Physical Chemistry , Ch 17 - Quan… · PHYSICAL CHEMISTRY Sixth Edition Ira N. Levine Chemistry...

PHYSICAL CHEMISTRY

Sixth Edition

Ira N. LevineChemistry Department

Brooklyn CollegeCity University of New York

Brooklyn, New York

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Quantum Mechanics

We now begin the study of quantum chemistry, which applies quantum mechanics tochemistry. Chapter 17 deals with quantum mechanics, the laws governing the be-havior of microscopic particles such as electrons and nuclei. Chapters 18 and 19 applyquantum mechanics to atoms and molecules. Chapter 20 applies quantum mechanicsto spectroscopy, the study of the absorption and emission of electromagnetic radiation.Quantum mechanics is used in statistical mechanics (Chapter 21) and in theoreticalchemical kinetics (Chapter 22).

Unlike thermodynamics, quantum mechanics deals with systems that are not partof everyday macroscopic experience, and the formulation of quantum mechanics isquite mathematical and abstract. This abstractness takes a while to get used to, and itis natural to feel somewhat uneasy when first reading Chapter 17.

In an undergraduate physical chemistry course, it is not possible to give a full pre-sentation of quantum mechanics. Derivations of results that are given without proofmay be found in quantum chemistry texts listed in the Bibliography.

Sections 17.1 to 17.4 give the historical background of quantum mechanics.Section 17.5 discusses the uncertainty principle, a key concept that underlies the dif-ferences between quantum mechanics and classical (Newtonian) mechanics. Quantummechanics describes the state of a system using a state function (or wave function) �.Sections 17.6 and 17.7 describe the meaning of � and the time-dependent and time-independent Schrödinger equations used to find �. Sections 17.8, 17.9, 17.10, 17.12,17.13, and 17.14 consider the Schrödinger equation, the wave functions, and theallowed quantum-mechanical energy levels for several systems. Sections 17.11 and17.16 discuss operators, which are used extensively in quantum mechanics. Section17.15 introduces some of the approximation methods used to apply quantum mechan-ics to chemistry.

Essentially all of chemistry is a consequence of the laws of quantum mechanics.If we want to understand chemistry at the fundamental level of electrons, atoms, andmolecules, we must understand quantum mechanics. Quantities such as the heat ofcombustion of octane, the 25°C entropy of liquid water, the reaction rate of N2 and H2gases at specified conditions, the equilibrium constants of chemical reactions, theabsorption spectra of coordination compounds, the NMR spectra of organic com-pounds, the nature of the products formed when organic compounds react, the shapea protein molecule folds into when it is formed in a cell, the structure and function ofDNA are all a consequence of quantum mechanics.

In 1929, Dirac, one of the founders of quantum mechanics, wrote that “The gen-eral theory of quantum mechanics is now almost complete . . . . The underlying phys-ical laws necessary for the mathematical theory of . . . the whole of chemistry are thuscompletely known, and the difficulty is only that the exact application of these lawsleads to equations much too complicated to be soluble.” After its discovery, quantummechanics was used to develop many concepts that helped explain chemical proper-ties. However, because of the very difficult calculations needed to apply quantummechanics to chemical systems, quantum mechanics was of little practical value in

C H A P T E R

17CHAPTER OUTLINE

17.1 Blackbody Radiation andEnergy Quantization

17.2 The Photoelectric Effectand Photons

17.3 The Bohr Theory of theHydrogen Atom

17.4 The de Broglie Hypothesis

17.5 The Uncertainty Principle

17.6 Quantum Mechanics

17.7 The Time-IndependentSchrödinger Equation

17.8 The Particle in a One-Dimensional Box

17.9 The Particle in a Three-Dimensional Box

17.10 Degeneracy

17.11 Operators

17.12 The One-DimensionalHarmonic Oscillator

17.13 Two-Particle Problems

17.14 The Two-Particle Rigid Rotor

17.15 Approximation Methods

17.16 Hermitian Operators

17.17 Summary

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Section 17.1Blackbody Radiation

and Energy Quantization

591

accurately calculating the properties of chemical systems for many years after its dis-covery. Nowadays, however, the extraordinary computational power of modern com-puters allows quantum-mechanical calculations to give accurate chemical predictionsin many systems of real chemical interest. As computers become even more powerfuland applications of quantum mechanics in chemistry increase, the need for allchemists to be familiar with quantum mechanics will increase.

17.1 BLACKBODY RADIATION AND ENERGY QUANTIZATIONClassical physics is the physics developed before 1900. It consists of classical me-chanics (Sec. 2.1), Maxwell’s theory of electricity, magnetism, and electromagneticradiation (Sec. 20.1), thermodynamics, and the kinetic theory of gases (Chapters 14and 15). In the late nineteenth century, some physicists believed that the theoreticalstructure of physics was complete, but in the last quarter of the nineteenth century, var-ious experimental results were obtained that could not be explained by classicalphysics. These results led to the development of quantum theory and the theory of rel-ativity. An understanding of atomic structure, chemical bonding, and molecular spec-troscopy must be based on quantum theory, which is the subject of this chapter.

One failure of classical physics was the incorrect CV,m values of polyatomic mol-ecules predicted by the kinetic theory of gases (Sec. 14.10). A second failure was theinability of classical physics to explain the observed frequency distribution of radiantenergy emitted by a hot solid.

When a solid is heated, it emits light. Classical physics pictures light as a waveconsisting of oscillating electric and magnetic fields, an electromagnetic wave. (SeeSec. 20.1 for a fuller discussion.) The frequency n (nu) and wavelength l (lambda) ofan electromagnetic wave traveling through vacuum are related by

(17.1)*

where c � 3.0 � 108 m/s is the speed of light in vacuum. The human eye is sensitive toelectromagnetic waves whose frequencies lie in the range 4 � 1014 to 7 � 1014 cycles/s.However, electromagnetic radiation can have any frequency (see Fig. 20.2). We shalluse the term “light” as synonymous with electromagnetic radiation, not restrictingit to visible light.

Different solids emit radiation at different rates at the same temperature. To sim-plify things, one deals with the radiation emitted by a blackbody. A blackbody is abody that absorbs all the electromagnetic radiation that falls on it. A good approxi-mation to a blackbody is a cavity with a tiny hole. Radiation that enters the hole is re-peatedly reflected within the cavity (Fig. 17.1a). At each reflection, a certain fraction

ln � c

Figure 17.1

(a) A cavity acting as a blackbody.(b) Frequency distribution ofblackbody radiation at twotemperatures. (The visible regionis from 4 � 1014 to 7 � 1014 s�1.)

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Chapter 17Quantum Mechanics

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of the radiation is absorbed by the cavity walls, and the large number of reflectionscauses virtually all the incident radiation to be absorbed. When the cavity is heated,its walls emit light, a tiny portion of which escapes through the hole. It can be shownthat the rate of radiation emitted per unit surface area of a blackbody is a function ofonly its temperature and is independent of the material of which the blackbody ismade. (See Zemansky and Dittman, sec. 4-14, for a proof.)

By using a prism to separate the various frequencies emitted by the cavity, one canmeasure the amount of blackbody radiant energy emitted in a given narrow frequencyrange. Let the frequency distribution of the emitted blackbody radiation be describedby the function R(n), where R(n) dn is the energy with frequency in the range n ton � dn that is radiated per unit time and per unit surface area. (Recall the discussionof distribution functions in Sec. 14.4.) Figure 17.1b shows some experimentally ob-served R(n) curves. As T increases, the maximum in R(n) shifts to higher frequencies.When a metal rod is heated, it first glows red, then orange-yellow, then white, thenblue-white. (White light is a mixture of all colors.) Our bodies are not hot enough toemit visible light, but we do emit infrared radiation.

In June 1900, Lord Rayleigh attempted to derive the theoretical expression for thefunction R(n). Using the equipartition-of-energy theorem (Sec. 14.10), he found thatclassical physics predicted R(n) � (2pkT/c2)n2, where k and c are Boltzmann’s con-stant and the speed of light. But this result is absurd, since it predicts that the amountof energy radiated would increase without limit as n increases. In actuality, R(n)reaches a maximum and then falls off to zero as n increases (Fig. 17.1b). Thus, clas-sical physics fails to predict the spectrum of blackbody radiation.

On October 19, 1900, the physicist Max Planck announced to the GermanPhysical Society his discovery of a formula that gave a highly accurate fit to the ob-served curves of blackbody radiation. Planck’s formula was R(n) � an3/(ebn/T � 1),where a and b are constants with certain numerical values. Planck had obtained thisformula by trial and error and at that time had no theory to explain it. On December14, 1900, Planck presented to the German Physical Society a theory that yielded theblackbody-radiation formula he had found empirically a few weeks earlier. Planck’stheory gave the constants a and b as a � 2ph/c2 and b � h/k, where h was a new con-stant in physics and k is Boltzmann’s constant [Eq. (3.57)]. Planck’s theoretical ex-pression for the frequency distribution of blackbody radiation is then

(17.2)

Planck considered the walls of the blackbody to contain electric charges thatoscillated (vibrated) at various frequencies [Maxwell’s electromagnetic theory of light(Sec. 20.1) shows that electromagnetic waves are produced by accelerated electriccharges. A charge oscillating at frequency n will emit radiation at that frequency.] Inorder to derive (17.2), Planck found that he had to assume that the energy of eachoscillating charge could take on only the possible values 0, hn, 2hn, 3hn, . . . , wheren is the frequency of the oscillator and h is a constant (later called Planck’s constant)with the dimensions of energy � time. This assumption then leads to Eq. (17.2). (ForPlanck’s derivation, see M. Jammer, The Conceptual Development of QuantumMechanics, McGraw-Hill, 1966, sec. 1.2.) Planck obtained a numerical value of h byfitting the formula (17.2) to the observed blackbody curves. The modern value is

(17.3)*

In classical physics, energy takes on a continuous range of values, and a system canlose or gain any amount of energy. In direct contradiction to classical physics, Planckrestricted the energy of each oscillating charge to a whole-number multiple of hn and

h � 6.626 � 10�34 J # s

R1n 2 �2ph

c2

n3

ehn>kT � 1

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Section 17.2The Photoelectric Effect and Photons

593

hence restricted the amount of energy each oscillator could gain or lose to an integralmultiple of hn. Planck called the quantity hn a quantum of energy (the Latin wordquantum means “how much”). In classical physics, energy is a continuous variable. Inquantum physics, the energy of a system is quantized, meaning that the energy can takeon only certain values. Planck introduced the idea of energy quantization in one case,the emission of blackbody radiation. In the years 1900–1926, the concept of energyquantization was gradually extended to all microscopic systems.

Planck was not very explicit in his derivation and several historians of sciencehave argued that Planck’s apparent introduction of energy quantization was solely formathematical convenience to allow him to evaluate a certain quantity needed in thederivation, and Planck was not actually proposing energy quantization as a physicalreality. See S. G. Brush, Am. J. Phys., 70, 119 (2002); O. Darrigol, Centaurus, 43, 219(2001)—available at www.mpiwg-berlin.mpg.de/Preprints/P150.PDF.

17.2 THE PHOTOELECTRIC EFFECT AND PHOTONSThe person who recognized the value of Planck’s idea was Einstein, who applied theconcept of energy quantization to electromagnetic radiation and showed that thisexplained the experimental observations in the photoelectric effect.

In the photoelectric effect, a beam of electromagnetic radiation (light) shining ona metal surface causes the metal to emit electrons; electrons absorb energy from thelight beam, thereby acquiring enough energy to escape from the metal. A practical ap-plication is the photoelectric cell, used to measure light intensities, to prevent elevatordoors from crushing people, and in smoke detectors (light scattered by smoke parti-cles causes electron emission, which sets off an alarm).

Experimental work around 1900 had shown that (a) Electrons are emitted onlywhen the frequency of the light exceeds a certain minimum frequency n0 (the thresh-old frequency). The value of n0 differs for different metals and lies in the ultravioletfor most metals. (b) Increasing the intensity of the light increases the number of elec-trons emitted but does not affect the kinetic energy of the emitted electrons. (c) Increas-ing the frequency of the radiation increases the kinetic energy of the emitted electrons.

These observations on the photoelectric effect cannot be understood using theclassical picture of light as a wave. The energy in a wave is proportional to its inten-sity but is independent of its frequency, so one would expect the kinetic energy of theemitted electrons to increase with an increase in light intensity and to be independentof the light’s frequency. Moreover, the wave picture of light would predict the photo-electric effect to occur at any frequency, provided the light is sufficiently intense.

In 1905 Einstein explained the photoelectric effect by extending Planck’s conceptof energy quantization to electromagnetic radiation. (Planck had applied energy quan-tization to the oscillators in the blackbody but had considered the electromagneticradiation to be a wave.) Einstein proposed that in addition to having wavelike proper-ties, light could also be considered to consist of particlelike entities (quanta), each quan-tum of light having an energy hn, where h is Planck’s constant and n is the frequency ofthe light. These entities were later named photons, and the energy of a photon is

(17.4)*

The energy in a light beam is the sum of the energies of the individual photons and istherefore quantized.

Let electromagnetic radiation of frequency n fall on a metal. The photoelectriceffect occurs when an electron in the metal is hit by a photon. The photon disappears,and its energy hn is transferred to the electron. Part of the energy absorbed by theelectron is used to overcome the forces holding the electron in the metal, and the

Ephoton � hn

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remainder appears as kinetic energy of the emitted electron. Conservation of energytherefore gives

(17.5)

where the work function � is the minimum energy needed by an electron to escape themetal and mv2 is the kinetic energy of the free electron. The valence electrons in met-als have a distribution of energies (Sec. 23.11), so some electrons need more energy thanothers to leave the metal. The emitted electrons therefore show a distribution of kineticenergies, and mv2 in (17.5) is the maximum kinetic energy of emitted electrons.

Einstein’s equation (17.5) explains all the observations in the photoelectric effect.If the light frequency is such that hn � �, a photon does not have enough energy toallow an electron to escape the metal and no photoelectric effect occurs. The minimumfrequency n0 at which the effect occurs is given by hn0 � �. (The work function �differs for different metals, being lowest for the alkali metals.) Equation (17.5) showsthe kinetic energy of the emitted electrons to increase with n and to be independent ofthe light intensity. An increase in intensity with no change in frequency increases theenergy of the light beam and hence increases the number of photons per unit volumein the light beam, thereby increasing the rate of emission of electrons.

Einstein’s theory of the photoelectric effect agreed with the qualitative observa-tions, but it wasn’t until 1916 that R. A. Millikan made an accurate quantitative test ofEq. (17.5). The difficulty in testing (17.5) is the need to maintain a very clean surfaceof the metal. Millikan found accurate agreement between (17.5) and experiment.

At first, physicists were very reluctant to accept Einstein’s hypothesis of photons.Light shows the phenomena of diffraction and interference (Sec. 17.5), and theseeffects are shown only by waves, not by particles. Eventually, physicists became con-vinced that the photoelectric effect could be understood only by viewing light as beingcomposed of photons. However, diffraction and interference can be understood onlyby viewing light as a wave and not as a collection of particles.

Thus, light seems to exhibit a dual nature, behaving like waves in some situationsand like particles in other situations. This apparent duality is logically contradictory,since the wave and particle models are mutually exclusive. Particles are localized inspace, but waves are not. The photon picture gives a quantization of the light energy, butthe wave picture does not. In Einstein’s equation Ephoton � hn, the quantity Ephoton is aparticle concept, but the frequency n is a wave concept, so this equation is, in a sense,self-contradictory. An explanation of these apparent contradictions is given in Sec. 17.4.

In 1907 Einstein applied the concept of energy quantization to the vibrations ofthe atoms in a solid, thereby showing that the heat capacity of a solid goes to zero asT goes to zero, a result in agreement with experiment but in disagreement with theclassical equipartition theorem. See Sec. 23.12 for details.

