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Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Mon Sep 12, 2016 1:40 AM

Post by s n on September 11 at 02:31:08 PM

As usual, fantastic lecture. Just wanted to point out one typo at 22:21. The h bar should be squared in the Kinetic Energy operator. Really clear explanations. Thank you so much.

1 answer

Last reply by: Professor Hovasapian
Wed Dec 30, 2015 12:49 AM

Post by bohdan schatschneider on December 26, 2015

Do you cover the idea of a well behaved wave function in other videos?  Its usually covered along with the 1st postulate.

The Postulates & Principles of Quantum Mechanics, Part I

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 0:00
  • Postulate I 0:31
    • Probability That The Particle Will Be Found in a Differential Volume Element
  • Example I: Normalize This Wave Function 11:30
  • Postulate II 18:20
    • Postulate II
    • Quantum Mechanical Operators: Position
    • Quantum Mechanical Operators: Kinetic Energy
    • Quantum Mechanical Operators: Potential Energy
    • Quantum Mechanical Operators: Total Energy
    • Quantum Mechanical Operators: Momentum
    • Quantum Mechanical Operators: Angular Momentum
    • More On The Kinetic Energy Operator
  • Angular Momentum 28:08
    • Angular Momentum Overview
    • Angular Momentum Operator in Quantum Mechanic
    • The Classical Mechanical Observable
    • Quantum Mechanical Operator
    • Getting the Quantum Mechanical Operator from the Classical Mechanical Observable
  • Postulate II, cont. 43:40
    • Quantum Mechanical Operators are Both Linear & Hermetical

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

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

We have just finished discussing the idea of a particle in a box and we have taken a look at some basic quantum mechanical systems.0006

And we have taken a look at some of the properties and integrals, and things like that.0013

In the next couple of lessons, what I'm going to do is I'm going to actually formalize what it is that we discussed in the last few lessons.0017

We are going to talk about the Postulates and principles of quantum mechanics.0024

Let us jump right on in.0029

When we talk about a postulate, you can also call it an axiom if you will.0032

What we are doing is we are taking a look at this entire body of data and we are saying let us take a look at how successful it has been.0038

Instead of developing it one way, this idea of quantum mechanics, 0050

let us just begin with some postulates that are confirmed by a whole many years of data and we will use that as our starting point.0053

For our first course in quantum mechanics, that is actually the best way to go.0063

I would not read more into that, this is just sort of a formality of quantum mechanics.0068

We lay these out as a bunch of axioms just like what we use in geometry, there was the axiom of parallel lines.0073

There was the axiom of the shortest distance between two points as a straight line.0080

Certain things that we need, in order to develop the rest of the theory.0084

These are going to be the postulate in quantum mechanics.0087

Number 1, the state of a quantum mechanical system is completely determined by its wave function, also called the state function.0093

Just like the PV= nrt, we call it an equation of state.0103

Ψ is the wave function but it is an equation of state.0107

It tells us what state the quantum mechanical system happens to be in.0109

This wave function is a function of the particles coordinates.0114

If we are talking about a particle in a 1 dimensional box, it is going to be a function of X.0117

A particle in a 2 dimensional box, it s XY.0121

A 3 dimensional box, it is XYZ and so one.0125

However many coordinates if you are talking about a 5 space quantum mechanical system.0128

Who knows?0133

It is going to be XYZST, something like that.0134

All information about the system can be extracted from the ψ.0138

This is very important.0141

All the information that we need about a system can be extracted from ψ.0146

The ψ * ψ conjugate × ψ DX DY DZ is the probability that the particle will be found 0151

in a differential volume element located at the point XYZ.0160

We have seen this before.0164

Remember when we said that ψ conjugate × ψ , we call this,0166

when we multiply these functions together we get something called the probability density.0170

When we multiply the probability density × either a length or an area or a volume, we get the probability.0176

This DX DY DZ is just a differential volume element.0185

At a given point XYZ, let us just say in the center of the box, this DX DY.0194

I will do the DY here and DZ here, this is just a differential cube.0202

It is the probability that the particle will be found in that particular little region in space or 0207

in the interval or in the square, whatever it happens to be.0214

Once again, ψ * × ψ × DX DY DZ is the C at the probability that 0219

the particle will be found in a differential volume element DX DY DZ located at that particular point XYZ.0225

A lot of this will make sense when we start doing the problems.0234

A lot of these lessons are going to be a lot of theoretical discussion.0239

We are going to lay out the theory and then we are going to do the problems in one big swoop.0242

