ ) dθ ] ] + l(l + ) m dµ dµ µ Θ = 0. (4.) consider first the case of m = 0 where there is no azimuthal
More informationVANDERBILT UNIVERSITY. MATH 3120 INTRO DO PDES The Schrödinger equation
VANDERBILT UNIVERSITY MATH 31 INTRO DO PDES The Schrödinger equation 1. Introduction Our goal is to investigate solutions to the Schrödinger equation, i Ψ t = Ψ + V Ψ, 1.1 µ where i is the imaginary number
More informationSolutions to Laplace s Equations- II
Solutions to Laplace s Equations- II Lecture 15: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Laplace s Equation in Spherical Coordinates : In spherical coordinates
More informationPhysics 228 Today: Ch 41: 1-3: 3D quantum mechanics, hydrogen atom
Physics 228 Today: Ch 41: 1-3: 3D quantum mechanics, hydrogen atom Website: Sakai 01:750:228 or www.physics.rutgers.edu/ugrad/228 Happy April Fools Day Example / Worked Problems What is the ratio of the
More informationThe Hydrogen atom. Chapter The Schrödinger Equation. 2.2 Angular momentum
Chapter 2 The Hydrogen atom In the previous chapter we gave a quick overview of the Bohr model, which is only really valid in the semiclassical limit. cf. section 1.7.) We now begin our task in earnest
More informationCHEM-UA 127: Advanced General Chemistry I
1 CHEM-UA 127: Advanced General Chemistry I Notes for Lecture 11 Nowthatwehaveintroducedthebasicconceptsofquantummechanics, wecanstarttoapplythese conceptsto build up matter, starting from its most elementary
More informationThe Hydrogen Atom. Chapter 18. P. J. Grandinetti. Nov 6, Chem P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, / 41
The Hydrogen Atom Chapter 18 P. J. Grandinetti Chem. 4300 Nov 6, 2017 P. J. Grandinetti (Chem. 4300) The Hydrogen Atom Nov 6, 2017 1 / 41 The Hydrogen Atom Hydrogen atom is simplest atomic system where
More informationThe Hydrogen Atom. Dr. Sabry El-Taher 1. e 4. U U r
The Hydrogen Atom Atom is a 3D object, and the electron motion is three-dimensional. We ll start with the simplest case - The hydrogen atom. An electron and a proton (nucleus) are bound by the central-symmetric
More information2m r2 (~r )+V (~r ) (~r )=E (~r )
Review of the Hydrogen Atom The Schrodinger equation (for 1D, 2D, or 3D) can be expressed as: ~ 2 2m r2 (~r, t )+V (~r ) (~r, t )=i~ @ @t The Laplacian is the divergence of the gradient: r 2 =r r The time-independent
More informationVelocities in Quantum Mechanics
Velocities in Quantum Mechanics Toshiki Shimbori and Tsunehiro Kobayashi Institute of Physics, University of Tsukuba Ibaraki 305-8571, Japan Department of General Education for the Hearing Impaired, Tsukuba
More informationThe 3 dimensional Schrödinger Equation
Chapter 6 The 3 dimensional Schrödinger Equation 6.1 Angular Momentum To study how angular momentum is represented in quantum mechanics we start by reviewing the classical vector of orbital angular momentum
More informationQuantum Orbits. Quantum Theory for the Computer Age Unit 9. Diving orbit. Caustic. for KE/PE =R=-3/8. for KE/PE =R=-3/8. p"...