17.3 THE BOHR THEORY OF THE HYDROGEN ATOMThe next major application of energy quantization was the Danish physicist NielsBohr’s 1913 theory of the hydrogen atom. A heated gas of hydrogen atoms emits elec-tromagnetic radiation containing only certain distinct frequencies (Fig. 20.36). During1885 to 1910, Balmer, Rydberg, and others found that the following empirical formulacorrectly reproduces the observed H-atom spectral frequencies:

(17.6)

where the Rydberg constant R equals 1.096776 � 105 cm�1. There was no explana-tion for this formula until Bohr’s work.

n

c�

1

l� R a 1

n2b

�1

n2a

b nb � 1, 2, 3, . . . ; na � 2, 3, . . . ; na 7 nb

12

12

hn � £ � 12 mv2

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Section 17.4The de Broglie Hypothesis

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If one accepts Einstein’s equation the fact that only certain frequen-cies of light are emitted by H atoms indicates that contrary to classical ideas, a hydro-gen atom can exist only in certain energy states. Bohr therefore postulated that theenergy of a hydrogen atom is quantized: (1) An atom can take on only certain distinctenergies E1, E2, E3, . . . . Bohr called these allowed states of constant energy the sta-tionary states of the atom. This term is not meant to imply that the electron is at restin a stationary state. Bohr further assumed that (2) An atom in a stationary state doesnot emit electromagnetic radiation. To explain the line spectrum of hydrogen, Bohrassumed that (3) When an atom makes a transition from a stationary state with energyEupper to a lower-energy stationary state with energy Elower, it emits a photon of light.Since Ephoton � hn, conservation of energy gives

(17.7)*

where Eupper � Elower is the energy difference between the atomic states involved in thetransition and n is the frequency of the light emitted. Similarly, an atom can make atransition from a lower-energy to a higher-energy state by absorbing a photon of fre-quency given by (17.7). The Bohr theory provided no description of the transitionprocess between two stationary states. Of course, transitions between stationary statescan occur by means other than absorption or emission of electromagnetic radiation. Forexample, an atom can gain or lose electronic energy in a collision with another atom.

Equations (17.6) and (17.7) with upper and lower replaced by a and b give Ea �Eb � Rhc(1/nb

2 � 1/na2), which strongly indicates that the energies of the H-atom sta-

tionary states are given by E � �Rhc/n2, with n � 1, 2, 3, . . . . Bohr then introducedfurther postulates to derive a theoretical expression for the Rydberg constant. Heassumed that (4) The electron in an H-atom stationary state moves in a circle aroundthe nucleus and obeys the laws of classical mechanics. The energy of the electron isthe sum of its kinetic energy and the potential energy of the electron–nucleus electro-static attraction. Classical mechanics shows that the energy depends on the radius ofthe orbit. Since the energy is quantized, only certain orbits are allowed. Bohr used onefinal postulate to select the allowed orbits. Most books give this postulate as (5) Theallowed orbits are those for which the electron’s angular momentum mevr equalsnh/2p, where me and v are the electron’s mass and speed, r is the radius of the orbit,and n � 1, 2, 3, . . . . Actually, Bohr used a different postulate which is less arbitrarythan 5 but less simple to state. The postulate Bohr used is equivalent to 5 and is omit-ted here. (If you’re curious, see Karplus and Porter, sec. 1.4.)

With his postulates, Bohr derived the following expression for the H-atom energylevels: E � �mee

4/8e02h2n2, where e is the proton charge and the electric constant e0

occurs in Coulomb’s law (13.1). Therefore, Bohr predicted that Rhc � mee4/8e0

2h2 andR � mee

4/8e02h3c. Substitution of the values of me, e, h, e0, and c gave a result in good

agreement with the experimental value of the Rydberg constant, indicating that theBohr model gave the correct energy levels of H.

Although the Bohr theory is historically important for the development of quan-tum theory, postulates 4 and 5 are in fact false, and the Bohr theory was superseded in1926 by the Schrödinger equation, which provides a correct picture of electronicbehavior in atoms and molecules. Although postulates 4 and 5 are false, postulates 1,2, and 3 are consistent with quantum mechanics.

17.4 THE DE BROGLIE HYPOTHESISIn the years 1913 to 1925, attempts were made to apply the Bohr theory to atoms withmore than one electron and to molecules. However, all attempts to derive the spectraof such systems using extensions of the Bohr theory failed. It gradually became clear

Eupper � Elower � hn

Ephoton � hv,

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that there was a fundamental error in the Bohr theory. The fact that the Bohr theoryworks for H is something of an accident.

A key idea toward resolving these difficulties was advanced by the French physi-cist Louis de Broglie (1892–1987) in 1923. The fact that a heated gas of atoms or mol-ecules emits radiation of only certain frequencies shows that the energies of atoms andmolecules are quantized, only certain energy values being allowed. Quantization ofenergy does not occur in classical mechanics; a particle can have any energy in clas-sical mechanics. Quantization does occur in wave motion. For example, a string heldfixed at each end has quantized modes of vibration (Fig. 17.2). The string can vibrateat its fundamental frequency n, at its first overtone frequency 2n, at its second over-tone frequency 3n, etc. Frequencies lying between these integral multiples of n are notallowed.

De Broglie therefore proposed that just as light shows both wave and particle as-pects, matter also has a “dual” nature. As well as showing particlelike behavior, anelectron could also show wavelike behavior, the wavelike behavior manifesting itselfin the quantized energy levels of electrons in atoms and molecules. Holding the endsof a string fixed quantizes its vibrational frequencies. Similarly, confining an electronin an atom quantizes its energies.

De Broglie obtained an equation for the wavelength l to be associated with amaterial particle by reasoning in analogy with photons. We have Ephoton � hn.Einstein’s special theory of relativity gives the energy of a photon as Ephoton � pc,where p is the momentum of the photon and c is the speed of light. Equating these twoexpressions for Ephoton, we get h � pc. But � c /l, so hc/l � pc and l � h /p for aphoton. By analogy, de Broglie proposed that a material particle with momentum pwould have a wavelength l given by

(17.8)

The momentum of a particle with a speed v much less than the speed of light isp � mv, where m is the particle’s rest mass.

The de Broglie wavelength of an electron moving at 1.0 � 106 m/s is

This wavelength is on the order of magnitude of molecular dimensions and indicatesthat wave effects are important in electronic motions in atoms and molecules. For amacroscopic particle of mass 1.0 g moving at 1.0 cm/s, a similar calculation givesl � 7 � 10 �27 cm. The extremely small size of l (which results from the smallnessof Planck’s constant h in comparison with mv) indicates that quantum effects areunobservable for the motion of macroscopic objects.

De Broglie’s bold hypothesis was experimentally confirmed in 1927 by Davissonand Germer, who observed diffraction effects when an electron beam was reflectedfrom a crystal of Ni; G. P. Thomson observed diffraction effects when electrons werepassed through a thin sheet of metal. See Fig. 17.3. Similar diffraction effects havebeen observed with neutrons, protons, helium atoms, and hydrogen molecules, indi-cating that the de Broglie hypothesis applies to all material particles, not just electrons.An application of the wavelike behavior of microscopic particles is the use of electrondiffraction and neutron diffraction to obtain molecular structures (Secs. 23.9 and23.10).

Electrons show particlelike behavior in some experiments (for example, thecathode-ray experiments of J. J. Thomson, Sec. 18.2) and wavelike behavior in otherexperiments. As noted in Sec. 17.2, the wave and particle models are incompatiblewith each other. An entity cannot be both a wave and a particle. How can we explain

l �6.6 � 10�34 J s

19.1 � 10�31 kg 2 11.0 � 106 m>s 2 � 7 � 10�10 m � 7 Å

l � h>p

nn

Figure 17.2

Fundamental and overtonevibrations of a string.

Figure 17.3

Diffraction rings observed whenelectrons are passed through a thinpolycrystalline metal sheet.

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Section 17.5The Uncertainty Principle

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the apparently contradictory behavior of electrons? The source of the difficulty is theattempt to describe microscopic entities like electrons by using concepts developedfrom our experience in the macroscopic world. The particle and wave concepts weredeveloped from observations on large-scale objects, and there is no guarantee that theywill be fully applicable on the microscopic scale. Under certain experimental condi-tions, an electron behaves like a particle. Under other conditions, it behaves like awave. However, an electron is neither a particle nor a wave. It is something that can-not be adequately described in terms of a model we can visualize.

A similar situation holds for light, which shows wave properties in some situa-tions and particle properties in others. Light originates in the microscopic world ofatoms and molecules and cannot be fully understood in terms of models visualizableby the human mind.

Although both electrons and light exhibit an apparent “wave–particle duality,”there are significant differences between these entities. Light travels at speed c in vac-uum, and photons have zero rest mass. Electrons always travel at speeds less than cand have a nonzero rest mass.

17.5 THE UNCERTAINTY PRINCIPLEThe apparent wave–particle duality of matter and of radiation imposes certain limita-tions on the information we can obtain about a microscopic system. Consider a mi-croscopic particle traveling in the y direction. Suppose we measure the x coordinate ofthe particle by having it pass through a narrow slit of width w and fall on a fluorescentscreen (Fig. 17.4). If we see a spot on the screen, we can be sure the particle passedthrough the slit. Therefore, we have measured the x coordinate at the time of passingthe slit to an accuracy w. Before the measurement, the particle had zero velocity vx andzero momentum px � mvx in the x direction. Because the microscopic particle haswavelike properties, it will be diffracted at the slit. Photographs of electron-diffractionpatterns at a single slit and at multiple slits are given in C. Jönsson, Am. J. Phys., 42,4 (1974).

Diffraction is the bending of a wave around an obstacle. A classical particlewould go straight through the slit, and a beam of such particles would show a spreadof length w in where they hit the screen. A wave passing through the slit will spreadout to give a diffraction pattern. The curve in Fig. 17.4 shows the intensity of thewave at various points on the screen. The maxima and minima result from con-structive and destructive interference between waves originating from various partsof the slit. Interference results from the superposition of two waves travelingthrough the same region of space. When the waves are in phase (crests occurringtogether), constructive interference occurs, with the amplitudes adding to give astronger wave. When the waves are out of phase (crests of one wave coinciding withtroughs of the second wave), destructive interference occurs and the intensity isdiminished.

The first minima (points P and Q) in the single-slit diffraction pattern occur atplaces on the screen where waves originating from the top of the slit travel one-halfwavelength less or more than waves originating from the middle of the slit. Thesewaves are then exactly out of phase and cancel each other. Similarly, waves originat-ing from a distance d below the top of the slit cancel waves originating a distance dbelow the center of the slit. The condition for the first diffraction minimum is then

in Fig. 17.4, where C is located so that . Becausethe distance from the slit to the screen is much greater than the slit width, angle APCis nearly zero and angles PAC and ACP are each nearly 90°. Hence, angle ACD isessentially 90°. Angles PDE and DAC each equal 90° minus angle ADC. These

CP � APDP � AP � 12l � CD

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two angles are therefore equal and have been marked u. We have sin u � /l/ w � l/w. The angle u at which the first diffraction minimum occurs is given by

sin l/w.For a microscopic particle passing through the slit, diffraction at the slit will

change the particle’s direction of motion. A particle diffracted by angle u and hittingthe screen at P or Q will have an x component of momentum px � p sin u at the slit(Fig. 17.4), where p is the particle’s momentum. The intensity curve in Fig. 17.4shows that the particle is most likely to be diffracted by an angle lying in the range�u to �u, where u is the angle to the first diffraction minimum. Hence, the positionmeasurement produces an uncertainty in the px value given by p sin u � (�p sin u) �2p sin u. We write px � 2p sin u, where px gives the uncertainty in our knowledgeof px at the slit. We saw in the previous paragraph that sin u � l/w, so px � 2pl/w.The de Broglie relation (17.8) gives l � h/p, so px � 2h/w. The uncertainty inour knowledge of the x coordinate is given by the slit width, so x � w. Therefore,x px � 2h.

Before the measurement, we had no knowledge of the particle’s x coordinate,but we knew that it was traveling in the y direction and so had px � 0. Thus, beforethe measurement, x � q and px � 0. The slit of width w gave the x coordinateto an uncertainty w (x � w) but introduced an uncertainty px � 2h/w in px. Byreducing the slit width w, we can measure the x coordinate as accurately as weplease, but as x � w becomes smaller, px � 2h/w becomes larger. The more weknow about x, the less we know about px. The measurement introduces an uncon-trollable and unpredictable disturbance in the system, changing px by an unknownamount.

Although we have analyzed only one experiment, analysis of many other experi-ments leads to the same conclusion: the product of the uncertainties in x and px of aparticle is on the order of magnitude of Planck’s constant or greater:

(17.9)*

This is the uncertainty principle, discovered by Heisenberg in 1927. A generalquantum-mechanical proof of (17.9) was given by Robertson in 1929. Similarly wehave y py h and z pz h.

The small size of h makes the uncertainty principle of no consequence formacroscopic particles.

¢x ¢px g h

u �

12

12

AD � DC

Figure 17.4

Diffraction at a slit.

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Section 17.6Quantum Mechanics

599

17.6 QUANTUM MECHANICSThe fact that electrons and other microscopic “particles” show wavelike as well as par-ticlelike behavior indicates that electrons do not obey classical mechanics. Classicalmechanics was formulated from the observed behavior of macroscopic objects anddoes not apply to microscopic particles. The form of mechanics obeyed by micro-scopic systems is called quantum mechanics, since a key feature of this mechanicsis the quantization of energy. The laws of quantum mechanics were discovered byHeisenberg, Born, and Jordan in 1925 and by Schrödinger in 1926. Before discussingthese laws, we consider some aspects of classical mechanics.

Classical MechanicsThe motion of a one-particle, one-dimensional classical-mechanical system is governedby Newton’s second law F � ma � m d2x/dt2. To obtain the particle’s position x as afunction of time, this differential equation must be integrated twice with respect to time.The first integration gives dx/dt, and the second integration gives x. Each integration in-troduces an arbitrary integration constant. Therefore, integration of F � ma gives anequation for x that contains two unknown constants c1 and c2; we have x � f (t, c1, c2),where f is some function. To evaluate c1 and c2, we need two pieces of information aboutthe system. If we know that at a certain time t0, the particle was at the position x0 andhad speed v0, then c1 and c2 can be evaluated from the equations x0 � f (t0, c1, c2) andv0 � f �(t0, c1, c2), where f � is the derivative of f with respect to t. Thus, provided we knowthe force F and the particle’s initial position and velocity (or momentum), we can useNewton’s second law to predict the position of the particle at any future time. A similarconclusion holds for a three-dimensional many-particle classical system.

The state of a system in classical mechanics is defined by specifying all the forcesacting and all the positions and velocities (or momenta) of the particles. We saw in thepreceding paragraph that knowledge of the present state of a classical-mechanicalsystem enables its future state to be predicted with certainty.

The Heisenberg uncertainty principle, Eq. (17.9), shows that simultaneous speci-fication of position and momentum is impossible for a microscopic particle. Hence,the very knowledge needed to specify the classical-mechanical state of a system isunobtainable in quantum theory. The state of a quantum-mechanical system musttherefore involve less knowledge about the system than in classical mechanics.

Quantum MechanicsIn quantum mechanics, the state of a system is defined by a mathematical function �(capital psi) called the state function or the time-dependent wave function. (As partof the definition of the state, the potential-energy function V must also be specified.)� is a function of the coordinates of the particles of the system and (since the statemay change with time) is also a function of time. For example, for a two-particle sys-tem, � � �(x1, y1, z1, x2, y2, z2, t), where x1, y1, z1 and x2, y2, z2 are the coordinates ofparticles 1 and 2, respectively. The state function is in general a complex quantity; thatis, � � f � ig, where f and g are real functions of the coordinates and time and

. The state function is an abstract entity, but we shall later see how � isrelated to physically measurable quantities.