Let us see what we have got.0247

This thing that we just we wrote, the ψ *, ψ, DX, DY, DZ, we will often be written this way.0254

I often do it this way, we will just call it DV, the differential volume element.0262

This DV and the DX DY DZ, they are the same.0268

This is just the theoretical description, differential volume element, DX DY DZ when we actually have to integrate this function, 0272

we integrate one variable at a time with respect to X, with respect to Y, with respect to Z.0280

When we integrate this, this is going to be triple integral.0285

If we have a DX DY it is going to be a double integral.0288

We use the breaking down of the individual quarters when we have to run a calculation.0292

The two dimensional version is going to be, just for completeness, that 2D version is and we do DA,0297

Where DA is a differential area element and that is equal to DX DY.0315

That is what is going on here.0321

A little differential area where this is DX and this is DY, if we happen to be talking about 2 dimensions.0323

And of course, the 1 dimensional case.0329

The one dimensional case, what we have already seen several ×.0332

This is just DX, where DX is some differential length element.0337

Now, because this ψ * ψ is a probability density or if you want to just refer to it as probability, that is fine.0345

As a math function, it must satisfy the following properties.0370

We are dealing with mathematical functions here, and certain things have to be the case.0399

The first one, the integral of this ψ * ψ DV has to equal 1, when we integrate it over the entire space that we happen to be concerned about.0403

If it is some 3 dimensional box that is length 1 ×2 × 3, that entire space is integrating from 0 to 1 in the X coordinate, 0417

0 to 2 on the Y coordinate, 0 to 3 on the Z coordinate, our entire space.0426

When we integrate that, it has to satisfy that the integral of this thing is equal to 1.0432

Here is why, we said that this thing right here is the probability of finding it in a particular differential element.0438

If I'm integrating over the entire space, I know that I'm going to find it somewhere in that space, I know that.0447

Because I am 100% sure that I'm going to find it somewhere in that space, 0457

when I add up all the probabilities it is going to end up being 1 is just the same as 100% or 1.00.0461

When you find the probability you are always going to get some decimal which is going to be less than 1.0469

When you add up all the probabilities for a particular situation, you are going to get 1.0473

That is all these means.0477

Since this is a probability, adding, integrating all the probabilities gives you 1.0479

This is called the normalization condition.0484

This is the case because the probability of finding the particle somewhere in our region of interest is 100% or 1.00, that is all that means.0486

A little bit of a notation here.0528

When we are discussing theory, we are going to use a single integral as the symbol.0531

Now of course, this is a differential volume element which means DX DY DZ.0539

In practice, again, that is going to be a triple integral.0543

But in theory, when we are just discussing the theory of it, we are going to use a single integral symbol.0547

I hope it does not bother you.0552

I will just write that down.0559

When discussing theory, we will use a single integral symbol.0564

In 3 dimensions it will look like this thing, will look like this.0591

We have ψ * which is a function of XYZ × ψ which is a function of XYZ × DXDYDZ = 1.0598

This thing in 3 dimensions is actually is this thing is all we are saying.0613

Once again, wave functions, the ψ , that satisfied this property, that satisfy the integral of ψ * ψ = 1 are said to be normalized.0627

That is the normalization condition.0650

Recall, that we actually use this property of normalization to find the normalization constant.0656

If any function satisfies a particular equation, any constant × that function, will also satisfy the differential equation.0662

We want this to be true so we can multiply our function by any constant we want to make this true, to make it equal to 1.0672

We use this property to find that constant.0681

That is what we did in previous examples.0684

In fact, let us go ahead and do one now.0686

Let us see, so we shall see later that the wave function for a particle moving in a circle is given by the following.0691

Ψ of θ = B sub ML × E ⁺IM θ, where θ goes from 0 to 2 π particles moving in a circle.0697

Normalize this wave function.0707

We want to normalize this wave function.0708

We know the normalization condition is what we just wrote.0711

We know it is going to be this × that = 1.0715

Our ψ = B sub ML E ⁺IM sub L θ.0722

This implies that the ψ sub * which is a conjugate, in this particular case, it is not the same as ψ .0733

This is IME ⁺IM θ.0741

B sub M sub L E ⁺IM sub L θ.0745

That is the conjugate.0751

The conjugate is + I – I.0752

In this case, these are actually different.0754

Our normalization condition of ψ * × ψ is equal to,0759

When we are doing over the entire space,0765

In this particular case, our θ is going to go from 0 to 2 π.0766

When I multiply, the ψ * ψ , I’m just multiplying this function and this function,0771

