david ronis mcgill university · 2019. 11. 4. · wed. nov. 14the chemical bond ch10 sec 6-10 fri....

Albert Einstein

Max Planck Louis de Broglie

CHEMISTRY 345

Introduction to Quantum Chemistry

David Ronis

McGill University

Erwin Schrodinger

Arnold Sommerfeld Max Born

Fall Term, 2019

CHEMISTRY 345

Introduction to Quantum Chemistry

David Ronis

McGill University

© 2019

All rights reserved. In accordance with Canadian Copyright Law,

reproduction of this material, in whole or in part, without the prior

written consent the author is strictly prohibitied.

Last modified on 4 November 2019

1. General Information

Fall Term, 2019

Table of Contents

1. General Information . . . . . . . . . . . . . . . . . . . . . . 1

2. Wav e Packets: Group and Phase Velocity . . . . . . . . . . . . . . . 6

3. Operators and the Schrodinger Equation . . . . . . . . . . . . . . . 10

3.1. The Schrodinger Equation . . . . . . . . . . . . . . . . . . . 10

3.2. The Time Independent Schrodinger Equation . . . . . . . . . . . . . 11

4. Some Properties of the Schrodinger Eq. . . . . . . . . . . . . . . . 12

4.1. Particle Conservation . . . . . . . . . . . . . . . . . . . . . 12

4.2. Ehrenfest’s Theorem . . . . . . . . . . . . . . . . . . . . . 13

4.3. Nodes and Energy Levels . . . . . . . . . . . . . . . . . . . . 14

5. Inner Products and Uncertainty Relations . . . . . . . . . . . . . . . 16

5.2. The Schwarz Inequality . . . . . . . . . . . . . . . . . . . . 17

5.3. Generalized Uncertainty Relations . . . . . . . . . . . . . . . . . 17

6. Symmetry and Eigenvalues . . . . . . . . . . . . . . . . . . . . 19

6.1. Some Properties of Hermitian Operators . . . . . . . . . . . . . . . 19

6.2. Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . 19

7. The Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . 22

7.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . 22

7.2. Ladder Operators . . . . . . . . . . . . . . . . . . . . . . 23

7.3. The Frobenius Method . . . . . . . . . . . . . . . . . . . . . 25

8. Operators in Polar Coordinates . . . . . . . . . . . . . . . . . . 27

9. Legendre Functions . . . . . . . . . . . . . . . . . . . . . . 31

10. Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . 36

10.1. Problem Set 1 . . . . . . . . . . . . . . . . . . . . . . . 36

10.2. Problem Set 2 . . . . . . . . . . . . . . . . . . . . . . . 38

10.3. Problem Set 3 . . . . . . . . . . . . . . . . . . . . . . . 40

10.4. Problem Set 4 . . . . . . . . . . . . . . . . . . . . . . . 43

11. 2019 Midterm Exam . . . . . . . . . . . . . . . . . . . . . 44

12. 2019 Midterm Exam Solutions . . . . . . . . . . . . . . . . . . 46

13. Lecture Slides . . . . . . . . . . . . . . . . . . . . . . . 50

Fall Term, 2019

General Information -3- Chemistry 345

Chemistry 345: Introduction to Quantum Chemistry

An introduction to quantum chemistry covering the historicaldevelopment, wav e theory, methods of quantum mechanics, and

applications of quantum chemistry.

Lectures on Monday, Wednesday & Friday, 12:35-1:25 P.M. in Maass 217The midterm is on Thursday, October 17, 2019, from 6 to 9 P.M., in STBIO N2/2.

Professor E-Mail* Office Office Hours

David Ronis [email protected] Maass 426 Any timeLinda Reven [email protected] Maass 214 Any time* Please put Chem 345 in the subject field of any e-mails to make it easierfor us to find it.

We also have a tutor, Cameron Reid, (e-mail: [email protected]), whowill be holding tutorials focusing on the homework assignments. Meeting timeand place will be announced soon.

1.1. TEXTS

• D. A. McQuarrie, Quantum Chemistry, 2nd ed., (University Science Books, 2008).

• H. O. Leung and M. D. Marshall, Problems and Solutions to Accompany Donald A. McQuar-rie’s Quantum Chemistry, 2nd ed. (University Science Books, 2008).

1.1.1. Supplementary Texts

• M. Karplus and R.N. Porter, Atoms & Molecules: An Introduction for Students of PhysicalChemistry (On Reserve).

• James R. Barrante, Applied Mathematics for Physical Chemistry. (Not really a book on quan-tum chemistry, but is a nice refresher/review of some of the mathematics used).

1.1.2. General References

1.1.2.1. Old Classics

• Max Born, Atomic Physics (A good description of the early development of quantum mechan-ics with lots of references to experiments).

1.1.2.2. More Advanced References

• L. I. Schiff, Quantum Mechanics (On Reserve)

• Albert Messiah, Quantum Mechanics, (Dover--A bargain)

Fall Term, 2019

Chemistry 345 -4- General Information

1.2. GRADING SCHEME

The grade in this course will be computed as:

Grading Scheme: CHEM 345

Homework & Quizzes 20%Midterm 30%Final 50%

The midterm is on Thursday, October 17, 2019, from 6 to 9 P.M., in STBIO N2/2.

1.3. Notes:

• McGill University values academic integrity. Therefore, all students must understand themeaning and consequences of cheating, plagiarism and other academic offenses under theCode of Student Conduct and Disciplinary Procedures (see www.mcgill.ca/students/srr/honest/for more information). (Approved by Senate on 29 January 2003.)

• In accord with McGill University’s Charter of Students’ Rights, students in this course havethe right to submit in English or in French any written work that is to be graded. (approved bySenate on 21 January 2009)

• In the event of extraordinary circumstances beyond the University’s control, the content and/orevaluation scheme in this course is subject to change.

• No class on Monday, September 30, 2019.

Fall Term, 2019

General Information -5- Chemistry 345

1.4. Tentative Outline

Week Date Topic Quiz Reading1 Wed. Sept. 4 Introduction

Fri. Sept. 6 History of QM Ch 1 Sec 1-5

2 Mon. Sept. 9 History of QM Ch 1 Sec 6-14Wed. Sept. 11 The Wav e Equation Ch 2 Sec 1-3Fri. Sept.13 The Wav e Equation Ch 2 Sec 4-6HW1 Quiz

(Ch 1)3 Mon. Sept. 16 The Schrodinger Equation: Particle in a Box Ch 3 Sec 1-3

Wed. Sept.18 The Schrodinger Equation: Particle in a Box Ch 3 Sec 4-5Fri. Sept. 20 The Schrodinger Equation: Particle in a Box Ch 3 Sec 6-7HW2 Quiz

(Ch2)4 Mon. Sept. 23 The Schrodinger Equation: Particle in a Box Ch 3 Sec 8-9

Wed. Sept. 25 The Postulates of QM Ch 4 Sec 1-5Fri. Sept. 27 The Postulates of QM Ch 4 Sec 6-10HW3 Quiz

(Ch 3)5 Mon. Sept 30 Harmonic Oscillator: Vibrational Spectroscopy Ch 5 Sec 1-6

Wed. Oct. 2 Harmonic Oscillator: Vibrational Spect. Ch 5 Sec 7-12HW4 Quiz(Ch 4,5)

Fri. Oct. 4 The Rigid Rotor: Rotational Spectroscopy Ch 6 Sec 1-46 Mon. Oct.7 The Rigid Rotor: Rotational Spectroscopy HW5 Quiz

(Ch 6)Wed. Oct.9 The Hydrogen Atom Ch 6 Sec 5-8Fri. Oct.11 Review Ch 1-6 Ch 7 Sec 1-2

7 Mon. Oct. 14 Thanksgiving no classWed. Oct. 16 The Hydrogen Atom Ch 7 Sec 3-4Thurs. Oct 17 Ch 7 Sec 3-4Midterm exam Ch 1-6 Maass 217Fri. Oct. 18 The Hydrogen Atom Ch 7 Sec 4-5

8 Mon. Oct. 22 The Hydrogen Atom Ch 7 Sec 6-9Wed. Oct. 24 Approximation Methods Ch 8 Sec 1-3Fri. Oct. 26 Approximation Methods Ch 8 Sec 4-6HW6 Quiz

(Ch 7)9 Mon. Oct. 29 Many Electron Atoms Ch 9 Sec 1-2

Wed. Oct. 31 Many Electron Atoms Ch 9 Sec 3-4Fri. Nov. 2 Many Electron Atoms Ch 9 Sec 5-6HW7 Quiz

(Ch 8)

10 Mon. Nov. 5 Many Electron Atoms Ch 9 Sec 7-9Wed. Nov. 7 Many Electron Atoms Ch 9 Sec 10-13Fri. Nov. 9 The Chemical Bond Ch10 Sec 1-3HW8 Quiz

(Ch9 Sec 1-6)11 Mon. Nov. 12 The Chemical Bond Ch10 Sec 4-6

Wed. Nov. 14 The Chemical Bond Ch10 Sec 6-10Fri. Nov. 16 Qualitative Theory of Chemical Bonding Ch11 Sec 1-3HW9 Quiz

(Ch 9 Sec 7-13)12 Mon. Nov. 19 Qualitative Theory of Chemical Bonding Ch11 Sec 4-6

Wed. Nov. 21 Qualitative Theory of Chemical Bonding Ch 11 Sec 7Fri. Nov. 23 Computational Quantum Chemistry Ch 12 Sec 1-2HW10 Quiz

(Ch10)13 Mon. Nov.26 Computational Quantum Chemistry

Wed Nov. 28 Computational Quantum Chemistry Ch 12 Sec 5-7Fri. Nov. 30 Computational Quantum Chemistry HW11 Quiz

(Ch11)14 Mon. Dec. 3 Computational Quantum Chemistry

Tues. Dec. 4 Review

Fall Term, 2019

Chemistry 345 -6- General Information

2. Wav e Packets: Group and Phase Velocity

2.1. Fourier’s Theorem

Fourier proved a remarkable theorem, namely, that any function, f (x), on a domain [0, L]can be represented as a Fourier series:

f (x) =∞

n=−∞Σ fn e

ikn x

√ L, (2.1)

where the wav evector kn ≡ 2π n/L. (Actually the theorem shows that the sum will converge to theav erage of the limiting values of the function to the right and left of the point x, but this isn’t rel-evant for us). The Fourier coefficients or amplitudes can be obtained by noting that

1

L ∫L

0dx exp[i(kn − kn′)x] = δ n,n′, (2.2)

where

δ n,n′ ≡

1 if n = n′0 otherwise,

(2.3)

is referred to as a Kronecker-δ . By multiplying Eq. (2.1) by exp[−ikm x]/√ L, integrating bothsides from 0 to L, and using Eq. (2.2) it follows that

fm = ∫L

0dx

e−ikm x

√ Lf (x), (2.4)

which is sometimes referred to as the Fourier transform of f (x).