The state function changes with time. For an n-particle system, quantum mechan-ics postulates that the equation governing how � changes with t is

(17.10)�U 2

2mn

a 02°0x2

n

�02°0y2

n

�02°0z2

n

b � V °

�Ui

0°0t

� �U 2

2m1a 02°

0x21

�02°0y2

1

�02°0z2

1

b � . . .

i � 1�1

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600

In this equation, U (h-bar) is Planck’s constant divided by 2p,

(17.11)*

i is ; m1, . . . , mn are the masses of particles 1, . . . , n; x1, y1, z1 are the spatial co-ordinates of particle 1; and V is the potential energy of the system. Since the potentialenergy is energy due to the particles’ positions, V is a function of the particles’ coor-dinates. Also, V can vary with time if an externally applied field varies with time.Hence, V is in general a function of the particles’ coordinates and the time. V is de-rived from the forces acting in the system; see Eq. (2.17). The dots in Eq. (17.10) standfor terms involving the spatial derivatives of particles 2, 3, . . . , n � 1.

Equation (17.10) is a complicated partial differential equation. For most of theproblems dealt with in this book, it will not be necessary to use (17.10), so don’t panic.

The concept of the state function � and Eq. (17.10) were introduced by theAustrian physicist Erwin Schrödinger (1887–1961) in 1926. Equation (17.10) isthe time-dependent Schrödinger equation. Schrödinger was inspired by the deBroglie hypothesis to search for a mathematical equation that would resemble the dif-ferential equations that govern wave motion and that would have solutions giving theallowed energy levels of a quantum system. Using the de Broglie relation l� h/p andcertain plausibility arguments, Schrödinger proposed Eq. (17.10) and the related time-independent equation (17.24) below. These plausibility arguments have been omittedin this book. It should be emphasized that these arguments can at best make theSchrödinger equation seem plausible. They can in no sense be used to derive or provethe Schrödinger equation. The Schrödinger equation is a fundamental postulate ofquantum mechanics and cannot be derived. The reason we believe it to be true is thatits predictions give excellent agreement with experimental results. “One could arguethat the Schrödinger equation has had more to do with the evolution of twentieth-century science and technology than any other discovery in physics.” (JeremyBernstein, Cranks, Quarks, and the Cosmos, Basic Books, 1993, p. 54.)

In 1925, several months before Schrödinger’s work, Werner Heisenberg (1901–1976),Max Born (1882–1970), and Pascual Jordan (1902–1980) developed a form of quantummechanics based on mathematical entities called matrices. A matrix is a rectangular arrayof numbers; matrices are added and multiplied according to certain rules. The matrixmechanics of these workers turns out to be fully equivalent to the Schrödinger form ofquantum mechanics (which is often called wave mechanics). We shall not discuss matrixmechanics.

Schrödinger also contributed to statistical mechanics, relativity, and the theory ofcolor vision and was deeply interested in philosophy. In an epilogue to his 1944 book,What Is Life?, Schrödinger wrote: “So let us see whether we cannot draw the correct, non-contradictory conclusion from the following two premises: (i) My body functions as a puremechanism according to the Laws of Nature. (ii) Yet I know, by incontrovertible direct ex-perience, that I am directing its motions . . . . The only possible inference from these twofacts is, I think, that I—I in the widest meaning of the word, that is to say, every consciousmind that has ever said or felt ‘I’—am the person, if any, who controls the ‘motion of theatoms’ according to the Laws of Nature.” Schrödinger’s life and loves are chronicled inW. Moore, Schrödinger, Life and Thought, Cambridge University Press, 1989.

The time-dependent Schrödinger equation (17.10) contains the first derivative of� with respect to t, and a single integration with respect to time gives us �. Inte-gration of (17.10) therefore introduces only one integration constant, which can beevaluated if � is known at some initial time t0. Therefore, knowing the initial quantum-mechanical state �(x1, . . . , zn, t0) and the potential energy V, we can use (17.10) topredict the future quantum-mechanical state. The time-dependent Schrödinger equation

1�1

U � h>2p

lev38627_ch17.qxd 3/25/08 12:22 PM Page 600

Section 17.6Quantum Mechanics

601

Figure 17.5

An infinitesimal box in space.

is the quantum-mechanical analog of Newton’s second law, which allows the futurestate of a classical-mechanical system to be predicted from its present state. We shallsoon see, however, that knowledge of the state in quantum mechanics usually involvesa knowledge of only probabilities, rather than certainties, as in classical mechanics.

What is the relation between quantum mechanics and classical mechanics?Experiment shows that macroscopic bodies obey classical mechanics (provided theirspeed is much less than the speed of light). We therefore expect that in the classical-mechanical limit of taking h → 0, the time-dependent Schrödinger equation ought toreduce to Newton’s second law. This was shown by Ehrenfest in 1927; for Ehrenfest’sproof, see Park, sec. 3.3.

Physical Meaning of the State Function �Schrödinger originally conceived of � as the amplitude of some sort of wave that wasassociated with the system. It soon became clear that this interpretation was wrong.For example, for a two-particle system, � is a function of the six spatial coordinatesx1, y1, z1, x2, y2, and z2, whereas a wave moving through space is a function of onlythree spatial coordinates. The correct physical interpretation of � was given by MaxBorn in 1926. Born postulated that ��2 gives the probability density for findingthe particles at given locations in space. (Probability densities for molecular speedswere discussed in Sec. 14.4.) To be more precise, suppose a one-particle system hasthe state function �(x, y, z, t�) at time t�. Consider the probability that a measurementof the particle’s position at time t� will find the particle with its x, y, and z coordinatesin the infinitesimal ranges xa to xa � dx, ya to ya � dy, and za to za � dz, respectively.This is the probability of finding the particle in a tiny rectangular-box-shaped regionlocated at point (xa, ya, za) in space and having edges dx, dy, and dz (Fig. 17.5). Born’spostulate is that the probability is given by

(17.12)*

where the left side of (17.12) denotes the probability the particle is found in the boxof Fig. 17.5.

EXAMPLE 17.1 Probability for finding a particle

Suppose that at time t� the state function of a one-particle system is

[One nanometer (nm) � 10�9 m.] Find the probability that a measurement of theparticle’s position at time t� will find the particle in the tiny cubic region with itscenter at x � 1.2 nm, y � �1.0 nm, and z � 0 and with edges each of length0.004 nm.

The distance 0.004 nm is much less than the value of c and a change of 0.004 nmin one or more of the coordinates will not change the probability density ��2 sig-nificantly. It is therefore a good approximation to consider the interval 0.004 nmas infinitesimal and to use (17.12) to write the desired probability as

� 1.200 � 10�9

� 32> 14p nm2 2 4 3>2e�2311.222�1�122�024>410.004 nm 2 30° 0 2 dx dy dz � 12>pc2 2 3>2e�21x2�y2�z22>c2

dx dy dz

° � 12>pc2 2 3>4e�1x2�y2�z22>c2

where c � 2 nm

� 0° 1xa, ya, za, t¿ 2 0 2 dx dy dz

Pr1xa � x � xa � dx, ya � y � ya � dy, za � z � za � dz 2y

z

x

xa

ya

zadz

dydx

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602

Exercise(a) At what point is the probability density a maximum for the � of this exam-ple? Answer by simply looking at ��2. (b) Redo the calculation with x changedto its minimum value in the tiny cubic region and then with x changed to its max-imum value in the region. Compare the results with that found when the centralvalue of x is used. [Answers: (a) At the origin. (b) 1.203 � 10�9, 1.197 � 10�9.]

The state function � is a complex quantity, and �� is the absolute value of �. Let� � f � ig, where f and g are real functions and . The absolute value of �is defined by . For a real quantity, g is zero, and the absolute valuebecomes ( f 2)1/2, which is the usual meaning of absolute value for a real quantity. Thecomplex conjugate �* of � is defined by

(17.13)*

To get �*, we replace i by �i wherever it occurs in �. Note that

(17.14)

since i2 � �1. Therefore, instead of ��2, we can write �*�. The quantity ��2 ��*� � f 2 � g2 is real and nonnegative, as a probability density must be.

In a two-particle system, ��(x1, y1, z1, x2, y2, z2, t�)�2 dx1 dy1 dz1 dx2 dy2 dz2 is theprobability that, at time t�, particle 1 is in a tiny rectangular-box-shaped region locatedat point (x1, y1, z1) and having dimensions dx1, dy1, and dz1, and particle 2 is simulta-neously in a box-shaped region at (x2, y2, z2) with dimensions dx2, dy2, and dz2. Born’sinterpretation of � gives results fully consistent with experiment.

For a one-particle, one-dimensional system, ��(x, t)�2 dx is the probability that theparticle is between x and x � dx at time t. The probability that it is in the regionbetween a and b is found by summing the infinitesimal probabilities over the intervalfrom a to b to give the definite integral �a

b ��2 dx. Thus

(17.15)*

The probability of finding the particle somewhere on the x axis must be 1. Hence, �q

�q ��2 dx � 1. When � satisfies this equation, it is said to be normalized. The nor-malization condition for a one-particle, three-dimensional system is

(17.16)

For an n-particle, three-dimensional system, the integral of ��2 over all 3n coordinatesx1, . . . , zn, each integrated from �q to q, equals 1.

The integral in (17.16) is a multiple integral. In a double integral like �b

a �dc f (x, y) dx dy, one first integrates f (x, y) with respect to x (while treating y as a

constant) between the limits c and d, and then integrates the result with respect to y.For example, �1

0 �40 (2xy � y2) dx dy � �1

0 (x2y � xy2) �40 dy � �10 (16y � 4y2) dy �

28/3. To evaluate a triple integral like (17.16), we first integrate with respect to x whiletreating y and z as constants, then integrate with respect to y while treating z as con-stant, and finally integrate with respect to z.

The normalization requirement is often written

(17.17)*� 0° 0 2 dt � 1

�q

�q�

q

�q�

q

�q

0° 1x, y, z, t 2 0 2 dx dy dz � 1

Pr1a � x � b 2 � �b

a

0° 0 2 dx one-particle, one-dim. syst.

°*° � 1 f � ig 2 1 f � ig 2 � f 2 � i2g2 � f 2 � g2 � 0° 0 2°* � f � ig, where ° � f � ig

0° 0 � 1f 2 � g2 2 1>2 i � 1�1

lev38627_ch17.qxd 3/25/08 12:22 PM Page 602

where is a shorthand notation that stands for the definite integral over the fullranges of all the spatial coordinates of the system. For a one-particle, three-dimensionalsystem, � dt implies a triple integral over x, y, and z from �q to q for each coordi-nate [Eq. (17.16)].

By substitution, it is easy to see that, if � is a solution of (17.10), then so is c�,where c is an arbitrary constant. Thus, there is always an arbitrary multiplicative con-stant in each solution to (17.10). The value of this constant is chosen so as to satisfythe normalization requirement (17.17).

From the state function �, we can calculate the probabilities of the various possi-ble outcomes when a measurement of position is made on the system. In fact, Born’swork is more general than this. It turns out that � gives information on the outcome ofa measurement of any property of the system, not just position. For example, if � isknown, we can calculate the probability of each possible outcome when a measurementof px, the x component of momentum, is made. The same is true for a measurement ofenergy, or angular momentum, etc. (The procedure for calculating these probabilitiesfrom � is discussed in Levine, sec. 7.6.)

The state function � is not to be thought of as a physical wave. Instead � is anabstract mathematical entity that gives information about the state of the system.Everything that can be known about the system in a given state is contained in the statefunction �. Instead of saying “the state described by the function �,” we can just aswell say “the state �.” The information given by � is the probabilities for the possi-ble outcomes of measurements of the system’s physical properties.

The state function � describes a physical system. In Chapters 17 to 20, the sys-tem will usually be a particle, atom, or molecule. One can also consider the state func-tion of a system that contains a large number of molecules, for example, a mole ofsome compound; this will be done in Chapter 21 on statistical mechanics.

Classical mechanics is a deterministic theory in that it allows us to predict theexact paths taken by the particles of the system and tells us where they will be at anyfuture time. In contrast, quantum mechanics gives only the probabilities of finding theparticles at various locations in space. The concept of a path for a particle becomesrather fuzzy in a time-dependent quantum-mechanical system and disappears in atime-independent quantum-mechanical system.

Some philosophers have used the Heisenberg uncertainty principle and the nondeterminis-tic nature of quantum mechanics as arguments in favor of human free will.

The probabilistic nature of quantum mechanics disturbed many physicists, includingEinstein, Schrödinger, and de Broglie. (Einstein wrote in 1926: “Quantum mechanics . . .says a lot, but does not really bring us any closer to the secret of the Old One. I, at any rate,am convinced that He does not throw dice.” When someone pointed out to Einstein thatEinstein himself had introduced probability into quantum theory when he interpreted alight wave’s intensity in each small region of space as being proportional to the probabil-ity of finding a photon in that region, Einstein replied, “A good joke should not be repeatedtoo often.”) These scientists believed that quantum mechanics does not furnish a completedescription of physical reality. However, attempts to replace quantum mechanics by anunderlying deterministic theory have failed. There appears to be a fundamental random-ness in nature at the microscopic level.

SummaryThe state of a quantum-mechanical system is described by its state function �, whichis a function of time and the spatial coordinates of the particles of the system. The statefunction provides information on the probabilities of the outcomes of measurementson the system. For example, when a position measurement is made on a one-particlesystem at time t�, the probability that the particle’s coordinates are found to be in the

�dt Section 17.6Quantum Mechanics

603

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604

ranges x to x � dx, y to y � dy, z to z � dz is given by ��(x, y, z, t�)�2 dx dy dz. Thefunction ��2 is the probability density for position. Because the total probability offinding the particles somewhere is 1, the state function is normalized, meaning that thedefinite integral of ��2 over the full range of all the spatial coordinates is equal to 1.The state function � changes with time according to the time-dependent Schrödingerequation (17.10), which allows the future state (function) to be calculated from thepresent state (function).

17.7 THE TIME-INDEPENDENT SCHRÖDINGER EQUATIONFor an isolated atom or molecule, the forces acting depend only on the coordinates ofthe charged particles of the system and are independent of time. Therefore, the poten-tial energy V is independent of t for an isolated system. For systems where V is inde-pendent of time, the time-dependent Schrödinger equation (17.10) has solutions of theform �(x1, . . . , zn, t) � f (t)c(x1, . . . , zn), where c (lowercase psi) is a function of the3n coordinates of the n particles and f is a certain function of time. We shall demon-strate this for a one-particle, one-dimensional system.

For a one-particle, one-dimensional system with V independent of t, Eq. (17.10)becomes

(17.18)

Let us look for those solutions of (17.18) that have the form

(17.19)

We have �2�/�x2 � f (t) d2c/dx2 and ��/�t � c(x) df/dt. Substitution into (17.18)followed by division by fc � � gives

(17.20)

where the parameter E was defined as E � �(U/i) f �(t)/f (t).From the definition of E, it is equal to a function of t only and hence is indepen-

dent of x. However, (17.20) shows that E � �(U2/2m)c �(x)/c (x) � V(x), which is afunction of x only and is independent of t. Hence, E is independent of t as well as inde-pendent of x and must therefore be a constant. Since the constant E has the same di-mensions as V, it has the dimensions of energy. Quantum mechanics postulates that Eis in fact the energy of the system.

Equation (17.20) gives df/f � �(iE/U) dt, which integrates to ln f � �iEt/U � C.Therefore f � eCe�iEt/U � Ae�iEt/U, where A � eC is an arbitrary constant. The constantA can be included as part of the c(x) factor in (17.19), so we omit it from f. Thus

(17.21)

Equation (17.20) also gives

(17.22)

which is the (time-independent) Schrödinger equation for a one-particle, one-dimensional system. Equation (17.22) can be solved for c when the potential-energyfunction V(x) has been specified.

