Unit-IV: Wave Mechanics (Part-I) - ii ( unit... · 2020. 5. 18. · Concept of Modern Physics, by Arthur Beiser, Tata McGraw-Hill, 6th edition (2003). Quantum Physics for Scientists

Unit-IV: Wave Mechanics (Part-I)

Dr. Prashanta Kumar Khandai

Department of PhysicsEwing Christian College

Allahabad

Email: [email protected] Number: +91-9122008433, +91-7717756468.

May 17, 2020

Dr. Prashanta Kumar Khandai (ECC) Lecture May 17, 2020 1 / 27

Outline

1 Introduction

2 Wave functionAcceptable or well-behaved wave function

3 Schrodinger’s equationTime dependent and Time independent formsIts properties and short-comings

4 Different properties and theorems of wave functions

5 Expectation values

6 Equation of Continuity or Conservation of probability


Introduction

Reference Books and other sources

• Concept of Modern Physics, by Arthur Beiser, Tata McGraw-Hill, 6thedition (2003).

• Quantum Physics for Scientists and Technologists, by Paul Sanghera,John Wiley & Sons (2011).• Notes of Anjani Sir.


Introduction

Failures of Classical Physics

• Classical Physics on its own can not explain many natural/ observedphenomena.• There are two main drawbacks of Classical Physics:

1 Einstein’s theory of relativity: Newtonian mechanics failed at veryhigh speed phenomena, where a body moves with speed of light.

2 Classical Physics fails to work microscopic level: It fails to explain thestructure of atom and how light interacts with particles.


Introduction

Introduction

• The original formulation of quantum mechanics is generally credited toWerner Heisenberg and Erwin Schrodinger, who presented essentially thesame theory in two different mathematical formulations.

• Heisenberg developed quantum mechanics, in 1925, along with MaxBorn and Pascual Jordan, in the form of matrices, whereas Schrodingerdeveloped the quantum mechanics, in 1926, in the form of wavemechanics.

• Paul Dirac, Schrodinger, and others later demonstrated that bothformulations yield identical results even though they use a differentmathematical structure.


Wave function

Wave function: building block of quantum mechanics

• In wave mechanics, a particle (or a physical system) can be representedby a wave and it must have a wave function associated with it.

• In quantum mechanics, such a wave function, Ψ , represents theprobability of finding the particle at a given point in space and time.

• To be precise, we will learn later that Ψ, as being a complex number,does not have any direct physical significance. However, the square of itsabsolute magnitude |Ψ|2 = Ψ∗Ψ evaluated at a specific point in space andtime represents the probability density (P ) of experimentally finding theparticle at that point in space and time. Here Ψ∗ is the complex conjugateof Ψ.


Wave function

Normalization Condition

• Because the particle must be somewhere in the space, the integration of|Ψ|2 over total space and time must be equal to 1. This can berepresented mathematically as:

∫ +∞−∞ |Ψ|

2dV = 1⇒∫ +∞−∞ Ψ∗(~r, t)Ψ(~r, t)dV = 1⇒

∫ +∞−∞ PdV = 1

where dV = dxdydz = d3r represents infinitesimal volume in 3D at time tin which the particle can be found with the probability of |Ψ|2. This is alsocalled the normalization condition for the wave function. The wavefunction that meets this condition is called a normalized wave function.

• If∫ +∞−∞ |Ψ|

2dV = 0 particle does not exist there.


Wave function Acceptable or well-behaved wave function

Acceptable or Well-Behaved wave function

• Ψ must be finite, single-valued every where. Also Ψ is continuous everywhere (except in the region where P.E: U =∞).

• ∂Ψ∂x , ∂Ψ

∂y and ∂Ψ∂z must be finite, continuous and single valued everywhere.

• Ψ must be normalizable, which means that Ψ must go to 0 as x→ ±∞,y → ±∞ and z → ±∞ in order that

∫|Ψ|2dV over all space be a finite

constant.



Operator: information extractor

• We know that a wave function has all the information about the systemthat it represents. How do we extract information from it? That is whereoperators come into the picture.

