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Example Problems III

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
  • Example I: Good Candidate for a Wave Function 0:08
  • Example II: Variance of the Energy 7:00
  • Example III: Evaluate the Angular Momentum Operators 15:00
  • Example IV: Real Eigenvalues Imposes the Hermitian Property on Operators 28:44
  • Example V: A Demonstration of Why the Eigenfunctions of Hermitian Operators are Orthogonal 35:33

Transcription: Example Problems III

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

Today we are going to continue on with our example problems.0004

Let us jump right on in. 0007

Our first example is, is the function E ⁺to the power – X²/ 2 / the interval for - infinity to infinity is a good candidate for a wave function?0010

Recall that for a wave function to be viable, not any function can be a wave function, it has to satisfy certain properties.0024

In other words, it needs to be what we call well behaved.0045

The wave function to be viable must be well behaved.0047

Well behaved means, one, that the integral of ψ* ψ must converge.0062

By converge, we mean it must reach some finite number.0073

You must get actual number not infinity.0076

You must get 5, 7, 8, 6.2, π, 11 π, whatever it is, that is what we mean by converge.0079

It must converge to a number.0085

In other words, it must be finite.0089

I can get some value, it cannot be improper.0092

Two, that both the wave function and the derivative of the wave function must be finite.0096

The functions can go off into infinity over their particular interval of definition, over their domain.0106

Three, ψ and ψ prime must be continuous and must be smooth, no breaks in the graph.0114

Four, both the wave function ψ and the derivative of the wave function ψ prime, they must be single valued.0133

In other words, just a nice well defined function.0145

It have to be a function.0148

Let us go ahead and see if this particular function, E ⁻X²/ 2 over this interval, actually satisfies these properties.0149

Let us go ahead and talk about the first one.0160

We are going to have to integrate from -infinity to infinity.0161

We have to do this, ψ * × ψ.0168

In this particular case, ψ * is ψ because there is nothing complex in here.0172

There is no I involved.0176

It is E ⁻X²/ 2 × E ⁻X²/ 2 DX.0178

This is a symmetric interval and the function itself is an even function.0187

It is an even function, therefore, I can go ahead and write this as the integral from to infinity.0195

It is symmetric about the Y axis so it is not a problem.0200

I can just take half of it and just multiply by 2 to get the other half.0206

I do not have to do this -infinity to infinity thing.0209

E ⁻X² DX.0213

This definitely does converge.0227

Let us go ahead and write it out.0231

I will do this interval and I recommend you let your software do it for you.0243

It is going to be 2 × π/ 4 ^½.0248

The integral converges.0260

Again, we were able to do this thing because E ⁻X ^/ 2 is an even function.0265

Remember what an even function is.0280

It is where F of –X = F of X.0282

It is something that is going to be symmetric about the Y axis.0288

The same on the left it is on the right.0293

That takes care of one, that is good.0299

Now, we need to check to see whether it is finite.0303

Ψ is equal to E ⁻X²/ 2 and ψ prime = -X × E ⁻X²/ 2.0309

I hope you are checking my derivatives.0337

Yes, both of these are absolutely finite on their particular interval as X goes to + infinity or –infinity.0339

Here both are finite.0346

So far so good.0352

Now, we need to check continuity.0354

This is continuous and this is continuous, both are continuous.0357

There are no discontinuities, both are continuous, our domain of definition.0362

And last, both are single valued.0371

In other words, when you put 1X in there you will get 1Y value.0378

You are not going to get 2 different Y values.0381

Both are single valued.0383

Yes, this is a good candidate for a wave function.0385

This is not just any random function that can be a wave function.0402

It has to be well behaved.0405

It has to satisfy these properties.0406

I apologize for a little bit of sniffling and my voice being a little different.0412

I’m just getting over a bit of cold.0418

Example number 2, look at the solutions to the problem of a free particle in a 2 dimensional box, 0424

These are the solutions to the Eigen value problem, is the Schrodinger equation expressed as an Eigen value problem.0431

H is the hamiltonian operator.0439

It is just a shorthand notation and E is the energy Eigen value.0440

It shows that whenever the system is in one of its energy Eigen states, 0446

In other words whenever the system is one of the wave functions, 0449

One of the Eigen functions, that the variance of the energy is equal to 0.0453

The standard deviation is equal to 0.0459

In other words, that every time you measure the energy, you are going to get the same value.0462

