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

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

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### The Postulates & Principles of Quantum Mechanics, Part IV

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
- The Postulates & Principles of Quantum Mechanics, Part IV 0:09
- Operators can be Applied Sequentially
- Sample Calculation 1
- Sample Calculation 2
- Commutator of Two Operators
- The Uncertainty Principle
- In the Case of Linear Momentum and Position Operator
- When the Commutator of Two Operators Equals to Zero

### Physical Chemistry Online Course

### Transcription: The Postulates & Principles of Quantum Mechanics, Part IV

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

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

*Let us dive right on in.*0008

*We have already seen that operators can be applied sequentially.*0012

*Let us go ahead and write that down.*0019

*We have already seen that operators can be applied sequentially.*0027

*In other words, if I do A, B of some function this just means working from right to left.*0045

*Apply B first then apply it to what you got.*0056

*It means apply B to F first, then apply A to the result.*0059

*In general, AB of F does not equal BA of F.*0085

*In general, if you apply B first and apply A to what you got, it is not the same as applying A and applying B to what you got.*0100

*When it does happen that way, when AB of F, when it does equal BA of F, we say the operators AB commute.*0110

*In other words, A and B are commutative just like the numbers 2 × 4 = 4 × 2.*0145

*When numbers is fine with the operator, sometimes it is, sometimes it is not.*0155

*When it does happen that way, we call it commutative.*0158

*Let us go ahead and calculate a couple of things.*0161

*Let us calculate, pick a couple of operators.*0165

*Let us calculate the linear momentum that of F.*0170

*P ̂ X ̂ of F and then let us calculate X ̂ P ̂ of F.*0181

*Let us see what happens.*0190

*Well P ̂ X ̂ of F = - I H ̅ DDX of XF.*0193

*The X operator just means multiply the function by X.*0207

*When we do that, we get - I H ̅.*0213

*I will go ahead and use it.*0221

*DDX, this is X and this is a function of X so we have to use the product rule.*0222

*This × the derivative of that + that × the derivative of this.*0226

*This is going to be X DF DX, this × the derivative of this + that.*0229

*That is that one, that is P ̂ X ̂.*0239

*Let us go ahead and do X have P ̂.*0243

*X ̂ P ̂ of F is going to equal X ̂ apply to - I H ̅ DDX of F.*0246

*I will go ahead and just do it as a single.*0262

*This is going to be DF DX = - I H ̅ X DF DX, that is that one.*0266

*Notice, this and this are not the same.*0283

*P ̂ X ̂ of F does not equal X ̂ P had of F.*0292

*They are not commutative.*0302

*These two operators, in other words the linear momentum operator, and the position operator do not commute.*0304

*Let us calculate another pair.*0318

*Let us calculate the kinetic energy and linear momentum.*0328

*This is sub X because we are just doing in the X direction.*0337

*Let us do K ̂ P ̂ and let us do P ̂ K ̂.*0340

*I think I'm going to actually go ahead and drop the X.*0348

*I do not think it is going to be very confusing.*0351

*For right now, I’m going to stick with working in one dimension.*0353

*I’m realizing that it is actually too many symbols floating around and it is a little confusing.*0357

*K ̂ P ̂ and P ̂ K ̂, let us see what we get.*0362

*The K ̂ P ̂ of F is going to be – H ̅²/ 2 MA² DX² apply to - I H ̅ DF DX.*0366

*What is in the brackets is this one, P apply to F first and then I'm going to apply K to what I got.*0390

*When I do that, I end up with -I H ̅³/ 2M D³ F DX³.*0396

*That takes care of K ̂ P ̂ of F.*0410

*Let us do the other.*0413

*Let us do P ̂ K ̂ of F.*0415

*That is going to equal - I H ̅ DDX of –H ̅²/ 2 M D² F DX².*0419

*We apply K first to F, that is what is in the brackets and then we are going to apply P.*0435

*When we do this, we end up with I H ̅³/ 2 M D³ F DX³.*0439

*Notice, this and this are the same.*0464

*The kinetic energy operator and the linear momentum operator they do commute.*0468

*K ̂ P ̂ = P ̂ K ̂, this is commutative.*0477

*The kinetic energy operator and the linear momentum operator, they do commute.*0487

*We just solved the linear momentum operator and the position operators do not commute.*0491

*When A and B, when the two operators commute then we know that AB of F = BA of F.*0501

*I'm going to move this over to the left hand side.*0524

*This is going to be A ̂ B ̂ of F - B ̂ A ̂ of F.*0527

*I’m just working symbolically.*0534

*An operator is just the thing that you can treat like a number, = 0.*0535

*Let me go ahead and factor out the F.*0542

*I get A ̂ B ̂ - B ̂ and A ̂ of F = 0.*0545

*What I’m going to do is I’m going to introduce symbol here, 0 ̂ of F,*0564

*Where this 0 ̂ is just the 0 operator and it means multiply by 0.*0572

*Again, we just want to be mathematically consistent.*0583

*We have an operator operating on F, an operator operating on F.*0586

*We know it equal 0 but we want to symbolically be consistent.*0590

*Is the 0 operator meaning multiply by 0, multiply by the number 0.*0594

*This expression right here, let me go back to red.*0609

*This expression right here is called the commutator of A and B.*0614

*Commutator of A and B symbolizes as follows.*0637

*It is A, B and we put in brackets.*0641

*This symbol means AB – BA.*0646

*It is called the commutator of A and B.*0655

*We saw that the commutator of the kinetic energy and the linear momentum operator is equal to 0.*0660