• Operator is a rule which changes a function to another function. Forexample, derivative is an operator which shows: d

dx (sinx) = cosx.

• In quantum mechanics, a physical observable (e.g. position, momentum,energy etc.) is represented by an operator that operates on a wavefunction to predict the value of the measurements of the observable.That is AΨ = aΨ′, where A is the operator and a is the value of themeasurement of the observable.



Continued...

• Let’s Consider the wave function of a particle having momentum p andenergy E is given by: Ψ(x, t) = ei(kx−ωt), where k and ω are determinedfrom momentum: p = ~k and energy E = ~ω respectively.

Now we are going to extract information from the wave function using asimple operator, i.e., derivative. So differentiating Ψ with respect to x weget: ∂

∂x [Ψ(x, t)] = ∂∂x [ei(kx−ωt)] = ikΨ. ⇒ ~

i∂∂x [Ψ(x, t)] = ~kΨ(x, t).

⇒ p = ~i∂∂x .

As Ψ = Ψ(x, t), so again differentiating Ψ with respect to t we get:∂∂t [Ψ(x, t)] = ∂

∂t [ei(kx−ωt)] = −iωΨ. ⇒ i~ ∂∂t [Ψ(x, t)] = ~ωΨ(x, t).

⇒ E = i~ ∂∂t .

So the operator form of momentum: p = ~i∂∂x and energy: E = i~ ∂∂t .



Eigen value equation

• It is an expression in which an operator A acting on a wave function Ψ

representing the same wave function Ψ multiplied by a constant factor λ.

That is AΨ = λΨ , here λ is called the eigen value.

• For example: suppose ddx is an operator acting on the wave function e2x,

then it yields: ddx (e2x) = 2e2x, where 2 is the eigen value and it satifies theeigen value equation.

• The only possible results in the measurement of a physical observable Aare the eigenvalues λ of the corresponding operator A operating on thewave function Ψ.

• If two or more wave functions corresponds to same eigen value, thenthese wave functions are called degenerate wave functions or simplydegenearte states.



Problems

1 Select the acceptable wave functions from the following: (a) Ψ = xn,(b) Ψ = ex, (c) Ψ = e−x, (d) Ψ = e−x

2

, (e) Ψ = A sin(x).

2 Show that the wave function Ψ(x) = e−α2x2

2 is an eigen function ofoperator A = − d2

dx2 + α4x2 with an eigen value of α2.

3 Which of the following functions are eigen function of the operatord2

dx2 ? Find the eigen value in each case, if exists. (a) Ψ = A sin(mx),(b) Ψ = B cos(nx), (c) Ψ = Ae−mx, (d) Ψ = Bex

2

, (e) Ψ = ex, (f)Ψ = α

x .

4 The normalized state of free particle is given by wave function

ψ(x) = Ne−x2

2a2 eikx. (a) Find the normalization factor N , (b) In whatregion of space, the particle is most likely to found.


Schrodinger’s equation

Schrodinger’s equation

• It is a fundamental equation of wave mechanics having two forms: Timedepenedent and time independent Schrodinger equations.

• Time dependent Schrodinger equation: i~∂Ψ∂t = − ~2

2m∇2Ψ + UΨ , where

Ψ = Ψ(x, y, z; t) and ∇2 = ∂2

∂x2 + ∂2

∂y2 + ∂2

∂z2 .

• Time independent Schrodinger equation: ∇2Ψ + 2m~2 (E − V )Ψ = 0 ,

where Ψ = Ψ(x, y, z)

• The general form of Schrodinger equation: HΨ = EΨ, where H is theHamiltonian operator equal to the sum of K.E ( p

2

2m) and P.E (U) and E isthe energy eigen value or in some cases it is called as the energy operator.


Schrodinger’s equation Time dependent and Time independent forms

Time dependent Schrodinger equation (TDSE)

Let us consider a particle which is moving (with non-relativistic speed)freely in +ve x direction. Then the corresponding wave equation for theparticle can be written as:

Ψ = Aei(kx−ωt)

Since we know p = ~k and E = ~ω, then the above equation can bewritten as:

Ψ = Aei

~ (px−Et) (1)

Although the above equation is correct for a freely moving particle, but weare interested in the situations where the particle is bound by somerestrictions. Now we are going to obtain a differential equation for Ψ.



