Raffi Hovasapian

The Postulates & Principles of Quantum Mechanics, Part II

Slide Duration:

Table of Contents

Section 1: Classical Thermodynamics Preliminaries

The Ideal Gas Law

46m 5s

Intro

0:00

Course Overview

0:16

Thermodynamics & Classical Thermodynamics

0:17

Structure of the Course

1:30

The Ideal Gas Law

3:06

Ideal Gas Law: PV=nRT

3:07

Units of Pressure

4:51

Manipulating Units

5:52

Atmosphere : atm

8:15

Millimeter of Mercury: mm Hg

8:48

SI Unit of Volume

9:32

SI Unit of Temperature

10:32

Value of R (Gas Constant): Pv = nRT

10:51

Extensive and Intensive Variables (Properties)

15:23

Intensive Property

15:52

Extensive Property

16:30

Example: Extensive and Intensive Variables

18:20

Ideal Gas Law

19:24

Ideal Gas Law with Intensive Variables

19:25

Graphing Equations

23:51

Hold T Constant & Graph P vs. V

23:52

Hold P Constant & Graph V vs. T

31:08

Hold V Constant & Graph P vs. T

34:38

Isochores or Isometrics

37:08

Physical Chemistry The Postulates & Principles of Quantum Mechanics, Part II

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Section 12: Postulates and Principles of Quantum Mechanics: Lecture 2 | 39:28 min

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1 answer

Last reply by: Professor Hovasapian
Tue Apr 7, 2015 10:43 PM

Post by shashikanth sothuku on March 29, 2015

hello prof,
in average value, what is the significance of taking both modulus and squaring, as taking either of these can give a positive value.

1 answer

Last reply by: Josh Winfield
Sat Mar 19, 2016 6:53 PM

Post by Anhtuan Tran on February 1, 2015

Hi Professor Hovasapian,
On the first slide when you were finding the eigenvalue of the K^ operator, you included (2/a)^1/2 in your a sub n. However, since the (2/a)^1/2 is a part of your wave equation, I believe that a sub n shouldn't include that part.
By the way, thank you for your great lecture. I really enjoyed it.

The Postulates & Principles of Quantum Mechanics, Part II

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Postulate III

Postulate IV

Postulate IV

Postulate V

Postulate V

Average Value

Average Value

Intro 0:00
Postulate III 0:09

Postulate III: Part I
Postulate III: Part II
Postulate III: Part III
Postulate III: Part IV