For an n-particle, three-dimensional system, the same procedure that led toEqs. (17.19), (17.21), and (17.22) gives

(17.23)° � e�iEt>Uc1x1, y1, z1, . . . , xn, yn, zn 2

�U 2

2m

d2c1x 2dx2 � V1x 2c1x 2 � Ec1x 2

f 1t 2 � e�iEt>U

�U 2

2m

1

c 1x 2d2c

dx2 � V1x 2 � �Ui

1

f 1t 2df 1t 2

dt� E

° 1x, t 2 � f 1t 2c1x 2

�U 2

2m

02°0x2 � V1x 2° � �

Ui

0°0t

lev38627_ch17.qxd 3/27/08 11:59 AM Page 604

where the function c is found by solving

(17.24)*

The solutions c to the time-independent Schrödinger equation (17.24) are the (time-independent) wave functions. States for which � is given by (17.23) are called sta-tionary states. We shall see that for a given system there are many different solutionsto (17.24), different solutions corresponding to different values of the energy E. Ingeneral, quantum mechanics gives only probabilities and not certainties for the out-come of a measurement. However, when a system is in a stationary state, a measure-ment of its energy is certain to give the particular energy value that corresponds tothe wave function c of the system. Different systems have different forms for thepotential-energy function V(x1, . . . , zn), and this leads to different sets of allowedwave functions and energies when (17.24) is solved for different systems. All this willbe made clearer by the examples in the next few sections.

For a stationary state, the probability density ��2 becomes

(17.25)

where we used (17.19), (17.21) and the identity

(Prob. 17.19). Hence, for a stationary state, ��2 � �c�2, which is independent of time.For a stationary state, the probability density and the energy are constant with time.There is no implication, however, that the particles of the system are at rest in a sta-tionary state.

It turns out that the probabilities for the outcomes of measurements of any phys-ical property involve ��, and since �� c�, these probabilities are independent oftime for a stationary state. Thus, the e�iEt/U factor in (17.23) is of little consequence,and the essential part of the state function for a stationary state is the time-independentwave function c(x1, . . . , zn). For a stationary state, the normalization condition (17.17)becomes � �c�2 dt � 1, where � dt denotes the definite integral over all space.

The wave function c of a stationary state of energy E must satisfy the time-independent Schrödinger equation (17.24). However, quantum mechanics postulatesthat not all functions that satisfy (17.24) are allowed as wave functions for the system.In addition to being a solution of (17.24), a wave function must meet the followingthree conditions: (a) The wave function must be single-valued. (b) The wave functionmust be continuous. (c) The wave function must be quadratically integrable.Condition (a) means that c has one and only one value at each point in space. Thefunction of Fig. 17.6a, which is multiple-valued at some points, is not a possible wavefunction for a one-particle, one-dimensional system. Condition (b) means that cmakes no sudden jumps in value. A function like that in Fig. 17.6b is ruled out.Condition (c) means that the integral over all space � �c�2 dt is a finite number. Thefunction x2 (Fig. 17.6c) is not quadratically integrable, since �q

�q x4 dx � (x5/5)�q�q �q � (�q) � q. Condition (c) allows the wave function to be multiplied by a con-stant that normalizes it, that is, that makes � �c�2 dt � 1. [If c is a solution of theSchrödinger equation (17.24), then so is kc, where k is any constant; see Prob. 17.20.]A function obeying conditions (a), (b), and (c) is said to be well-behaved.

Since E occurs as an undetermined parameter in the Schrödinger equation (17.24),the solutions c that are found by solving (17.24) will depend on E as a parameter: c�c(x1, . . . , zn; E). It turns out that c is well-behaved only for certain particular valuesof E, and it is these values that are the allowed energy levels. An example is given inthe next section.

1 fc 2* � f *c*

0° 0 2 � 0 fc 0 2 � 1 fc 2*fc � f*c*fc � eiEt>Uc*e�iEt>Uc � e0c*c � 0c 0 2

�U 2

2m1a 02c

0x21

�02c

0y21

�02c

0z21

b � . . . �U 2

2mn

a 02c

0x2n

�02c

0y2n

�02c

0z2n

b � Vc � Ec

Section 17.7The Time-Independent Schrödinger Equation

605

Figure 17.6

(a) A multivalued function. (b) Adiscontinuous function. (c) Afunction that is not quadraticallyintegrable.

lev38627_ch17.qxd 3/25/08 12:22 PM Page 605


606

Figure 17.7

Potential-energy function for aparticle in a one-dimensional box.

We shall mainly be interested in the stationary states of atoms and molecules,since these give the allowed energy levels. For a collision between two molecules orfor a molecule exposed to the time-varying electric and magnetic fields of electro-magnetic radiation, the potential energy V depends on time, and one must deal withthe time-dependent Schrödinger equation and with nonstationary states.

SummaryIn an isolated atom or molecule, the potential energy V is independent of time and thesystem can exist in a stationary state, which is a state of constant energy and time-independent probability density. For a stationary state, the probability density for theparticles’ locations is given by �c�2, where the time-independent wave function c is afunction of the coordinates of the particles of the system. The possible stationary-statewave functions and energies for a system are found by solving the time-independentSchrödinger equation (17.24) and picking out only those solutions that are single-valued, continuous, and quadratically integrable.

17.8 THE PARTICLE IN A ONE-DIMENSIONAL BOXThe introduction to quantum mechanics in the last two sections is quite abstract. Tohelp make the ideas of quantum mechanics more understandable, this section exam-ines the stationary states of a simple system, a particle in a one-dimensional box. Bythis is meant a single microscopic particle of mass m moving in one dimension x andsubject to the potential-energy function of Fig. 17.7. The potential energy is zero forx between 0 and a (region II) and is infinite elsewhere (regions I and III):

This potential energy confines the particle to move in the region between 0 and a onthe x axis. No real system has a V as simple as Fig. 17.7, but the particle in a boxcan be used as a crude model for dealing with pi electrons in conjugated molecules(Sec. 19.11).

We restrict ourselves to considering the states of constant energy, the stationarystates. For these states, the (time-independent) wave functions c are found by solvingthe Schrödinger equation (17.24), which for a one-particle, one-dimensional system is

(17.26)

Since a particle cannot have infinite energy, there must be zero probability of findingthe particle in regions I and III, where V is infinite. Therefore, the probability density�c�2 and hence c must be zero in these regions: cI � 0 and cIII � 0, or

(17.27)

Inside the box (region II), V is zero and (17.26) becomes

(17.28)

To solve this equation, we need a function whose second derivative gives us the samefunction back again, but multiplied by a constant. Two functions that behave this wayare the sine function and the cosine function, so let us try as a solution

c � A sin rx � B cos sx

d2c

dx2 � �2mE

U 2 c for 0 � x � a

c � 0 for x 6 0 and for x 7 a

�U 2

2m

d2c

dx2 � Vc � Ec

V � e0 for 0 � x � a

q for x 6 0 and for x 7 a

lev38627_ch17.qxd 3/25/08 12:22 PM Page 606

where A, B, r, and s are constants. Differentiation of c gives d2c/dx2 � �Ar2 sin rx �Bs2 cos sx. Substitution of the trial solution in (17.28) gives

(17.29)

If we take r � s � (2mE)1/2 U�1, Eq. (17.29) is satisfied. The solution of (17.28) istherefore

(17.30)

A more formal derivation than we have given shows that (17.30) is indeed the generalsolution of the differential equation (17.28).

As noted in Sec. 17.7, not all solutions of the Schrödinger equation are acceptablewave functions. Only well-behaved functions are allowed. The solution of the particle-in-a-box Schrödinger equation is the function defined by (17.27) and (17.30), whereA and B are arbitrary constants of integration. For this function to be continuous, thewave function inside the box must go to zero at the two ends of the box, since c equalszero outside the box. We must require that c in (17.30) go to zero as x → 0 and as x →a. Setting x � 0 and c � 0 in (17.30), we get 0 � A sin 0 � B cos 0 � A � 0 � B � 1,so B � 0. Therefore

(17.31)

Setting x � a and c� 0 in (17.31), we get 0 � sin[(2mE)1/2U�1a]. The function sin wequals zero when w is 0, �p, �2p, . . . , �np, so we must have

(17.32)

Substitution of (17.32) in (17.31) gives c � A sin (�npx/a) � �A sin (npx/a), sincesin (�z) � �sin z. The use of �n instead of n multiplies c by �1. Since A is arbi-trary, this doesn’t give a solution different from the �n solution, so there is no need toconsider the �n values. Also, the value n � 0 must be ruled out, since it would makec � 0 everywhere (Prob. 17.26), meaning there is no probability of finding the parti-cle in the box. The allowed wave functions are therefore

(17.33)

The allowed energies are found by solving (17.32) for E to get

(17.34)*

where U � h/2pwas used. Only these values of E make c a well-behaved (continuous)function. For example, Fig. 17.8 plots c of (17.27) and (17.31) for E � (1.1)2h2/8ma2.Because of the discontinuity at x � a, this is not an acceptable wave function.

Confining the particle to be between 0 and a requires that c be zero at x � 0 andx � a, and this quantizes the energy. An analogy is the quantization of the vibrationalmodes of a string that occurs when the string is held fixed at both ends. The energylevels (17.34) are proportional to n2, and the separation between adjacent levelsincreases as n increases (Fig. 17.9).

The magnitude of the constant A in c in (17.33) is found from the normalizationcondition (17.17) and (17.25): � �c�2 dt � 1. Since c � 0 outside the box, we needonly integrate from 0 to a, and

1 � �q

�q

0c 0 2 dx � �a

0

0c 0 2 dx � 0A 0 2�a

0

sin2 a npxab dx

E �n2h2

8ma2 , n � 1, 2, 3, . . .

c � A sin 1npx>a 2 for 0 � x � a, where n � 1, 2, 3, . . .

12mE 2 1>2 ��1a � �np

c � A sin 3 12mE 2 1>2 ��1x 4 for 0 � x � a

c � A sin 3 12mE 21>2��1x 4 � B cos 3 12mE 2 1>2��1x 4 for 0 � x � a

�Ar2 sin rx � Bs2 cos sx � �2mE��2A sin rx � 2mE��2B cos sx

Section 17.8The Particle in a One-Dimensional Box

607

Figure 17.8

Plot of the solution to the particle-in-a-box Schrödinger equation forE � (1.1)2h2/8ma2. This solutionis discontinuous at x � a.

Figure 17.9

Lowest four energy levels of aparticle in a one-dimensional box.

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608

A table of integrals gives � sin2 cx dx � x/2 � (1/4c) sin 2cx, and we find �A� � (2/a)1/2.The normalization constant A can be taken as any number having absolute value(2/a)1/2. We could take A � (2/a)1/2, or A � �(2/a)1/2, or A � i(2/a)1/2 (where i �etc. Choosing A � (2/a)1/2, we get

(17.35)

For a one-particle, one-dimensional system, �c(x)�2 dx is a probability. Since prob-abilities have no units, c(x) must have dimensions of length�1/2, as is true for c in(17.35).

The state functions for the stationary states of the particle in a box are given by (17.19),(17.21), and (17.35) as � � e�iEt/U(2/a)1/2 sin (npx/a), for 0 � x � a, where E �n2h2/8ma2 and n � 1, 2, 3, . . . .

EXAMPLE 17.2 Calculation of a transition wavelength

Find the wavelength of the light emitted when a 1 � 10�27 g particle in a 3-Åone-dimensional box goes from the n � 2 to the n � 1 level.

The wavelength l can be found from the frequency n. The quantity h� is theenergy of the emitted photon and equals the energy difference between the twolevels involved in the transition [Eq. (17.7)]:

where (17.34) was used. Use of l � c/n and 1 Å � 10�10 m [Eq. (2.87)] gives

(The mass m is that of an electron, and the wavelength lies in the ultraviolet.)

Exercise(a) For a particle of mass 9.1 � 10�31 kg in a certain one-dimensional box, then � 3 to n � 2 transition occurs at n� 4.0 � 1014 s�1. Find the length of the box.(Answer: 1.07 nm.) (b) Show that the frequency of the n � 3 to 2 particle-in-a-one-dimensional-box transition is 5/3 times the frequency of the 2 to 1 transition.

Let us contrast the quantum-mechanical and classical pictures. Classically, theparticle can rattle around in the box with any nonnegative energy; Eclassical can be anynumber from zero on up. (The potential energy is zero in the box, so the particle’s en-ergy is entirely kinetic. Its speed v can have any nonnegative value, so can haveany nonnegative value.) Quantum-mechanically, the energy can take on only the val-ues (17.34). The energy is quantized in quantum mechanics, whereas it is continuousin classical mechanics.

Classically, the minimum energy is zero. Quantum-mechanically, the particle in abox has a minimum energy that is greater than zero. This energy, h2/8ma2, is the zero-point energy. Its existence is a consequence of the uncertainty principle. Suppose theparticle could have zero energy. Since its energy is entirely kinetic, its speed vx andmomentum mvx � px would then be zero. With px known to be zero, the uncertaintypx is zero, and the uncertainty principle x px h gives x � q. However, weknow the particle to be somewhere between x � 0 and x � a, so x cannot exceed a.Hence, a zero energy is impossible for a particle in a box.

12 mv2

l �8ma2c

3h�

811 � 10�30 kg 2 13 � 10�10 m 2 213 � 108 m>s 2316.6 � 10�34 J s 2 � 1 � 10�7 m

hn � Eupper � Elower � 22h2>8ma2 � 12h2>8ma2 and n � 3h>8ma2

c � a 2ab 1>2

sin npx

a for 0 � x � a, where n � 1, 2, 3, . . .

1�1 2 ,

lev38627_ch17.qxd 3/25/08 12:22 PM Page 608

Section 17.8The Particle in a One-Dimensional Box

609

Figure 17.10

Wave functions and probabilitydensities for the lowest threeparticle-in-a-box stationary states.

The stationary states of a particle in a box are specified by giving the value of theinteger n in (17.35). n is called a quantum number. The lowest-energy state (n � 1)is the ground state. States higher in energy than the ground state are excited states.

Figure 17.10 plots the wave functions c and the probability densities �c�2 for thefirst three particle-in-a-box stationary states. For n � 1, npx/a in the wave function(17.35) goes from 0 to p as x goes from 0 to a, so c is half of one cycle of a sinefunction.

Classically, all locations for the particle in the box are equally likely. Quantum-mechanically, the probability density is not uniform along the length of the box, butshows oscillations. In the limit of a very high quantum number n, the oscillations in�c�2 come closer and closer together and ultimately become undetectable; this corre-sponds to the classical result of uniform probability density. The relation 8ma2E/h2 �n2 shows that for a macroscopic system (E, m, and a having macroscopic magnitudes),n is very large, so the limit of large n is the classical limit.

A point at which c � 0 is called a node. The number of nodes increases by 1 foreach increase in n. The existence of nodes is surprising from a classical viewpoint. Forexample, for the n � 2 state, it is hard to understand how the particle can be found inthe left half of the box or in the right half but never at the center. The behavior ofmicroscopic particles (which have a wave aspect) cannot be rationalized in terms of avisualizable model.

The wave functions c and probability densities �c�2 are spread out over the lengthof the box, much like a wave (compare Figs. 17.10 and 17.2). However, quantum me-chanics does not assert that the particle itself is spread out like a wave; a measurementof position will give a definite location for the particle. It is the wave function c(which gives the probability density �c�2) that is spread out in space and obeys a waveequation.

EXAMPLE 17.3 Probability calculations

(a) For the ground state of a particle in a one-dimensional box of length a, findthe probability that the particle is within �0.001a of the point x � a/2. (b) Forthe particle-in-a-box stationary state with quantum number n, write down (but donot evaluate) an expression for the probability that the particle will be found

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610

between a/4 and a/2. (c) For a particle-in-a-box stationary state, what is theprobability that the particle will be found in the left half of the box?

(a) The probability density (the probability per unit length) equals �c�2.Figure 17.10 shows that �c�2 for n � 1 is essentially constant over the very smallinterval 0.002a, so we can consider this interval to be infinitesimal and take �c�2 dxas the desired probability. For n � 1, Eq. (17.35) gives �c�2 � (2/a) sin2 (px/a).With x � a/2 and dx � 0.002a, the probability is �c�2 dx � (2/a) sin2 (p/2) �0.002a � 0.004.

(b) From Eq. (17.15), the probability that the particle is between points c andd is �d

c ��2 dx. But ��2 � �c�2 for a stationary state [Eq. (17.25)], so the proba-bility is � d

c �c�2 dx. The desired probability is �aa

//24 (2/a) sin2 (npx/a) dx, where

(17.35) was used for c.(c) For each particle-in-a-box stationary state, the graph of �c�2 is symmet-

ric about the midpoint of the box, so the probabilities of being in the left andright halves are equal and are each equal to 0.5.