You are going to get some of that particular value and you will never get anything else.0468

In other words, there is no going to be deviation.0472

You are going to get the same point every time.0474

55555555 a million × and the average of that is 5.0478

The deviation is 0 because there is no other number except 5.0484

That is what this means.0488

Let us go ahead and find out what ψ is, look up the solutions.0492

For a 2 dimensional box, ψ N sub X N sub Y, we have 2 quantum numbers, 0497

Is going to equal 2/ AB¹/2 × the sin of B sub X π/ A × X × the sin of N sub Y π/ B × Y.0503

That is the wave function.0523

We want to show that this is equal to 0.0529

This is equal to the average value of E² – the average value of E².0534

Whenever that system is in one of its energy Eigen states, that means this is satisfied.0550

Its Eigen value is satisfied.0557

When the system is in one of its energy Eigen states, remember the Hamiltonian operator is the operator for the total energy of the system.0568

The E and H, when we talk about the total energy of the system, we are talking about the Hamiltonian operator.0586

It is one of its energy Eigen states, that means that the H of ψ N sub X N sub Y is equal to the energy N sub X N sub Y of ψ N sub X N sub Y.0595

We are just applying this, we are just working symbolically and formally.0611

N² ψ, just apply it twice.0617

N sub X N sub Y= E², N sub X N sub Y ψ, N sub X N sub Y.0629

It is just an application of this.0639

This is true.0642

The expectation value of the energy is going to equal the double integral.0646

Double integral because we are talking about 2 variables.0653

We have X and Y of ψ * N sub X N sub Y × the energy Hamiltonian operator are the same.0654

Ψ N sub X N sub Y.0671

This is the definition of the average value.0674

This is a double integral because it is a two variable problem.0683

This is going to equal the double integral.0691

This is this, this part this part is this part, I’m just going to substitute in.0698

It is going to be ψ* N sub X N sub Y × E N sub X N sub Y ψ N sub X N sub Y.0709

This is the definition, this thing is just this thing and I just substituted this in for here to get this.0727

E now is a scalar, it comes out of the integral.0740

This equals E N sub X N sub Y × the double integral of ψ*.0747

N sub X N sub Y × ψ N sub X N sub Y which is equal to E, because this integral =1.0757

This is a normalized wave function and this is the normalization condition.0767

Because the integral equals 1.0773

D² that is equal to the integral of ψ* N sub X N sub Y × the hamiltonian² ψ N sub X N sub Y.0786

I just substitute in, equals the double integral of ψ sub * N sub X N sub Y E N sub X N sub Y ψ N sub X N sub Y.0800

This comes out because it is a scalar.0815

This is², N sub X N sub Y × the double integral.0820

We are back to N sub X N sub Y ψ N sub X N sub Y is equal to E² N sub X N sub Y.0828

Our σ² of E which is equal to the expectation value of E² - the expectation value of E that is² is equal to E² N sub X N sub Y - E N sub X N sub Y² is equal to 0.0841

We did what we were supposed to do just by manipulating the basic definition of the Eigen value problem.0871

That and that were the case of the hamiltonian operator and hamiltonian operator of ψ is equal to the energy × ψ.0877

The Schrodinger equation and we just use this and the definition of the expectation value.0888

Evaluate L sub X L sub Y.0902

Very simple statement in the problem.0908

Let us see what we can do.0911

This is the angular momentum operator in the X direction.0915

Angular momentum in the Y direction.0919

The X component of the angular momentum, the Y component of the angular momentum, the commutator of those two.0922

Let us go ahead and write out what these are.0932

Let me go back to blue here.0934

You know what, let me go back to black, I think.0936

I have L sub X = - I H ̅ Y DDZ - Z DDY and the Y component of the angular momentum operator equals - I H ̅ Z DDX - X DDZ.0940

Trying to keep all of these symbolism straight is the hardest part of quantum mechanics, I promise you.0976

I really do not like these brackets.0986

I do not like physically actually drawing them out so I hope you will forgive me, I tend to use a slightly different notation.0990

I just tend to write comma, L sub X, L sub Y, commutators are not a big deal.0994

You can use whatever symbolism you want as long as you know what is being said.1002