W.G. Harter Coulomb Obits 6-1 Quantum Theory for the Computer Age Unit 9 Caustic for KE/PE =R=-3/8 F p' p g r p"... P F' F P Diving orbit T" T T' Contact Pt. for KE/PE =R=-3/8 Quantum Orbits W.G. Harter
More information20 The Hydrogen Atom. Ze2 r R (20.1) H( r, R) = h2 2m 2 r h2 2M 2 R
20 The Hydrogen Atom 1. We want to solve the time independent Schrödinger Equation for the hydrogen atom. 2. There are two particles in the system, an electron and a nucleus, and so we can write the Hamiltonian
More informationPhysics 221A Fall 1996 Notes 12 Orbital Angular Momentum and Spherical Harmonics
Physics 221A Fall 1996 Notes 12 Orbital Angular Momentum and Spherical Harmonics We now consider the spatial degrees of freedom of a particle moving in 3-dimensional space, which of course is an important
More informationWelcome back to PHY 3305
Welcome back to PHY 3305 Today s Lecture: Hydrogen Atom Part I John von Neumann 1903-1957 One-Dimensional Atom To analyze the hydrogen atom, we must solve the Schrodinger equation for the Coulomb potential
More informationProblem 1: Spin 1 2. particles (10 points)
Problem 1: Spin 1 particles 1 points 1 Consider a system made up of spin 1/ particles. If one measures the spin of the particles, one can only measure spin up or spin down. The general spin state of a
More informationSelf-consistent Field
Chapter 6 Self-consistent Field A way to solve a system of many electrons is to consider each electron under the electrostatic field generated by all other electrons. The many-body problem is thus reduced
More informationIV. Electronic Spectroscopy, Angular Momentum, and Magnetic Resonance
IV. Electronic Spectroscopy, Angular Momentum, and Magnetic Resonance The foundation of electronic spectroscopy is the exact solution of the time-independent Schrodinger equation for the hydrogen atom.
More information1 Commutators (10 pts)
Final Exam Solutions 37A Fall 0 I. Siddiqi / E. Dodds Commutators 0 pts) ) Consider the operator  = Ĵx Ĵ y + ĴyĴx where J i represents the total angular momentum in the ith direction. a) Express both
More informationLecture 10. Central potential
Lecture 10 Central potential 89 90 LECTURE 10. CENTRAL POTENTIAL 10.1 Introduction We are now ready to study a generic class of three-dimensional physical systems. They are the systems that have a central
More informationThe Central Force Problem: Hydrogen Atom
The Central Force Problem: Hydrogen Atom B. Ramachandran Separation of Variables The Schrödinger equation for an atomic system with Z protons in the nucleus and one electron outside is h µ Ze ψ = Eψ, r
More informationSpherical Coordinates and Legendre Functions
Spherical Coordinates and Legendre Functions Spherical coordinates Let s adopt the notation for spherical coordinates that is standard in physics: φ = longitude or azimuth, θ = colatitude ( π 2 latitude)
More informationQuantum Mechanics: The Hydrogen Atom
Quantum Mechanics: The Hydrogen Atom 4th April 9 I. The Hydrogen Atom In this next section, we will tie together the elements of the last several sections to arrive at a complete description of the hydrogen
More information(3.1) Module 1 : Atomic Structure Lecture 3 : Angular Momentum. Objectives In this Lecture you will learn the following
Module 1 : Atomic Structure Lecture 3 : Angular Momentum Objectives In this Lecture you will learn the following Define angular momentum and obtain the operators for angular momentum. Solve the problem
More informationAtoms 2012 update -- start with single electron: H-atom
Atoms 2012 update -- start with single electron: H-atom x z φ θ e -1 y 3-D problem - free move in x, y, z - easier if change coord. systems: Cartesian Spherical Coordinate (x, y, z) (r, θ, φ) Reason: V(r)
More informationLigand Field Theory Notes
Ligand Field Theory Notes Read: Hughbanks, Antisymmetry (Handout). Carter, Molecular Symmetry..., Sections 7.4-6. Cotton, Chemical Applications..., Chapter 9. Harris & Bertolucci, Symmetry and Spectroscopy...,
More information4/21/2010. Schrödinger Equation For Hydrogen Atom. Spherical Coordinates CHAPTER 8
CHAPTER 8 Hydrogen Atom 8.1 Spherical Coordinates 8.2 Schrödinger's Equation in Spherical Coordinate 8.3 Separation of Variables 8.4 Three Quantum Numbers 8.5 Hydrogen Atom Wave Function 8.6 Electron Spin
More informationSchrödinger equation for the nuclear potential
Schrödinger equation for the nuclear potential Introduction to Nuclear Science Simon Fraser University Spring 2011 NUCS 342 January 24, 2011 NUCS 342 (Lecture 4) January 24, 2011 1 / 32 Outline 1 One-dimensional
More informationTransformation Matrices; Geometric and Otherwise As examples, consider the transformation matrices of the C 3v
