Lecture 1 – Fundamental Concepts in Quantum Physicsyangtze.hku.hk/lecture/bk/CHEM3541_2019_Lecture_Notes.docx · Web viewCHEM 3541 Physical Chemistry Lecture Notes Textbooks Atkins’

CHEM 3541 Physical ChemistryLecture Notes

Textbooks

Atkins’ Physical Chemistry, 7th ed., pp. 304-316

Atkins’ Physical Chemistry, 10th ed., pp. 290, 292-305

Table of ContentsLecture 1 – Fundamental Concepts in Quantum Physics11.Schrödinger Equation12.Hermitian operator33.Expectation value54.Heisenberg’s uncertainty principle5Lecture 2 – Translational Motion81.One-dimensional particle-in-a-box model82.Two-dimensional model103.Three-dimensional model114.Tunnelling125.Particle in a finite square-well potential13Lecture 3 – Vibrational Motion151.One-dimensional harmonic oscillator152.The vibration of a diatomic molecule17Lecture 4 – Rotational Motion191.Two-dimensional rotational motion192.Compatibility theorem203.Three-dimensional rotational motion20Lecture 5 – Hydrogen Atom231.Separation of CM motion and relative motion232.Separation of radial motion and rotational motion24

Lecture 1 – Fundamental Concepts in Quantum Physics1. Schrödinger Equation

Time-independent S.E.

· : Hamiltonian operator

· : Wavefunction,

· : Kinetic energy operator and potential energy operator. In many situations, operator is simply a function.

Time-dependent S.E.

· If potential and thus wavefunction depend on time explicitly, one need to solve time-dependent S.E.

I. One-dimensional system with one particle

is Planck constant, whereas is called reduced Planck constant. operator is simply a function. The S.E. reads

Wavefunction contains all the information of this system. The simplest one is probability density , which tells that the probability to find the particle between and is .

Since total probability to find the particle between and should be 1, we require the wavefunction to be normalized:

For an unnormalized wavefunction, it can be normalized as

Example 1: constant potential

For , the S.E. is

If , the general solution is , where . The wave vector is defined as .

· Kinetic energy .

· Momentum . is de Broglie wavelength. To see the meaning of , let , then the period of is .

Example 2: normalization of wavefunction

Q. Given a wave function where is a given constant, try to normalize it. Noted that in the exponent is radial distance but not position vector.

A. , where .

Thus .

II. Three-dimensional system with one particle

i. Cartesian coordinates

Wavefunction is . The probability to find the particle in is .

, where is Laplacian operator. S.E. is . Wavefunction and energy are unknown variables.

ii. Spherical coordinates

Relationship to Cartesian coordinates

: radial distance

: polar angle

: azimuth angle

Wavefunction is . The probability to find the particle in is .

Laplacian operator is more complex in spherical coordinates

where .

S.E. is

III. Eigenequation, eigenfunction and eigenvalue

The solution of S.E. is a set of eigenenergy/value and eigenfunction , .

For any operator , if there exist some values satisfy , is then called eigen function of operator .

For example, since , is an eigenfunction of operator with corresponding eigen value . But is not an eigenfunction of .

For one-dimensional system with one particle, we can choose two wavefunctions with different eigenvalues and ,

· ,

· ,

Generally, is NOT a solution of S.E. . Only when , .

2. Hermitian operator

For any two functions, if the following equation holds, then is called Hermitian operator.

For example, is not a Hermitian operator since

But is a Hermitian operator

I. Eigenvalues of Hermitian operator

· Theorem: Any eigenvalue of Hermitian operator is a real number.

· Proof:

Given a Hermitian operator with eigenvalue and normalized eigenfunction , i.e., . Applying the definition of Hermitian operator, we have

Thus must be real.

Q.E.D.

All operators corresponding to physical observables (properties that can be measured) are Hermitian operators.

II. Eigenfunctions of Hermitian operator

· Theorem: Eigenfunctions of Hermitian operator with different eigenvalues are orthogonal.