There are some simple properties of Fourier transforms that are listed here that you canprove for yourself:

1. If the function f (x) is real then f−n = f *n2. If g(x) = f (x − x0) then gn = e−ikn x0 fn

In order to see how this all works, consider the Gaussian function

f (x) ≡ exp[−(x − x0)2/2σ 2]/√ 2π σ 2, (2.5)

where σ

Wa ve Packets: Group and Phase Velocity -7- Chemistry 345

Fig. 2.1. Partial sums, ΣNn=−N fneikn x /√ L for a Gaussian wav e packet (L=20, σ =1) for severalvalues of N. The dashed line is the exact answer. For this case, the difference between theexact and partial answer is indistinguishable for N ≥ 10.

Finally, note that in the limit L → ∞ the difference between successive kn’s becomes infini-tesimal and the Fourier sums become integrals, specifically

nΣ → ∫ dn → L ∫

dk

2π. (2.7)

2.2. Group and Phase Velocity

The obvious conclusion of the preceding bit of mathematics is that we can represent local-ized (i.e., particle-like) objects as a superposition of wav es. This is intriguing and plays a keyrole in the quantum theory. Consider the following time-dependent wav e packet:

Ψ(x, t) = ∫ dk ei[kx−ω (k)t] A(k). (2.8)The component wav es are dispersive (i.e., the frequency depends on the wav e vector). Clearlyeach wav e has a velocity vk = ω (k)/k; this is called the phase velocity (e.g., for light in vacuum,ω (k) = kc, and the phase velocity is simply the speed of light). For Einstein-de Broglie wav es ofnon-relativistic particles of mass m and momentum p, we hav e

E = h− ω = p2

2mand k =

p

h− , (2.9)

Fall Term, 2019

Chemistry 345 -8- Wav e Packets: Group and Phase Velocity

and hence, the phase velocity. vp of such a wav e is p/2m, not quite what we might haveexpected, but then again, a wav e isn’t a particle.

Matters become worse if relativistic corrections are considered. In Einstein’s theory of spe-cial relativity, a particle’s energy and momentum are

E =m0c

2

√ 1 − (v /c)2and p =

m0v

√ 1 − (v /c)2, (2.10)

where m0 is the rest mass of the particle. If Eq. (2.10) is used to compute the phase velocity weobtain vp = c2/v, which exceeds the speed of light and is problematic to say the least. Of course,if you think about it for a bit, you’ll agree that it is basically impossible to measure the phasevelocity of a single plane wav e; there is no way to "mark" the position of one of the crests andsee where it goes without using another wav e to interfere with it. Once you do this you are con-sidering the properties of a wav e packet, perhaps like the ones considered above, and the realquestion becomes does the wav e packet behave as expected?

The velocity of the wav e packet is called the group velocity, vg. Consider a modulatedwave of the form

Ψ(x, t) = ei(k0 x−ω0t) B(x, t), (2.11a)

where

B(x, t) ≡ ∫ dk1 ei[k1 x−ω1(k1)t] A(k0 + k1), (2.11b)is called the modulation factor, and is obtained by letting ω ≡ ω0 + ω1(k1) and k ≡ k0 + k1 in Eq.(2.8). The modulation factor is assumed to be wider and slower than the base wav elength andperiod, 2π /k0 and 2π /ω0, respectively, i.e., |ω1(k1)|

Wa ve Packets: Group and Phase Velocity -9- Chemistry 345

and

∂B(x, t)∂x

t

= ∫ dk1 ik1ei[k1 x−ω1(k1)t] A(k0 + k1). (2.14b)

Now the assumption about shape of the modulation factor means only a narrow range of k1,around k1 = 0, contribute to the integrals; hence, we can approximate the behavior of ω1(k1) bywriting the Taylor expansion ω1(k1) ≈ (dω (k)/dk)k=k0 k1 + . . ., which, when used in Eqs. (2.14a)and (2.14b), shows that

vg =

dω (k)dk

k=k0. (2.15)

(Remember that the total wav evector of each of the components of the wav e packet is k1 + k0, cf.Eq. (2.11)). By using the Einstein-de Broglie relations for the energy and momentum, it is easyto verify that the wav e packet’s velocity is exactly the velocity of the particle (both classicallyand relativistically).

Fall Term, 2019

Chemistry 345 -10- Operators and the Schrodinger Equation

3. Operators and the Schrodinger Equation

3.1. The Schrodinger Equation

As we’ve seen in the preceding chapter, the group velocity, together with the Planck and deBroglie hypotheses, allow us to rewrite Eqs. (2.14) as

ih− ∂Ψ(x, t)∂t

= ∫ dk1 E(k1)ei[k1 x−ω1(k1)t] A(k1), (3.1.1a)

P xΨ(x, t) ≡h−i

∂Ψ(x, t)∂x

= ∫ dk1 h− k1ei[k1 x−ω1(k1)t] A(k1) (3.1.1b)and

P2xΨ(x, t) = ∫ dk1 (h− k1)2ei[k1 x−ω1(k1)t] A(k1), (3.1.1c)where E(k) ≡ h− ω (k) = p2(k)/2m is the energy and

P x ≡h−i

∂∂x

(3.1.2)

is the x component of the momentum operator. It has some important properties, for now, sufficeit to note that it is linear and thus allows for superposition. i.e.,

P x[λ1 f (x) + λ2g(x)] = λ1P x f (x) + λ2P x g(x). (3.1.3)

By comparing Eqs. (3.1.1a) and (3.1.1c), we see that

ih− ∂Ψ∂t

=P2x

2mΨ = −

h−2

2m

∂2Ψ∂x2

. (3.1.4)

By repeating our analysis here and in the last chapter we see that Eq. (3.1.4) becomes

ih− ∂Ψ∂t

= −P2x + P2y + P2z

2mΨ = −

h−2

2m∇2Ψ, (3.1.5)

where

∇2 ≡∂2

∂x2+

∂2

∂y2+

∂2

∂z2, (3.1.6)

is known as the Laplacian, while, Eqs. (3.1.4) and (3.1.5) are known as the time dependentSchrodinger equation for a single, free particle. Note that this is not the only way the Schroding-er equation could be obtained. Since each time derivative brings down a factor of ω1(k1) andeach space derivative brings one of k1, a second order, in time, equation, consistent with Planckand de Broglie hypotheses could have been obtained (it also can be obtained by taking a time

Fall Term, 2019

Operators and the Schrodinger Equation -11- Chemistry 345

derivative of Eq. (3.1.4)).

The preceding discussion doesn’t include the effects of interactions. For example, supposethere is a potential energy, V (x, t) acting on the system, as is the case if electric or magneticfields are present. Since, p2x /(2m) is the kinetic energy, T , of the particle, we’ll postulate that allthat needs to be done is to include the potential energy; i.e.,

P2x

2m→

P2x

2m+ V (x, t) ≡ H , (3.1.7)

where H is the so-called Hamiltonian (a function of momentum and position classically) or theHamiltonian operator, quantum mechanically. With this, we can rewrite Eq. (3.1.4) as

ih− ∂Ψ(x, t)∂t

= HΨ(x, t), (3.1.8)

where the generalizations to systems in 2D or 3D or that include more than one particle, areobvious.

3.2. The Time Independent Schrodinger Equation

As is often the case, the potential energy, V , doesn’t explicitly depend on time and Schro-dinger’s equation becomes separable; specifically, we can write Ψ(x, t) ≡ e−iEt/h−u(x). When thisis used in the time dependent Schrodinger equation, we see that

Hu(x) = Eu(x). (3.2.1)

This has a form similar to an eigenequation encountered in linear algebra, where the energy, E, isthe eigenvalue and u(x) is the eigenfunction. As matrix eigenequations can have multiple eigen-values and eigenvectors, so too can Eq. (3.2.1).

Fall Term, 2019

Chemistry 345 -12- Some Properties of the Schrodinger Eq.

4. Some Properties of the Schrodinger Eq.

4.1. Particle Conservation

Now that we’ve constructed an equation that governs Ψ(r,t), the obvious next question isjust what is Ψ(r, t)? From Eqs. (2.11a&b), we see that Ψ is complex, which disqualifies it frombeing measurable directly. Worse, the prefactor ei(k0r−ω0t) is hard for detectors to resolve (thisactually is less true today) given how short the oscillation period is.