I’m just making the brackets for some odd reason just that bothers me.1006

That is equal to L sub X L sub Y of F – L sub Y L sub X of F.1010

Basic definition of commutator.1023

Let us go ahead and do this first one.1027

I’m going to blue, let us go ahead and do this first one.1029

L sub X L sub Y of F = - IH Y DDZ - Z DDY of - I H ̅ Z DF DX - X DF DZ.1033

We have to multiply this out.1069

Let me go ahead and take care of the constants first.1075

This and this, - IH and - IH ends up becoming a -H ̅².1078

An operator, you can treat it just like a binomial or a polynomial.1087

You have a binomial operator.1092

A binomial operator is when I do this × this, this × that, this × this, this × that.1093

We are just multiplying, we are operating so we have to be careful here.1099

This one here, let us do this first.1102

This and this, this is the DDZ differential operator.1106

Here we have Z × a function of XYZ so this is going to be a product rule.1112

This one is going to end up being , let us see here.1118

YZ is going to be this Y and then the derivative of this is going to be this × the derivative of that.1124

It is going to be D² F DZ DX + that × the derivative of this.1144

The Y and Z part, we just multiply that.1158

Y the derivative of this thing is this time the derivative of that which is why I get the Y × Z.1161

D² of DZ DX.1168

It is going to be the derivative of this.1171

This × the derivative of this which is 1.1173

We are going to get Y DF DX.1176

I hope that made sense because this is Z and this is a function of XY and Z.1180

This is the product rule so when I apply this differential operator to this term, I have to do the product rule.1187

The rest are easy.1195

This × this, we will do this × this.1197

This is going to be - XY D² F DZ².1204

We will do this × this, not a problem these are just straight because there is no function of that variable.1217

I'm not taking the derivative with respect to a variable of, you know something that includes that variable like this one.1228

They are the normal product rules.1234

This we get - Z² D² F DY DX +,1236

I will do it this right here.1248

It is going to be XZ D² F DY DZ.1250

Let us go ahead and do L of Y, L of X of F.1263

That is going to be - I H ̅ Z DDX - X DDZ that is the operator of LY, × - I H ̅ of Y DF DX - X DF DY.1271

Again, you notice I use an F, I did not just work with the operators.1302

For this one, I included the F.1308

For this part here, I include the F.1311

I have working on some function, I do not want to lose my way.1315

This is going to equal to – H ̅².1318

It is going to be this in this, it is going to be a YZ D² X.1329

Let me see if I got this here.1350

DF DZ this is XYZ.1354

We have to be careful here.1364

Notice, the LX operator, X is going to be YZ ZY.1366

This is the Y operator which is going to be ZX XZ.1374

There we go, now we are good.1379

We have YZ D² F DX DZ.1381

I will do this one, this is going to be –Z² D² F DX DY.1390

Let me have this one which is going to be –XY D² F DZ².1401

I have Z and Z, I’m going to get two terms.1414

This is going to be product rule, again.1418

It is going to be + XZ D² F DZ DY.1420

In other words, this × the derivative of this and it is going to be + X.1429

This × the derivative of this DF DY.1440

Everything actually ends up canceling out.1451

YZ D² of DZ DX YZ D² F DX DZ.1454

Remember, mixed partials are equal so that cancels with that.1459

That stays and that stays.1469

XY D² of DZ² XY D² of DZ².1471

This - - becomes +, so that cancels.1477

Z² D² of DY DX D² of DX DY cancels.1483

X of Z D² of DY DZ D² of DZ DY cancels.1490

What I’m left with is the following.1498

I’m left with LX LY F – LY LX of F = -H ̅² × Y DF DX – X DF DY.1501

I'm going to go ahead and switch this around.1535

I’m going to go ahead and write this as H ̅², flip that.1537

Basically pull out a -1 from here I get X DF DY - Y DF DX.1543

I’m going to write this as - I H ̅ × - I H ̅.1557

- and – is a + here, X D FDY - Y DF DX.1567

If you notice this thing right here, this part right here happens to be L of Z.1582

We get I H ̅ L of Z of F and we can drop the F.1594

Our final operator notation I H ̅ L of Z.1605

I will just go ahead and use the XLY, it equals that.1616

Notice that this does not equal to 0 operator.1624

It is not possible.1633

These operators the LX and LY do not commute.1635

It is not possible to measure any two components of the angular momentum simultaneously to an arbitrary degree of precision.1639