I'm going to get this B sub M sub L² and this is going to be E ⁻IM sub L θ,0778

That is this one × E ⁺I M sub L θ that is this one, right this one × this one D θ,0788

We want it to equal 1, that is the normalization condition.0801

I'm sorry, E ⁻IM θ × E ⁺IM θ cancels and becomes E⁰.0812

It is going to be this + this which is E⁰ which is going to be 1.0818

What we are left with is the integral of B sub M sub L² D θ.0822

I will pull the constant out, BM sub L² × the integral D θ.0832

Let me not forget my upper and lower limit of integration.0838

This is equal to B sub M sub L² × θ 0 to 2 π and this is equal to B sub M of L² × 2 π.0844

And I know that this equals 1, that is my normalization condition.0862

I set up the integral, I just set this up by literally put this in.0866

Ψ is equal to this, ψ * is equal to that, I multiply them together, I put them in and I solved the integral and I just set it equal to 1.0870

I just solved for the B sub M sub L.0877

Do I have another page here?0880

I can go ahead and do it on this page.0885

When I solve for the B sub M sub L, I get the following.0888

B sub M sub L, the normalization constant is equal to √1/ 2 π or to the ½ power, 0891

this is my normalization constant.0904

Let us go ahead and do it on the next page.0921

Ψ sub M sub L is equal to,0925

We have a normalized wave function = 1/ 2 π ^½ E ⁺I M sub L whatever that happens to be. 0928

Θ is now a normalize wave function.0941

You always do the same thing, just use the normalize condition.0954

It is a normalize wave function that satisfies, that equal to 1.0956

I just happen to use that particular condition to find B.0967

Once I found B, if I put this as a function itself into here, I'm going to get 1.0970

That is what I did.0978

That was the first property that the wave function has to satisfy.0981

The other properties, I’m just going to go ahead and list it.0985

They are just for mathematical completeness.0987

Both the ψ and ψ *, they must be continuous functions, at least over our main of interest.0990

Let us go ahead and do it this way.1010

Actually ψ and ψ prime, the function and its derivative, they must be continuous.1023

And ψ and its derivative must be finite.1029

At a particular region of interest, the function cannot go to infinity.1035

That is not going to work when we can integrate something like that.1044

For ψ and ψ prime, they must be single valued.1048

You cannot have some inverse trigonometric function where are you going to get this the sort of periodic behavior, 1063

or it is going to be multiple valued.1072

Remember, inverse trigonometric functions like the inverse sin function, that is going to end up looking like this, XY axis.1073

It is a multiple value, that is not going to work.1083

These have to satisfy these conditions in order for it to be a viable wave function.1087

Let us go ahead and go on to Postulate number 2.1100

For every observable quantity in classical mechanics that correspond a linear hermitian operator in quantum mechanics.1103

Here is what happening, in classical mechanics we have some particle.1115

That particle has a velocity, that particle has a momentum, that particle has a kinetic energy, 1119

that particle has an angular momentum, whatever it happens to be.1126

That particle has a potential energy.1130

We have certain formulas for that, like the kinetic energy ½ the mass × the velocity².1133

For its momentum, we have its mass × its velocity.1139

These are the observables in classical mechanics, things that we observe.1143

You know we can see it has this energy.1147

We can see that it has this momentum.1149

In quantum mechanics, to each observable kinetic energy, momentum, angular momentum, position, whatever it is,1151

They correspondence an operator in quantum mechanics.1158

That operator happens to be linear and also happens to be hermitian.1162

Because here is what we are doing.1167

What we are doing is we are taking a look at particles not as particles.1168

We are taking a look at them as waves.1172

Because they are waves, we find a wave equation and that is what we do.1174

We find the Schrӧdinger equation for the system.1178

We solve the Schrӧdinger equation, the differential equation and we end up with this function that represents the particle.1180

An electron in this particular state satisfies this wave equation.1186

I have this wave equation, I do not have the particle anymore like in classical mechanics and classical physics.1190

I have this wave function that represents the particle.1197

It is just a mathematical function.1200

In order to find things like the kinetic energy of the particle, the momentum of a particle, the angular momentum, the position of the particle,1203

Whatever it happens to be, I cannot do it directly.1211

What I have to do is I have to operate on the wave function that represents the particle and that is what we are doing,1216