Consider

ρ(r, t) ≡ Ψ*(r, t)Ψ(r, t) = |Ψ(r, t)|2. (4.1.1)

This clearly avoids the two criticisms raised above. From the time dependent Schrodinger equa-tion, Eq. (3.1.8), we see that

∂ρ(r, t)∂t

=Ψ*(r, t)HΨ(r, t) − Ψ(r, t)HΨ*(r, t)

ih− (4.1.2a)

= −h−

2mi

Ψ*(r, t)∇2Ψ(r, t) − Ψ(r, t)∇2Ψ*(r, t)

(4.1.2b)

= −∇ ⋅ J(r, t), (4.1.2c)

where

J(r, t) ≡1

2m

Ψ*(r, t)PΨ(r, t) + Ψ(r, t)[PΨ(r, t)]*

=

1

mRe

Ψ*(r, t)PΨ(r, t)

(4.1.3)

is known as the probability flux, while P is the vector momentum operator; i.e., P ≡ h− ∂/∂ri .The overall divergence, ∇, has an interesting consequence, namely,

d

dtV

∫ dr |Ψ(r, t)|2 = −∂V∫ dA n ⋅ J(r, t), (4.1.4)

where V is an arbitrary volume, ∂V is its surface, and n is unit vector pointing out of V . Equa-tion (4.1.4) shows that the only way the you can change the total amount of |Ψ(r, t)|2 is by "flow-ing" it in at the surface. In particular, if the flux vanishes at the surface of V then ρ(r, t) is con-served. This is analogous to what arises in non-reacting diffusive systems. Based on this, Schro-dinger and others initially viewed |Ψ(r, t)|2 as a mass density for the particle under consideration.

This interpretation is not consistent with experiment. Consider the two slit electron dif-fraction experiment. In a diffusing system or classical electromagnetic wav es, reducing the con-centration of the diffusing species or the intensity of the electric and magnetic fields, results in aproportionate reduction of the intensity of the diffraction pattern. Not so in the electron

Fall Term, 2019

Some Properties of the Schrodinger Eq. -13- Chemistry 345

diffraction measurements, where one observes single hits at the detector. If the single hits arerecorded, then after enough time the quantum mechanical diffraction pattern is obtained!

As such, we interpret |Ψ(r, t |2 as the probability density for finding the particle in somespecified region, dr. This interpretation was controversial when it was first introduced (and forsome it still is).

4.2. Ehrenfest’s Theorem

All this discussion of probabilities etc., is very nice, but where does classical mechanicsarise in our discussion? After all, a serious study of classical mechanics arguably starts withNewton (1642-1726) was a very very mature field by the early 20’th century. Consider, the aver-age values of the position and momentum; i.e.,

< r > (t) = ∫ dr Ψ*(r, t)rΨ(r, t) (4.2.1a)and

 (t) = ∫ dr Ψ*(r, t)PΨ(r, t), (4.2.1b)where < O >=< Ψ|O|Ψ > is known as the expectation (average) value of the operator O. Notethat the position of the operator O matters. By using the Schrodinger Equation to obtain thetime derivatives, we see that

d < r > (t)

dt= ∫

dr

ih− Ψ*(r, t)rHΨ(r, t) − rΨ(r, t)HΨ*(r, t) (4.2.2a)

= −h−

2mi ∫ dr r[Ψ*(r, t)∇ ⋅ ∇Ψ(r, t) − Ψ(r, t)∇ ⋅ ∇Ψ*(r, t)] (4.2.2b)

= −h−

2mi ∫ dr Ψ(r, t)∇ ⋅ ∇[rΨ*(r, t)] − rΨ(r, t)∇ ⋅ ∇Ψ*(r, t)] (4.2.2c)

= ∫dr

mΨ*(r, t)PΨ(r, t) =

 (t)

m, (4.2.2d)

where Eq. (4.2.2c) is obtained by integrating the first term in Eq.(4.2.2b) by parts twice, whileEq. (4.2.2d) has an additional integration by parts to move P from Ψ* to Ψ. Turning to themomentum, cf. Eq. (4.2.1b), it follows that

d (t)

dt= ∫

dr

ih− Ψ*(r, t)PHΨ(r, t) − [HΨ*(r, t)]PΨ(r, t) (4.2.3a)

= − ∫dr

ih− 2m Ψ*(r, t)PP2Ψ(r, t) − [P2Ψ*(r, t)]PΨ(r, t)

Fall Term, 2019

Chemistry 345 -14- Some Properties of the Schrodinger Eq.

+ ∫dr

ih− Ψ*(r, t)PV (r, t)Ψ(r, t) − [V (r, t)Ψ*(r, t)]PΨ(r, t), (4.2.3b)

where we’ve decomposed H into kinetic and potential energies. By using the explicit form ofthe momentum operator, i.e., P = h− ∂/i∂r, and integrating by parts twice in the second term in thefirst integral in Eq. (4.2.3b) it follows that the two terms in the integrand cancel and the integralvanishes. Moreover, by using the explicit form of P in the second integral, the latter can berewritten as

d (t)

dt= − ∫ dr Ψ*(r, t)

∂V (r, t)∂r

Ψ(r, t) = < F > , (4.2.4)

where F(r, t) ≡ −∂V /∂r is the force exerted on the particle. This is known as Ehrenfest’s Theo-rem. Basically it replaces ma = F in Newton’s Laws by a force that is averaged over the proba-bility density, and in particular, it will reduce to the classical result if the probability density issharply peaked at the position of the particle.

4.3. Nodes and Energy Levels

There is an interesting link between energy levels and the number of nodes (zeros) in thewave-function. In one dimension, the time independent Schrodinger’s equation becomes

−h−2

2m

∂2Ψi(x)∂x2

+ V (x)Ψi(x) = EiΨi(x). (4.3.1)

For what follows, consider two eigenstates, 1 and 2, assume that E2 ≥ E1 and that Ψ1(x) has apair of adjacent nodes at a and b but is otherwise positive on (a, b).‡

‡The overall sign of the wav e function is immaterial and we can simply multiply Ψi(x) by -1 if it is neg a-tive on the interval (a, b).

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Some Properties of the Schrodinger Eq. -15- Chemistry 345

Fig. 4.3.1. Tw o states with different wav e-functions, one with nodes andlower energy and the other having higher energy and assuming that it isnonzero on the interval (a, b). Notice that ∂Ψ1(x)/∂x is positive (negative) atx = a(b).

Suppose that Ψ2(x) > 0 on (a, b) and consider

E1 ∫b

adx Ψ2(x)Ψ1(x) = ∫

b

adx Ψ2(x)HΨ1(x) (4.3.2)

= −h−2

2m

Ψ2(x)

∂Ψ1(x)∂x

−∂Ψ2(x)

∂xΨ1(x)

b

a

+ ∫b

adx HΨ2(x)Ψ1(x)

= −h−2

2mΨ2(x)

∂Ψ1(x)∂x

b

a

+ E2 ∫b

adx Ψ2(x)Ψ1(x), (4.3.3)

where we’ve integrated by parts twice and remembered that Ψ1(a) = Ψ1(b) = 0. This can berearranged to give

0 = −h−2

2mΨ2(x)

∂Ψ1(x)∂x

b

a

+ (E2 − E1) ∫b

adx Ψ2(x)Ψ1(x), (4.3.4)

It is easy to show that the boundary terms are positive (cf. Fig. 4.3.1) as is the integral. Clearlythis doesn’t add up to zero and the only way to resolve the contradiction is to have one or morenodes in Ψ2(x) in the interval (a, b).

Fall Term, 2019

Chemistry 345 -16- Inner Products and Uncertainty Relations

5. Inner Products and Uncertainty Relations

5.1. Inner Products

An inner product is a generalization of the dot-product used in vector spaces. It is denotedas (A, B) and maps pairs of quantities (e.g., vectors, operators, functions) to a single complexnumber. For this mapping to be an inner product several properties must be satisfied:

(5.1.1) (A, B) = (B, A)*, where * denotes the complex conjugate,

(5.1.2) For any complex scalar α , (A, α B) = α (A, B),

(5.1.3) (A, B + C) = (A, B) + (A, C),

and

(5.1.4) ||A||2 ≡ (A, A) ≥ 0. The equality holds if and only if A = 0. ||A|| is called the norm(length) of A.

As an example, consider complex vectors in 3D and let

(A, B) ≡ A* ⋅ B = A*1 B1 + A*2 B2 + A*3 B3, (5.1.5)

where ⋅ denotes a dot product. This can be extended to arbitrary dimensions by letting

(A, B) ≡N

i=1Σ A*i Bi . (5.1.6)

This sort of inner product can be extended to matrices as A final exmaple, one which is moreuseful in quantum mechanics, an inner product can be defined for the space of complex squarematrices; specifically,

(A, B) ≡ Tr(A† B) =i, jΣ A*j,i B j,i , (5.1.7)

where Tr denotes the trace and † the Hermitian conjugate, i.e., for any square matrix, with com-plex matrix elements Ai, j ,

(A†)i, j = A*j,i (5.1.8)

A final exmaple, one which is more useful in quantum mechanics, consider an inner prod-uct for complex functions, f (q1, . . . , qN ) and g(q1, . . . , qN ) defined as

( f , g) ≡ ∫ dq1, . . . , dqN f *(q1, . . . , qN )g(q1, . . . , qN ). (5.1.9)It is easy to show that these inner products satisfy the conditions given in Eqs. (5.1.1)−(5.1.4).