Postulate IV 23:57

Postulate IV

Postulate V 27:02

Postulate V

Average Value 36:38

Average Value

Physical Chemistry Online Course

Section 1: Classical Thermodynamics Preliminaries
	The Ideal Gas Law	46:05
	Math Lesson 1: Partial Differentiation	46:02
Section 2: Energy
	Energy & the First Law I	1:06:45
	Energy & the First Law II	1:06:33
	Energy & the First Law III	1:02:17
	Changes in Energy & State: Constant Volume	1:04:39
	Joule's Experiment	16:50
	Changes in Energy & State: Constant Pressure	43:40
	The Relationship Between Cp & Cv	32:23
	The Joule Thompson Experiment	39:15
	Adiabatic Changes of State	35:52
Section 3: Energy Example Problems
	1st Law Example Problems I	42:40
	1st Law Example Problems II	1:00:23
	1st Law Example Problems III	44:34
	Thermochemistry Example Problems	59:07
Section 4: Entropy
	Entropy	49:16
	Math Lesson II	33:59
	Entropy As a Function of Temperature & Volume	54:37
	Entropy as a Function of Temperature & Pressure	31:18
	Summary of Entropy So Far	23:06
	Entropy Changes for an Ideal Gas	25:42
Section 5: Entropy Example Problems
	Entropy Example Problems I	43:39
	Entropy Example Problems II	56:44
	Entropy Example Problems III	57:06
Section 6: Entropy and Probability
	Entropy & Probability I	54:35
	Entropy & Probability II	35:05
Section 7: Spontaneity, Equilibrium, and the Fundamental Equations
	Spontaneity & Equilibrium I	28:42
	Spontaneity & Equilibrium II	34:38
	The Fundamental Equations of Thermodynamics	30:50
	The General Thermodynamic Equations of State	34:06
	Properties of the Helmholtz & Gibbs Energies	39:18
	The Entropy of the Universe & the Surroundings	19:40
Section 8: Free Energy Example Problems
	Free Energy Example Problems I	54:16
	Free Energy Example Problems II	31:17
	Free Energy Example Problems III	45:00
	Three Miscellaneous Example Problems	58:05
Section 9: Equation Review for Thermodynamics
	Looking Back Over Everything: All the Equations in One Place	25:20
Section 10: Quantum Mechanics Preliminaries
	Complex Numbers	34:25
	Probability & Statistics	59:57
	SchrÓ§dinger Equation & Operators	42:05
	SchrÓ§dinger Equation as an Eigenvalue Problem	30:26
	The Plausibility of the SchrÓ§dinger Equation	21:34
Section 11: The Particle in a Box
	The Particle in a Box Part I	56:22
	The Particle in a Box Part II	45:24
	The Particle in a Box Part III	48:43
Section 12: Postulates and Principles of Quantum Mechanics
	The Postulates & Principles of Quantum Mechanics, Part I	46:18
	The Postulates & Principles of Quantum Mechanics, Part II	39:28
	The Postulates & Principles of Quantum Mechanics, Part III	35:32
	The Postulates & Principles of Quantum Mechanics, Part IV	29:55
Section 13: Postulates and Principles Example Problems, Including Particle in a Box
	Example Problems I	54:25
	Example Problems II	46:58
	Example Problems III	44:11
Section 14: The Harmonic Oscillator
	The Harmonic Oscillator I	35:33
	The Harmonic Oscillator II	43:04
	The Harmonic Oscillator III	26:30
Section 15: The Rigid Rotator
	The Rigid Rotator I	41:10
	The Rigid Rotator II	30:32
	The Rigid Rotator III	35:19
Section 16: Oscillator and Rotator Example Problems
	Example Problems I	33:48
	Example Problems II	46:43
Section 17: The Hydrogen Atom
	The Hydrogen Atom I	40:00
	The Hydrogen Atom II	35:58
	The Hydrogen Atom III	36:18
	The Hydrogen Atom IV	33:55
	The Hydrogen Atom V: Where We Are	51:53
	The Hydrogen Atom VI	51:53
	The Hydrogen Atom VII	34:29
Section 18: Hydrogen Atom Example Problems
	Hydrogen Atom Example Problems I	43:49
	The Hydrogen Atom Example Problems II	1:01:57
	The Hydrogen Atom Example Problems III	48:33
	The Hydrogen Atom Example Problems IV	48:33
	The Hydrogen Atom Example Problems V	48:33
Section 19: Spin Quantum Number and Atomic Term Symbols
	Spin Quantum Number: Term Symbols I	59:18
	Spin Quantum Number: Term Symbols II	34:54
	Spin Quantum Number: Term Symbols III	38:03
	Term Symbols & Atomic Spectra	57:49
Section 20: Term Symbols Example Problems
	Example Problems I	1:01:20
	Example Problems II	56:34
Section 21: Equation Review for Quantum Mechanics
	Quantum Mechanics: All the Equations in One Place	18:24
Section 22: Molecular Spectroscopy
	Spectroscopic Overview: Which Equation Do I Use & Why	50:02
	Vibration-Rotation	59:47
	Vibration-Rotation Interaction	46:22
	The Non-Rigid Rotator	29:24
	The Anharmonic Oscillator	30:53
	Electronic Transitions	1:01:33
Section 23: Molecular Spectroscopy Example Problems
	Example Problems I	33:47
	Example Problems II	1:01:05
	Example Problems III	33:31
Section 24: Statistical Thermodynamics
	Statistical Thermodynamics: The Big Picture	1:01:15
	Statistical Thermodynamics: The Various Partition Functions I	47:23
	Statistical Thermodynamics: The Various Partition Functions II	54:09
Section 25: Statistical Thermodynamics Example Problems
	Example Problems I	48:32
	Example Problems II	57:30

