For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

### 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
- Postulate I
- Example I: Normalize This Wave Function
- Postulate II
- 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
- 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.

- 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

### Physical Chemistry Online Course

### 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 crossed*1767

*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

1 answer

Last reply by: Professor Hovasapian

Thu Feb 23, 2017 1:00 AM

Post by Kerwin Clement on February 15 at 09:18:43 PM

Is there an Eigenfuction for the position and momentum operator that would generate a Gaussian function.

1 answer

Last reply by: Professor Hovasapian

Mon Sep 12, 2016 1:40 AM

Post by s n on September 11, 2016

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.