That is what these operators are.1222

They are the wave function that is representing the particle.1225

Therefore, if I want to extract information about that particle, whatever it is that I want, I have to operate on it.1227

Every single observable kinetic energy and potential energy, momentum, in classical mechanics that we know, 1235

that we are familiar with, in quantum mechanics they are represented by operators.1241

And those operators happen to be linear and hermitian.1245

Let us go through... I will list these operators just to familiarize ourselves and we will of course discuss them further.1249

Here is a table of quantum mechanical operators.1255

The first one, the position operator.1258

The position operator, operator always have that little hat symbol.1260

This is symbolized by X hat, this is the one dimensional version.1263

The operator in the action, in other words what the operator does when you see X.1267

For example, if you see X of ψ it means do this to ψ.1275

In this particular case, it is multiply ψ by X.1281

That is what we mean by the operator in the action.1285

The action itself is multiplied by X.1287

In a 3 dimensional case, the operator is represented by R, the position vector.1290

And the same thing, the action of this operator is just multiply the wave function by R.1298

In other words, if you had ψ out here, if you are operating on a function ψ, you are going to do this to it.1305

Remember, operators distribute.1313

We have already seen that in previous lessons and we will see more of it.1314

The kinetic energy, in classical mechanics, you know that kinetic energy is ½ the mass × the velocity².1318

In quantum mechanics, the operator kinetic energy is just K hat on it.1326

And what I'm doing to the ψ is this thing.1330

I'm taking the second partial with respect to X.1334

I’m adding the second partial with respect to Y.1336

I'm adding the second partial with respect to Z.1338

Multiplying by –H ̅/ 2M, where M is the mass of the particle.1341

This is the operator, this is the operator.1346

The action is the operator itself.1349

What you are doing is you are just a differentiating the function and you are multiplying.1352

That is what operators is.1357

It says do this to the wave function.1358

The potential energy, symbolized by V means just multiply the wave function by V, whatever V happens to be at that particular XYZ.1363

The total energy function, the total energy you know in classical mechanics, 1378

total energy is just equal to the kinetic energy + the potential energy.1382

That is the same thing here.1386

Take the kinetic energy operator and add the potential energy operator.1389

When you are applying this to some wave function, you do all of this and then you add, multiplying it by V, that is all you are doing.1393

The momentum operator, the momentum is mass × the velocity, that is the classical mechanics.1403

In quantum mechanics, I have momentum.1410

Notice, momentum is a vector, this is bolded out.1413

This is a vector operator.1416

I'm going to just apply this operator to the wave function and whatever it is I get, that is what I get.1418

The angular momentum, I’m going to discuss the angular momentum a little bit further 1427

but I want you to actually see what it is here in the table.1433

Angular momentum is also a vector operator.1437

You remember the angular momentum L in classical mechanics is equal to the position vector 1441

with a cross product of the linear momentum, R crossed P.1446

A cross product of two vectors gives you another vector.1451

This is an actual vector operator.1455

Basically, because they are a little involved we decided to break them up into components.1458

The X component of the angular momentum is this.1464

The Y component of the angular momentum is this and the Z component of the angular momentum is this.1468

Again, this is just a reference for you to see these operators.1473

All this will make sense when we start doing the problems.1476

Let us go ahead and move forward here.1482

We go back to blue.1489

In classical mechanics, the potential energy is ½ the mass × the velocity of the particles².1494

This is observable.1504

When you take a measurement, you observe something.1507

In quantum mechanics, the kinetic energy operator happens to be –H ̅/ 2M Del².1511

This is a linear operator, a linear hermitian operator.1524

I'm going to talk about hermitian a little bit later.1527

For right now, I will just call it a linear operator.1532

If I want information about the kinetic energy of a quantum mechanical system, the process begins by operating on the wave function.1537

But operating on the wave function which is ψ with the kinetic energy operator.1573

Again, I cannot stress this enough.1588

The biggest problem the kids have in quantum mechanics is conceptually, 1589

I mean I understand that the biggest problem that you are going to have in quantum mechanics is going to be the biggest problem that you have 1596

with any of these particular math or science courses is going to be the mathematic.1600

Sometimes, it just gets involved.1603

In quantum mechanics, the mathematics tends to be really daunting.1605

It is not difficult mathematics.1608

It just tends to be symbolically intense, that is all.1610

It has a lot of symbols on the page and that tends to be very intimidating.1613

Conceptually, quantum mechanics is not altogether that difficult, it has this history.1617