Transcription: The Postulates & Principles of Quantum Mechanics, Part II

Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.0000

Today, we are going to continue our discussion of the postulates and principles of Quantum Mechanics.0004

Postulate number 3, for a quantum mechanical system described by the wave function ψ,0011

whenever a measurement is taken of the observable quantity associated with the operator A,0018

the only values ever observed will be the Eigen values of A.0024

That is the only values observed will be the A sub N that satisfy the equation A of ψ = A sub B of ψ.0029

In other words, if I want to measure the kinetic energy of a particle, the kinetic energy operator.0041

The only energies that I will ever going to find when I take the measurement, by observe, are going to be the Eigen values of the operator.0049

It is very important postulate.0058

Let us go ahead and work a little bit with it.0061

Let us see, I will stick with black.0065

Let us try to measure the kinetic energy of a particle in a 1 dimensional box.0070

What are we going to find?0077

Let us find out.0079

Our kinetic energy operator is – H ̅/ 2 M D² DX².0081

We are dealing with a 1 dimensional box, this is actually just a regular derivative.0099

It is not a partial derivative.0104

I hope you forgive me if I just keep that partial notation.0106

Our ψ sub N for a 1 dimensional box is equal to 2/ A¹/2.0111

You remember sin N π/ A × X, those are the wave functions for the particle in a 1 dimensional box.0119

Let us see what we have got.0128

Let us operate on it.0129

The ψ sub N is equal to.0132

That is fine, let us go ahead and write it out.0145

H ̅/ 2 M and I'm operating on the function 2/ A¹/2.0149

We want to write as much as possible and when you are doing these quantum mechanical problems and0157

calculations because there are these symbols all over the place.0163

You definitely want to write out as much as you can.0166

You will lose your way, that is what everybody does, that is just the nature of the game.0169

Sin N π/ A × X.0175

I’m going to go ahead and differentiate.0186

I’m going to take ψ.0188

I’m not concentrating here so let me do this again.0191

We have D² DX² and we are operating on the function 2/ A ^½ × the sin of N π/ AX.0197

We are going to differentiate this function twice and then multiply by this thing.0221

When I differentiate the first time, I’m going to get N π/ A × 2/ A¹/2 cos N π/ A × X.0225

When I differentiate it the second time, I'm going to end up with N² π² / A² × 2/ A¹/2 and0237

it is going to be negative because the derivative of cos is negative sin.0247

It is going to be sin N π/ A × X.0252

What I end up getting here is, this is going to be equaling - H/ 2 M × N² π²/ A² × 2/ A¹/2.0257

This is actually a negative because that is negative.0280

Sin N π/ A × X and what I end up with is, the negatives cancel, this is H ̅.0283

I get H ̅ N² π²/ 2 MA² × 2/ A¹/2 × the sin of N π/ A × X.0292

And that is equal to the operator.0310

When I operated on it, what I end up with is.0311

This thing right here, that is my ψ and that thing right there that is my A sub N, that is my Eigen value.0315

In this particular case, this wave function it happens to be an Eigen function of the kinetic energy operator.0329

In other words, when I operate on this function with a kinetic energy operator, I end up getting the function back × some constant.0340

This constant is an Eigen value so when I get this, that means this is an Eigen function.0350

Ψ sub N is an Eigen function of the kinetic energy operator.0360

Any measurement that I make of the kinetic energy of the particle in a box is going to end up giving me,0374

What I get is going to be the Eigen value.0382

Any measurement I make of the kinetic energy of the system will give me H ̅ N² π²/ 2 MA² 2 / A ^½,0390

I hope that I have not forgotten some exponents somewhere.0430

Depending on what N is, depending on N.0434

N is the state of the system.0441

Depending on the state of the system whatever N happens to be, remember N can be 123456.0446

Those are the quantum numbers for the particle in a 1 dimensional box.0451

The measurement that I make is going to be different.0455

But whatever measurement I do make, the only measurement that I see,0457

the only thing I'm going to observe is going to be an Eigen value of that particular operator,0461

Because the wave function itself happens to be an Eigen function of the operator.0468