ExerciseFor the n � 2 state of a particle in a box of length a, (a) find the probability theparticle is within �0.0015a of x � a/8; (b) find the probability the particle isbetween x � 0 and x � a/8. (Answers: (a) 0.0030; (b) 1/8 � 1/4p � 0.0454.)

If ci and cj are particle-in-a-box wave functions with quantum numbers ni and nj,one finds (Prob. 17.29) that

(17.36)

where ci � (2/a)1/2 sin (nipx/a) and cj � (2/a)1/2 sin (njpx/a). The functions f and gare said to be orthogonal when � f*g dt � 0, where the integral is a definite integralover the full range of the spatial coordinates. One can show that two wave functionsthat correspond to different energy levels of a quantum-mechanical system are or-thogonal (Sec. 17.16).

17.9 THE PARTICLE IN A THREE-DIMENSIONAL BOXThe particle in a three-dimensional box is a single particle of mass m confined toremain within the volume of a box by an infinite potential energy outside the box. Thesimplest box shape to deal with is a rectangular parallelepiped. The potential energyis therefore V � 0 for points such that 0 � x � a, 0 � y � b, and 0 � z � c and V �q elsewhere. The dimensions of the box are a, b, and c. In Secs. 20.3 and 21.6, thissystem will be used to give the energy levels for translational motion of ideal-gas mol-ecules in a container.

Let us solve the time-independent Schrödinger equation for the stationary-statewave functions and energies. Since V � q outside the box, c is zero outside the box,just as for the corresponding one-dimensional problem. Inside the box, V � 0, and theSchrödinger equation (17.24) becomes

(17.37)

Let us assume that solutions of (17.37) exist that have the form X(x)Y(y)Z(z),where X(x) is a function of x only and Y and Z are functions of y and z. For an arbi-trary partial differential equation, it is not in general possible to find solutions in whichthe variables are present in separate factors. However, it can be proved mathematically

��2

2ma 02c

0x2 �02c

0y2 �02c

0z2 b � Ec

�a

0

c*i cj dt � 0 for ni � nj

lev38627_ch17.qxd 3/25/08 12:22 PM Page 610

that, if we succeed in finding well-behaved solutions to (17.37) that have the formX(x)Y(y)Z(z), then there are no other well-behaved solutions, so we shall have foundthe general solution of (17.37). Our assumption is then

(17.38)

Partial differentiation of (17.38) gives

, ,

Substitution in (17.37) followed by division by X(x)Y(y)Z(z) � c gives

(17.39)

Let Ex � �(U2/2m)X �(x)/X(x). Then (17.39) gives

(17.40)

From its definition, Ex is a function of x only. However, the relation Ex � E �U2Y�/2mY � U2Z�/2mZ in (17.40) shows Ex to be independent of x. Therefore Ex is aconstant, and we have from (17.40)

(17.41)

Equation (17.41) is the same as the Schrödinger equation (17.28) for a particle ina one-dimensional box if X and Ex in (17.41) are identified with c and E, respectively,in (17.28). Moreover, the condition that X(x) be continuous requires that X(x) � 0 atx � 0 and at x � a, since the three-dimensional wave function is zero outside the box.These are the same requirements that c in (17.28) must satisfy. Therefore, the well-behaved solutions of (17.41) and (17.28) are the same. Replacing c and E in (17.34)and (17.35) by X and Ex, we get

(17.42)

where the quantum number is called nx.Equation (17.39) is symmetric with respect to x, y, and z, so the same reasoning

that gave (17.42) gives

(17.43)

(17.44)

where, by analogy to (17.40),

(17.45)

We assumed in Eq. (17.38) that the wave function c is the product of separate fac-tors X(x), Y(y), and Z(z) for each coordinate. Having found X, Y, and Z [Eqs. (17.42),(17.43), and (17.44)], we have as the stationary-state wave functions for a particle ina three-dimensional rectangular box

(17.46)

Outside the box, c � 0.

c � a 8

abcb 1>2

sin nxpx

a sin

nypy

b sin

nzpz

c inside the box

Ey � �U 2

2m

Y– 1y 2Y1y 2 , Ez � �

U 2

2m

Z– 1z 2Z1z 2

Z1z 2 � a 2cb 1>2

sin nzpz

c, Ez �

nz2h2

8mc2, nz � 1, 2, 3, p

Y1y 2 � a 2

bb 1>2

sin nypy

b, Ey �

ny2h2

8mb2, ny � 1, 2, 3, p

X1x 2 � a 2ab 1>2

sin nxpx

a, Ex �

nx2h2

8ma2, nx � 1, 2, 3, p

�1 U 2>2m 2X– 1x 2 � ExX1x 2 for 0 � x � a

Ex � �U 2

2m

X– 1x 2X1x 2 � E �

U 2

2m

Y– 1y 2Y1y 2 �

U 2

2m

Z– 1z 2Z1z 2

��2

2m

X– 1x 2X1x 2 �

�2

2m

Y– 1y 2Y1y 2 �

�2

2m

Z– 1z 2Z1z 2 � E

02c>0z2 � X1x 2Y1y 2Z– 1z 202c>0y2 � X1x 2Y– 1y 2Z1z 202c>0x2 � X– 1x 2Y1y 2Z1z 2

c � X1x 2Y1y 2Z1z 2

Section 17.9The Particle in a

Three-Dimensional Box

611

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612

Equations (17.39), (17.40), and (17.45) give E � Ex � Ey � Ez, and use of (17.42)to (17.44) for Ex, Ey, and Ez gives the allowed energy levels as

(17.47)

The quantities Ex, Ey, and Ez are the kinetic energies associated with motion in the x,y, and z directions.

The procedure used to solve (17.37) is called separation of variables. The con-ditions under which it works are discussed in Sec. 17.11.

The wave function has three quantum numbers because this is a three-dimensionalproblem. The quantum numbers nx, ny, and nz vary independently of one another. Thestate of the particle in the box is specified by giving the values of nx, ny, and nz. Theground state is nx � 1, ny � 1, and nz � 1.

The Particle in a Two-Dimensional BoxFor a particle in a two-dimensional rectangular box with sides a and b, the same pro-cedure that gave (17.46) and (17.47) gives

(17.48)

and E � (h2/8m) (nx2/a2 � ny

2/b2). For a two-dimensional box with b � 2a, Fig. 17.11shows the variation of the probability density �c�2 in the box for three states. Thegreater the density of dots in a region, the greater the value of �c�2. Figure 17.12 showsthree-dimensional graphs of �c�2 for the lowest two states. The height of the surfaceabove the xy plane gives the value of �c�2 at point (x, y). Figure 17.13 is a three-dimensional graph of c for the nx � 1, ny � 2 state; c is positive in half the box, neg-ative in the other half, and zero on the line that separates these two halves. Figure 17.14shows contour plots of constant �c� for the nx � 1, ny � 2 state; the contours shown arethose for which �c�/�c�max � 0.9 (the two innermost loops), 0.7, 0.5, 0.3, and 0.1, where�c�max is the maximum value of �c�. These contours correspond to �c�2/�c2�max � 0.81,0.49, 0.25, 0.09, and 0.01.

17.10 DEGENERACYSuppose the sides of the three-dimensional box of the last section have equal lengths:a � b � c. Then (17.46) and (17.47) become

(17.49)c � 12>a 2 3>2 sin 1nxpx>a 2 sin 1nypy>a 2 sin 1nzpz>a 2

c � 14>ab 2 1>2 sin 1nxpx>a 2 sin 1nypy>b 2 for 0 � x � a, 0 � y � b

E �h2

8ma nx

2

a2 �ny

2

b2 �nz

2

c2 b

Figure 17.11

Probability densities for threestates of a particle in a two-dimensional box whosedimensions have a 2:1 ratio. Thestates are the c11, c12, and c21states, where the subscripts givethe nx and ny values.

Figure 17.12

Three-dimensional plot of �c�2 forthe c11 and c12 states of a two-dimensional box with b � 2a.

Figure 17.13

Three-dimensional plot of c12 fora particle in a two-dimensionalbox with b � 2a.

lev38627_ch17.qxd 3/25/08 12:22 PM Page 612

(17.50)

Let us use numerical subscripts on c to specify the nx, ny, and nz values. The lowest-energy state is c111 with E 3h2/8ma2. The states c211, c121, and c112 each have en-ergy 6h2/8ma2. Even though they have the same energy, these are different states. Withnx 2, ny 1, and nz � 1 in (17.49), we get a different wave function than with nx1, ny 2, and nz 1. The c211 state has zero probability density of finding the parti-cle at x a/2 (see Fig. 17.10), but the c121 state has a maximum probability densityat x a/2.

The terms “state” and “energy level” have different meanings in quantum mechan-ics. A stationary state is specified by giving the wave function c. Each different c isa different state. An energy level is specified by giving the value of the energy. Eachdifferent value of E is a different energy level. The three different particle-in-a-boxstates c211, c121, and c112 belong to the same energy level, 6h2/8ma2. Figure 17.15shows the lowest few stationary states and energy levels of a particle in a cubic box.

An energy level that corresponds to more than one state is said to be degenerate.The number of different states belonging to the level is the degree of degeneracy ofthe level. The particle-in-a-cubic-box level 6h2/8ma2 is threefold degenerate. Theparticle-in-a-box degeneracy arises when the dimensions of the box are made equal.Degeneracy usually arises from the symmetry of the system.

17.11 OPERATORSOperatorsQuantum mechanics is most conveniently formulated in terms of operators. An oper-ator is a rule for transforming a given function into another function. For example, theoperator d/dx transforms a function into its first derivative: (d/dx) f (x) f �(x). Let symbolize an arbitrary operator. (We shall use a circumflex to denote an operator.) If transforms the function f (x) into the function g(x), we write f (x) g(x). If isthe operator d/dx, then g(x) (x). If is the operator “multiplication by 3x2,” theng(x) 3x2f (x). If log, then g(x) log f (x).

The sum of two operators and is defined by

(17.51)*

For example, (ln � d/dx) f(x) � ln f(x) � (d/dx) f(x) � ln f(x) � f�(x). Similarly,( � ) f(x) � f(x) � f(x).

The square of an operator is defined by 2f (x) � [ f (x)]. For example,

Therefore, (d/dx)2 � d2/dx2.The product of two operators is defined by

(17.52)*

The notation [ f (x)] means that we first apply the operator to the function f(x) toget a new function, and then we apply the operator to this new function.

Two operators are equal if they produce the same result when operating on anarbitrary function: � if and only if f � f for every function f.

EXAMPLE 17.4 Operator algebra

Let the operators and be defined as x and d/dx. (a) Find ( )(x3 cos x). (b) Find f (x) and f (x). Are the operators and equal? (c) Find .AB � BA

BAABBAAB�A � B�B#�ABA

CBCB

ABBA

1AB 2 f 1x 2 � A 3 B f 1x 2 4

1d>dx 22 f 1x 2 � 1d>dx 2 3 1d>dx 2 f 1x 2 4 � 1d>dx 2 3 f ¿1x 2 4 � f –1x 2 � 1d2>dx2 2 f 1x 2AAA

BABA

1A � B 2 f 1x 2 � Af 1x 2 � Bf 1x 2BA��A�

Af ¿�A�AA

A�

��

��

�

E � 1nx2 � ny

2 � nz2 2h2>8ma2 Section 17.11

Operators

613

Figure 17.14

Contour plot of constant �c� for thestate of Fig. 17.13.

Figure 17.15

Lowest seven stationary states(and lowest three energy levels)for a particle in a cubic box. Thenumbers are the values of thequantum numbers nx, ny, and nz.

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614

(a) Using the definition (17.51) of the sum of operators, we have

(b) The definition (17.52) of the operator product gives

In this example, and produce different results when they operate on f (x),so and are not equal in this case. In multiplication of numbers, the orderdoesn’t matter. In multiplication of operators, the order may matter.

(c) To find the operator � , we examine the result of applying itto an arbitrary function f(x). We have ( � ) f(x) � f � f � xf��(xf� � f) � �f, where the definition of the difference of operators and the resultsof (b) were used. Since ( � ) f(x) � �1� f(x) for all functions f(x), the defin-ition of equality of operators gives

where the multiplication sign after the �1 is omitted, as is customary.The operator � is called the commutator of and and is sym-

bolized by [ , ];

ExerciseLet � x2 and � d2/dx2. (a) Find ( � )(x 4 � 1/x). (b) Find f (x) and

f (x). (c) Find [ , ]. [Answers: (a) x6 � 12x2 � x � 2/x3; (b) x2f �(x), 2f (x) �4xf �(x) � x2f �(x); (c) �2 � 4x (d/dx).]

Operators in Quantum MechanicsIn quantum mechanics, each physical property of a system has a correspondingoperator. The operator that corresponds to px, the x component of momentum of a par-ticle, is postulated to be (U/i)( / x), with similar operators for py and pz:

(17.53)*

where x is the quantum-mechanical operator for the property px and . Theoperator that corresponds to the x coordinate of a particle is multiplication by x, andthe operator that corresponds to f (x, y, z), where f is any function, is multiplication bythat function. Thus,

(17.54)*

To find the operator that corresponds to any other physical property, we writedown the classical-mechanical expression for that property as a function of cartesiancoordinates and corresponding momenta and then replace the coordinates and mo-menta by their corresponding operators (17.53) and (17.54). For example, the energyof a one-particle system is the sum of its kinetic and potential energies:

E � K � V � 12 m1vx

2 � vy2 � vz

2 2 � V1x, y, z, t 2

x � x � , y � y � , z � z � , f 1x, y, z 2 � f 1x, y, z 2 �

i � 1�1p

px �Ui

00x

, py �Ui

00y

, pz �Ui

00z

SRRSSRSRSR

3 A, B 4 � AB � BA

BABABAAB

AB � BA � �1

BAAB

BAABBAABBAAB

BAABBAAB

BA f 1x 2 � B 3 A f 1x 2 4 � 1d>dx 2 3x f 1x 2 4 � x f ¿1x 2 � f 1x 2AB f 1x 2 � A 3B ˆ f 1x 2 4 � x 3 1d>dx 2 f 1x 2 4 � x 3 f ¿ 1x 2 4 � x f ¿ 1x 2

� x4 � x cos x � 3x2 � sin x

� x1x3 � cos x 2 � 1d>dx 2 1x3 � cos x 21A � B 2 1x3 � cos x 2 � 1x � d>dx 2 1x3 � cos x 2

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Section 17.11Operators

615

To express E as a function of the momenta and coordinates, we note that px mvx,py � mvy , pz � mvz. Therefore,

(17.55)*

The expression for the energy as a function of coordinates and momenta is called thesystem’s Hamiltonian H [after W. R. Hamilton (1805–1865), who reformulatedNewton’s second law in terms of H ]. The use of Eq. (17.53) and i2 � �1 gives

so � �U2 2/ x2 and /2m � �(U2/2m) 2/ x2. From (17.54), the potential-energyoperator is simply multiplication by V(x, y, z, t). (Time is a parameter in quantum me-chanics, and there is no time operator.) Replacing px

2, py2, pz

2, and V in (17.55) by theiroperators, we get as the energy operator, or Hamiltonian operator, for a one-particlesystem

(17.56)

To save time in writing, we define the Laplacian operator � 2 (read as “delsquared”) by �2 � 2/ x2 � 2/ y2 � 2/ z2 and write the one-particle Hamiltonianoperator as

(17.57)

where the multiplication sign after V is understood.For a many-particle system, we have x,1 � (U/i) / x1 for particle 1, and the

Hamiltonian operator is readily found to be

(17.58)*

(17.59)*

with similar definitions for �22, . . . , �n

2. The terms in (17.58) are the operators for thekinetic energies of particles 1, 2, . . . , n and the potential energy of the system.