Fall Term, 2019

Inner Products and Uncertainty Relations -17- Chemistry 345

5.2. The Schwarz Inequality

Our discussion of inner products and, in particular, the criterion, that for any operator A,

(A, A) ≥ 0, (5.2.1)

where the equality holds if and only if A = 0. By taking A ≡ X + λY , where X and Y are Hermit-ian operators and λ is a c-number, in Eq. (5.2.1) it follows that

||X ||2 + λ(X , Y ) + λ*(Y , X) + |λ |2||Y ||2 ≥ 0, (5.2.2)

where ||X || ≡ (X , X)1/2 is called the norm of X . Note that ||X || becomes the usual length in a vec-tor space. By writing complex quantities in polar form, i.e.,

(X , Y ) ≡ eiφ |(X , Y )| and λ ≡ eiδ |λ |, (5.2.3)

we rewrite Eq. (5.2.2) as

||X ||2 + 2|λ | cos(φ + δ )|(X , Y )| + |λ |2||Y ||2 ≥ 0. (5.2.4)

The left hand side of Eq. (5.2.4) is a quadratic polynomial in |λ | who’s roots cannot be realif Eq. (5.2.1) is to hold. This will be the case if the discriminant is negative; i.e.,

cos2(φ + δ )|(X , Y )|2 ≤ ||X ||2||Y ||2. (5.2.5)

Since the phase of λ is arbitrary, we obtain the maximum lower bound by choosing a δ thatmakes the cosine 1 and thus

|(X , Y )| ≤ ||X || ||Y ||. (5.2.6)

which is the Schwarz inequality.

5.3. Generalized Uncertainty Relations

To make the connection to quantum mechanics we reintroduce bra and ket notation anddefine the inner product as

(X , Y ) ≡ 〈Ψ|XY |Ψ〉 = ∫ dqN Ψ*(qN )XY Ψ(qN ) = ∫ dqN XΨ(qN )*

Y Ψ(qN ), (5.3.1)

where qN ≡ (q1, . . . , qN ) are generalized coordinates and Ψ(qN ) is a normalized wav e functionfor a system having N degrees of freedom. Next, let

|ψ Z 〉 ≡Z − 〈Z 〉Ψ

|

Ψ〉 Z = X , Y . (5.3.2)

The Schwarz inequality becomes

||∆X ||2||∆Y ||2 ≥ | < ψ X |ψY > |2, (5.3.3)

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Chemistry 345 -18- Inner Products and Uncertainty Relations

where

||∆Z ||2 ≡ ∫ dqN Ψ*(qN )(Z− < Z >Ψ)2Ψ(qN ), Z ≡ X , Y , (5.3.4a)

= ∫ dqN |(Z − 〈Z 〉)Ψ(qN )|2 = 〈ψ Z |ψ Z 〉 (5.3.4b)is a variance (standard deviation squared) and is a measure of the uncertainty. With this, we re-write the right hand side of Eq. (5.3.3) as

| < ψ X |ψY > |2 =

< Ψ|X̂Ŷ |Ψ >

2

=1

4< Ψ|{X̂ , Ŷ }|Ψ > + < Ψ|[X̂ , Ŷ ]|Ψ >

2

, (5.3.5)

where the ˆ denotes the deviation from the average, i.e., Ẑ ≡ Z − 〈Z 〉Ψ. In addition,{A, B} ≡ AB + BA and [A, B] ≡ AB − BA, are known as an anticommutator or a commutator ofA and B, respectively. For Hermitian X and Y it is easy to show that 〈Ψ|{X̂ , Ŷ }|Ψ〉 is real while< Ψ|[X̂ , Ŷ ]|Ψ > is purely imaginary. Finally, note that [X̂ , Ŷ ] = [X , Y ]. Thus Eq. (5.3.3) can berewritten as

||∆X ||2||∆Y ||2 ≥1

4< Ψ|{X̂ , Ŷ }|Ψ >2 +

1

4C2 ≥

C2

4, (5.3.6)

where iC ≡< Ψ|[X , Y ]|Ψ >. Note that the term involving the anticommutator often vanishes, e.g.,when Ψ is an eigenstate of X or Y . This is a generalization of the Heisenberg uncertainty rela-tions.

Consider the usual momentum-position case.

[ px , x] =h−i

(5.3.7)

which, when used in Eq. (5.3.6), shows that

∆ px∆X ≥h−2

, (5.3.8)

as expected.

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Symmetry and Eigenvalues -19- Chemistry 345

6. Symmetry and Eigenvalues

6.1. Some Properties of Hermitian Operators

Consider any of the energy eigenstates, specifically one that solves

Hψ n = Enψ n (6.1.1)

where < ψ n |ψ n >= 1. It follows that

En = (ψ n Hψ n) = (Hψ n,ψ n) = E*n, (6.1.2)

where the second to last equality follows from the fact that H is hermitian. Thus we have justshown that En = E*n; i.e., En is real. Since we’ve used no properties of H other that it be hermit-ian, any hermitian operator has real eigenvalues.

Next consider any two energy eigenfunctions, ψ n and ψ n′

(ψ n, Hψ n′) = En′ < n|n′ >= En < n|n′ > (6.1.3)

which can be rearranged to show that

(En′ − En) < n|n′ >= 0. (6.1.4)

Therefore, if the two states n and n′ are nondegenerate we see that they must be orthogonal, i.e.,< n|n′ >= 0. If some of the states are degenerate, their eigenfunctions can be orthonormalizedusing the Grahm-Schmidt method.

6.2. Symmetry

Consider two Hermitian operators, H and P, which commute, i.e.,

[H , P] ≡ HP − PH = 0. (6.2.1)

If we have a solution to the Schrodinger eigenequation,

HΨ = EΨ, (6.2.2)

then the fact that H and P commute implies that PΨ is also an eigenfunction of H with the sameeigenvalue E. If the eigenvalue is nondegenerate, i.e., there is only one eigenfunction corre-sponding to it, not counting trivial rescalings like changing it’s sign or normalization, then

PΨ∝Ψ or PΨ = pΨ. (6.2.3)

Thus, Ψ is also an eigenfunction of P corresponding to eigenvalue p.

What happens if the eigenvalue E is degenerate? That is, we find several acceptable solu-tions to Eq. (6.2.2),

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Chemistry 345 -20- Symmetry and Eigenvalues

Hun = Eun, n = 1, . . . ,ν , (6.2.4)

where ν is called the degeneracy of the eigenvalue E. For example, for the hydrogen atomicorbitals, the degeneracy of the n = 2 orbitals is 4 (2s, 2px 2py and 2pz). As before, given that Pand H commute, it follows that Pun is still an eigenfunction of H corresponding to eigenvalueE, howev er we don’t know which eigenfunction it is. All we can really say is that it is some lin-ear combination of the eigenfunctions. Hence, for each n,

Pun =ν

i=1Σ cn,iui , (6.2.5)

where the cn,i’s are constants (remember that the ui’s are functions of position). Note that wecan always assume that the ui eigenfunctions have been made orthonormal by the Gram-Schmidtprocedure, and thus

< ui |u j >= (ui , u j) = ∫ dr u*i u j = δ i, j . (6.2.6)By multiplying Eq. (6.2.5) on the left by u*i , integrating, and using Eq. (6.2.6) it follows that

cn,i =< ui |P|un >= (< un |P|ui >)* = c*i,n, (6.2.7)

where the second equality follows from the assumed Hermiticity of P. This sort of inner prod-uct, < ui |P|u j > is called the ij matrix element of the operator P.

We can view the coefficients, ci, j as the matrix elements of a ν × ν square matrix, C(hence, the name). The last equality in Eq. (6.2.6) says this matrix is special, namely,

C = (CT )* ≡ C†, (6.2.8)

where T denotes the transpose of the matrix. The combination of the operations of transposing amatrix and complex conjugating the matrix elements is called Hermitian conjugation and isdenoted by the superscript †.

It turns out that Hermitian matrices are very special. The property we need (this is linearalgebra which most of you don’t hav e) is that a Hermitian matrix can be diagonalized by a uni-tary transformation (isn’t that a mouthful). This means that we can find a matrix D with the fol-lowing properties

DCD† = Λ, (6.2.9a)

where Λ is diagonal, namely

Λ ≡

λ10

⋅0

0

0

λ2⋅0

0

⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅⋅ ⋅ ⋅

0

0

⋅λν −1

0

0

0

⋅0

λν

,

and where D is unitary, that is

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Symmetry and Eigenvalues -21- Chemistry 345

D† = D−1. (6.2.9b)

By organizing the ν eigenfunctions un into a column vector U , i.e.,

U ≡

u1

u2

⋅⋅

uν

, (6.2.10)

we can rewrite Eq. (6.2.5) as

PU = CU . (6.2.11)

Finally by multiplying on the left by D and using Eqs. (6.2.9a) and (6.2.9b) we find that

DPU = PDU = DCU = DCD−1 DU = ΛDU . (6.2.12)

Now each of the elements of V ≡ DU , vi , is just some linear combination of the original eigen-functions of H , and are thus still eigenfunctions corresponding to the eigenvalue E. Moreover, itis not too hard to show using Eqs. (6.2.6) and (6.2.9b) that

< vi |v j >= δ i, j (6.2.13)

i.e., these new eigenfunctions are still orthonormal (assuming the original ones were). Finally, ifwe examine the last equality in Eq. (6.2.11) component-wise we see that

Pvn = λ nvn. (6.2.14)

Hence, vn is the eigenfunction of P corresponding to eigenvalue λ n (and is still an eigenfunctionof H corresponding to eigenvalue E).