Transformation Matrices; Geometric and Otherwise As examples, consider the transformation matrices of the v group. The form of these matrices depends on the basis we choose. Examples: Cartesian vectors:
More informationChapter 6: Quantum Theory of the Hydrogen Atom
Chapter 6: Quantum Theory of the Hydrogen Atom The first problem that Schrödinger tackled with his new wave equation was that of the hydrogen atom. The discovery of how naturally quantization occurs in
More informationQuantum Mechanics-I Prof. Dr. S. Lakshmi Bala Department of Physics Indian Institute of Technology, Madras. Lecture - 21 Square-Integrable Functions
Quantum Mechanics-I Prof. Dr. S. Lakshmi Bala Department of Physics Indian Institute of Technology, Madras Lecture - 21 Square-Integrable Functions (Refer Slide Time: 00:06) (Refer Slide Time: 00:14) We
More informationSpin Dynamics Basic Theory Operators. Richard Green SBD Research Group Department of Chemistry
Spin Dynamics Basic Theory Operators Richard Green SBD Research Group Department of Chemistry Objective of this session Introduce you to operators used in quantum mechanics Achieve this by looking at:
More informationTime part of the equation can be separated by substituting independent equation
Lecture 9 Schrödinger Equation in 3D and Angular Momentum Operator In this section we will construct 3D Schrödinger equation and we give some simple examples. In this course we will consider problems where
More informationSingle Electron Atoms
Single Electron Atoms In this section we study the spectrum and wave functions of single electron atoms. These are hydrogen, singly ionized He, doubly ionized Li, etc. We will write the formulae for hydrogen
More information1 r 2 sin 2 θ. This must be the case as we can see by the following argument + L2
PHYS 4 3. The momentum operator in three dimensions is p = i Therefore the momentum-squared operator is [ p 2 = 2 2 = 2 r 2 ) + r 2 r r r 2 sin θ We notice that this can be written as sin θ ) + θ θ r 2
More informationWe now turn to our first quantum mechanical problems that represent real, as
84 Lectures 16-17 We now turn to our first quantum mechanical problems that represent real, as opposed to idealized, systems. These problems are the structures of atoms. We will begin first with hydrogen-like
More informationCollection of formulae Quantum mechanics. Basic Formulas Division of Material Science Hans Weber. Operators
Basic Formulas 17-1-1 Division of Material Science Hans Weer The de Broglie wave length λ = h p The Schrödinger equation Hψr,t = i h t ψr,t Stationary states Hψr,t = Eψr,t Collection of formulae Quantum
More informationquantization condition.
/8/016 PHYS 34 Modern Physics Atom II: Hydrogen Atom Roadmap for Exploring Hydrogen Atom Today Contents: a) Schrodinger Equation for Hydrogen Atom b) Angular Momentum in Quantum Mechanics c) Quantum Number
More informationLecture 4 Quantum mechanics in more than one-dimension
Lecture 4 Quantum mechanics in more than one-dimension Background Previously, we have addressed quantum mechanics of 1d systems and explored bound and unbound (scattering) states. Although general concepts
More informationPHYSICS 304 QUANTUM PHYSICS II (2005) Assignment 1 Solutions
PHYSICS 04 QUANTUM PHYSICS II 200 Assignment Solutions. The general state of a spin half particle with spin component S n = S n = be shown to be given by 2 h can S n = 2 h = cos 2 θ S z = 2 h + eiφ sin
More informationBrief review of Quantum Mechanics (QM)
Brief review of Quantum Mechanics (QM) Note: This is a collection of several formulae and facts that we will use throughout the course. It is by no means a complete discussion of QM, nor will I attempt
More informationRepresentation theory and quantum mechanics tutorial Spin and the hydrogen atom
Representation theory and quantum mechanics tutorial Spin and the hydrogen atom Justin Campbell August 3, 2017 1 Representations of SU 2 and SO 3 (R) 1.1 The following observation is long overdue. Proposition
More informationLecture 5: Harmonic oscillator, Morse Oscillator, 1D Rigid Rotor
Lecture 5: Harmonic oscillator, Morse Oscillator, 1D Rigid Rotor It turns out that the boundary condition of the wavefunction going to zero at infinity is sufficient to quantize the value of energy that
More informationCrystal field effect on atomic states
Crystal field effect on atomic states Mehdi Amara, Université Joseph-Fourier et Institut Néel, C.N.R.S. BP 66X, F-3842 Grenoble, France References : Articles - H. Bethe, Annalen der Physik, 929, 3, p.
More informationFun With Carbon Monoxide. p. 1/2
Fun With Carbon Monoxide p. 1/2 p. 1/2 Fun With Carbon Monoxide E = 0.25 ± 0.05 ev Electron beam results p. 1/2 Fun With Carbon Monoxide E = 0.25 ± 0.05 ev Electron beam results C V (J/K-mole) 35 30 25
More informationReading: Mathchapters F and G, MQ - Ch. 7-8, Lecture notes on hydrogen atom.