This reputation, I should say of being this bizarre way of looking at the world.1624

It is not, all you are doing is you are saying instead of treating a particle like this object that I can touch and see,1629

I'm just taking that particle and representing it with a mathematical function which happens to be a wave.1634

I'm just looking at the particle as a wave.1639

Because now I'm not dealing with a particle itself, I’m dealing with a wave, in order to get information 1643

about that particle I have to do something to the wave function.1648

That is what these operators are.1653

They are just things that we are doing to the wave function to extract information about what it is that we want.1655

In the case of a kinetic energy, while I’m treating a particle like a wave I want to find out the kinetic energy of the particular particle.1663

Since, I have a wave function what I have to do is I have to operate on that wave function with a kinetic energy operator.1670

If I want to know something about the linear momentum, 1676

I operate on the wave function with a linear momentum operator and I get whatever information that I need.1679

That is all that is going on with quantum mechanics, that is all.1685

Let us go ahead and talk a little bit about angular momentum.1691

Angular momentum.1702

Any particle moving along the curve path relative to a fixed point has an angular momentum.1708

For our purposes, mostly it is going to be some particle moving in a circle.1738

It has an angular momentum.1741

Angular momentum is a vector.1749

Again, I’m not saying anything here that you have not seen in your previous physics course.1754

Angular momentum is a vector and it just might have been awhile since you have seen it.1761

L, it is the position vector crossed with the linear momentum which happens to be the position vector crossed1767

and linear momentum is just the mass × the linear velocity of that particle.1777

I will go ahead and write a couple more equations here.1784

The magnitude of the vector is equal to the magnitude of the position vector × the magnitude 1786

of the momentum vector × the sin of the angle in between them.1798

I will draw a picture in just a second.1803

And of course for P, this thing right here, I will go ahead and write P itself.1808

It is equal to the momentum in the X direction I, the momentum of the Y direction J, the momentum in the Z direction × K.1813

That is it, we are just breaking a vector up into its components XYZ, IJK.1824

You have seen all of this before, I hope.1830

In this particular case, what we have is let us say this is our fixed point, let us say this is our R, this is our object.1834

Let us say at a particular moment it happens to be moving in this direction.1845

This is our fixed point and let us say this particle is actually moving along some curved path.1850

This is our R, this is our distance from a fixed point.1856

This is going to be its momentum which is equal to the mass × the linear velocity.1863

Θ is this angle right here.1871

Carry out this and drop this here, that is what the θ is.1876

All of these equations apply to this particular system.1881

Again, the angular momentum happens to be the R cross the P.1887

The angular momentum operator in quantum mechanics.1896

The angular momentum operator is exactly the same thing, operator in quantum mechanics.1904

It is the exact same thing except now it is an operator.1915

The symbol is has a little hat, R that cross P.1917

We have the R operator, multiply by R, we have the momentum operator 1925

which that linear momentum operator thing that happens to be the angular momentum operator.1931

Let us go ahead and next go through the process of finding the classical mechanical observable.1941

What this R crossed P, let us go through the process of finding it in terms of its components and 1949

we will go ahead and do the same thing for the angular momentum operator.1955

We are going to do the classical mechanical version and we are going to do the quantum mechanical version, the operator version.1959

I just want you to go ahead and see it.1965

Let me go back to black.1967

The classical mechanical observable which is the angular momentum.1981

In vector calculus, in fact in calculus the cross product of two vectors is symbolically represented as follows.1992

We say symbolically because this just gives us a symbolic way of actually calculating what it is.2012

It is symbolically represented as follows.2020

Whenever we have a vector A cross a vector B that is going to equal the determinant A sub X A sub Y A sub Z, B sub X B sub Y B sub Z.2030

It is this determinant.2048

We take care of that 3 × 3 determinant and that is how we actually find what A cross B is.2051

In the case of L which is equal to R cross P, that is going to equal IJK and then this is going to be just XYZ.2060

R is just the vector of the position and the momentum is just PX PY PZ.2074

We do that determinant and then when we work this out, we get the following.2084

We expand that way so we end up with YPZ – ZPY, that is the I version.2090

We expand this way.2106

Remember, it is alternating + - + - , this is going to be - XPZ - ZPX that is going to be J.2110