From (17.58), we see that the time-dependent Schrödinger equation (17.10) canbe written as

(17.60)

and the time-independent Schrödinger equation (17.24) can be written as

(17.61)*

where V in (17.61) is independent of time. Since there is a whole set of allowedstationary-state wave functions and energies, (17.61) is often written as cj � Ejcj,where the subscript j labels the various wave functions (states) and their energies.

When an operator applied to the function f gives the function back again butmultiplied by the constant c, that is, when

one says that f is an eigenfunction of with eigenvalue c. (However, the functionf � 0 everywhere is not allowed as an eigenfunction.) The wave functions c in (17.61)are eigenfunctions of the Hamiltonian operator , the eigenvalues being the allowedenergies E.

H

B

Bf � cf

B

H

Hc � Ec

�Ui

0°0t

� H °

§12 �

02

0x12 �

02

0y12 �

02

0z12

H � �U 2

2m1§1

2 �U 2

2m2§2

2 � p �U 2

2mn

§n2 � V1x1, p , zn, t 2

p

H � �1 U 2>2m 2§2 � V

E � H � �U 2

2ma 02

0x2 �02

0y2 �02

0z2 b � V1x, y, z, t 2 �

px2px

2

px2 f 1x, y, z 2 � 1 U>i 2 10>0x 2 3 1 U>i 2 10>0x 2 f 4 � 1 U2>i2 2 02f>0x2 � �U 2 02f>0x2

E �1

2m1px

2 � py2 � pz

2 2 � V1x, y, z, t 2 � H

�

lev38627_ch17.qxd 3/25/08 12:22 PM Page 615


616

Operator algebra differs from ordinary algebra. From c Ec [Eq. (17.61)], onecannot conclude that � E. is an operator and E is a number, and the two are notequal. Note, for example, that (d/dx)e2x � 2e2x, but d/dx 2. In Example 17.4, wefound that ( � ) f (x) � �1 f (x) (for � x and � d/dx) and concluded that

� � �1. Because this equation applies to all functions f (x), it is valid todelete the f (x) here. However, the relation (d/dx)e2x � 2e2x applies only to the functione2x, and this function cannot be deleted.

EXAMPLE 17.5 Eigenfunctions

Verify directly that c � Ec for the particle in a one-dimensional box.Inside the box (Fig. 17.7), V � 0 and Eq. (17.56) gives � �(U2/2m) d2/dx2

for this one-dimensional problem. The wave functions are given by (17.35) asc � (2/a)1/2 sin (npx/a). We have, using (1.27) and (17.11):

since E � n2h2/8ma2 [Eq. (17.34)].

ExerciseVerify that the function Aeikx, where A and k are constants, is an eigenfunctionof the operator x. What is the eigenvalue? (Answer: kU.)

The operators that correspond to physical quantities in quantum mechanics arelinear. A linear operator is one that satisfies the following two equations for allfunctions f and g and all constants c:

The operator �/�x is linear, since (�/�x)(f � g) � �f/�x � �g/�x and (�/�x)(cf ) �c �f/�x. The operator is nonlinear, since .

If the function satisfies the time-independent Schrödinger equation then so does the function cc, where c is any constant. Proof of this follows from thefact that the Hamiltonian operator is a linear operator. We have (cc) � c c �cEc � E(cc). The freedom to multiply c by a constant enables us to normalize c.

MeasurementMultiplication of c � Ec [Eq. (17.61)] by e�iEt/U gives e�iEt/U c � Ee�iEt/Uc. For astationary state, does not involve time and e�iEt/U c � (e�iEt/Uc). Using � �e�iEt/Uc [Eq. (17.23)], we have

so � is an eigenfunction of with eigenvalue E for a stationary state. A stationarystate has a definite energy, and measurement of the system’s energy will always givea single predictable value when the system is in a stationary state. For example, for then � 2 particle-in-a-box stationary state, measurement of the energy will always givethe result 22h2/8ma2 [Eq. (17.34)].

What about properties other than the energy? Let the operator correspond to theproperty M. Quantum mechanics postulates that if the system’s state function � hap-pens to be an eigenfunction of with eigenvalue c (that is, if � � c�), then aMM

M

H

H° � E°

HHHHH

HHH

Hc � Ec,c1f � g 1f � 1g1

L1 f � g 2 � Lf � Lg and L1cf 2 � cLf

L

p

�n2h2

8ma2 a 2ab 1>2

sin npx

a� Ec

Hc � �U 2

2m

d2

dx2 c a 2ab 1>2

sin npx

ad � �

h2

4p212m 2 a2ab 1>2 a�n2p2

a2 b sin npx

a

HH

BAABBABAAB

HH�H

lev38627_ch17.qxd 3/27/08 11:59 AM Page 616

measurement of M is certain to give the value c as the result. (Examples will be givenwhen we consider angular momentum in Sec. 18.4). If � is not an eigenfunction of

then the result of measuring M cannot be predicted. (However, the probabilities ofthe various possible outcomes of a measurement of M can be calculated from �, butdiscussion of how this is done is omitted.) For stationary states, the essential part of �is the time-independent wave function c, and c replaces � in the italicized statementin this paragraph.

Average ValuesFrom (14.38), the average value of x for a one-particle, one-dimensional quantum-mechanical system equals �q

�q xg(x) dx, where g(x) is the probability density forfinding the particle between x and x � dx. But the Born postulate (Sec. 17.6) gives g(x) � ��(x)�2. Hence, �x� � �q

�q x��(x)�2 dx. Since ��2 � �*�, we have �x� ��q

�q �*x� dx � �q�q �* � dx, where (17.54) was used.

What about the average value of an arbitrary physical property M for a generalquantum-mechanical system? Quantum mechanics postulates that the average value ofany physical property M in a system whose state function is � is given by

(17.62)

where is the operator for the property M and the integral is a definite integral overall space. In (17.62), operates on � to produce the result �, which is a function.The function � is then multiplied by �*, and the resulting function �* � is inte-grated over the full range of the spatial coordinates of the system. For example, Eq. (17.53) gives the px operator as x � (U/i) / x, and the average value of px for a one-particle, three-dimensional system whose state function is � is � px� �(U/i) �q

�q �q�q �

q�q �*( �/ x) dx dy dz.

The average value of M is the average of the results of a very large number ofmeasurements of M made on identical systems, each of which is in the same state �just before the measurement.

If � happens to be an eigenfunction of with eigenvalue c, then � � c� and(17.62) becomes �M� � � �* � dt � � �*c� dt � c� �*� dt � c, since � isnormalized. This result makes sense since, as noted in the last subsection, c is the onlypossible result of a measurement of M if � � c�.

For a stationary state, � equals e�iEt/Uc [Eq. (17.23)]. Since doesn’t affect thee�iEt/U factor, we have

Therefore, for a stationary state,

(17.63)*

EXAMPLE 17.6 Average value

For a particle in a one-dimensional-box stationary state, give the expressionfor �x2�.

For a one-particle, one-dimensional problem, dt � dx. Since 2 � x2 �, wehave

8x2 9 � �q

�q

c*x2c dx � �0

�q

x2 0c 0 2 dx � �a

0

x2 0c 0 2 dx � �q

a

x2 0c 0 2 dx

x

8M 9 � �c*Mc dt

°*M° � eiEt>Uc*Me�iEt>Uc � eiEt>Ue�iEt>Uc*Mc � c*Mc

MM

MMM

p

MMMM

M

8M 9 � �°*M° dt

x

M,

Section 17.11Operators

617

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618

since c*c � �c�2 [Eq. (17.14)]. For x � 0 and x � a, we have c � 0[Eq. (17.27)] and inside the box c� (2/a)1/2 sin (npx/a) [Eq. (17.35)]. Therefore

Evaluation of the integral is left as a homework problem (Prob. 17.42).

ExerciseEvaluate �px� for a particle in a one-dimensional-box stationary state. [Answer:(2npU/ia2) �a

0 sin (npx/a) cos (npx/a) dx � 0.]

Separation of VariablesLet q1, q2, . . . , qr be the coordinates of a system. For example, for a two-particle sys-tem, q1 � x1, q2 � y1, . . . , q6 � z2. Suppose the Hamiltonian operator has the form

(17.64)

where the operator involves only q1, the operator involves only q2, etc. An ex-ample is the particle in a three-dimensional box, where one has � � �with x � �(U2/2m) 2/ x2, etc. We saw in Sec. 17.9 that, for this case, c �X(x)Y(y)Z(z) and E � Ex � Ey � Ez, where X(x) � ExX(x), Y(y) � EyY(y), zZ(z) �EzZ(z) [Eqs. (17.41) and (17.45)].

The same type of argument used in Sec. 17.9 shows (Prob. 17.43) that when isthe sum of separate terms for each coordinate, as in (17.64), then each stationary-statewave function is the product of separate factors for each coordinate and each stationary-state energy is the sum of energies for each coordinate:

(17.65)*

(17.66)*

where E1, E2, . . . and the functions f1, f2, . . . are found by solving

(17.67)

The equations in (17.67) are, in effect, separate Schrödinger equations, one for eachcoordinate.

Noninteracting ParticlesAn important case where separation of variables applies is a system of n noninteract-ing particles, meaning that the particles exert no forces on one another. For such a sys-tem, the classical-mechanical energy is the sum of the energies of the individual par-ticles, so the classical Hamiltonian H and the quantum-mechanical Hamiltonianoperator have the forms H � H1 � H2 � � � � � Hn and � � � � � � �where involves only the coordinates of particle 1, involves only particle 2, etc.Here, by analogy to (17.65) to (17.67), we have

(17.68)*

(17.69)*

(17.70)*

For a system of noninteracting particles, there is a separate Schrödinger equation foreach particle, the wave function is the product of wave functions of the individual

H1 f1 � E1 f1, H2 f2 � E2 f2, p , Hn fn � En fn

E � E1 � E2 � p � En

c � f11x1, y1, z1 2 f21x2, y2, z2 2 p fn1xn, yn, zn 2H2H1

Hn,H2H1HH

H1 f1 � E1 f1, H2 f2 � E2 f2, . . . , Hr fr � Er fr

E � E1 � E2 � p � Er

c � f11q1 2 f21q2 2 p fr 1qr 2

H

HHyHx

HHz,HyHxH

H2H1

H � H1 � H2 � . . . � Hr

8x2 9 �2a

�a

0

x2 sin2 npx

a dx

lev38627_ch17.qxd 3/25/08 12:22 PM Page 618

particles, and the energy is the sum of the energies of the individual particles. (Fornoninteracting particles, the probability density �c�2 is the product of probability den-sities for each particle: �c�2 � f1�2� f2�2 � � � � fn�2. This is in accord with the theoremthat the probability that several independent events will all occur is the product of theprobabilities of the separate events.)

17.12 THE ONE-DIMENSIONAL HARMONIC OSCILLATORThe one-dimensional harmonic oscillator is a useful model for treating the vibrationof a diatomic molecule (Sec. 20.3) and is also relevant to vibrations of polyatomicmolecules (Sec. 20.8) and crystals (Sec. 23.12).

Classical TreatmentBefore examining the quantum mechanics of a harmonic oscillator, we review theclassical treatment. Consider a particle of mass m that moves in one dimension and isattracted to the coordinate origin by a force proportional to its displacement from theorigin: F � �kx, where k is called the force constant. When x is positive, the forceis in the �x direction, and when x is negative, F is in the �x direction. A physicalexample is a mass attached to a frictionless spring, x being the displacement from theequilibrium position. From (2.17), F � �dV/dx, where V is the potential energy.Hence �dV/dx � �kx, and . The choice of zero of potential energy isarbitrary. Choosing the integration constant c as zero, we have (Fig. 17.16)

(17.71)

Newton’s second law F � ma gives m d2x/dt2 � �kx. The solution to this differ-ential equation is

(17.72)

as can be verified by substitution in the differential equation (Prob. 17.52). In (17.72),A and b are integration constants. The maximum and minimum values of the sine func-tion are �1 and �1, so the particle’s x coordinate oscillates back and forth between�A and �A. A is the amplitude of the motion.

The period t (tau) of the oscillator is the time required for one complete cycle ofoscillation. For one cycle of oscillation, the argument of the sine function in (17.72)must increase by 2p, since 2p is the period of a sine function. Hence the periodsatisfies (k/m)1/2t � 2p, and t � 2p(m/k)1/2. The frequency n is the reciprocal of theperiod and equals the number of vibrations per second (n � 1/t); thus

(17.73)*

The energy of the harmonic oscillator is . The use of(17.72) for x and of leads to (Prob. 17.52)

(17.74)

Equation (17.74) shows that the classical energy can have any nonnegative value. Asthe particle oscillates, its kinetic energy and potential energy continually change, butthe total energy remains constant at .

Classically, the particle is limited to the region When the particlereaches x A or x A, its speed is zero (since it reverses its direction of motionat and ) and its potential energy is a maximum, being equal to . If theparticle were to move beyond x A, its potential energy would increase above

. This is impossible for a classical particle. The total energy is and the12 kA21

2 kA2� ;

12 kA2�A�A

� ��A � x � A.

12 kA2

E � 12 kA2

vx � dx/dt � 1k/m 2 1/2A cos 3 1k/m 2 1/2t � b 4E � K � V � 12 mv2

x � 12 kx2

n �1

2pa k

mb 1>2

x � A sin 3 1k>m 2 1>2t � b 4

V � 12 kx2

V � 12 kx2 � c

�

Section 17.12The One-Dimensional Harmonic

Oscillator

619

Figure 17.16

The potential-energy function fora one-dimensional harmonicoscillator.

lev38627_ch17.qxd 3/25/08 12:22 PM Page 619


620

kinetic energy is nonnegative, so the potential energy (V E K ) cannot exceedthe total energy.

Quantum-Mechanical TreatmentNow for the quantum-mechanical treatment. Substitution of V � in (17.26) givesthe time-independent Schrödinger equation as

(17.75)

Solution of the harmonic-oscillator Schrödinger equation (17.75) is complicated andis omitted in this book (see any quantum chemistry text). Here, we examine the results.One finds that quadratically integrable (Sec. 17.7) solutions to (17.75) exist only forthe following values of E:

(17.76)*

where the vibrational frequency n is given by (17.73) and the quantum number v takeson nonnegative integral values. [Don’t confuse the typographically similar symbols n(nu) and v (vee).] The energy is quantized. The allowed energy levels (Fig. 17.17) areequally spaced (unlike the particle in a box). The zero-point energy is . (For a col-lection of harmonic oscillators in thermal equilibrium, all the oscillators will fall to theground state as the temperature goes to absolute zero; hence the name zero-pointenergy.) For all values of E other than (17.76), one finds that the solutions to (17.75)go to infinity as x goes to �q, so these solutions are not quadratically integrable andare not allowed as wave functions.

The well-behaved solutions to (17.75) turn out to have the form

where a � 2pnm/U. The polynomial that multiplies e�ax2/2 contains only even powersof x or only odd powers, depending on whether the quantum number v is even or odd.The explicit forms of the lowest few wave functions c0, c1, c2, and c3 (where the sub-script on c gives the value of the quantum number v) are given in Fig. 17.18, whichplots these c’s. As with the particle in a one-dimensional box, the number of nodes in-creases by one for each increase in the quantum number. Note the qualitative resem-blance of the wave functions in Figs. 17.18 and 17.10.

The harmonic-oscillator wave functions fall off exponentially to zero as x → �q.Note, however, that even for very large values of x, the wave function c and the prob-ability density �c�2 are not zero. There is some probability of finding the particle at anindefinitely large value of x. For a classical-mechanical harmonic oscillator withenergy , Eq. (17.74) gives , and A � [(2v � 1)hn/k]1/2. Aclassical oscillator is confined to the region �A � x � A. However, a quantum-mechanical oscillator has some probability of being found in the classically forbiddenregions x � A and x � �A, where the potential energy is greater than the particle’stotal energy. This penetration into classically forbidden regions is called tunneling.Tunneling occurs more readily the smaller the particle’s mass and is most importantin chemistry for electrons, protons, and H atoms. Tunneling influences the rates ofreactions involving these species (see Secs. 22.3 and 22.4). Electron tunneling is thebasis for the scanning tunneling microscope, a remarkable device that gives picturesof the atoms on the surface of a solid (Sec. 23.10). Tunneling makes possible thefusion of hydrogen nuclei to helium nuclei in the sun, despite the electrical repulsionbetween two hydrogen nuclei.