Chemistry 356 017: Problem set No. 6; Reading: Mathchapters F and G, MQ - Ch. 7-8, Lecture notes on hydrogen atom. The H atom involves spherical coordinates and angular momentum, which leads to the shapes
More informationSolving the Schrödinger Equation for the 1 Electron Atom (Hydrogen-Like)
Stockton Univeristy Chemistry Program, School of Natural Sciences an Mathematics 101 Vera King Farris Dr, Galloway, NJ CHEM 340: Physical Chemistry II Solving the Schröinger Equation for the 1 Electron
More information1.6. Quantum mechanical description of the hydrogen atom
29.6. Quantum mechanical description of the hydrogen atom.6.. Hamiltonian for the hydrogen atom Atomic units To avoid dealing with very small numbers, let us introduce the so called atomic units : Quantity
More informationEigenfunctions on the surface of a sphere. In spherical coordinates, the Laplacian is. u = u rr + 2 r u r + 1 r 2. sin 2 (θ) + 1
Eigenfunctions on the surface of a sphere In spherical coordinates, the Laplacian is u = u rr + 2 r u r + 1 r 2 [ uφφ sin 2 (θ) + 1 sin θ (sin θ u θ) θ ]. Eigenfunctions on the surface of a sphere In spherical
More informationFORMULA SHEET FOR QUIZ 2 Exam Date: November 8, 2017
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II November 5, 207 Prof. Alan Guth FORMULA SHEET FOR QUIZ 2 Exam Date: November 8, 207 A few items below are marked
More informationInfinitesimal Rotations
Universit of Connecticut DigitalCommons@UConn Chemistr Education Materials Department of Chemistr Januar 007 Infinitesimal Rotations Carl W. David Universit of Connecticut, Carl.David@uconn.edu Follow
More informationSummary: angular momentum derivation
Summary: angular momentum derivation L = r p L x = yp z zp y, etc. [x, p y ] = 0, etc. (-) (-) (-3) Angular momentum commutation relations [L x, L y ] = i hl z (-4) [L i, L j ] = i hɛ ijk L k (-5) Levi-Civita
More informationChem 442 Review for Exam 2. Exact separation of the Hamiltonian of a hydrogenic atom into center-of-mass (3D) and relative (3D) components.
Chem 44 Review for Exam Hydrogenic atoms: The Coulomb energy between two point charges Ze and e: V r Ze r Exact separation of the Hamiltonian of a hydrogenic atom into center-of-mass (3D) and relative
More informationAngular momentum & spin
Angular momentum & spin January 8, 2002 1 Angular momentum Angular momentum appears as a very important aspect of almost any quantum mechanical system, so we need to briefly review some basic properties
More informationRelevant sections from AMATH 351 Course Notes (Wainwright): Relevant sections from AMATH 351 Course Notes (Poulin and Ingalls):
Lecture 5 Series solutions to DEs Relevant sections from AMATH 35 Course Notes (Wainwright):.4. Relevant sections from AMATH 35 Course Notes (Poulin and Ingalls): 2.-2.3 As mentioned earlier in this course,
More informationSpherical Harmonics on S 2
7 August 00 1 Spherical Harmonics on 1 The Laplace-Beltrami Operator In what follows, we describe points on using the parametrization x = cos ϕ sin θ, y = sin ϕ sin θ, z = cos θ, where θ is the colatitude
More informationOh, the humanity! David J. Starling Penn State Hazleton PHYS 214
Oh, the humanity! -Herbert Morrison, radio reporter of the Hindenburg disaster David J. Starling Penn State Hazleton PHYS 24 The hydrogen atom is composed of a proton and an electron with potential energy:
More informationAngular Momentum. Classically the orbital angular momentum with respect to a fixed origin is. L = r p. = yp z. L x. zp y L y. = zp x. xpz L z.
Angular momentum is an important concept in quantum theory, necessary for analyzing motion in 3D as well as intrinsic properties such as spin Classically the orbital angular momentum with respect to a
More informationSection 9 Variational Method. Page 492
Section 9 Variational Method Page 492 Page 493 Lecture 27: The Variational Method Date Given: 2008/12/03 Date Revised: 2008/12/03 Derivation Section 9.1 Variational Method: Derivation Page 494 Motivation
More information1 Schroenger s Equation for the Hydrogen Atom
Schroenger s Equation for the Hydrogen Atom Here is the Schroedinger equation in D in spherical polar coordinates. Note that the definitions of θ and φ are the exact reverse of what they are in mathematics.
