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Course: CHEM 444, Fall 2011
School: Delaware
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Mechanics: Quantum Commutation 5 april 2010 I. Commutators: Measuring Several Properties Simultaneously In classical mechanics, once we determine the dynamical state of a system, we can simultaneously obtain many dierent system properties (i.e., velocity, position, momentum, acceleration, angular/linear momentum, kinetic and potential energies, etc.). The uncertainty is governed by the resolution and precision...

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Mechanics: Quantum Commutation 5 april 2010 I. Commutators: Measuring Several Properties Simultaneously In classical mechanics, once we determine the dynamical state of a system, we can simultaneously obtain many dierent system properties (i.e., velocity, position, momentum, acceleration, angular/linear momentum, kinetic and potential energies, etc.). The uncertainty is governed by the resolution and precision of the instruments at our disposal. In quantum mechanics, the situation is dierent. Consider the following: 1. We would like to measure several properties of a particle represented by a wavefunction. 2. Properties of a q.m. system can be measured experimentally. Theoretically, the measurement process corresponds to an operator acting on the wavefunction. The outcomes of the measurement are the eigenvalues that correspond to the operator. The operator is taken to be acting on a wavefunction that is either a pure eigenfunction of the operator of interest, or an expansion in the basis of eigenfunctions. In order to measure, for instance, 2 properties simultaneously, the wavefunction of the particle must be an eigenstate of the two operators that corespond to the properties we would like to measure simultaneously. How do we formalize this mathematically? Consider: We have to operators, A and B . Each operator acting on its eigenstate gives back the corresponding eigenvalues, A i and Bj , respectively. A Ai = A i Ai B Bi = B i Bi If we have a wavefunction that is an eigenstate of both operators, then: AAi ,Bj = Ai Ai ,Bj 1 BAi ,Bj = Bj Ai ,Bj Thus, B AAi ,Bj = BAi Ai ,Bj ABAi ,Bj = ABj Ai ,Bj So, using the fact that Ai ,Bj is an eigenfunction of A and B : B AAi ,Bj = Bj Ai Ai ,Bj ABAi ,Bj = Ai Bj Ai ,Bj Subtracting the equations, we realize a compact notation for dening what is called a commutator: [A, B ] = AB B A The commutator is itself either zero or an operator The order of operations is important and will give unique commutators depending on this ordering For two physical properties to be simultaneously observable, their operator representations must commute. Thus, A Bf (x) B Af (x) = 0 2 operators that commute Example Problem 17.1: whether Determine the momentum operator commutes with the a) kinetic energy and b) total energy operators. a). To determine whether the two operators commute (and importantly, to determine whether the two observables associated with those operators can be known simultaneously), one considers the following: 2 momentum and kinetic energy ih d dx h2 d2 2m dx2 f (x) i h3 d3 2m dx3 h2 d2 2m dx2 ih f (x) i h3 d3 2m dx3 d f (x) dx f (x) 0 Thus, the momentum and kinetic energy operators commute b). momentum and total energy ih d dx h2 d2 h2 d2 + V (x) f (x) + V (x) 2m dx2 2m dx2 ih d f (x) dx (1) Since the momentum and kinetic energy operators commute from part a, we can write d V (x)f (x)) + ihV (x) dx d d ihV (x) f (x) ihf (x) V (x) + ihV (x) dx dx d f (x) dx d f (x) dx d ihf (x) V (x) dx ih (2) Thus, the commutator for the momentum and total energy reduces as follows: d d d = V (x), ih = ih V (x) H, ih dx dx dx The last equation does not equal zero identically, and thus we see two things: 1. the momentum and total energy do not commute 2. the commutator reduces to a unique operation (we will see this again with respect to angular momentum) Heisenberg Uncertainty Principle Recall the discussion of the free particle. For that system, we determined that the energy (and momentum) spectrum is continuous since there were no boundary conditions imposed on the wavefunction (thus we arrive at plan-wave representations of the particle-wave entity). The important point for the present dicussion is the relation between our knowledge of the momentum of the particle and its position. We note a few things: 3 The Heisenberg Uncertainty principle is stated as: px h 2 For a quantum mechanical description of a particles dynamics, we cannot know exactly and simultaneously both the particles position and momentum. We must accept an uncertainty in measurements of these quantities as given by the inequality. To relate the uncertainty principle to variances and statistical measures, the relation: x p h 2 can be used in conjunction with wavefunctions and denitions of average properties: 2 p =< p2 > < p >2 4 (likewise f or x )