· Proof:

Using the definition of Hermitian operator, we have

Also,

Subtracting the first equation from the second one gives

For , we have

Q.E.D.

Example1: one-dimensional momentum operator

For particle in constant potential, take the wavefunction as . Define the momentum operator as

which has been proved to be Hermitian. Apply it to above wavefunction,

By definition, , thus , and is indeed the momentum operator.

Example2: three-dimensional momentum operator

Momentum operator is Hermitian since all components of it are Hermitian.

3. Expectation value

For a Hamiltonian with normalized wavefunctions , , we construct a wavefunction where . The expectation value, or the average value of energy is then

Generally, for any operator in a quantum state , its expectation value is

4. Heisenberg’s uncertainty principle

Suppose we have measured an observable times (), then the uncertainty of is defined as

For position and momentum, their uncertainties satisfy the following Heisenberg’s uncertainty principle

Apply to an arbitrary wavefunction gives

Thus, . Since is arbitrary, we have

Define commutator of two operators and as , the above equation can be rewritten as

If , quantum mechanics will degenerate to classical mechanics.

· Heisenberg’s uncertainty principle: If the commutator of two operators and is , then their uncertainties satisfy the following inequality

· Proof:

Given an arbitrary wave function , constructing a non-negative integral with real variable

Noted that for a complex number 𝑐, . Expanding above integral, we have

To ensure above quadratic function of is non-negative, there must be

i.e.

Define , , apparently,

Thus

Substitute , with and respectively in ,

Noted that and , finally we have

Q.E.D.

Examples

· , . , . .

· has definite momentum , so that . must be , which means the particle is diffused in the whole space.

· If , we have , thus . This means we need infinite energy to constrain one quantal particle to a certain position.

· If two operators commute, i.e. , , we have . So and can be measured precisely at the same time.

·

· A particle with mass moves along direction subjected to a potential . , . , i.e.

· If and only if is equal to some constant will and commute, thus energy and momentum can be measured precisely at the same time.

· From time-dependent S.E. , we define . Since , .

· If a quantum state has definite energy, i.e. , then the lifetime of this state will be .

· In reality, energy level is broadened, and is regarded as lifetime of the energy level.

Lecture 2 – Translational Motion1. One-dimensional particle-in-a-box model

Suppose we have one particle with mass confined in a box , then its Hamiltonian is

where the potential is

I. Wavefunctions and energy levels

Within , this S.E. has solution

Outside

Since wavefunction should be continuous, we impose following boundary conditions

Plug and into , we have

So that after normalization, within ,

Energy levels corresponding to each is

,

Figure 1 First five normalized wavefunctions

II. Orthogonality of wavefunctions

For ,

III. Uncertainty principle for ground state

Noted that and , we can rewrite uncertainty as

and are needed to calculate .

Thus

· Calculation details

Let

So

Let

So

· Example 8A.2 (pp. 321)

· Problem: β-Carotene is a linear polyene in which 10 single and 11 double bonds alternate along a chain of 22 carbon atoms. If we take each C-C bond length to be about 140 pm, then the length L of the molecular box in β-carotene is L = 2.94 nm. Estimate the wavelength of the light absorbed by this molecule from its ground state to the next higher excited state.

· Answer:

2. Two-dimensional model

m

y=L2

y=0

x=0

x=L1

Hamiltonian operator

S.E. is .

Within and , S.E. is

Boundary conditions are

To solve this multivariable equation, we perform separation of variables. Let and plug this equation into S.E., we get

Divide both sides by

To ensure , each term in the LHS should be some constant, viz.

,

with .

Now the 2D S.E. of has been decomposed into two 1D S.E., thus its solution is just the product of two separated equations,

Its energy levels are

where .

3. Three-dimensional model

Hamiltonian:

S.E.: , within , , and

Boundary conditions

Now we perform similar procedures as 2D model. Let and plug this equation into S.E., we get

Divide above equation by

Each term in the LHS should be some constant, viz.