Thus, we can always find a suitable linear combination of the original degenerate eigenval-ues that results in a simultaneous orthonormal eigenfunctions of both H and P. In particular, if Pis some symmetry of the problem, e.g., a reflection or rotation, then this implies that we can clas-sify the energy eigenstates of the system according to how they transform under the symmetryoperation; i.e., to what eigenvalue of the operator do they correspond. This was what was donein our simple one-dimensional examples with reflection symmetry where we characterized thesolutions according to whether they were even or odd functions. More generally, this can beapplied to molecules with symmetry, the operators in that case being the symmetry operations ofthe point group appropriate to the molecule.

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Chemistry 345 -22- The Harmonic Oscillator

7. The Harmonic Oscillator

7.1. Introduction

Consider the interaction potential for a non-reactive pair of neutral molecules; theLennard-Jones 6-12 potential is shown in Fig. 7.1.1.

Fig. 7.1.1. The Lennard-Jones 6-12 potential, cf. Eq. (7.1.1) (black solid line) , and its har-monic oscillator approximation (blue dashed line). The minimum energy, −ε , occurs atr = 21/6σ . The force constant (curvature) at the minimum is 28/39ε /σ 2

The 6-12 potential is characterized by two parameters, ε and σ , corresponding to the well depthand place where the potential is zero, respectively.. We can write the potential as

u(r12) ≡ 4ε

σr12

12

−

σr12

6, (7.1.1)

cf. Fig. 7.1.1. The harmonic approximation is reasonable for energies near the minimum, andthere the Schrodinger’s equation becomes

−

h−2

2m

∂2

∂x2+

K

2x2

Ψn(x) = EnΨn(x), (7.1.2)

where x ≡ r − rmin, K is the force constant, and where the energy is measured relative to the

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The Harmonic Oscillator -23- Chemistry 345

minimum potential (−ε for the 6-12 potential).

The calculation becomes slightly less complicated if we transform Eq. (7.1.2) by introduc-ing dimensionless quantities; specifically, we reexpress the momentum and position operators as

p =h−i

∂∂x

≡ √ mh− ω0P and x ≡ h−

mω0

1/2

Q, (7.1.3)

where ω0 ≡ √ K /m is the oscillator frequency, both classically or quantum mechanically, cf.Eherenfest’s theorem. The dimensionless operators, P and Q, are related as

P =1

i

∂∂Q

. (7.1.4)

As expected, P and Q are Hermitian and don’t commute; specifically

[P, Q] =1

i(7.1.5)

When used in the Schrodinger equation, Eq. (7.1.2), the latter becomes

HΨn(Q) ≡1

2

P2 + Q2

Ψn(Q) = ε nΨn(Q), (7.1.6)

where ε n ≡ En/(h− ω0) is the energy of the nth state in units of h− ω0. There are many ways tosolve the harmonic oscillator Schrodinger equation. Is the next sections we’ll examine some ofthem.

7.2. Ladder Operators

Perhaps the most elegant way to solve the harmonic oscillator Schrodinger equation is byusing ladder operators. Consider

a ≡1

21/2(Q + iP) and a† ≡

1

21/2(Q − iP), (7.2.1)

which implies that

Q =a + a†

21/2and P =

a − a†

i21/2, (7.2.2)

and that

[a, a†] = 1. (7.2.3)

Equation (7.2.1) implies that a and a† are Hermitian conjugates of each other, but are not Her-mitian themselves.

By using Eqs. (7.2.1) in Eq. (1.6) the Hamiltonian becomes

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H =1

4[−(a − a†)2 + (a + a†)2] =

1

2(aa† + a†a), (7.2.4)

which is Hermitian as expected. Next, we use the commutation relation, Eq. (7.2.3), to show thataa† = a†a + 1 and this allows us to rewrite Eq. (7.2.4) as

H = N + 1/2, (7.2.5)

where N ≡ a†a is Hermitian.

Next consider the eigenequation for N ; i.e.,

N |ν > = ν |ν > , (7.2.6)

where ν is the eigenvalue and |ν > is the eigenfunction. Clearly,

Na|ν > = a†aa|ν > = (aa† − 1)a|ν > = (ν − 1)a|ν > ; (7.2.7)

i.e., a|ν > is an eigenfunction for the eigenvalue ν − 1. It is for this reason that a is referred to asa lowering operator. Similarly,

Na†|ν > = a†aa†|ν > = a†(a†a + 1)|ν > = (ν + 1)a†|ν > , (7.2.8)

and hence, a†|ν > is an eigenfunction for eigenvalue ν + 1 (a† is called a raising operator.)

By repeated applications of a or a† we obtain a ladder of eigenvalues, each separated by 1from its neighbors and in principle extending from −∞ to ∞ .

It seems strange that the eigenvalues are unbounded below, especially if we consider

< ν |a†a|ν > = ν < ν |ν > = < aν |aν > (7.2.9)

or

ν =< aν |aν >

< ν |ν >≥ 0. (7.2.10)

The only way out of this contradiction is that the ladder terminates; i.e., when we apply the low-ering operator to the lowest energy state, |νmin > the result vanishes. When this is used in Eq.(7.2.10), we see that νmin = 0. Hence, by returning to the Hamiltonian, cf. Eq. (7.2.5), we seethat

En = h− ω0(n + 1/2), n = 0, 1, 2, . . . (7.2.11)

The extra 1/2 is called the zero point energy and wasn’t known by Planck.

Note that the eigenfunctions are not normalized; nonetheless the norms are easily found.Consider

< a†ν |a†ν > = < ν |aa†|ν > = < ν |a†a + 1|ν > = (ν + 1) < ν |ν > . (7.2.12)

If < ν |ν > = 1, then |ν + 1 >= a†|ν > /√ ν + 1 is also normalized.

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The Harmonic Oscillator -25- Chemistry 345

Finally we need to know the ground state wav e function, which solves Eq. (1.6) forε = 1/2; i.e.,

−

∂2

∂Q2+ Q2 − 1

Ψ0(Q) = 0. (7.2.13)

By trying a solution of the form e−γ Q2

in Eq. (7.2.13) we see that

−(2γ − 1)[Q2(2γ + 1) − 1]e−Q2γ = 0

or γ = 1/2. Hence, the ground state wav e-function Ψ0(Q) = e−Q2/2. The first excited state wav e-

function is obtained by operating with a†, cf. Eq. (7.2.2.1), which gives Ψ1(Q) = 2Qe−Q2/2. Note

that neither wav e-function is normalized.

7.3. The Frobenius Method

The Frobenius method is a way to find general solutions to ordinary differential equations.Before using it, it is useful to rewrite the harmonic oscillator wav e-function as

Ψn(Q) = e−Q2/2Fn(Q), (7.3.1)

which, when used in Eq. (7.3.1.6), shows that

−

∂2

∂Q2+ Q2 − 2ε n

e−Q

2/2Fn(Q) = 0. (7.3.2)

After evaluating the derivatives we find that

∂2

∂Q2− 2Q

∂∂Q

+ (2ε n − 1)Fn(Q) = 0. (7.3.3)

In the Frobenius method, we assume a series solution, i.e.,

Fn(Q) =∞

i=0Σ ciQi+s, (7.3.4)

where c0 ≠ 0. By using Eq. (7.3.4) in (7.3.3) the latter becomes

∞

i=0Σ ci(s + i)(s + i − 1)Qs+i−2 − ciQi+s[2(i + s) + 1 − 2ε n] = 0.

This can be rearranged giving

∞

i=−2Σ Qs+i{ci+2(s + i + 2)(s + i + 1) − ci[2(i + s) + 1 − 2ε n]} = 0, (7.3.5)

where we have assumed that ci = 0 for i < 0. By equating the coefficients of like powers of Q tozero we find that

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c0s(s − 1) = 0, for i = −2, (7.3.6a)

c1s(s + 1) = 0, for i = −1, (7.3.6b)

and

ci+2 =2(i + s) + 1 − 2ε n

(s + i + 2)(s + i + 1)ci for i ≥ 0. (7.3.6c)

From Eq. (7.3.6a) we see that s = 0 or 1 which, if c1 = 0 correspond to even or odd Fn(Q),respectively. Note that c1 = 0 implies that ci = 0 for all odd i’s.

Next, consider the recursion relation, Eq. (7.3.6c). For large i it becomes

ci+2 =2

ici or

ci+2

ci=

2

i, (7.3.7)

which is the recursion relation for eQ2

. This leads to a non-normalizable wav e function, evenwith the extra factor of exp(−Q2/2) introduced in Eq. ( 7.3.1). The only way out of this problemis to have the recursion terminate; from Eq. (7.3.6c) this means that

ε n = n + 1/2 for n = 1, 2, . . . , (7.3.8)

where n = i + s at the termination point. Since i is even, the even or odd n’s arise whens = 0 or 1, respectively. This is is the same result as found using the ladder operators. Theresulting polynomials are, up to normalization, known as Hermite polynomials.* See Fig. 5.10 inMcQuarrie for images of the wav e-functions. Note that they tunnel into classically forbiddenregions of the potential. They also are consistent with the node counting theorem discussed ear-lier in the course.

*See, e.g., M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, (Dover), ch. 22.

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Operators in Polar Coordinates -27- Chemistry 345

8. Operators in Polar Coordinates

aθ

ar

aϕ

xy

z

r

ϕ

θ

Fig. 8.1. Polar coordinates.