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Delaware - CHEM - 444
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Quantum Mechanics: The Hydrogen Atom12th April 2010I. The Hydrogen AtomIn this next section, we will tie together the elements of the last severalsections to arrive at a complete description of the hydrogen atom. This willculminate in the denition of
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Quantum Mechanics: The Hydrogen Atom12th April 2010I. The Hydrogen Atom Wavefunctions: Orbital Building Blocks1The Radial FunctionLets consider s, p , and dtype orbitals and arrive at some generaldescriptions of the spatial variation of these functi
Delaware - CHEM - 444
Helium Atom, Many-Electron Atoms, VariationalPrinciple, Approximate Methods, Spin20th April 2010I. The Helium Atom and Variational Principle: ApproximationMethods for Complex Atomic SystemsThe hydrogen atom wavefunctions and energies, we have seen, a
Delaware - CHEM - 444
Many-Electron Atomic States, Terms, and Levels27th April 2010I. Hartree-Fock with Antisymmetrized WavefunctionsRecall the earlier discussion of the Hartree-Fock self-consistent method. Forthe purposes of that introduction, we used trial wavefunctions
Delaware - CHEM - 444
Atomic Terms,Hunds Rules, Atomic Spectroscopy27th April 2010I. Atomic Terms, Hunds Rules, Atomic SpectroscopyHaving dened ways to determine atomic terms (which group various quantum microstates of similar energy) we need to specify a protocol to allow
Delaware - CHEM - 444
Many-Electron Atomic States, Terms, and Levels5th May 2009I. Hartree-Fock with Antisymmetrized Wavefunctions Recall the earlier discussion of the Hartree-Fock self-consistent method. For the purposes of that introduction, we used trial wavefunctions tha
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Atomic Terms,Hund's Rules, Atomic Spectroscopy5th May 2009I. Atomic Terms, Hund's Rules, Atomic Spectroscopy Having defined ways to determine atomic terms (which group various quantum microstates of similar energy) we need to specify a protocol to allow
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Diatomic Molecules7th May 20091Hydrogen Molecule: Born-Oppenheimer ApproximationIn this discussion, we consider the formulation of the Schrodinger equation for diatomic molecules; this can be extended to larger molecules. First we will consider the se
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Diatomic Molecules12th May 20091Hydrogen Molecule: Born-Oppenheimer ApproximationIn this discussion, we consider the formulation of the Schrodinger equation for diatomic molecules; this can be extended to larger molecules. First we will consider the s
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Delaware - CHEM - 444
Averages: A Simple Treatment16th February 2010Consider the following series of 10 (N data = 10) numbers (we are beingdiscrete:10, 30, 20, 80, 80, 70, 70, 60, 50, 30What is the average of these discrete numbers? That is, what is theaverage value of x
Delaware - CHEM - 444
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Delaware - CHEM - 444
NAME:Natalie Portman201110Circle Section Number:CHEMISTRY 444, SPRING,118081Examination 1, March 5, 2011Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILYunderstand
Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
Reaction Kinetics: Elementary Ideas18th February 20101Rate LawsReturning to our hypothetical model reaction:i i + j j k k + l lIn practice, the rate of reaction is treated as follows, thus introducingthe concept of the rate constant, k :N[i]iiR
Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
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Delaware - CHEM - 444
Delaware - CHEM - 444
Diatomic Molecules12th May 20091Hydrogen Molecule: Born-Oppenheimer ApproximationIn this discussion, we consider the formulation of the Schrodinger equation for diatomic molecules; this can be extended to larger molecules. First we will consider the s
