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Lecture Comments (3)

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?

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

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