, ,

where . So, the solution is

with energy levels

where .

4. Tunnelling

In this quantum tunnelling model, potential is set to be zero in or and be constant in . The energy of incident wavefunction is and . Denote the amplitude of incident and exit wave functions as and respectively and define transmission coefficient as .

V

A

A’

A

A’

In zero-potential region, S.E. and wavefunction are

For incident region , we choose , where two terms stand for incident and reflection wavefunctions respectively. For , stands for tunnelling wavefunction.

In potential barrier region, . The general solution is

where .

At the two interfaces, wavefunction shall be smooth, i.e.

thus we can obtain four equations

, and can all be expressed in as

: , :

:

Then we try to solve transmission coefficient.

denote ,

Finally,

· If , i.e. : .

· If , i.e. high, wide barrier: .

· The heavier the mass, the smaller the .

5. Particle in a finite square-well potential

Potential is constant in or and zero in . The energy of particle is and . Denote and .

Similar to above tunnelling model, we use following wavefunctions

:

:

:

At infinity, wavefunction should vanish, thus . At , applying boundary conditions and , we get and , i.e.

,

Similarly, at we have and , i.e.

,

Then can be expressed in terms of as or . Use the first expression, we have

At , should be continuous

i.e.

Real parts of LHS and RHS are identical, and the imaginary parts should be equal

i.e.

· When , we have , i.e. .

· When and , we have and where . For a given , if there is no such satisfy these two equations, then this state is quantum forbidden.

Lecture 3 – Vibrational Motion1. One-dimensional harmonic oscillator

The force of spring is , thus its potential can be calculated as . The Hamiltonian is then , and its corresponding S.E. is

We first change variable from to where . Under this operation, changes to . After some algebra, we have

where and .

Now we take a look at asymptotic behaviour of above equation. When ,

thus . Rewrite , we have following Hermite equation

Expand as and plug this expansion into Hermite equation,

But when , as fast as , causing as fast as . To ensure the expansion of must be truncated, i.e.

Thus we get quantized energy as

It is worthwhile noting that the ground state energy, also called zero-point energy, is non-zero.

I. Wavefunction

Solutions of Hermite equation are called Hermite polynomials. They satisfy following recursive and orthonormal relations

First three Hermite polynomials are listed below.

Now we need to normalize wavefunction .

Thus and the normalized result is

II. Uncertainty of position

Thus .

III. Potential energy

IV. Tunnelling – classically forbidden region

Classically forbidden region is where . Quantal oscillator can reach classically forbidden region with some tunnelling probability.

For ground state, , . Denote and as the negative and positive solutions of respectively. The tunnelling probability is then

2. The vibration of a diatomic molecule

Consider a diatomic molecule moving along axis. The coordinates of two atoms with mass and are and respectively. The force constant is .

Classically, the kinetic energy of this system is

where . Define the coordinate of centre-of-mass as

and the distance between two atoms as

It is easy to find the inverse transformations as

Thus the kinetic energy can be expressed in as

Here is the total mass and is the reduced mass.

Define the momentum of centre-of-mass and vibration as

,

The kinetic energy is then

Its corresponding quantum Hamiltonian is

The Schrödinger equation is now separable. Assuming that

we have

Divide both sides by ,

Thus

The solution is

Appendix: Direct coordinate transformations of Hamiltonian

It’s straightforward to find that

Thus

Finally,

Lecture 4 – Rotational Motion

1. Two-dimensional rotational motion

A particle of mass moves in a ring of radius in the -plane with zero potential. We use cylindrical coordinates for convenience. Laplace operator has following form

r

m

Since are fixed, it can be simplified to .

S.E. is . Denote as moment of inertia, we have

The solution is where . We then apply cyclic boundary condition , i.e. . leads to , thus

,

After normalization, we have and

,

I. Energy

, ,

· Ground state energy is zero.