Delaware - CHEM - 444
Quantum Mechanics: The Hydrogen Atom13th April 2011I. The Hydrogen AtomIn this next section, we will tie together the elements of the last severalsections to arrive at a complete description of the hydrogen atom. This will culminate in the denition of
Delaware - CHEM - 444
Helium Atom, Many-Electron Atoms, VariationalPrinciple, Approximate Methods, Spin21st April 2011I. The Helium Atom and Variational Principle: ApproximationMethods for Complex Atomic SystemsThe hydrogen atom wavefunctions and energies, we have seen, a
Delaware - CHEM - 444
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Delaware - CHEM - 444
Waves and the Schroedinger Equation5 april 20101The Wave EquationWe have seen from previous discussions that the wave-particle duality of matter requires we describe entities through some wave-form based represenation. The most natural consideration a
Delaware - CHEM - 444
CHEMISTRY 444.11QUIZ 3Spring, 2008 (08S)March 13, 2008NAME:Score _/151. (10 points)Consider the following reaction mechanism:k2A + A A *+Ak2A *+A A + A'k2A+ M A * + M'k2A * + M A+ Mk1A *B + Ca. Write an expression for the rate of react
Delaware - CHEM - 444
1Problem 1Do the one-dimensional kinetic energy and momentum operators commute?If not, what operator does their commutator represent?h2 d2KE =2m dx21.1dP = ihdxSolutionThis question requires calculating the commutator of the operators given.K
Delaware - CHEM - 444
1Problem 1The d orbitals have the nomenclature d z 2 , dxy , dxz ,dyz , and dx2 y2 . Showhow the d orbital given below can be written in the form y z F (r ).3dyz1.12=81 1ao3/2r 2 r/3aoesin( ) cos( ) sin()a2oSolutionz = r cos ( )x = r si
Delaware - CHEM - 444
1Problem 1In previous discussion, we have considered approximate methods for obtaining energies, wavefunctions, or both for quantum systems more complicatedthan the Hydrogen atom. In this exercise, consider the quantum harmonicoscillator and use the v
Delaware - CHEM - 444
1Problem 1For the atomic term 3 P , what are the energy dierences between the levelsarising due to spin-orbit coupling.1.1SolutionFor the atomic term 3 P , show the energy dierences between the levelsarising due to spin-orbit coupling.Answer For t
Delaware - CHEM - 444
CHEMISTRY 444.10/80QUIZ 1NAME:Spring, 2011February 18, 2011EMIL JANNINGSScore _/20[Numbers without decimal points are considered infinitely precise. Show reasonable significant figuresand proper units. Answers should be in reasonable units for the
Delaware - CHEM - 444
1Problem 1The rotational constant, Be , for IBr determined from microwave spectroscopyis 0.2241619 cm1 . Approximate the bond length of this molecule.1.1SolutionFrom the Handbook:Be =h28 2 ReRearranging gives the required form for solving for t
Delaware - CHEM - 444
Quiz Solutions: S10 (Section 011, CHEM444)27th February 20101Quiz 11.1Problem 1What is the ratio of the following values for a particle of Argon and Kryptonat 298K?AraveKraveArave=KraveArmostprobableArmostprobableKrmost83.798= 1.
Delaware - CHEM - 444
CHEMISTRY 444.11QUIZ 1Spring, 2011 (11S)Feb. 17, 2011NAME:Score _/20Problem 1 (8 points)Starting from the continuous probability distribution for one-dimensional diffusion, show how one woulddetermine the average squared displacement in a model de
Delaware - CHEM - 444
Quiz Solutions: S10 (Section 011, CHEM444)27th February 201011.1Quiz 1Problem 1What is the ratio of the following values for a particle of Argon and Krypton at 298K? Ar ave Kr aveAr ave = Kr avemKr = mAr83.798 = 1.45 39.948Ar most Kr mostproba
Delaware - CHEM - 444
Transport Properties: Momentum Transport,Viscosity13th February 20111IntroductionMuch as mass(material) is transported within uids (gases and liquids),linear momentum is also associated with transport, in this case, due togradients in velocity.See
Delaware - CHEM - 419
CORRECTIONS TO PHYSICAL CHEMISTRY FOR THE LIFE SCIENCES BYENGEL, DROBNY, AND REIDPage 337: The third sentence from the top of the page should read For E&lt; V(x) (insidethe box), the solutions have.Page 622: In Example Problem 24.1, the equation in the s
Delaware - CHEM - 419
CorrectionsEssential Data and Equations for a Course in PhysicalChemistryCecil Dybowski and Andrew V. TeplyakovPage 3-2, Table 3.41 joule = 9.47810-4 BTUPage 3-2, Table 3.61 Pa = 7.5006210-3 torr1 torr = 1.3157910-3 atm1 psi = 6.804610-2 atmPage