That is what that particular postulate that we just saw says.0473

In this case, let us go to blue.0478

In this case, our wave function for the system ψ happened to be an Eigen function of the operator that we chose,0485

which is the kinetic energy operator, in this case.0509

In this case, our wave function ψ happen to be an Eigen function of our operator of interest which was K.0512

This does not have to be the case.0538

This need not be the case.0542

In other words, you can have a wave function for a particular system and then try to measure the linear momentum of that system.0548

You are going to have to apply the linear momentum operator to the wave function.0558

But when you do that, you are going to discover that the wave function for the system is not an Eigen function of that operator.0563

The question becomes when you measure it, what value you are going to get?0572

We have already seen that when it is an Eigen function that we are just going to get the Eigen values, that is what we are going to see.0576

We are going to deal with the case where it is not necessarily the case.0582

This may not be the case.0585

For simplicity, we just want to deal with some simple functions so we actually see what is going on0589

rather than complicating it where the function is going to end up getting in the way of our understanding.0594

Let me go back to black here.0601

Let us take our wave function ψ equal to the sin of KX and let us choose our linear momentum operator in the X direction.0612

Let us choose that as our operator of interest.0628

We have sin of KX and we have our operator so let us go ahead and write down what it is that we have.0638

We have ψ is equal to sin of K × X and we have the linear momentum operator which happens to be – I H ̅ DDX that is the operator.0646

We are going to operate on this function.0661

Operating on the ψ that is the same as taking the –I H DDX of this sin K of X.0667

This happens to be - I H ̅ K cos of KX.0680

Operating on the function sin of KX gives me - IHK cos KX.0699

Cos is not sin, you did not end up operating on this function and getting some constant × the function itself.0712

You ended up with a different function.0721

Clearly, the sin X function, our wave function here is not an Eigen function of the linear momentum operator.0723

Precisely, because it ended up not being equal to some constant × the function itself.0734

I will write this separately.0756

I do not what to do this here.0759

Let me go to the next page.0763

Clearly, ψ which is equal to sin K of X is not an Eigen function of our linear momentum operator.0769

The question is what value do we get when we measure the linear momentum of this particular wave function of this particular particle?0788

What value do we observe when we measure the momentum of the system, that is the question.0798

Let us see what we can do here.0822

We are going to start playing around with some mathematics here.0825

Here is how we deal with this.0828

There is a function E ⁺IKX, E ⁺I θ.0835

The Euler’s formula E ⁺IKX.0844

The exponential function actually factors into the cos and sin function.0847

This is cos of KX + I × the sin of KX.0851

There is also another function E ^- IKX that is equal to cos KX - I × the sin of KX.0858

These are just basic mathematical relations that exist.0870

Here is what is interesting.0873

The wave function that we have, our wave function sin of KX, it can be written as a linear combination of E ⁺IKX and E ^- IKX.0875

In other words, I can take E ⁺IKX which is this thing and E ⁻IKX which is this thing and0916

I can end up writing our original wave function which a sin of KX as some combination of these two, multiplied by some constants.0923

It is going to turn out that these two functions are Eigen functions of the linear momentum operator0930

and because that is the case, I can extract some information.0937

The wave function of KX can be written as a linear combination of E ⁺IKX and E ⁻IKX.0940

Let us go ahead and see how.0946

By the way, a linear combination of these two functions, a linear combination means some constant ×0951

the first function + some constant × the second function.0960

You might have 4 functions, a linear combination of 5 functions.0966

It is just going to be some constant × that function + another constant × that function + another constant × that function,0969

That is what linear combination means.0977

You are combining them linearly, adding them up.0978

Let us go ahead and see how it is represented.0985

Our sin of KX, I can actually write it as 1/ 2 I × E ⁺IKX - 1/ 2 I × E ^- IKX.0989

In this particular case, my coefficients are 1/2 I and -1 / 2 I, that is my C1 and C2.1008