More informationNotes: Most of the material presented in this chapter is taken from Jackson, Chap. 2, 3, and 4, and Di Bartolo, Chap. 2. 2π nx i a. ( ) = G n.
Chapter. Electrostatic II Notes: Most of the material presented in this chapter is taken from Jackson, Chap.,, and 4, and Di Bartolo, Chap... Mathematical Considerations.. The Fourier series and the Fourier
More informationRecall that any inner product space V has an associated norm defined by
Hilbert Spaces Recall that any inner product space V has an associated norm defined by v = v v. Thus an inner product space can be viewed as a special kind of normed vector space. In particular every inner
More informationClassical Field Theory: Electrostatics-Magnetostatics
Classical Field Theory: Electrostatics-Magnetostatics April 27, 2010 1 1 J.D.Jackson, Classical Electrodynamics, 2nd Edition, Section 1-5 Electrostatics The behavior of an electrostatic field can be described
More informationCh120 - Study Guide 10
Ch120 - Study Guide 10 Adam Griffith October 17, 2005 In this guide: Symmetry; Diatomic Term Symbols; Molecular Term Symbols Last updated October 27, 2005. 1 The Origin of m l States and Symmetry We are
More informationName Solutions to Test 3 November 7, 2018
Name Solutions to Test November 7 8 This test consists of three parts. Please note that in parts II and III you can skip one question of those offered. Some possibly useful formulas can be found below.
More informationThe Rigid Rotor Problem: A quantum par7cle confined to a sphere Spherical Harmonics. Reading: McIntyre 7.6
The Rigid Rotor Problem: A quantum par7cle confined to a sphere Spherical Harmonics Reading: McIntyre 7.6 Legendre s equa7on (m = 0) The series must be finite! If the series is not finite, the polynomial
More information1.4 Solution of the Hydrogen Atom
The phase of α can freely be chosen to be real so that α = h (l m)(l + m + 1). Then L + l m = h (l m)(l + m + 1) l m + 1 (1.24) L l m = h (l + m)(l m + 1) l m 1 (1.25) Since m is bounded, it follow that
More information(Refer Slide Time: 1:20) (Refer Slide Time: 1:24 min)
Engineering Chemistry - 1 Prof. K. Mangala Sunder Department of Chemistry Indian Institute of Technology, Madras Lecture - 5 Module 1: Atoms and Molecules Harmonic Oscillator (Continued) (Refer Slide Time:
More informationB2.III Revision notes: quantum physics
B.III Revision notes: quantum physics Dr D.M.Lucas, TT 0 These notes give a summary of most of the Quantum part of this course, to complement Prof. Ewart s notes on Atomic Structure, and Prof. Hooker s
More informationChem 467 Supplement to Lecture 19 Hydrogen Atom, Atomic Orbitals
Chem 467 Supplement to Lecture 19 Hydrogen Atom, Atomic Orbitals Pre-Quantum Atomic Structure The existence of atoms and molecules had long been theorized, but never rigorously proven until the late 19
More informationSheet 06.6: Curvilinear Integration, Matrices I
Fakultät für Physik R: Rechenmethoden für Physiker, WiSe 5/6 Doent: Jan von Delft Übungen: Benedikt Bruognolo, Dennis Schimmel, Frauke Schwar, Lukas Weidinger http://homepages.physik.uni-muenchen.de/~vondelft/lehre/5r/
More information1. (3) Write Gauss Law in differential form. Explain the physical meaning.