It turns out to be very convenient to use polar coordinates to deal with problems withspherical symmetry. As is well known, polar coordinates are defined by

r =

x

y

z

=

r cos(φ ) sin(θ )r sin(φ ) sin(θ )

r cos(θ )

, (8.1)

as shown in Fig. 8.1. Alternately we can express the polar coordinates (r,θ , φ ) in terms of theCartesian ones. From Eq. (8.1) it follows that

r = (x2 + y2 + z2)1/2, θ = arctan

√ x2 + y2z

, and φ = arctan

y

x

. (8.2)

The question we will deal with here is how to express some of the differential operators we’veencountered in terms of polar coordinates, e.g., ∂/∂x, etc. The chain rule is our basic tool, andwe have

∂∂x

=

∂r∂x

∂∂r

+

∂θ∂x

∂∂θ

+

∂φ∂x

∂∂φ

, (8.3)

etc.. The derivatives in Eq. (8.3) can be done using Eq. (8.2), recalling that

darctan(x)

dx=

1

1 + x2, (8.4)

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Chemistry 345 -28- Operators in Polar Coordinates

with the result:

∂r∂x

=x

(x2 + y2 + z2)1/2= cos(φ ) sin(θ ), (8.5a)

∂θ∂x

=xz

(x2 + y2 + z2)(x2 + y2)1/2=

cos(θ ) cos(φ )r

, (8.5b)

and

∂φ∂x

= −y

x2 + y2= −

sin(φ )r sin(θ )

. (8.5c)

Thus, by using Eqs. (8.5a-c) in Eq. (8.3) we find that

∂∂x

= cos(φ ) sin(θ )∂∂r

+cos(θ ) cos(φ )

r

∂∂θ

−sin(φ )

r sin(θ )∂

∂φ. (8.6a)

The derivatives in y and x can be performed in a similar fashion, giving

∂∂y

= sin(φ ) sin(θ )∂∂r

+cos(θ ) sin(φ )

r

∂∂θ

+cos(φ )r sin(θ )

∂∂φ

, (8.6b)

and

∂∂z

= cos(θ )∂∂r

−sin(θ )

r

∂∂θ

. (8.6c)

It turns out to be convenient to write the final result in vector form for the gradient operator,namely,

→∇ = ar

∂∂r

+aθ

r

∂∂θ

+aφ

r sin(θ )∂

∂φ, (8.7)

where

ar ≡

cos(φ ) sin(θ )sin(φ ) sin(θ )

cos(θ )

, aθ ≡

cos(φ ) cos(θ )sin(φ ) cos(θ )

− sin(θ )

, and aφ ≡

− sin(φ )cos(φ )

0

(8.8)

are 3 orthonormal unit vectors and are shown in Fig. 8.1.

Of course, we can’t really stop here since we also need the Laplacian in order to writedown Schrodinger’s equation in polar coordinates. Since, in 3D,

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2, (8.9)

we have to consider operators like

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Operators in Polar Coordinates -29- Chemistry 345

∂2

∂z2=

cos(θ )

∂∂r

−sin(θ )

r

∂∂θ

cos(θ )

∂∂r

−sin(θ )

r

∂∂θ

(8.10)

= cos2(θ )∂2

∂r2−

2 sin(θ ) cos(θ )r

∂2

∂θ ∂r+

sin2(θ )r2

∂2

∂θ 2

+cos(θ ) sin(θ )

r2

∂∂θ

+sin2(θ )

r

∂∂r

+sin(θ ) cos(θ )

r2

∂∂θ

, (8.11)

where, as usual, we have assumed that the order of differentiation on whatever function is oper-ated on doesn’t matter.

The expressions for the y and z second derivatives can be obtained in a similar manner*

and when added together give

∇2 =1

r2

∂∂r

r2

∂∂r

+1

r2 sin(θ )∂

∂θsin(θ )

∂∂θ

+1

r2 sin2(θ )∂2

∂φ 2. (8.12)

which is the result stated in class. Notice that all the mixed derivatives hav e canceled!

Next, consider the angular momentum operator, which is defined as

L ≡ r × p = −ih− rar × ∇, (8.13)

where the second equality is obtained from the explicit definition of the momentum operator andEqs. (8) and (9). By noting that ar × aθ = aφ and that ar × aφ = −aθ it follows that

L = −ih− aφ

∂∂θ

−aθ

sin(θ )∂

∂φ. (8.14)

The individual components become:

L x = −ih−− sin(φ )

∂∂θ

− cot (θ ) cos(φ )∂

∂φ, (8.15a)

L y = −ih−cos(φ )

∂∂θ

− cot (θ ) sin(φ )∂

∂φ, (8.15b)

and

Lz = −ih−∂

∂φ. (8.15c)

*As McQuarrie says (p. 206 in the 1st edition): "One can see that this is a fairly lengthy algebraic process,and the conversion of the Laplacian operator from Cartesian coordinates to spherical coordinates is a long,tedious, standard problem that arises in every quantum-mechanics course and should be done once (andonly once!) by any serious student." I agree.

Fall Term, 2019

Chemistry 345 -30- Operators in Polar Coordinates

Equation (8.15c) and the fact that our functions must be periodic functions of φ with period 2πleads to very simple expressions for the z-component angular momentum eigenfunctions (deter-

mined by Lzuz = λuz). It is easily shown that uz(φ ) = exp(imφ )/√ 2π , m = 0, ±1, ±2, . . . are thenormalized eigenfunctions, corresponding to eigenvalue, λ m = mh−.

If Eqs. (8.15a-c) are used to compute the operator corresponding to the squared magnitudeof the angular momentum , L2 ≡ L2x + L2y + L2z , one obtains

L2 = −h−2

1

sin(θ )∂

∂θsin(θ )

∂∂θ

+1

sin2(θ )∂2

∂φ 2, (8.16)

which, when used in Eq. (8.12), allows us to write the kinetic energy operator, KE = −h−2 ∇2/2mas

KE = −h−2

2mr2∂∂r

r2

∂∂r

+L2

2mr2. (8.17)

Finally, remember that if you need to do integrals in polar coordinates, you must includethe Jacobian for the transformation, i.e.,

∫ dr → ∫∞0

r2dr ∫π

0sin(θ )dθ ∫

2π

0dφ . (8.18)

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Legendre Functions -31- Chemistry 345

9. Legendre Functions

We hav e shown that the the angular momentum contribution to the kinetic energy involvedthe square of the angular momentum operator, which is simply:

L2 = −h−2

1

sin(θ )∂

∂θsin(θ )

∂∂θ

+1

sin2(θ )∂2

∂φ 2, (9.1)

By writing the eigenfunctions of L2 as

Yl,m(θ , φ )∝Pml (x)eimφ (9.2)

where x = cos(θ ), m = 0, ±1, . . ., and where for convenience later, we will write the eigenvalue ash−2 l(l + 1),we see that Pml (x) satisfies:*

d

dx

(1 − x2)

dPml (x)

dx

+ l(l + 1) −

m2

1 − x2Pml (x) = 0. (9.3)

This is a well known differential equation called Legendre’s equation and was studied in detailby Legendre in the late 18th century.

We first consider the special case of m = 0 and write

P0l (x) ≡ Pl(x) =∞

n=0Σ an xn+s. (9.4)

By using Eq. (9.4) in (9.3) (for m = 0) and equating like powers of x, we find that s = 0, 1 andobtain the following recursion relation:

an+2 =(n + s)(n + s + 1) − l(l + 1)

(n + 2 + s)(n + 1 + s)an n ≥ 0. (9.5)

Note that even and odd n terms decouple, and since we already have the possibility of s = 0 ors = 1 we can simply drop all the odd terms, as they can be reconstructed from the s = 1 solution.Now for large enough n, the terms in l are negligible and Eq. (9.5) becomes

an+2 ≈(n + s)

(n + 2 + s)an, (9.6)

which in turn implies that an ≈ 1/n for large enough n. The series

∞

n=1Σ x

2n

2n= −

1

2log(1 − x2), (9.7)

*Note that

sin(θ )d

dθ= − sin2(θ )

d

dx= −(1 − x2)

d

dx

Fall Term, 2019

Chemistry 345 -32- Legendre Functions

which is divergent at x = ±1 (i.e., at θ = 0, π ). Clearly this is unacceptable, and the only way toavoid this is to have the series terminate (e.g., as was the case for the harmonic oscillator). FromEq. (9.5) we see that this will happen if l(l + 1) = (n + s)(n + s + 1) for some n, or equivalently ifl is a nonnegative integer. The resulting polynomials are called Legendre polynomials, and thefirst few are listed in the following table

Table 1: Legendre Polynomials

l Pl(x)

0 11 x2 (3x2 − 1) / 23 (5x3 − 3x)/2

The book lists several properties of Legendre polynomials, the most useful for us probablybeing

∫1

−1dx Pl(x)Pl′(x) =

2

2l + 1δ l,l′, (9.8)

which can be used to normalize the wav efunction.

Finally, returning to the more general case of m ≠ 0 we simply note that the general solu-tion (at least the ones that do not diverge at x = ±1) can written as

P|m|l = (1 − x

2)|m|/2d |m|

dxmPl(x), (9.9)

where remember that Eq. (9.(E#) only depends on m2, i.e., on the magnitude of m. These func-tions are called the Associated Legendre polynomials and several are listed in the book. SincePl(x) is a polynomial of degree l, Eq. (9.9) implies that |m| ≤ l.