This sin of KX can be written this way.1021

Let us go ahead and show that that is the case.1023

Let us expand it.1026

This is equal to 1/ 2 I E ⁺IKX.1027

E ⁺IKX is equal to this thing.1031

I’m going to write it all out, multiply out.1034

1/2 I × the cos of KX + 1/2 I × I sin of KX.1038

Let me put it over here.1059

-1/ 2 I × this which is this so we get 1/2 I × the cos of KX - 1/2 I × I × the sin of KX.1061

This and this, they cancel, the I's cancel.1090

I'm left with ½ KX -- + ½ sin KX, I end up with sin KX which is my wave function.1096

The wave function ψ equals the sin of KX can be written as a linear combination of E ⁺IKX and E ⁻IKX.1112

And we just saw that it was.1144

We said that these functions happen to be Eigen functions of the linear momentum operator.1153

We have taken a function that is not an Eigen function of the linear momentum operator but1162

we have expressed it as a linear combination of functions that are Eigen functions of the linear momentum operator.1167

Let us go ahead and prove that that is the case first, before we continue.1174

The linear momentum operator operating on E ⁺IKX is equal to – I H ̅ DDX of E ⁺IKX is equal to - I H ̅ IK E ⁺IKX which is equal to,1179

I × I is -1, - -1 ends up being H ̅ KE ⁺IKX.1196

Clearly, linear momentum operator operating on IKX equals some constant × IKX.1219

E ⁺IKX is an Eigen function of the linear momentum operator and the same thing the other way around.1230

The same thing with when I apply this to E ^- IKX.1236

I’m just demonstrating that these are Eigen functions.1240

- I H ̅ DDX of E ^- IKX and what we end up with is - I H ̅ × - I K × E ^- IKX that equals.1245

I and I is -1.1265

-1 is negative so we end up with –H ̅ KE ^- IKX.1273

Clearly, the linear momentum operator operating on this function gives me some constant × that function.1284

E ⁺IKX and E ⁻IKX are Eigen functions of the linear momentum operator.1296

Even though, ψ = KX is not an Eigen function of the linear momentum operator,1308

it can be written as a linear combination of functions that are Eigen functions.1335

When we measure the linear momentum, sometimes we get H ̅ K, one of the Eigen values, the Eigen value,1362

one of the functions that is an Eigen function.1398

Sometimes, we get – HK.1402

HK and -HK happen to be the Eigen values of the functions that are Eigen functions of the linear momentum operator.1414

That is all that we have done.1423

I hope that makes sense.1429

Let us see what we have got next.1432

Let us go ahead.1434

Postulate 4, this is what we are leading to this particular discussion.1441

When a measurement is being made on an observable quantity corresponding to an operator and1451

the wave function of the system is expressed as a linear combination, also called the super position of Eigen functions of the operator like what we just did.1455

If the wave function itself is not an Eigen function of the operator but1466

it can be expressed as a linear combination of functions that are Eigen functions of the operator,1469

where the ψ sub I or expansion coefficients and they can be complex,1477

The F sub I or the Eigen functions of the operator and each measurement will give an Eigen value of the operator.1481

This is the definition of Eigen function Eigen value.1490

That is the Eigen function of the operator, that is the Eigen value.1495

When you end up taking the operator operating on it and getting some constant × that function itself.1499

In the previous example, we had HK and – HK.1506

There were two of them.1511

The coefficients were the same, 1/ 2I 1/ 2 I.1513

Because the coefficients are the same, when you take the measurement,1518

half the time you are going to end up with HK and the other half of the measurements you get are going to be – HK.1522

In general, if you have more than two functions and if these coefficients are not necessarily equal,1529

there is going to be a distribution of how many times you get an Eigen value.1538

The probability of a given Eigen value appearing as a measure of value is going to be proportional to square of the modulus of the actual coefficient.1542

If the wave function is normalized, then the probability is exactly that squared.1554

This is an expression of what it is that we have actually done.1563

Let me say it again.1567

When a measurement is being made of observable quantity corresponding to an operator and the wave function ψ1568

is expressed as a linear combinations of Eigen functions of the operator or the ψ sub I1574

are the coefficients and the N sub I are the Eigen functions of the operator.1579

Each measurement will give an Eigen value of the operator A corresponding to the particular function.1583