Continued...

Now let us begin by Differentiating eq-1 twice with respect to x, which

gives: ∂2Ψ∂x2 = −p

2

~2 Ψ. ⇒ p2Ψ = −~2 ∂2Ψ∂x2 .........(2)

Again differentiating eq-1 once wrt t, we get: ∂Ψ∂t = − iE~ Ψ.

EΨ = i~∂Ψ∂t ...............(3)

As we know that the total energy E of a particle is equal to the sum of itsK.E ( p

2

2m) and P.E (U(x, t)), so we can write E = p2

2m + U(x, t).By multiplying Ψ on both sides of the equation we get:EΨ = p2

2mΨ + U(x, t)Ψ........(4)Substituting the eq-(2) and eq-(3) in eq-(4) it becomes the time

dependent Schrodinger equation in 1-D: i~∂Ψ∂t = − ~2

2m∂2Ψ∂x2 + UΨ .

Schrodinger’s equation in 3D is:

i~∂Ψ∂t = − ~2

2m (∂2Ψ∂x2 + ∂2Ψ

∂y2 + ∂2Ψ∂z2 ) + U(x, y, z, t)Ψ .

⇒ i~∂Ψ∂t = − ~2

2m∇2Ψ + U(x, y, z, t)Ψ .



Time independent Schrodinger equation (TISE)

Here we will prove the TISE from TDSE using the separation of variablestechnique in 1D.When the potential energy (U) is independent of time, then the wavefunction can be written as:Ψ(x, t) = ψ(x)φ(t).We begin by taking the wave function of a particle moving in +x directionas: Ψ(x, t) = Ae

i

~ (px−Et) = Aei

~pxe−i

~Et = ψe−i

~Et.Substituting the Ψ in TDSE, we get:i~ ∂∂t (ψe

− i

~Et) = − ~2

2m∂2

∂x2 (ψe−i

~Et) + Uψe−i

~Et

Eψe−i

~Et = − ~2

2me− i

~Et ∂2Ψ∂x2 + Uψe−

i

~Et.Dividing through by the common exponential factor we get the TISE in

1D: ∂2ψ∂x2 + 2m

~2 (E − U)ψ = 0 . It is also called as the steady-state form of

Schrodinger’s equation. In three dimension, it becomes:∂2ψ∂x2 + ∂2ψ

∂y2 + ∂2ψ∂z2 + 2m

~2 (E − U)ψ = 0 . ⇒ ∇2ψ + 2m~2 (E − U)ψ = 0 .


Schrodinger’s equation Its properties and short-comings

Short-comings and properties of Schrodinger equation

• Properties of Schrodinger equation:

1 The differential equation is first order in time. This means that for aninitial condition it suffices to know the wavefunction completely atsome initial time t0 and the Schrodinger equation then determines thewave function for all times.

2 Schrodinger equation is linear in the wave function Ψ. If Ψ1 and Ψ2

are two solutions of Schrodinger’s equation, then Ψ = Ψ1 + Ψ2 is alsoa solution, where a1 and a2 are constants. Thus the wave functionsΨ1 and Ψ2 obey the superposition principle that other waves do.

• Short-comings of Schrodinger equation:

1 Time dependent Schrodinger’s equation is not symmetric with respectto space and time. Because it contains second order derivative ofspace and first order derivative of time.



Physical interpretation of wave function

• Properties of Schrodinger’s Interpretation:

1 Schrodinger thought that Ψ represents particles that disintegrate. Youhave a wave function and the wave function is spread all over space,so the particle has disintegrated completely. And wherever you findmore Ψ, more of the particle is there. That was his interpretation.

2 In 1926, Schrdinger believed that electron waves were always spreadout across all of space and that the square of the wave function gavethe charge density of the electron wave in any particular location.This was a reasonable assumption since the wave appeared to bedensest in the places where Bohr’s theory predicted electrons wouldbe. Yet Schrdinger’s interpretation could not explain quantumtunnelling.