Finally, putting the θ and φ pieces back together gives the angular momentum eigenfunc-tions:

Yl,m(θ , φ ) = Nl,m P|m|l (cos(θ ))e

imφ , (9.10)

where the normalization constant turns out to be

Nl,m =

(2l + 1)4π

(l − |m|)!(l + |m|)!

1/2

. (9.11)

These functions are called Spherical Harmonics and are orthonormal in the sense that

∫π

0dθ ∫

2π

0dφ sin(θ )Y *l,m(θ , φ )Yl′,m′(θ , φ ) = δ l,l′δ m,m′

They are eigenvalues of the total angular momentum squared operator corresponding to eigenval-ue h−2 l(l + 1), l = 0, 1, 2. . . , and are also eigenfunctions of the Lz operator (cf. the last section)corresponding to eigenvalue mh−, m = 0, ±1, . . . ±l. Note that the square of the angular

Fall Term, 2019


momentum is degenerate, with degeneracy 2l + 1.

Some examples are shown on the following pages.*

*From M. Karplus and R.N. Porter, Atoms and Molecules (W.A. Benjamine, Inc.)

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Chemistry 345 -34- Legendre Functions

Hydrogen orbitals for l = 0, 1

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Hydrogen orbitals for l = 2

Fall Term, 2019

Chemistry 345 -36- Problem Sets

10. Problem Sets

Note that the due dates are last year’s. This year’s will be announced in class and on the web

site.

10.1. Problem Set 1

DUE: Monday, September 16, 2019

L

L’

d

R

Θ

P

Fig. 10.1. Diffraction/Interference pat-tern in the double slit experiment.

1. For the geometry shown in the figure, derive the Bragg formula for constructive interfer-ence:

d sin(Θ) = nλ ,

where the notation is as in class. (Remember that L, L′ >> d).

2. McQuarrie, Quantum Chemistry, Problem 1-5.

3. McQuarrie, Quantum Chemistry, Problem 1-12.

4. McQuarrie, Quantum Chemistry, Problem 1-17 [Note that the book uses different units(factors of 4π ε0) in Coulomb’s law, use whatever units you prefer].

5. Complete the analysis of the compton effect started in class; namely, giv en that energy andmomentum conservation imply that

h− ω + mec2 = h− ω ′ +mec

2

[1 − (v /c)2]1/2

and

h−→k = h−

→k ′ +

me→v

[1 − (v /c)2]1/2,

where me and→v are the electron’s rest mass and final velocity, respectively, and where the

Fall Term, 2019

Problem Sets -37- Chemistry 345

prime denotes the photon’s final properties. Finally, note, that we have assumed that theelectron was at rest initially.

Show that

∆λ = 2h

mecsin2(θ /2)

[HINT: you can use the fact that the shifts in the photon’s frequency and wav elength aresmall].

6. By looking at the Planck Blackbody distribution law at low frequency, show how theRayleigh-Jeans formula is obtained.

Fall Term, 2019

Chemistry 345 -38- Problem Set 2

10.2.


1. In the class notes it was stated that

By using the Einstein-deBroglie relations for the energy and momentum,it is easy to verify that the wav e packet’s group velocity is exactly thevelocity of the particle (both classically and relativistically).

Prove this.

2. On the domain 0 ≤ x ≤ L consider the function:

f (x) ≡

ε −1, for(L − ε )

2≤ x ≤

(L + ε )2

0 otherwise.

Derive a general expression for the Fourier amplitudes and investigate how the series con-verges by plotting partial sums as a function of x.

3. In class, we considered the group velocity of an quantum mechanical wav e packet by con-sidering the behavior of the frequencies, ω (k), near k = 0 in a linear approximation. Usethe gaussian wav e packet we considered in the notes and use the resultω (k) = E/h− = h− k2/2m to show that the width of the wav e packet grows in time; i.e., thewave packet spreads. (HINT: assume that L is large and approximate the Fourier sum byan integral.) What are the implications of your answer for measuring the position of a freeparticle?

4. Quantum dots can be viewed as particle (the electron) in a box systems. Here we willinvestigate the properties of the model one-dimensional square-well quantum dot (consid-ered in class), of side length L and energy depth −V0 as shown in the figure.

−V0

L/2−L/2

x

U(x)

a) Derive the quantization conditions. (HINT: The wav efunction will be either anev en or odd function of x).

Fall Term, 2019

Problem Set 2 -39- Chemistry 345

b) Consider the graphical solution of your quantization conditions and estimate theminimum value of V0 needed to have at least two bound states.

c) Show how your energies go over to usual particle in a box energies (2D) as V0 → ∞.d) Finally, what must L be (when V0 = ∞) in order that the dot absorbs light in the

blue? What color would the dot appear?

5. McQuarrie (2nd ed.), Problem 3-35.

6. McQuarrie (2nd ed.), Problem 3-27.

Fall Term, 2019

10.3.


V

U(x)

0

xL−L −a aFig. 10.3.2. A particle in a box with an internal bar-rier.

1. Consider the multiple square-well potential shown in the figure.

a) By assuming that the energy is less than the middle barrier height, i.e., E < V0, showthat the general form of the stationary solutions is:

Ψ(x) =

A sin((k(x − L))) for a ≤ x ≤ LB sinh(κ x) + C cosh(κ x) for − a ≤ x ≤ aD sin((k(x + L))) for − L ≤ x ≤ −a,

where

k ≡ √ 2mEh−2 and κ ≡ √ 2m(V0 − E)h−2 ,b) Write out the boundary conditions at x = ±a and show that the states can be charac-

terizes as either being even or odd under reflection through x = 0; i.e., for the evenstates A = −D and B = 0, while for the odd ones A = D and C = 0. (As we will seelater in the course, being able to do this is no accident, and is a consequence of thereflection symmetry of the potential, and can be used to simplify, characterize thestates for other symmetries as well)

c) Show that the energies are determined by solving

tan((k(L − a)))k

+coth(κ a)

κ= 0 (1)

for the even states, while

Fall Term, 2019


tan((k(L − a)))k

+tanh(κ a)

κ= 0 (2)

for the odd ones.

d) Show that the solutions become the usual particle in a box energies for a box oflength 2L when a → 0. Also show that they become the particle in a box energies(doubly degenerate) for a box of length L − a when V0 → ∞. Finally, show that theenergies of the even and odd states become equal when a, L → ∞ keeping L − afixed. Why? Note that the energies in this last limit are less than the correspondingenergy for particle in a box with side L − a. Why?

e) Examine the graphical solutions to the equations for a/L = 0. 1 and2mV0 L

2/h−2 = 300. (HINT: Note that (kL)2 + (κ L)2 = 2mV0 L2/h−2).f) For cases like those shown in part e), the lower even and odd states have almost the

same energies. We can obtain an approximate expression for the energy difference

for large V0 by assuming that κ ≈ √ 2mV0/h−2 in Eqs. (1) and (2). Moreover, for thelowest energy states ( ke ≈ ko ≈ π /(L − a)), use these approximations to show that

tan((ko(L − a))) − tan((ke(L − a))) ≈4π e−2κ a

κ (L − a). (3)

Show that this also gives:

ko − ke ≈4π e−2κ a

κ (L − a)2(4)

and an energy splitting, ∆E,

∆E =h−2

2m(k2o − k2e) ≈

h2e−2κ a

mκ (L − a)3. (5)

This is known as the tunneling splitting, and ∆E/h is the tunneling frequency.

g) To understand the nature of the tunneling splitting better, sketch the wav e functions(don’t worry about normalization) for the lowest even and odd states and for thestates

Ψ±(x) ≡Ψe(x) ± Ψo(x)

√2.

If we started with a wav efunction of the form of either Ψ± what does it physicallycorrespond to? What happens to it as a function of time? Over what time-scale?This effect is observed in NH3 in the so-called umbrella motion of the hydrogensand of the lone-pair. What, if any, implications are there for things like the D- andL- forms of the proteins in your body. What must be the case if you aren’t to loosesleep over this?

h) For L = 5. 0Ao

and m the mass of a proton, use Eq. (5) to plot the log of the character-

istic time versus V0 for a = 1. 0 and 0. 1Ao. (Remember that Eq. (5) is an

Fall Term, 2019


approximation that breaks down when V0 is too small).

Fall Term, 2019


10.4.

DUE: Tuesday, October 8, 2019

1. McQuarrie, Problem 3-38.





McQuarrie refers to the 2nd edition of his Quantum Chemistry. The solutions on my website are for the 1st edition!

Fall Term, 2019

Chemistry 345 -44- 2019 Midterm Exam

11. 2019 Midterm Exam

Possibly Useful Information

h = 6. 62619 × 10−34J ⋅ s h− = 1. 05459 × 10−34J ⋅ scosh(x) ≡

ex + e−x

2cosh2(x) − sinh2(x) = 1 sinh(x) ≡

ex − e−x

2’

d cosh(x)

dx= sinh(x)

d sinh(x)

dx= cosh(x)

cos(x) ≡eix + e−ix

2cos2(x) + sin2(x) = 1 sin(x) ≡

eix − e−ix

2i’

d cos(x)

dx= − sin(x)

d sin(x)

dx= cos(x)

sin(x ± y) = sin(x) cos(y) ± cos(x) sin(y) cos(x ± y) = cos(x) cos(y) −+ sin(x) sin(y)

∫b

adx f (x)

dg(x)

dx= f (x)g(x)|ba − ∫

b

adx g(x)

df (x)

dx(Integration by parts)

This is a closed book exam. There are 4 problems, printed on the reverseside of this sheet, and each is worth 25%. Calculators are not necessary nor permitted.