*Because PP - PK was equal to 0, KP = PK.*0679

*We also saw that the linear momentum and the position operator does not equal that*0692

*because the commutator does not equal 0.*0705

*The linear momentum position operator of F - the position operator linear momentum of F equals the following.*0711

*That was - I H ̅ X DF DX - I H ̅ F -,*0725

*When I did this, I got some value.*0748

*When I did that, I got another value.*0749

*When I subtract these two, which is the commutator of that, I'm going to just put those values and then subtract.*0752

*- and - I H ̅ X DF DX, this is a + this is a -, these cancel.*0759

*I'm just left with this = - I H ̅ F.*0773

*I hope that makes sense.*0782

*I have P X bar of F - X ̂ P ̂ of F = this.*0786

*We can write this as, let me factor out the F.*0798

*P ̂ X ̂ - X ̂ P ̂ of F = - I H ̅ I ̂ of F.*0806

*Where I ̂ was being symbolically consistent.*0821

*I ̂ is called the identity operator and it just means multiply by 1.*0824

*It is the identity operator and means multiply by 1.*0839

*In other words, take this function and multiply by 1.*0852

*It is just the way of operator, you have some value.*0854

*Operator operating on a function, operator operating on function.*0859

*We are trying to establish some consistency, multiply by 1.*0863

*This is the commutator.*0871

*P ̂ X ̂ is equal to - I H ̅ identity operator.*0874

*All I have done is move the X out of the way, now I’m just dealing in operator notation.*0886

*The function itself is not irrelevant but from the mathematical perspective, you do not have to deal with it because any function will do.*0893

*It is the operator that is actually important.*0902

*Let me go back to black here.*0910

*We already know that σ, the standard deviation of the position of a particle × the standard deviation of the momentum of a particle is ≥ H ̅/ 2.*0913

*This is the Heisenberg uncertainty principle.*0934

*Standard deviation, uncertainty, that is what uncertainty is.*0937

*A standard deviation gives me a degree of uncertainty as far as quantum mechanics is concerned.*0942

*It represents the uncertainty in the measurement that I make.*0946

*If my standard deviation is 0 that means that every measurement that I make is the same number over and over and over again.*0952

*If I take 1000 measurements, not all the measurements are going to be the same, it is going to be some leeway.*0957

*We have already seen that the uncertainty in a position × the uncertainty in the momentum is going to be ≥ H ̅/ 2.*0963

*In other words, I cannot simultaneously measure the position and*0974

*the momentum of a particle to an arbitrary degree of the precision or accuracy.*0979

*I’m limited.*0985

*As if I'm better if I measure that one really well, then the other I’m not going to know.*0986

*If I measure the other one really well, the other one I'm not going to know.*0992

*That is what this means.*0995

*That is the position and momentum cannot simultaneously be measured to arbitrary degree of accuracy.*0998

*The non commutativity of two operators like the position of linear momentum operator, we saw that they are non commutative.*1045

*Those operators do not commute.*1072

*The non commutativity of two operators and we also saw that the uncertainty in the linear momentum*1073

*and the uncertainty in the position is ≥ some number.*1082

*The non commutativity of two operators and the uncertainty relationship between them is not coincidental.*1088

*We already knew that the uncertainty in the position × the uncertainty in the linear momentum is going to be≥ H ̅/ 2.*1111

*We also saw that those two linear operators, the linear momentum operator and the position operator do not commute.*1121

*That is not coincidental, it is always going to be like that.*1128

*I'm going to write down the general uncertainty relationship, the broadest one.*1135

*This is the uncertainty principle.*1142

*What we found before was actually a special case and we are going to go through it and derive it in a second from the general principle.*1155

*The general uncertainty principle, given two operators A and B, the uncertainty in the measurement of A × the uncertainty*1164

*in the measurement of B is going to be ≥ ½ × the absolute value of the integral of C sub *.*1186

*I cannot draw brackets anymore.*1200

*A B C the absolute value.*1206

*Where σ A and σ B are the standard deviations uncertainties in the measure quantities, corresponding to their respective operators.*1214

*This is the general uncertainty principle.*1263

*It says that if I have a particular wave function and let us say I go ahead and measure some property of that wave function A,*1266

*and then I measure some property of that wave function B,*1278

*I make multiple measurements and end up getting uncertainties.*1283

*I’m going to end up getting standard deviations for those particular measurements.*1289

*If I apply the commutator, in other words if I apply AB of ψ - BA of ψ and I multiply that by the conjugate of ψ.*1294