Electrodynamics I Midterm Exam - Part A - Closed Book KSU 204/0/23 Name Instructions: Use SI units. Where appropriate, define all variables or symbols you use, in words. Try to tell about the physics involved,
More informationThe Runge0Lenz Vector (continued)
University of Connecticut DigitalCommons@UConn Chemistry Education Materials Department of Chemistry 009 The Runge0Lenz Vector continued) Carl W. David University of Connecticut, Carl.David@uconn.edu Follow
More informationSolved radial equation: Last time For two simple cases: infinite and finite spherical wells Spherical analogs of 1D wells We introduced auxiliary func
Quantum Mechanics and Atomic Physics Lecture 16: The Coulomb Potential http://www.physics.rutgers.edu/ugrad/361 h / d/361 Prof. Sean Oh Solved radial equation: Last time For two simple cases: infinite
More informationwhich implies that we can take solutions which are simultaneous eigen functions of
Module 1 : Quantum Mechanics Chapter 6 : Quantum mechanics in 3-D Quantum mechanics in 3-D For most physical systems, the dynamics is in 3-D. The solutions to the general 3-d problem are quite complicated,
More informationChemistry 432 Problem Set 4 Spring 2018 Solutions
Chemistry 4 Problem Set 4 Spring 18 Solutions 1. V I II III a b c A one-dimensional particle of mass m is confined to move under the influence of the potential x a V V (x) = a < x b b x c elsewhere and
More informationQuantum Physics II (8.05) Fall 2002 Assignment 11
Quantum Physics II (8.05) Fall 00 Assignment 11 Readings Most of the reading needed for this problem set was already given on Problem Set 9. The new readings are: Phase shifts are discussed in Cohen-Tannoudji
More informationPhysics 504, Spring 2010 Electricity and Magnetism
Resonant cavities do not need to be cylindrical, of course. The surface of the Earth R E 6400 m the ionosphere R = R E + h, h 100 m form concentric spheres which are sufficiently good conductors to form
More informationLecture 4 Quantum mechanics in more than one-dimension
Lecture 4 Quantum mechanics in more than one-dimension Background Previously, we have addressed quantum mechanics of 1d systems and explored bound and unbound (scattering) states. Although general concepts
More informationSchrödinger equation for central potentials
Chapter 2 Schrödinger equation for central potentials In this chapter we will extend the concepts and methods introduced in the previous chapter ifor a one-dimenional problem to a specific and very important
More informationECEN 5005 Crystals, Nanocrystals and Device Applications Class 20 Group Theory For Crystals
ECEN 5005 Crystals, Nanocrystals and Device Applications Class 20 Group Theory For Crystals Laporte Selection Rule Polarization Dependence Spin Selection Rule 1 Laporte Selection Rule We first apply this
More informationCurvilinear coordinates
C Curvilinear coordinates The distance between two points Euclidean space takes the simplest form (2-4) in Cartesian coordinates. The geometry of concrete physical problems may make non-cartesian coordinates
More informationQuantum Numbers. principal quantum number: n. angular momentum quantum number: l (azimuthal) magnetic quantum number: m l
Quantum Numbers Quantum Numbers principal quantum number: n angular momentum quantum number: l (azimuthal) magnetic quantum number: m l Principal quantum number: n related to size and energy of orbital
More informationAtomic Units, Revisited
Atomic Units, evisited C. W. David (Dated: September, 004) I. INTODUCTION We assume that the reader has gone through an initial discussion of the quantum mechanics of atoms and molecules, and seeks an
More informationeff (r) which contains the influence of angular momentum. On the left is
1 Fig. 13.1. The radial eigenfunctions R nl (r) of bound states in a square-well potential for three angular-momentum values, l = 0, 1, 2, are shown as continuous lines in the left column. The form V (r)
More informationTotal Angular Momentum for Hydrogen
Physics 4 Lecture 7 Total Angular Momentum for Hydrogen Lecture 7 Physics 4 Quantum Mechanics I Friday, April th, 008 We have the Hydrogen Hamiltonian for central potential φ(r), we can write: H r = p
More informationQuantum Mechanics for Scientists and Engineers
Quantum Mechanics for Scientists and Engineers Syllabus and Textbook references All the main lessons (e.g., 1.1) and units (e.g., 1.1.1) for this class are listed below. Mostly, there are three lessons
More information(a) Determine the general solution for φ(ρ) near ρ = 0 for arbitary values E. (b) Show that the regular solution at ρ = 0 has the series expansion
Problem 1. Curious Wave Functions The eigenfunctions of a D7 brane in a curved geometry lead to the following eigenvalue equation of the Sturm Liouville type ρ ρ 3 ρ φ n (ρ) = E n w(ρ)φ n (ρ) w(ρ) = where
More informationQuantum Mechanics Solutions
Quantum Mechanics Solutions (a (i f A and B are Hermitian, since (AB = B A = BA, operator AB is Hermitian if and only if A and B commute So, we know that [A,B] = 0, which means that the Hilbert space H
More information