And of course, I’m going to expand third column that way.2124

This is going to be + XPY -YPX and that is going to be K.2127

When I put this all together, I get the following.2136

This vector is equal to this thing.2139

I have a YPZ - ZP sub Y, this is the I.2145

I'm going to do +, I’m going to switch these.2154

I'm going to distribute this negative sign with this and that.2156

I’m going to switch these.2158

It is going to end up being ZPX – XPZ, this is going to be the J.2160

This is going to be + the third one which is XPY - YPX and this is going to be K.2169

This is the component of my angular momentum in the X direction.2179

This is my component of angular momentum in the Y direction.2184

I will do this in red.2188

This is my component of angular momentum in the Z direction.2192

This is the classical mechanical representation.2197

If I were to expand everything, that is this.2199

If I want to find the angular momentum vector, this is my angular momentum vector based on the components of A and B.2205

Let us take a look at a quantum mechanical operator.2222

I just want you to see where it came from.2224

For the quantum mechanical operator.2231

L is equal to R cross P.2237

R cross P is equal to IJK.2247

This time we have the position operator X, position operator Y, position operator Z.2254

Again, it is the same exact thing symbolically for the classical mechanical version.2260

Except now, they are just operators, little hats.2265

And I have the X component operator, the Y component operator of the momentum, and I have the Z component operator of the momentum.2269

I will go ahead and I take that determinant.2280

This determinant happens to equal IJ and K.2283

This is the same XYZ, now we have a linear momentum operator for the X direction.2293

You already know what that is, that is - I H ̅ DDX.2301

For the Y it is - I H ̅ DDY and this is - I H ̅ DDZ.2308

When I expand this out, our first column, second column, third column, I get the following.2319

I get L is equal to - I H ̅ Y DDZ – Z DDY I + -I H ̅ Z DDX - X DDZ J + - I H ̅ × X DDY – Y DDX.2325

If you notice this is just the entry, this is just the X component of the linear operator in the table.2368

This is the Y component in the table and this is the Z component of the angular momentum operator in the table.2385

We just listed in the table that way because it is along that way.2396

We just decided to list them as the individual component.2399

It is important to remember that it is a vector operator which is why we have the IJK.2404

One final thing about this.2418

We will say notice, you could have gotten the quantum mechanical operator directly from the classical mechanical version observable 2419

by simply using the operator version in place of the actual X component of the momentum.2466

You could have gotten the quantum mechanical operator directly from the classical mechanical observable 2491

by simply replacing the PX in the classical mechanical version with the operator PX of the quantum mechanical version.2496

That said operator in place the same for Y and Z.2507

Remember, for the X component of the angular momentum classical mechanical, we had YPZ - ZPY in the I direction.2519

We just convert everything.2539

We just put the quantum mechanical operators right in this classical mechanical equation which is what we actually do all the time.2540

We have LX operator = Y × the PZ operator - Z × PY operator in the I direction.2548

The PZ operator is just – IH DDC and the PY operator is – I H ̅ DDY.2566

When I pull out the – H, I get - I H ̅ Y DDZ – Z DDY.2585

That is the quantum mechanical operator I do the same thing for the Y and Z.2605

Again, if you happen to have a classical mechanical version of equation, just replace everything with 2610

a quantum mechanical operator and everything should fall out.2615

Let us finish off.2623

Postulate 2, says that quantum mechanical operators are both linear and hermitian.2625

Let us recall the definition of linear.2658

Linear says that if I take an operator and if I operate on some function + another function,2670

In other words, if I add the two functions and then operate on it, 2677

I'm going to get the same answer as if I operate on the first one and add having operated on the second one.2682

The second part of that is, this ψ × ψ 1.2691

I will go ahead and just write it out and explain what it means, ψ L ψ 1.2696

Here L does not represent the angular momentum operator.2709

Here L just represents a generic operator.2712

Linear means, if I add two functions first and then operate on what I get when I add them, 2715

I end up getting the same answer as if I operate on the first and then add then operate on the second and add them.2723

You can add them first then operate or I can operate on them and then add.2729

You get the same answer.2733

The second part of that, if I multiply some function by a constant and operate on it, 2736

it is the same as operating on the function and the multiplying by the constant.2741

You are accustomed to seeing this just in your experience but by no means this is generally true.2746

There are many operators that are not linear.2752

In our case for quantum mechanics, all of our operators are going to be linear and this property is going to be a profoundly useful.2755

As far as the hermitian part is concerned, I'm going to end up talking about hermitian in the next couple of lessons.2763

I will leave that alone for right now.2769

I will go ahead and stop this lesson here and we will continue on next time.2772

Thank you so much for joining us here at www.educator.com.2775

See you next time, bye. 2777