1v � 12 2hn � 1

2kA21v � 12 2hn

cv � e e�ax2>2 1c0 � c2x2 � p � cvx

v 2 for v even

e�ax2>2 1c1x � c3x3 � p � cvx

v 2 for v odd

12 hn

E � 1v � 12 2hn where v � 0, 1, 2, p

�U 2

2m

d 2c

dx2 � 12 kx2c � Ec

12 kx2

��

Figure 17.17

Energy levels of a one-dimensional harmonic oscillator.

lev38627_ch17.qxd 3/25/08 12:22 PM Page 620

17.13 TWO-PARTICLE PROBLEMSConsider a two-particle system where the coordinates of the particles are x1, y1, z1 andx2, y2, z2. The relative (or internal) coordinates x, y, z are defined by

(17.77)

These are the coordinates of particle 2 in a coordinate system whose origin is attachedto particle 1 and moves with it.

In most cases, the potential energy V of the two-particle system depends only onthe relative coordinates x, y, and z. For example, if the particles are electricallycharged, the Coulomb’s law potential energy of interaction between the particlesdepends only on the distance r between them, and r � (x2 � y2 � z2)1/2. Let us assumethat V � V(x, y, z). Let X, Y, and Z be the coordinates of the center of mass of the sys-tem; X is given by (m1x1 � m2x2)/(m1 � m2), where m1 and m2 are the masses of theparticles (Halliday and Resnick, sec. 9-1). If one expresses the classical energy (thatis, the classical Hamiltonian) of the system in terms of the internal coordinates x, y,and z and the center-of-mass coordinates X, Y, and Z, instead of x1, y1, z1, x2, y2, andz2, it turns out (see Prob. 17.55) that

(17.78)

where M is the system’s total mass (M � m1 � m2), the reduced mass m is defined by

(17.79)*m �m1m2

m1 � m2

H � c 1

2m1px

2 � py2 � pz

2 2 � V1x, y, z 2 d � c 1

2M1pX

2 � pY2 � pZ

2 2 d

x � x2 � x1, y � y2 � y1, z � z2 � z1

Section 17.13Two-Particle Problems

621

� 0�

1

�/(�/�)1/4� � �

�2 2�1/2x�

� 1�

1

�1

�2 2

� 3�

1

�1

�2 2

� 2�

1

�1

�2 2

�0 � (�/�)1/4e��x2/2

�1 � (4�3/�)1/4xe��x2/2

�2 � (�/4�)1/4(2�x2 � 1)e��x2/2

�3 � (�/9�)1/4(2�3/2x3 � 3�1/2x)e��x2/2

� � 2�vm/�

�

�

�

�

�

�

�

�

�

�

�

�

��

�

�

�

�

�

�

�

Figure 17.18

Wave functions for the lowest fourharmonic-oscillator stationarystates.

lev38627_ch17.qxd 3/25/08 12:22 PM Page 621


622

and the momenta in (17.78) are defined by

(17.80)

where vx � dx/dt, etc., and vX � dX/dt, etc.Equation (17.55) shows that the Hamiltonian (17.78) is the sum of a Hamiltonian

for a fictitious particle of mass m and coordinates x, y, and z that has the potential en-ergy V(x, y, z) and a Hamiltonian for a second fictitious particle of mass M � m1 �m2 and coordinates X, Y, and Z that has V � 0. Moreover, there is no term for anyinteraction between these two fictitious particles. Hence, Eqs. (17.69) and (17.70)show that the quantum-mechanical energy E of the two-particle system is given byE � Em � EM, where Em and EM are found by solving

The Hamiltonian operator m is formed from the terms in the first pair of brackets in(17.78), and M is formed from the terms in the second pair of brackets.

Introduction of the relative coordinates x, y, and z and the center-of-mass coordi-nates X, Y, and Z reduces the two-particle problem to two separate one-particleproblems. We solve a Schrödinger equation for a fictitious particle of mass m movingsubject to the potential energy V(x, y, z), and we solve a separate Schrödinger equa-tion for a fictitious particle whose mass is M (� m1 � m2) and whose coordinates arethe system’s center-of-mass coordinates X, Y, and Z. The Hamiltonian M involvesonly kinetic energy. If the two particles are confined to a box, we can use the particle-in-a-box energies (17.47) for EM. The energy EM is translational energy of the two-particlesystem as a whole. The Hamiltonian m involves the kinetic energy and potential energyof motion of the particles relative to each other, so Em is the energy associated with thisrelative or “internal” motion.

The system’s total energy E is the sum of its translational energy EM and its in-ternal energy Em. For example, the energy of a hydrogen atom in a box is the sum ofthe atom’s translational energy through space and the atom’s internal energy, which iscomposed of potential energy of interaction between the electron and the proton andkinetic energy of motion of the electron relative to the proton.

17.14 THE TWO-PARTICLE RIGID ROTORThe two-particle rigid rotor consists of particles of masses m1 and m2 constrained toremain a fixed distance d from each other. This is a useful model for treating the rota-tion of a diatomic molecule; see Sec. 20.3. The system’s energy is wholly kinetic, andV � 0. Since V � 0 is a special case of V being a function of only the relative coordi-nates of the particles, the results of the last section apply. The quantum-mechanical en-ergy is the sum of the translational energy of the system as a whole and the energy ofinternal motion of one particle relative to the other. The interparticle distance is con-stant, so the internal motion consists entirely of changes in the spatial orientation ofthe interparticle axis. The internal motion is a rotation of the two-particle system.

Solution of the Schrödinger equation for internal motion is complicated, so weshall just quote the results without proof. (For a derivation, see, for example, Levine,sec. 6.4.) The allowed rotational energies turn out to be

(17.81)*

where the rotor’s moment of inertia I is

(17.82)*I � md2

Erot � J1J � 1 2 U 2

2I where J � 0, 1, 2, . . .

H

H

HH

Hmcm1x, y, z 2 � Emcm1x, y, z 2 and HMcM1X, Y, Z 2 � EMcM1X, Y, Z 2

pX � MvX, pY � MvY, pZ � MvZ

px � mvx, py � mvy, pz � mvz

lev38627_ch17.qxd 3/25/08 12:22 PM Page 622

with m m1m2/(m1 m2). The spacing between adjacent rotational energy levels in-creases with increasing quantum number J (Fig. 17.19). There is no zero-point rota-tional energy.

The rotational wave functions are most conveniently expressed in terms of the an-gles u and f that give the spatial orientation of the rotor (Fig. 17.20). One finds crot ��JMJ

(u)�MJ(f), where �JMJ

is a function of u whose form depends on the two quantumnumbers J and MJ, and �MJ

is a function of f whose form depends on MJ. These func-tions won’t be given here but will be discussed in Sec. 18.3.

Ordinarily, the wave function for internal motion of a two-particle system is afunction of three coordinates. However, since the interparticle distance is held fixed inthis problem, crot is a function of only two coordinates, u and f. Since there are twocoordinates, there are two quantum numbers, J and MJ. The possible values of MJ turnout to range from �J to J in steps of 1:

(17.83)*

For example, if J � 2, then MJ � �2, �1, 0, 1, 2. For a given J, there are 2J � 1 val-ues of MJ. The quantum numbers J and MJ determine the rotational wave function, butErot depends only on J. Hence, each rotational level is (2J � 1)-fold degenerate. Forexample, the value J � 1 corresponds to one energy level (Erot � U2/I) and correspondsto the three MJ values �1, 0, 1. Therefore for J � 1 there are three different crot func-tions, that is, three different rotational states.

EXAMPLE 17.7 Rotational energy levels

Find the two lowest rotational energy levels of the 1H35Cl molecule, treating itas a rigid rotor. The bond distance is 1.28 Å in HCl. Atomic masses are listed ina table inside the back cover.

The rotational energy [Eqs. (17.81) and (17.82)] depends on the reducedmass m of Eq. (17.79). The atomic mass m1 in m equals the molar mass M1 di-vided by the Avogadro constant NA. Using the table of atomic masses, we have

The two lowest rotational levels have J � 0 and J � 1, and (17.81) gives EJ�0 � 0and

ExerciseThe separation between the two lowest rotational levels of 12C32S is3.246 � 10�23 J. Calculate the bond distance in 12C32S. (Answer: 1.538 Å.)

17.15 APPROXIMATION METHODSFor a many-electron atom or molecule, the interelectronic repulsion terms in thepotential energy V make it impossible to solve the Schrödinger equation (17.24)exactly. One must resort to approximation methods.

EJ�1 �J1J � 1 2 U 2

2I�

112 2 16.63 � 10�34 J s 2 2212p 2 212.67 � 10�47 kg m2 2 � 4.17 � 10�22 J

I � md2 � 11.63 � 10�27 kg 2 11.28 � 10�10 m 2 2 � 2.67 � 10�47 kg m2

� 1.63 � 10�24 g

m �m1m2

m1 � m2�3 11.01 g>mol 2 >NA 4 3 135.0 g>mol 2 >NA 43 11.01 g>mol 2 � 135.0 g>mol 2 4 >NA

�0.982 g>mol

6.02 � 1023>mol

MJ � �J, �J � 1, p , J � 1, J

�� Section 17.15Approximation Methods

623

J � 3

J � 2

J � 1

J � 0

Figure 17.19

Lowest four energy levels of atwo-particle rigid rotor. Eachenergy level consists of 2J � 1states.

Figure 17.20

A two-particle rigid rotor.

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624

The Variation MethodThe most widely used approximation method is the variation method. From the pos-tulates of quantum mechanics, one can deduce the following theorem (for the proof,see Prob. 17.68). Let be the time-independent Hamiltonian operator of a quantum-mechanical system. If f is any normalized, well-behaved function of the coordinatesof the particles of the system, then

(17.84)

where Egs is the system’s true ground-state energy and the definite integral goes overall space. (Do not confuse the variation function f with the angle f in Fig. 17.20.)

To apply the variation method, one takes many different normalized, well-behaved functions f1, f2, . . . , and for each of them one computes the variationalintegral � f* f dt. The variation theorem (17.84) shows that the function giving thelowest value of � f* f dt provides the closest approximation to the ground-stateenergy (Fig. 17.21). This function can serve as an approximation to the true ground-state wave function and can be used to compute approximations to ground-statemolecular properties in addition to the energy (for example, the dipole moment).

Suppose we were lucky enough to guess the true ground-state wave function cgs.Substitution of f � cgs in (17.84) and the use of (17.61) and (17.17) give the varia-tional integral as � c*gs cgs dt � � c*gs Egscgs dt � Egs � c*gs cgs dt � Egs. We wouldthen get the true ground-state energy.

If the variation function f is not normalized, it must be multiplied by a normal-ization constant N before being used in (17.84). The normalization condition is 1 �� Nf�2 dt � �N�2 � �f�2 dt. Hence,

(17.85)

Use of the normalized function Nf in place of f in (17.84) gives � N*f* (Nf) dt��N�2 � f* f dt � Egs, where we used the linearity of (Sec. 17.11) to write (Nf)� N f. Substitution of (17.85) into the last inequality gives

(17.86)*

where f need not be normalized but must be well behaved.

EXAMPLE 17.8 Trial variation function

Devise a trial variation function for the particle in a one-dimensional box and useit to estimate Egs.

The particle in a box is exactly solvable, and there is no need to resort to anapproximate method. For instructional purposes, let’s pretend we don’t knowhow to solve the particle-in-a-box Schrödinger equation. We know that the trueground-state wave function is zero outside the box, so we take the variation func-tion f to be zero outside the box. Equations (17.84) and (17.86) are valid onlyif f is a well-behaved function, and this requires that f be continuous. For f tobe continuous at the ends of the box, it must be zero at x � 0 and at x � a, wherea is the box length. Perhaps the simplest way to get a function that vanishes at 0and a is to take f� x(a � x) for the region inside the box. As noted above, f� 0

� f*Hf dt

� f*f dt� Egs

HHHH

H

0N 0 2 �1

� 0f 0 2 dt

H

HH

�f*Hf dt � Egs for f normalized

HE

—— W[fa]

—— W[fc]

—— W[fd]

—— W[fb]

—— Egs

Figure 17.21

The variational integral cannot beless than the true ground-stateenergy Egs. The quantities W[fa],W[fb], W[fc], and W[fd] are thevalues of the variational integral in(17.84) for the normalizedfunctions fa, fb, fc, and fd. Ofthese functions, fb gives thelowest W and so its W is closest to Egs.

lev38627_ch17.qxd 3/25/08 12:22 PM Page 624

outside the box. Since we did not normalize f, Eq. (17.86) must be used. For theparticle in a box, V � 0 and � �(U2/2m) d2/dx2 inside the box. We have

Also, � f*f dt � �a0 x2(a � x)2 dx � a5/30. The variation theorem (17.86) be-

comes (U2a3/6m) � (a5/30) � Egs, or

From (17.34), the true ground-state energy is Egs � h2/8ma2 � 0.125h2/ma2. Thevariation function x(a � x) gives a 1.3% error in Egs.

Figure 17.22 plots the normalized variation function (30/a5)1/2x(a � x) andthe true ground-state wave function (2/a)1/2 sin (px/a). Figure 17.22 also plots thepercent deviation of the variation function from the true wave function versus x.

ExerciseWhich of the following functions could be used as trial variation functions forthe particle in a box? All functions are zero outside the box and the expressiongiven applies only inside the box. (a) �x2(a�x)2; (b) x2; (c) x3; (d ) sin (px/a); (e) cos (px/a); ( f ) x(a�x) sin (px/a). [Answer: (a), (d), ( f ).]

If the normalized variation function f contains the parameter c, then the vari-ational integral W � �f*Hfdt will be a function of c, and one minimizes W bysetting 0W/0c � 0.

EXAMPLE 17.9 Variation function with a parameter

Apply the variation function to the harmonic oscillator, where c is a param-eter whose value is chosen to minimize the variational integral.

The harmonic-oscillator potential energy (17.71) is kx2 and theHamiltonian operator (17.56) is (U2/2m) (d2/dx2) kx2. We have

where Table 14.1 of Sec. 14.4 was used to evaluate the integrals. Also,

W �� f*Hf dt

� f*f dt�

U 2c

2m�

k

8c

�f*f dt � �q

�q

e�2cx2

dx � a p2cb 1>2

�U 2

mapc

8b 1>2

�k

4a p

8c3 b1>2

�f*Hf dt � �q

�q

c� U 2

2m 14c2x2 � 2c 2e�2cx2

�1

2kx2e�2cx2 d dx

Hf � �U 2

2m

d21e�cx2 2dx2 �

1

2 kx2e�cx2

� �U 2

2m14c2x2 � 2c 2e�cx2

�1

2 kx2e�cx2

12�� H

12

e�cx2

Egs � 5h2>4p2ma2 � 0.12665h2>ma2

��U 2

2m �a

0

x1a � x 2 1�2 2 dx �U 2a3

6m

�f*Hf dt � �a

0

x1a � x 2 a�U 2

2mb

d2

dx2 3x1a � x 2 4 dx

H

Section 17.15Approximation Methods

625

Figure 17.22

The upper figure plots thevariation function f �(30/a5)1/2x(a � x) and the trueground-state wave function cgs forthe particle in a one-dimensionalbox. The lower figure plots thepercent deviation of this f fromthe true cgs.

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626

We now find the value of c that minimizes W:

We have c2 � mk/4U2 and c � �(mk)1/2/2U. The negative value for c would givea positive exponent in the variation function f� e�cx2

; f would go to infinity asx goes to �q and f would not be quadratically integrable. We therefore rejectthe negative value of c. With c � (mk)1/2/2U, the variational integral W becomes

where n � (1/2p)(k/m)1/2 [Eq. (17.73)] was used. The value hn/2 is the trueground-state energy of the harmonic oscillator [Eq. (17.76)], and with c �(mk)1/2/2U � pnm/U, the trial function is the same as the unnormalizedground-state wave function of the harmonic oscillator (Fig. 17.18).