Continued.....

• Max-Born Interpretation:

1 Max Born proposed a different interpretation in the same year. Bornstated that the square of the wave function does not represent thephysical density of electron waves, but their probability density. Thisis the probability of finding an electron in any particular state, that is,with any particular position, momentum, or energy, at any particulartime. The de Broglie model of the atom was now replaced with theidea that electrons exist in a superpositional ’probability cloud’.

2 Born’s statistical interpretation of the wave function, which says that|Ψ(x, t)|2 gives the probability of finding the particle at point x, attime t,or more precisely, |Ψ(x, t)|2dx = Probability of finding the particlebetween x and (x+ dx), at time t.


Different properties and theorems of wave functions

Stationary states

• The state of the system for which the probability density is independentof time is called as stationary states (i.e., dP

dt = 0). Such systems areknown as stationary systems.

• Example: for a wave function which satisfies the Schrodinger equation,its probability density is always independent of time.

The solution of Schrodinger equation is: Ψ(x, t) = ψ(x)e−i

~Et and theprobability density is:P = Ψ∗(x, t).Ψ(x, t) = ψ∗(x)e

i

~Et.ψ(x)e−i

~Et = ψ∗(x).ψ(x) independent oftime.



Orthogonal and orthonormal wave function

• Two wave functions Ψm(x, t) and Ψn(x, t) are said to be mutuallyorthogonal if

∫ +∞−∞ Ψ∗m(x, t).Ψn(x, t)dx = 0 when m 6= n.

• The normalized and the orthogonal wave functions are called orthonormalwave functions. Mathematically

∫ +∞−∞ Ψ∗m(x, t).Ψn(x, t)dx = δmn. Where

δmn is called as Kronecker delta and is defined as follows:

δmn =

{1, m = n

0, m 6= n



Complete (closed) wave function

• It is a wave function whose value at one point is independent of its valueat any other point.∫ +∞

−∞ψ∗m(x).ψn(x′)dx =

{0, x 6= x′

1, x = x′

• If∫ +∞−∞ ψ∗m(x)ψn(x′)dx = 0 for m 6= n and x = x′, then the wave function

ψ is said to be complete as well as orthonormal.



Complete (closed) wave function

According to this theorem, a single-valued, continuous, finite andnormalizable orthogonal wave function Φ can be expanded as series oforthogonal wave functions ψn of same variable.Mathematically: Φ =

∑anψn(x), where an are known as co-efficients and

ψn are known as basis set.Φ = a1ψ1 + a2ψ2 + .........+ anψn.Multiplying the above equation by ψ∗n and integrating over the entirespace, we get∫ +∞−∞ ψ∗n.Φdx =

a1∫ +∞−∞ ψ∗n.ψ1dx+ a2

∫ +∞−∞ ψ∗n.ψ2dx+ ............+ an

∫ +∞−∞ ψ∗n.ψndx

Again we know that:∫ +∞

−∞ψ∗m(x).ψn(x)dx =

{1, m = n

0, m 6= n.So an =

∫ +∞−∞ ψ∗n.Φdx


Expectation values

Expectation Values

• When we perform an experiment, we get a definite value and not aprobability. But the predictions of quantum mechanics are in terms ofprobabilities. So, then, how do we associate these predictions withexperimental measurements?

• This is where the concept of expectation values comes into the picture.We will derive the expectation value from the wave function and compareit to experimental results.• Expectation value of an observable (< A >) is defined as the average oflarge number of measurements on the observables (A) and it is

mathematically written as: < A >=

∫ +∞−∞ Ψ∗ A Ψ dx∫ +∞−∞ Ψ∗.Ψ dx

.

If Ψ is normalized, then∫ +∞−∞ Ψ∗.Ψ dx = 1, then the above equation

becomes: < A >=∫ +∞−∞ Ψ∗ A Ψ dx .