In other words, this is the same as the linear momentum position operator.1684

If you measure one really well, you have to lose the measurement of the other.1689

You cannot measure both to any degree of accuracy or precision that you want.1693

In the case of the angular momentum, any 2 components of the angular momentum cannot be measured simultaneously to any degree of precision.1697

The same holds for LX LZ and LY LZ.1704

LX LY and LY LZ, LX LZ, you are always going to get something like this.1709

A little messy but nice.1717

This is the stuff that you have to do.1720

You just have to do it through it.1721

Example 4, this is going to be a not so much of a problem, this is going to be a demonstration.1728

It is a demonstration of why the necessity of a real Eigen values imposes the hermitian property on operators.1733

Remember what we said about when we take our measurements in quantum mechanics, 1740

What we are measuring, we need the measurements to be real numbers.1743

In other words, we need the Eigen values to be real numbers.1748

Operators and Eigen functions can be complex.1751

They can be real or they can be complex.1753

There has to be a certain property of the operator that guarantees that the Eigen value is always real because what we measure has to be a real number.1757

It cannot be a complex, we cannot measure complex numbers.1766

We need a real number.1768

Here we are going to demonstrate why this is the case, before we just threw it out there and say that it is the case.1771

We are going to demonstrate why the necessity of having a real Eigen value forces the operator to be hermitian.1776

Let us go ahead and do that.1787

Should I work in red or black?1789

Let us go ahead and work with red.1794

We have A ψ = λ ψ, where λ is a real number.1797

It is in a real number system.1809

This is the general Eigen value problem.1811

The operator × ψ = ψ × some constant.1814

You are not getting ψ back when you operate on it but you end up getting it back multiplied by some real value.1819

What I'm going to do is I'm going to go ahead and take this and multiply on the left by ψ* both sides.1828

Whatever you do to both sides, you retain the equality.1837

Ψ * A ψ = ψ * λ ψ.1841

All of them have taken the basic definition and I have done this to it.1849

I’m going to integrate both sides.1853

The integral of that side is equal to the integral of this side.1857

This integral over here on the right, this is a real number so I could pull it out of the integral side.1862

This is going to be λ × the integral of ψ * ψ.1869

Ψ * ψ is the normalization condition and it equals 1.1874

This equal to λ × 1 which equals λ.1877

That is it, that is the first part.1883

Let us take the complex conjugate of this thing, start with that and see what we can do.1886

Let us start again with A ψ = λ × ψ.1897

Let us go ahead and take the complex conjugate.1903

This is going to be this conjugate, ψ conjugate = λ conjugate ψ conjugate.1905

We are going to manipulate this.1914

I’m going to multiply on the left by ψ.1915

I have ψ, I’m just manipulating.1919

All I’m doing is a mathematical manipulation and see where it takes me.1922

This is ho we did it.1926

Ψ λ * ψ *.1929

I’m going to integrate both sides.1938

When I integrate the side, I integrate this side.1941

Λ * is just a scalar so it comes out as λ * × ψ ψ *.1947

This is just the normalization condition so this is equal to λ *,but Λ is a real number.1957

Therefore, the conjugate of a real number is the number itself.1968

This equals λ.1972

Both integrals equal λ.1976

Let me go to blue.1978

This integral equals λ, this integral equals λ.1980

If they are both equal λ, then they are both equal to each other.1994

Therefore, I have the integral of ψ* A of ψ equals the integral of ψ A *ψ * which is the definition of the hermitian property.2002

In other words, we arrived at this λ being real forces these integrals to be equal.2035

We use this final step in our derivation as the definition of hermitian property.2043

In other words, their operator has to satisfy this condition.2048

That integral has to equal this integral, that is the definition.2053

The necessity of then being real forces to operator to have this property.2056

This property we call hermitian.2060

There you go.2063

If some operator satisfies the hermitian property regardless of whether 2067

The operator or the wave function is complex or not, the Eigen values λ sub N are always real.2092

This is very profound on so many levels.2116

Those of you who are going to continue on with some higher mathematics or go further with quantum mechanics, 2120

It is quite extraordinary what is going on here mathematically.2124

Let us see something else.2132

We said that the hermitian property guarantees that the Eigen values are real.2136

We say why, that is the case.2139

We also said that it guarantees that the Eigen functions are orthogonal.2141

In other words, the integral is equal to 0 but they are perpendicular to each other in some sense.2145