KEEP THE EXAM

October 17, 2019

2019 Midterm Exam -45- Chemistry 345

1. (25%) Define or give a defining equation for:

a) The deBroglie hypothesis b) The group velocityc) An inner product (give an equation) d) A hermitian operator

e) The time dependent Schrodinger equation (give the equation).

2. (25%) For Hermitian operators show:

a) That their eigenvalues are real.

b) If two Hermitian operators H and P commute, show that the eigenfunctions of oneare eigenfunctions of the other (again assume no degeneracy).

3. (25% From the quizzes) Eherenfest’s theorem in 1 dimension says that

∂ < x > (t)∂t

= (t)

mand

∂ (t)∂t

= < F(x) > (t), (3.1)

where x and p are the position and momentum operators, respectively, m is the particlemass, and F(x) is the classical force exerted on the particle at position x.

a) Prove Eherenfest’s theorem for a one-dimensional system, where −∞ < x < ∞ andin which the wav e function Ψ(x, t) → 0 as x → ±∞.

b) What does Eherenfest’s theorem become for a harmonic oscillator?

c) More generally, when does Eherenfest’s theorem reduce to classical Newtoniandynamics?

−V0

L/2−L/2

x

U(x)

Fig. 11.1. The potential energy for problem 4. Note that U(x) = 0 for|x| > L/2.

4. (25% From the homework) For the potential depicted in the figure and assuming the energyE < 0 derive or state:

a) The forms of the energy eigenfunctions, u(x) for |x| > L/2 and |x| < L/2.

b) The equations (boundary conditions) connecting the solutions at x = ±L/2 and theresulting transcendental equations that determine the eigenvalues, E.

c) Show how your expression gives the particle in a box energies (for a box of sidelength L) as V0 → ∞.

October 17, 2019

Chemistry 345 -46- 2019 Midterm Exam

12. 2019 Midterm Exam Solutions

1. (25%) Define or give a defining equation (each worth 5%) for:

a) The deBroglie hypothesis:

λ =h

p. (1.1)

b) The group velocity: The velocity of a wav e packet

vg =

∂ω (k)∂k

k=k0. (1.2)

c) An inner product (give an equation):

(A, B) = ∫ dx A*(x)B(x) (1.3)

d) A hermitian operator:

(Ψ1, OΨ2) = (OΨ1, Ψ2) = (Ψ1, OΨ2)*. (1.4)

e) The time dependent Schrodinger equation (give the equation).

ih− ∂Ψ∂t

= HΨ. (1.5)

2. (25%) For Hermitian operators show:

a) (15%) That their eigenvalues are real. By using Eq. (1.4) it follows that

(un, Oun) = λ n(un, un) = (Oun, un) = λ*n(un, un), (2.1)

where Oun = λ nun. By comparing the first and last equalities, we see thatλ n(un, un) = λ*n(un, un), i.e., λ n = λ*n, which implies that λ n is real.

b) (10%) If two Hermitian operators H and P commute, show that the eigenfunctions of oneare eigenfunctions of the other (again assume no degeneracy).

Suppose that un is an eigenfunction of H for eigenvalue En; i.e., Hun = Enun. By actingwith P on the left on each side of the equality, we see that

PHun = En Pun = HPun, (2.1)

where the last equality if valid when P and H commute. Hence, Pun is an unnormalizedeigenfunction of H with eigenvalue En. Howev er, we’ve assumed that the eigenfunctionsare non-degenerate and so Pun∝un. By calling the proportionality constant p, it follows

October 17, 2019

2019 Midterm Exam Solutions -47- Chemistry 345

that Pun = pun; i.e., un is also an eigenfunction of P with eigenvalue p.

3. (25% From the quizzes) Eherenfest’s theorem in 1 dimension says that

∂ < x > (t)∂t

= (t)

mand

∂ (t)∂t

= < F(x) > (t), (3.1)

where x and p are the position and momentum operators, respectively, m is the particle mass,

and F(x) is the classical force exerted on the particle at position x.

a) (15%) Prove Eherenfest’s theorem for a one-dimensional system, where −∞ < x < ∞ andin which the wave function Ψ(x, t) → 0 as x → ± ∞.

∂ < x > (t)∂t

= ∫ dx Ψ*(x, t)xΨ̇(x, t) + Ψ̇*(x, t)xΨ(x, t) (3.2)

= ∫dx

ih− Ψ*(x, t)x[HΨ(x, t)] − [HΨ(x, t)]* xΨ(x, t), (3.3)

=1

m ∫ dx Ψ*(x, t)

h−i

∂∂x

Ψ(x, t) =< px > (t)

m(3.4)

where

H ≡ −h−2

2m

∂2

∂x2+ V (x), (3.5)

and where Eq. (3.3) is obtained by using the time dependent Schrodinger equation, Ψ̇ = HΨ/ih−,while Eq. (3.4) is obtained by integrating Eq. (3.3) by parts twice.

The second part of Eherenfest’s Theorem follows in a similar manner; i.e.,

∂ < px > (t)∂t

= ∫ dx Ψ*(x, t) pxΨ̇(x, t) + Ψ̇*(x, t) pxΨ(x, t) (3.6)

= ∫dx

ih− Ψ*(x, t) pxV (x)Ψ(x, t) − Ψ*(x, t)V (x) pxΨ(x, t) (3.7)

= − ∫ dx Ψ*(x, t)∂V (x)

∂xΨ(x, t) =< F(x) > (t), (3.8)

where F(x) ≡ −∂V (x)/∂x. Note that px and the kinetic energy, p2x /2m, commute.

b) (5%) What does Eherenfest’s theorem become for a harmonic oscillator?

For the harmonic oscillator, F(x) = −Kx, where K is the force constant. By using this in Eq.(3.1) it follows that

October 17, 2019

Chemistry 345 -48- 2019 Midterm Exam Solutions

∂ < x > (t)∂t

= (t)

mand

∂ (t)∂t

= − K < x > (t), (3.9)

which is the classical (Newtonian) equation of motion. This is an exact result, and thus, simplyobserving the average trajectory of the oscillator will not distinguish between classical and quan-tum dynamics.

c) (5%)) More generally, when does Eherenfest’s theorem reduce to classical Newtoniandynamics?

Eherenfest’s theorem reduces to classical mechanics when < F(x) > ≈ F(< x > (t)); i.e., when|Ψ(x, t)|2 is sharply peaked around x =< x > (t).

−V0

L/2−L/2

x

U(x)

Fig. 12.1. The potential energy for problem 4. Note that U(x) = 0 for|x| > L/2.

4. (25% From the homework) For the potential depicted in the figure and assuming the energy

E < 0 derive or state:

a) (10% The forms of the energy eigenfunctions, u(x) for |x| > L/2 and |x| < L/2.

Schrodinger’s equation becomes

−h−2

2m

∂2u(x)∂x2

=

(E + V0)u(x), for |x| < L/2Eu(x), otherwise.

(4.1)

Since the problem has reflection symmetry (x → −x), we expect solutions that are even or oddunder reflection. The even solutions can be written as

u(x) =

A cos(kx) for |x| < L/2

Be−κ |x| otherwise,(4.2a)

where k ≡ √ 2m(E + V0)/h−2 and κ ≡ √ 2m|E |/h−2. Similarly, the odd solutions can be written as

u(x) =

A sin(kx) for |x| < L/2

±Be−κ |x| for |x| > L/2,(4.2b)

where the +(−) sign should be used for x > (

2019 Midterm Exam Solutions -49- Chemistry 345

The boundary conditions require that u(x) and du(x)/dx be continuous at x = ±L/2. For the evensolutions this becomes

A cos(kL/2) = Be−κ L/2 and Ak sin(kL/2) = Bκ e−κ L/2. (4.3)

By eliminating B we see that

k tan(kL/2) = κ . (4.4)

The factors of A cancel (nonetheless, |A| can be determined by normalizing the wav e-function).

The same boundary conditions apply for the odd solutions and thus Eq. (4.2b) gives

A sin(kL/2) = Be−κ L/2 and Ak cos(kL/2) = −Bκ e−κ L/2. (4.5)

Again eliminating B we find that

κ tan(kL/2) = −k. (4.6)

c) (5%) Show how your expression gives the particle in a box energies (for a box of sidelength L) as V0 → ∞.

First note that

(kL)2 + (κ L)2 =2mL2V0

h−2 ≡ γ2, (4.7)

where −V0 < E < 0. Next, let x ≡ kL and rewrite Eqs. (4.4) and (4.6) as

x tan(x/2) = √ γ 2 − x2 (even) (4.8a)

and

x cot (x/2) = −√ γ 2 − x2 (odd), (4.8b)

respectively. These are easy to solve graphically. As γ → ∞, e.g., for L or V0 → ∞, the righthand sides of Eqs. (4.8a) or (4.8b) diverge, and thus, x/2 will approach one of the points wheretan or cot diverge, namely, at x = (2n + 1)π (even solutions) or x = 2nπ (odd solutions), where nis an integer. Since, x ≡ kL and k ≡ √ 2m(V0 + E)/h−2 , we see that

E = −V0 +h−2

2mL2x2 = −V0 +

h2

8mL2n2, (4.9)

where n is a positive integer. Note that n is odd (even) for the even (odd) solutions. The lastresult is simply the particle in a box energy, shifted by −V0.

October 17, 2019

Chemistry 345 -50- 2019 Midterm Exam Solutions

13. Lecture Slides

Fall Term, 2019

david ronis mcgill university · 2019. 11. 4. · wed. nov. 14the chemical bond ch10 sec 6-10 fri....

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