The probability that a given Eigen value appearing as a measured value is going to be proportional1596

to the square of the coefficient, or the square of the model.1601

I just say the coefficient because in case it is a complex number, it is going to be the modulus, the length.1605

If the wave function happens to be already normalized, then the probability is exactly the square of the modulus of the coefficient.1610

Let us go to our next postulate.1621

When we take a bunch of individual measurements, we are going to get this value.1624

What if we want an average value?1632

Let us say we end up taking 1000 measurements?1633

There is going to be some average especially if the values are not the same over and over again.1635

If they have different values, there is going to be some average.1639

Again, sometimes, like we said the probabilities are not necessarily uniform or equal, it is not going to be half and half.1642

Sometimes, it will be 75% of the time you are going to get one value and 25% of the time you are going to get another Eigen value.1648

What is the average going to be when we want to find the average value of many measurements?1655

If ψ is a normalize wave function of a quantum mechanical system in a given state and the average value1662

also called an expectation value of the observable quantity associated with the operator is given by this thing right here.1667

In other words, what we do is we take the function.1676

We operate on a function and then we multiply.1679

What are those that we have got when we have operated on it by the complex conjugate of the function itself and then1682

we integrate that function over our region of interest because that is going to give us a numerical value of the average.1688

We already saw that the wave function for a system need not be an Eigen function of the operator of interest.1695

We already saw that with our sin KX.1701

Something interesting happens with the average values.1706

When that the ψ, the wave function of a given quantum mechanical system is the Eigen value of the specified operator.1711

We began this lesson by taking a look at what individual value we are going to get when we take one measurement.1717

We are talking about an average value.1723

Let us say I do not want to worry about individual measurement, I just want to calculate an average value if I were to take 10,000 measurements.1726

This integral will allow me to do that.1733

Something interesting happens when the average values, when the actual wave function itself is an Eigen function of the operator.1735

It is exactly what you think.1744

It is what we already saw in the beginning.1746

We are going to end up with one value over and over again.1747

We are going to confirm that mathematically.1751

Ψ is the wave function of a given system and we said that the ψ also satisfies, is an Eigen function of a particular operator.1763

It also satisfies the following relation.1783

A of ψ = λ × ψ.1787

Then, our postulate 5, the average value of A we said was equal this, the ψ conjugate × having operated on ψ, this is going to equal to,1795

A of ψ is this so I’m going to put that in here.1818

I’m going to get the ψ conjugate × the λ of ψ.1823

This is now just a constant, it is our scalar so I can pull it out.1828

It equals λ × the integral of ψ* and ψ.1832

Wave function is normalized so this is the normalization condition.1840

This is just λ × 1.1844

I end up with λ.1847

The average value is the actual Eigen value itself.1848

Furthermore, this is saying that the average value, if I take 10,000 measurements I’m going to end up with the Eigen value.1856

Earlier in this lesson, we talked about the individual measurements that we make.1870

An average value, if I take if I take the measurement and if I get 1, 10, 1, 10, 1, 10, 1, 10, 1, 10.1874

The average value is going to be 5.1882

If I do another experiment and I get 5555555, the average value is going to be 5.1884

What is the difference between those two situations?1890

The average value is the same but in one of them, I only got 5 and the other one I got 1, 10, 1, 10, 1, 10 or I could have gotten 110, 29, 38, 47,1893

I still end up with an average of 5.1907

This alone does not really tell me anything just yet.1909

This is gives me an average value.1913

What I need to find is I need to find out the actual variance, the standard deviation and1915

if that equals 0 then that means that this λ, that 5 is going to be the only value that I always get.1921

That is what is going on here.1928

Let us go ahead and calculate some other things.1931

Let us calculate the second moment A sub².1934

That is equal to ψ sub * A² of ψ.1938

We know that the square of an operator is just the operator applied twice.1946

It is going to be ψ sub * A, applied to A of ψ which equals the integral of ψ sub * A hat.1952

This is going to be λ × ψ.1970

The λ comes out so we end up with λ × the integral of ψ sub * A of ψ which equals λ × the integral of ψ*,1975