In this problem, it is going to be a demonstration of why the Eigen functions of hermitian operators are orthogonal.2151

Let us go ahead and start with A ψ sub M = λ sub M ψ sub M.2159

Let us start with A of ψ sub M= λ sub M ψ sub M.2176

I’m going to go ahead and do the same thing that I did before.2187

We are going to multiply on the left by the conjugate and take the integral.2190

I got ψ conjugate AN, ψ sub N = ψ N conjugate × λ sub N × ψ of N.2195

I’m going to integrate those two.2213

I’m going to call this integral 1 and I’m going to call this integral 1 prime.2217

Over here, I’m going to get the conjugate first.2225

I’m going to take A conjugate ψ sub M conjugate = λ sub M conjugate ψ sub M conjugate.2229

I’m going to multiply both sides on the left.2242

I’m sorry this is going to be ψ sub M not ψ sub N.2250

Ψ sub M and over here I’m going to multiply on the left by ψ sub N.2255

I got ψ sub N × A conjugate ψ sub M conjugate = ψ sub N × λ sub M conjugate ψ sub M conjugate.2264

Let me see if I got this right.2286

I’m going to call this integral 2 and I’m going to call this integral 2 prime.2290

I’m going to take 1 prime, this integral – 2 prime.2298

I’m going to go to red.2306

I got 1 prime – 2 prime = the integral ψ sub M conjugate × λ sub N ψ sub N – 2 prime which is.2310

I forgot to once I get to this point I’m going to integrate both sides.2327

-2 prime which is ψ sub N λ sub M conjugate ψ sub M conjugate.2335

That is going to equal, we are going to pull the λ out.2350

Λ N – λ M conjugate × the integral of ψ sub M conjugate ψ sub N.2354

Ψ sub N ψ sub N conjugate is the same thing.2365

I’m just going to write it as ψ sub M conjugate ψ sub N.2369

That is going to be that one.2379

I’m going to take 1-2, that integral.2382

1-2 is going to equal the integral of ψ sub M conjugate A ψ sub N – the integral of ψ sub N A conjugate ψ sub M conjugate.2386

This integral is going to equal 0.2415

The reason it is equal to 0 is because the operator we said was hermitian.2418

This is the definition of hermitian.2424

This thing and this thing, they are hermitian.2426

They are the same integral.2430

Therefore, this integral is equal to this integral.2431

This - that is equal to 0.2436

This is the case because A is hermitian.2438

This is the definition of hermitian ψ sub M * A ψ sub N is ψ sub N A * ψ sub M*.2446

So far so good.2464

I have got this thing and I got this thing.2466

The 1 prime -2 prime is equal to this.2474

1 -2 is equal to 0.2477

1 is 1 prime, 2 is equal to 2 prime.2482

1 prime -2 prime is equal to 1 -2.2492

Therefore, my 1 prime -2 prime which we said was λ sub N – λ sub M conjugate × 2498

The integral of ψ sub M conjugate ψ sub N is equal to 1 - 2 which we said is equal to 0.2509

I have something × something is equal to 0.2521

That means this is 0 or this is 0.2529

If M does not equal to N, then this λ sub N – λ sub M conjugate do not equal each other.2533

They are different Eigen values.2551

A different Eigen values are not going to equal each other.2552

These are not going to be equal to each other.2555

If they are not equal to each other, this thing is not 0, which means that this has to be 0.2558

Let me say it again.2565

If M does not equal N, then λ sub N and λ sub M do not equal each other.2565

Which means that λ sub N – λ sub M does not equal 0.2573

This equation =0 so that means this one has to equal 0, this term.2577

Therefore, the integral of ψ sub M conjugate × ψ sub N does equal 0 which we take as the definition orthogonality.2582

We already use the definition and now the justification for why it works, for why it is, what it is.2603

Which we take as the definition of orthogonality.2611

A hermitian operator, all the operators in quantum mechanics are linear.2625

They are hermitian, they guarantee that the Eigen values are going to be real and 2629

They guarantee that the Eigen functions are going to be orthogonal with respect to each other.2632

The integral is going to equal 0.2638

It is actually quite extraordinary.2642

With that, we will go ahead and end it here.2646

Thank you so much for joining us here at www.educator.com.2648

We will see you next time, bye. 2651