· First excited states are degenerated. with degeneracy 2.

II. Linear momentum

,

Thus are also eigenfunctions of .

III. Angular momentum

Classically, . In quantum mechanics, for a particle in a circle

,

Thus are also eigenfunctions of .

These co-eigenfunction phenomenon are described by compatibility theorem next section.

2. Compatibility theorem

· Theorem: Giving two Hermitian operators and , if and are commuting, viz , we can conclude that and have a common eigen basis, i.e. we can find a set of satisfying and

· Examples: , so are their common eigenfunctions.

3. Three-dimensional rotational motion

A particle of mass moves on the surface of a sphere of radius with zero potential. Now we use spherical coordinates where

Since is fixed, Laplace operator is simplified to . The S.E. is

i.e. with the same definition for as in 2D motion.

Let , we have

Dividing two sides by and rearranging the equation,

Thus,

and

should hold, where is a constant.

For , the solution is

,

For , let , this equation can be rewritten as associated Legendre equation

Its solutions are called associated Legendre functions where and .

The overall solutions are called spherical harmonics which satisfy following equation

Hereafter we will drop out subscript from for simplicity. First few of them are listed below.

·

· :

·

· :

· :

·

· :

· :

· :

I. Energy and square of angular momentum

The Hamiltonian ,

gives out energy levels

For square of angular momentum, , classically we have and quantum mechanically .

thus

II. Angular momentum

Classically, angular momentum is defined as . In quantum mechanics, we change it into , where , and for . Here ‘’ means cross product. Some basic properties of cross product are shown below

, , ,

And for bases, their cross products are

, ,

Using these relations, we have

The components of are then

, ,

respectively.

We then calculate commutators between these operators since the commutation relation is a key feature of angular momentum. Take and as example,

Similarly, we have , . For , we first decompose it into . Then it is straightforward to write

Thus , and . , and are mutual commuting, which confirms that are the common eigenfunctions for them, i.e.

Lecture 5 – Hydrogen Atom

In this last lecture, we will try to solve a real system – hydrogen atom. The system is composed of two particles – one electron and one positron, and thus its total degree of freedom (DoF) is 6. Three of DoFs belong to translational motion, two of them rotational motion, and the last one is the relative radial motion.

electron

O

positron

To begin with, we define following notations.

· Mass: nucleus , electron , total , reduced

· Position vector: nucleus , electron , centre of mass (CM) , electron relative to nucleus

· , (similar for )

· Classical momentum: nucleus , electron , CM , electron relative to nucleus

· Vector without arrow means modulus , etc

Then we can separate out CM motion and relative motion in classical energy expression as

Thus quantum Hamiltonian can also be written as sum of two parts

S.E.

1. Separation of CM motion and relative motion

Let ,

Divide both sides by :

Thus

and

where .

Since , . Roughly, and are nuclear and electron wavefunctions respectively.

I. CM motion

This is just a free particle moving in 3D space. Its solution is plane wave

Its wave vector has modulus and is parallel to .

II. Relative motion

can be expressed in spherical coordinate system located at nucleus

From now on, we will omit the subscript of for simplicity.

2. Separation of radial motion and rotational motion

Let ,

Divide both sides by ,

Thus we have

I. Rotational motion

Rearrange as . The solution is apparently spherical harmonic functions, with eigenvalues . Thus,

II. Radial motion

i.e.

Remember , here can be regarded as effective potential due to angular momentum.

Solution for above equation is

· , .

· , called Bohr radius, is the unit length in atomic unit.

· : associated Laguerre polynomial.

· Normalization factor .

· ; .

· Energy (or in SI ).

Overall, electronic wavefunction is

It depends on three quantum numbers,

· Principal quantum number: .

· Azimuthal quantum number: .

· Magnetic quantum number: .

But its energy levels only depend on principal quantum number . For ground state, , it is non-degenerate. For first excited state it is 4-fold degenerated.

First few electronic wavefunctions are listed below

1