ExerciseVerify the integration results in this example.

A common form for variational functions in quantum mechanics is the linearvariation function

where f1, . . . , fn are functions and c1, . . . , cn are variational parameters whose valuesare determined by minimizing the variational integral. Let W be the left side of(17.86). Then the conditions for a minimum in W are 0W/0c1 � 0, 0W/0c2 � 0, . . . ,0W/0cn � 0. These conditions lead to a set of equations that allows the c’s to be found.It turns out that there are n different sets of coefficients c1, . . . , cn that satisfy 0W/0c1 �� 0W/0cn � 0, so we end up with n different variational functions f1, . . . , fn andn different values for the variational integral W1, . . . , Wn, where W1 � � f 1* f1 dt/� f1*f1 dt, etc. If these W’s are numbered in order of increasing energy, it can beshown that W1 � Egs, W2 � Egs�1, etc., where Egs, Egs�1, . . . are the true energies ofthe ground state, the next-lowest state, etc. Thus, use of the linear variation functionc1 f1 � � � � � cn fn gives us approximations to the energies and wave functions of thelowest n states in the system. (In using this method, one deals separately with wavefunctions of different symmetry.)

H

f � c1 f1 � c2 f2 � . . . � cn fn

e�cx2

W �U 2c

2m�

k

8c�

Uk1>24m1>2 �

Uk1>24m1>2 �

hk1>24pm1>2 �

hn

2

0 �0W

0c�

U 2

2m�

k

8c2

lev38627_ch17.qxd 3/25/08 12:22 PM Page 626

Section 17.16Hermitian Operators

627

17.16 HERMITIAN OPERATORSSection 17.11 noted that operators in quantum mechanics are linear. Quantum-mechanical operators that correspond to a physical property must have another prop-erty besides linearity, namely, they must be Hermitian. This section discussesHermitian operators and their properties. The material of this section is important fora thorough understanding of quantum mechanics, but is not essential to understandingthe material in the remaining chapters of this book and so may be omitted if time doesnot allow its inclusion. The abstract material of this section can induce dizziness insusceptible individuals and is best studied in small doses.

Hermitian OperatorsThe quantum-mechanical average value �M� of the physical quantity M must be a realnumber. To take the complex conjugate of a number, we replace i by �i wherever itoccurs. A real number does not contain i, so a real number equals its complex conjugate:z � z* if z is real. Hence �M� � �M�*. We have �M� � � �* � dt [Eq. (17.62)] andM

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628

�M�* � � (�* �)* dt � � (�*)*( �)* dt � � �( �)* dt, where the result ofProb. 17.19 was used. Therefore

(17.91)

Equation (17.91) must hold for all possible state functions �, that is, for all functionsthat are continuous, single-valued, and quadratically integrable. A linear operator thatobeys (17.91) for all well-behaved functions is called a Hermitian operator. If isa Hermitian operator, it follows from (17.91) that (Prob. 17.63)

(17.92)*

where f and g are arbitrary well-behaved functions (not necessarily eigenfunctions ofany operator) and the integrals are definite integrals over all space. Although (17.92)looks like a more stringent requirement than (17.91), it is actually a consequence of(17.91). Thus a Hermitian operator obeys (17.92). The Hermitian property (17.92) isreadily verified for the quantum-mechanical operators x � and (U/i)(0/0x) (Prob. 17.64).

Eigenvalues of Hermitian OperatorsSection 17.11 noted that when � is an eigenfunction of with eigenvalue c, a mea-surement of M will give the value c. Since measured values are real, we expect c to bea real number. We now prove that the eigenvalues of a Hermitian operator are realnumbers. To prove the theorem, we take the special case of (17.92) where f and g arethe same function and this function is an eigenfunction of with eigenvalue b. Withf � g and f � bf, (17.92) becomes

Using (bf )* � b*f * and taking the constants outside the integrals, we get b � f *f dt� b* � ff * dt or

(17.93)

The quantity � f �2 is never negative. The only way the definite integral � � f �2 dt (whichis the infinite sum of the nonnegative infinitesimal quantities � f �2 dt) could be zerowould be if the function f were zero everywhere. However, the function f � 0 is notallowed as an eigenfunction (Sec. 17.11). Therefore (17.93) requires that b � b* � 0and b � b*. Only a real number is equal to its complex conjugate, so the eigenvalueb must be real.

Orthogonality of EigenfunctionsWe noted in Eq. (17.36) that the particle-in-a-one-dimensional-box stationary-statewave functions, which are eigenfunctions of , are orthogonal, meaning that � ci*cj dt � 0 when i � j. This is an example ofthe theorem that two eigenfunctions of a Hermitian operator that correspond to dif-ferent eigenvalues are orthogonal. The proof is as follows.

The Hermitian property (17.92) holds for all well-behaved functions. In particu-lar, it holds if we take f and g as two of the eigenfunctions of the Hermitian operator

. With f � bf and g � cg, the Hermitian property � f * g dt � � g( f )* dtbecomes

c� f *g dt � �g1bf 2* dt � �gb*f * dt � b�g f * dt

MMMMM

H

1b � b* 2 � 0 f 0 2 dt � 0

� f*bf dt � � f 1bf 2* dt

MM

M

� f *Mg dt � �g1Mf 2* dt

M

�°*M° dt � �° 1M° 2* dt

MMM

lev38627_ch17.qxd 3/25/08 12:22 PM Page 628

since a Hermitian operator has real eigenvalues. We have

If the eigenvalues c and b are different (c b), then � f *g dt � 0, and the theorem isproved.

If the eigenvalues b and c happen to be equal, then orthogonality need not neces-sarily hold. Recall that we saw examples of different eigenfunctions of having thesame eigenvalue when we discussed the degenerate energy levels of the particle in athree-dimensional cubic box and the rigid two-particle rotor (Secs. 17.10 and 17.14).Because the quantum-mechanical operator is linear, one can show (Prob. 17.65) thatif the functions f1 and f2 are eigenfunctions of with the same eigenvalue, that is, if

f1 bf1 and f2 bf2, then any linear combination c1 f1 c2 f2 (where c1 and c2 areconstants) is an eigenfunction of with eigenvalue b. This freedom to take linearcombinations of eigenfunctions with the same eigenvalue enables us to choose theconstants c1 and c2 so as to give orthogonal eigenfunctions (Prob. 17.66). From hereon, we shall assume that this has been done, so that all eigenfunctions of a Hermitianoperator that we deal with will be orthogonal.

Let the set of functions g1, g2, g3, . . . be the eigenfunctions of a Hermitian operator.Since these functions are (or can be chosen to be) orthogonal, we have � gj*gk dt � 0when j k (that is, when gj and gk are different eigenfunctions). We shall always nor-malize eigenfunctions of operators, so � gj*gj dt � 1. These two equations expressingorthogonality and normalization can be written as the single equation

(17.94)

where the Kronecker delta djk is a special symbol defined to equal 1 when j � k andto equal 0 when j and k differ:

(17.95)

A set of functions that are orthogonal and normalized is an orthonormal set.

Complete Sets of EigenfunctionsA set of functions g1, g2, g3, . . . is said to be a complete set if every well-behavedfunction that depends on the same variables as the g’s and obeys the same boundaryconditions as the g’s can be expressed as the sum �i cigi, where the c’s are constantswhose values depend on the function being expressed. The sets of eigenfunctions ofmany of the Hermitian operators that occur in quantum mechanics have been provedto be complete, and quantum mechanics assumes that the set of eigenfunctions of aHermitian operator that represents a physical quantity is a complete set. If F is a well-behaved function and the set g1, g2, g3, . . . is the set of eigenfunctions of the Hermitianoperator that corresponds to the physical property R, then

(17.96)

and one says that F has been expanded in terms of the set of g’s.How do we find the coefficients ck in the expansion (17.96)? Multiplication of

(17.96) by g j* gives g j*F � �k ck g j*gk. Integration of this equation over the full rangeof all the coordinates gives

where the orthonormality of the eigenfunctions of a Hermitian operator [Eq. (17.94)]and the fact that the integral of a sum equals the sum of the integrals were used. The

�gj* F dt � � ak

ckgj*gk dt � ak

�ckgj*gk dt � ak

ck�gj*gk dt � ak

ckdjk

F � ak

ckgk

R

djk � 1 when j � k, djk � 0 when j k

�gj*gk dt � djk

M��M�M

MM

H

1c � b 2 � f*g dt � 0

Section 17.16Hermitian Operators

629

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630

Kronecker delta djk is always zero except when k equals j. Therefore every term inthe sum �k ck djk is zero except for the one term where k becomes equal to j: thus,�k ck djk � cjdjj � cj [Eq. (17.95)]. Therefore

Changing j to k in this equation and substituting in the expansion (17.96), we have

(17.97)

where the definite integrals � gj*F dt are constants. Equation (17.97) shows how toexpand any function F in terms of a known complete set of functions g1, g2, g3, . . . .

Suppose we are unable to solve the Schrödinger equation for a system we are in-terested in. We can express the unknown ground-state wave function as cgs � �k ckgk,where the g’s are a known complete set of functions. We then use the linear variationmethod (Sec. 17.15) to solve for the coefficients ck, thereby obtaining cgs. The diffi-culty with this approach is that a complete set of functions usually contains an infinitenumber of functions. We are therefore forced to limit ourselves to a finite number offunctions in the expansion sum, thereby introducing error into our determination ofcgs. Most methods of calculating wave functions for molecules use expansions, as weshall see in Chapter 19.

Consider an example. Let the function F be defined as F � x2(a � x) for x between0 and a and F � 0 elsewhere. Could we use the particle-in-a-box stationary-state wavefunctions cn � (2/a)1/2 sin (npx/a) [Eq. (17.35)] to expand F? The function F is well-behaved and satisfies the same boundary conditions as cn, namely, F is zero at the ends of the box. The functions cn are the eigenfunctions of a Hermitian operator (theparticle-in-a-box Hamiltonian H) and so are a complete set. Therefore we can expressF as F � �q

n�1cncn, where the coefficients cn are given in (17.97) as

(17.98)

Problem 17.67 evaluates cn and shows how the sum �n cncn becomes a more and moreaccurate representation of F as more terms are included in the sum.

The proof of the variation theorem (17.84) uses an expansion of the variation func-tion , as in Eq. (17.96). The proof is outlined in Prob. 17.68.

SummaryQuantum-mechanical operators that correspond to physical properties are Hermitian,meaning that they satisfy (17.92) for all well-behaved functions f and g. The eigenval-ues of a Hermitian operator are real. The eigenfunctions of a Hermitian operator are (orcan be chosen to be) orthogonal. The eigenfunctions of a Hermitian operator form acomplete set, meaning that any well-behaved function can be expanded in terms of them.

17.17 SUMMARYElectromagnetic waves of frequency n and wavelength l travel at speed c � ln in vac-uum. Processes involving absorption or emission of electromagnetic radiation (for ex-ample, blackbody radiation, the photoelectric effect, spectra of atoms and molecules)can be understood by viewing the electromagnetic radiation to be composed of pho-tons, each photon having an energy hn, where h is Planck’s constant. When an atomor molecule absorbs or emits a photon, it makes a transition between two energy lev-els Ea and Eb whose energy difference is hn; Ea � Eb � hn.

De Broglie proposed that microscopic particles such as electrons have wavelikeproperties, and this was confirmed by observation of electron diffraction. Because

f

cn � �cn*F dt � �a

0

a 2ab 1>2

sin npx

ax21a � x 2 dx

F � ak

a �gk*F dt bgk

cj � �gj*F dt

lev38627_ch17.qxd 3/25/08 12:22 PM Page 630

of this wave–particle duality, simultaneous measurement of the precise position andmomentum of a microscopic particle is impossible (the Heisenberg uncertaintyprinciple).

The state of a quantum-mechanical system is described by the state function �,which is a function of the particles’ coordinates and the time. The change in � withtime is governed by the time-dependent Schrödinger equation (17.10) [or (17.60)],which is the quantum-mechanical analog of Newton’s second law in classical me-chanics. The probability density for finding the system’s particles is ��2. For exam-ple, for a two-particle, one-dimensional system, ��(x1, x2, t)�2 dx1 dx2 is the probabil-ity of simultaneously finding particle 1 between x1 and x1 � dx1 and particle 2 betweenx2 and x2 � dx2 at time t.

When the system’s potential energy V is independent of time, the system can existin one of many possible stationary states. For a stationary state, the state function is� � e�iEt/Uc. The (time-independent) wave function c is a function of the particles’coordinates and is one of the well-behaved solutions of the (time-independent) Schrödinger equation c � Ec, where E is the energy and the Hamiltonian operator

is the quantum-mechanical operator that corresponds to the classical quantity E. Tofind the operator corresponding to a classical quantity M, one writes down the classical-mechanical expression for M in terms of cartesian coordinates and momenta and thenreplaces the coordinates and momenta by their corresponding quantum-mechanicaloperators: 1 � x1 � , x,1 � (U/i) 0/0x1, etc. For a stationary state, ��2 � �c�2 and theprobability density and energy are independent of time.

In accord with the probability interpretation, the state function is normalized tosatisfy � ��2 dt� 1, where � dt denotes the definite integral over the full range of theparticles’ coordinates. For a stationary state, the normalization condition becomes � �c�2 dt � 1.

The average value of property M for a system in stationary state c is �M� �� c* c dt, where is the quantum-mechanical operator for property M.

The stationary-state wave functions and energies were found for the followingsystems.

(a) Particle in a one-dimensional box (V � 0 for x between 0 and a; V � q else-where): E � n2h2/8ma2, c � (2/a)1/2 sin (npx/a), n � 1, 2, 3, . . . .

(b) Particle in a three-dimensional rectangular box with dimensions a, b, c: E �(h2/8m) � (nx

2/a2 � ny2/b2 � nz

2/c2). (c) One-dimensional harmonic oscillator (V � kx2): E � (v � )hn, n �

(1/2p)(k/m)1/2, v � 0, 1, 2, . . . . (d ) Two-particle rigid rotor (particles at fixed distance d and energy entirely

kinetic): E � J(J � 1)U2/2I, I � md2, J � 0, 1, 2, . . . ; m K m1m2/(m1 � m2) is thereduced mass.

When more than one state function corresponds to the same energy level, that en-ergy level is said to be degenerate. There is degeneracy for the particle in a cubic boxand for the two-particle rigid rotor.

For a system of noninteracting particles, the stationary-state wave functions areproducts of wave functions for each particle and the energy is the sum of the energiesof the individual particles.

The variation theorem states that for any well-behaved trial variation function f,one has � f* f dt/� f*f dt � Egs, where is the system’s Hamiltonian operatorand Egs is its true ground-state energy.

Important kinds of calculations discussed in this chapter include:

• Use of ln� c to calculate the wavelength of light from the frequency, and vice versa.• Use of Eupper � Elower � hn to calculate the frequency of the photon emitted or

absorbed when a quantum-mechanical system makes a transition between twostates.

HH

12

12

MM

px

HH

Section 17.17Summary

631

lev38627_ch17.qxd 3/25/08 12:22 PM Page 631


632

• Use of energy-level formulas such as E � n2h2/8ma2 for the particle in a box orE � (v � )hn for the harmonic oscillator to calculate energy levels of quantum-mechanical systems.

• For a one-particle, one-dimensional, stationary-state system, use of �c�2 dx to cal-culate the probability of finding the particle between x and x � dx and of �b

a �c�2 dxto calculate the probability of finding the particle between a and b.

• Use of �M� � � c* c dt to calculate average values.• Use of the variation theorem to estimate the ground-state energy of a quantum-

mechanical system.

FURTHER READING

Hanna, chap. 3; Karplus and Porter, chap. 2; Levine, chaps. 1–4, 8, 9; Lowe andPeterson, chaps. 1–3, 7; McQuarrie (1983), chaps. 1–5; Atkins and Friedman, chaps.1–3, 6.

M

12

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