This is λ ψ.1991

This λ also comes out and ends up being λ² × the integral of that.1994

This is just 1 by the normalization condition because the wave function is normalized.2000

We end up with the λ².2006

Our variance of this particular operators, particular measurement.2009

The variance by definition is the average value of the second moment - the square of the average value, that is equal to λ² - λ² which equals 0,2014

Which means that our standard deviation is equal to 0.2032

The standard deviation is the square root of the variance.2035

A standard deviation of 0, it is a measure.2039

Remember what a standard deviation was, it is a measure of how far in general any individual measurement is from the mean.2043

The fact that the standard deviation is 0 tells you that every measurement is one number.2049

You keep getting λ over and over again, when you take, when you operate.2054

When the wave function itself is an Eigen function of the operator of interest, for an average value you end up just getting the Eigen value.2061

But more than that, it says that is the only value you get every time you take the measurement over and over again.2074

This standard deviation of 0 says there are no other values.2082

There is no average, you are just getting 5555555 over and over again.2085

It is just 5.2091

Let me go ahead and write that down.2096

Let me go back to blue and write this.2113

A variance equal to 0 or a standard deviation equal to 0 means every measurement of an observable,2121

whatever the observable happens to be, whatever operator,2147

every measurement of an observable associated with a generic operator A will give only λ, the Eigen value of A.2150

In other words, there is no average per say of multiple Eigen values.2188

It is the average of the same number over and over again.2192

Let us go ahead and finish off.2197

For systems whose wave functions are not Eigen functions of a given operator that can be expressed as2202

a linear combination of Eigen functions of the operator which is this thing, the average value is this thing.2214

Basically, all I do if I want to calculate an average value, I take the coefficients,2222

I find the modulus, I square them, and I multiply it by the particular Eigen value of that function.2230

And then add to it the square of the next coefficient × it Eigen value.2236

The square of the next coefficient × its Eigen value.2241

That will give me the average value.2244

This shows that the average value of an observable is a weighted average of the Eigen values of the operator,2248

Where the weights are equal to the square of the modulus of the expansion coefficients.2254

If I have some function which is not an Eigen function of my operator of interest and2261

let us say 75% of the time I end up with 2 and 25% of the time I end up with 5.2267

My average value is going to be the coefficient squared of the first function × 2.2278

The coefficients squared of the second function × 5.2287

That is going to end up giving me some average value.2291

It is going to be between 2 and 5 somewhere.2294

That is what this is saying.2296

This postulates that we have discussed in this lesson, they have to do with2299

what the individual values are going to be when I take individual measurements.2304

And what the average values are going to be when I want to calculate an average value and2309

see what it is that I expect to get in general, when I take 1000, 10000, 10,000,000 measurements.2314

That is what is going on here.2321

I do apologize that these individual lessons have not really included a lot of the example problems.2325

Again, I want to present the theory in a continuous way.2330

I want to go ahead and do the example problems in bulk, because when I do that,2338

it is going to allow us a chance to actually review the material rather than presenting some concept, doing example,2342

presenting some concept, doing an example.2349

If some of this is not exactly entirely clear, do not worry about it because we are going to be revisiting all of these and2352

writing things over and over again when we do the examples.2358

Terrific, we will go ahead and stop this lesson here.2363

Thank you so much for joining us here at www.educator.com.2365

We will see you next time, bye.2368

Related Books

Physical Chemistry: A Molecular Approach

Authors: Donald A. McQuarrie, John D. Simon

ISBN: 0935702997

Publisher: University Science Books

Year: 1997

"As the first modern physical chemistry textbook to cover quantum mechanics before thermodynamics and kinetics, this book provides a contemporary approach to the study of physical chemistry. By beginning with quantum chemistry, students will learn the fundamental principles upon which all modern physical chemistry is built. The text includes a special set of ""MathChapters"" to review and summarize the mathematical tools required to master the material Thermodynamics is simultaneously taught from a bulk and microscopic viewpoint that enables the student to understand how bulk properties of materials are related to the properties of individual constituent molecules. This new text includes a variety of modern research topics in physical chemistry as well as hundreds of worked problems and examples."

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