*And if I integrate that over my region of interest, whatever number I get, the absolute value of that divided by 2, this is the relationship.*1306

*All certainties of those measurements multiplied by each other is going to be ≥ that.*1315

*For operators that do not commute like the linear momentum and position operators, that happened to be H ̅/ 2.*1322

*In other words, I cannot know the position really well and the linear momentum really well.*1331

*I have to choose which one I know or I have to strike a balance between them.*1339

*But for operators that do commute, we saw that the commutator is equal to 0,*1342

*which means that this here, the σ A × the σ B is going to be equal to 0,*1348

*which means that I can measure each to an arbitrary degree of accuracy.*1354

*I can find out the kinetic energy and the linear momentum of a particle as closely as I want to.*1359

*That is what is going on here, that is the uncertainty principle here.*1365

*Once again, the uncertainty of a measurement for one operator × the uncertainty in the measurement for another operator*1371

*is going to be ≥ ½ the absolute value of this integral, where this thing is the commutator.*1380

*This commutator just says apply AB – BA to ψ, that is all that is going on here.*1386

*Let us go back.*1399

*In the case of this linear momentum and position operator.*1405

*Let us take a look of this one, this is a special case.*1411

*Let us use the definition that we just got.*1414

*The σ of P, the σ of X, it says that is ≥ ½ the absolute value of the integral of ψ* ×,*1417

*Apply the commutator 2 ψ DX.*1431

*This is going to equal the absolute value, the integral of ψ sub *.*1442

*We have already found what the commutator of this is.*1450

*It is equal to - I H ̅ × the operator of ψ DX, we found this already.*1454

*The commutator is what we just substituted in.*1465

*It is ½, I forgot the ½ here.*1470

*½ × the absolute value of - I H ̅ and I'm going to go ahead and pull the constant out.*1475

*The integral of ψ sub * I of ψ DX.*1488

*This right here, this identity operator just mean multiply by 1.*1500

*This integral here ends up becoming the integral of ψ* ψ DX.*1506

*The normalization condition guarantees that this is equal to 1.*1513

*This goes to 1.*1518

*What we have is ½ × the absolute value of - I H ̅ × 1.*1520

*This is just the absolute value of –I × the absolute value of H ̅.*1534

*The absolute value of –I is just 1.*1542

*Remember, the complex plane, here is 1, here is I, here is -1, here is –I.*1545

*The absolute value, the distance from 0 to -I is 1.*1552

*Therefore, this is just equal to H ̅/ 2,*1556

*Which means that σ of P × the σ of X is ≥ H ̅/ 2, but we knew that already.*1565

*But we ended up deriving it from the general expression of the uncertainty principle based on the commutator.*1577

*That is all we have done.*1587

*Let me go ahead and close this out.*1591

*Let us go back to black.*1593

*When the commutator of two operators = the 0 operator, then this integral,*1597

*I will go ahead and put the absolute values in.*1608

*Ψ sub * of ABC will equal 0, the commutator = 0.*1612

*It was a 0 operator which means this is just going to be the integral of ψ sub * × the 0 operator of ψ.*1623

*The 0 operator × operator of ψ is just 0.*1631

*This ends up being the integral of 0, it just equal 0.*1634

*The σ of A × the σ of B is going to be ≥ 0 or = 0.*1642

*Your σ A can be 0, your σ B can be 0.*1652

*The two things multiply together it equal 0.*1663

*One is 0 and the other is 0, or both can be 0.*1666

*When you have a standard deviation which is 0, when you have an uncertainty that is 0,*1677

*that means that you know exactly what the value is.*1683

*There is no uncertainty, there is no deviation of the data.*1687

*If you make 1000 measurements, those 1000 measurements are all going to be the same number.*1691

*If they all end up with the same number, your standard deviation is going to be 0.*1695

*If your standard deviation is 0, there is no uncertainty in that measurement.*1700

*I'm not uncertain, I know exactly what it is, it is that number.*1703

*In the case of operators that commute, I can measure each one to any degree of accuracy that I want.*1707

*In the case of the kinetic energy and the linear momentum, I can know the kinetic energy very well.*1717

*I can know the linear momentum very well.*1722

*In the case of operators that do not commute, these uncertainties have to be a certain minimum value.*1724

*They have to be bigger than that.*1732

*In the case of the σ X σ P, they have to be bigger than that.*1734

*In other words, I’m limited.*1741

*If I know the momentum really well, I do not know the position.*1743

*If I know the position really well and accurately, I do not know the momentum.*1747

*They have to have a strike of balance, that is what is going on here.*1751

*In this case, A and B, as observables, can be measured to any degree of accuracy.*1755

*That is the general expression for the uncertainty principle and*1779

*it is based on this thing called the commutator of those two operators.*1785

*Thank you so much for joining us here at www.educator.com.*1792

*We will see you next time, bye.*1794

2 answers

Last reply by: David Llewellyn

Mon Jan 5, 2015 2:21 PM

Post by David Llewellyn on January 2, 2015

Can you explain why |-i| = 1 and not i?