Expectation values

Expectation Value of known operators

• The expectation value of the position of a single particle is:< x >=

∫ +∞−∞ Ψ∗ x Ψ dx =

∫ +∞−∞ x |Ψ|2 dx.

• The expectation value of momentum of a particle in 1D is:< p >=

∫ +∞−∞ Ψ∗ p Ψ dx =

∫ +∞−∞ Ψ∗(~i

∂∂x ) Ψ dx = ~

i

∫ +∞−∞ Ψ∗ ∂Ψ

∂x dx, since

the operator form of p = ~i∂∂x .

• The expectation value of energy of a particle is:< E >=

∫ +∞−∞ Ψ∗ E Ψ dx =

∫ +∞−∞ Ψ∗ (i~ ∂∂t ) Ψ dx = i~

∫ +∞−∞ Ψ∗ ∂Ψ

∂t dx, since

the operator form of E = i~ ∂∂t . Here in all cases we assume that Ψ isnormalized.

• Problems: If Ψ = eikx√b−a then over the limit a to b calculate the

expectation value of (i) position (< x >), (ii) momentum (< p >) and (iii)kinetic energy (< T >).


Equation of Continuity or Conservation of probability

Conservation of Probability density or Equation ofContinuity

Probability density is defined as: P (r, t) = Ψ∗(r, t)Ψ(r, t). Then taking theintegration of the equation over all space, we get⇒

∫V P dV =

∫V Ψ∗(r, t)Ψ(r, t) dV . Then differentiating this equation

with respect to t we get:∫V

∂P∂t dV =

∫V (Ψ∗ ∂Ψ

∂t + Ψ∂Ψ∗

∂t ) dV ......(1)Now we are going to simplify the RHS of eq-(1) using Schrodingerequation.The time dependent Schrodinger equation is:i~∂Ψ

∂t = − ~2

2m∇2Ψ + UΨ............(2a).

Multiplying Ψ∗ on both sides of above equation:i~Ψ∗ ∂Ψ

∂t = − ~2

2mΨ∗∇2Ψ + Ψ∗UΨ............(2b).Taking complex conjugate of eq-(2a) we get:−i~∂Ψ∗

∂t = − ~2

2m∇2Ψ∗ + UΨ∗............(3a).

Multiplying Ψ on both sides of above equation, we get:


Equation of Continuity or Conservation of probability

Continued.....

−i~Ψ∂Ψ∗

∂t = − ~2

2mΨ∇2Ψ∗ + ΨUΨ∗............(3b).Then subtracting eq-(3b) from eq-(2b) and then doing somesimplification, we get:(Ψ∗ ∂Ψ

∂t + Ψ∂Ψ∗

∂t ) = i~2m (Ψ∗∇2Ψ−Ψ∇2Ψ∗).

Then eq-(1) implies:∫V

∂P∂t dV =

∫V

i~2m (Ψ∗∇2Ψ−Ψ∇2Ψ∗)dV

Using the Green’s theorem:Ψ∗∇2Ψ−Ψ∇2Ψ∗ = ∇.(Ψ∗∇Ψ−Ψ∇Ψ∗), theabove equation becomes:

∫V

∂P∂t dV = − ~

2im

∫V [∇.(Ψ∗∇Ψ−Ψ∇Ψ∗)] dV

⇒∫V

∂P∂t dV = −

∫V (∇.J) dV =

∮S J.dS ............(4) using Gauss

divergence theorem.where J = ~

2im (Ψ∗∇Ψ−Ψ∇Ψ∗) is called as current density.

⇒∫V [∂P∂t + (∇.J)] dV = 0 So we can also write ∂P

∂t + (∇.J) = 0 which is

the equation of continuity. As from the normalization condition, it is clearthat:

∫V

∂P∂t dV = 0. Here dV is arbitrary and can not be zero, so ∂P

∂t = 0

and that implies P is conserved.


Documents

Unit-IV: Wave Mechanics (Part-I) - ii ( unit... · 2020. 5. 18. · Concept of Modern Physics, by Arthur Beiser, Tata McGraw-Hill, 6th edition (2003). Quantum Physics for Scientists