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The Hydrogen Atom Example Problems II

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: Normalization & Pair-wise Orthogonal 0:13
    • Part 1: Normalized
    • Part 2: Pair-wise Orthogonal
  • Example II: Show Explicitly That the Following Statement is True for Any Integer n 27:10
  • Example III: Spherical Harmonics 29:26
  • Angular Momentum Cones 56:37
    • Angular Momentum Cones
    • Physical Interpretation of Orbital Angular Momentum in Quantum mechanics

Transcription: The Hydrogen Atom Example Problems II

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

Today, we are going to continue with our example problems for the hydrogen atom.0004

Let us get started here.0008

Example number 1, show that S1,0, S1, 1 and S1, -1 are normalized.0014

And that they are pair wise orthogonal.0022

These are the spherical harmonics.0024

They are the angular portion of the hydrogen atom wave function.0027

Let us go ahead and write down what they are so we have them.0033

I can go ahead and work in blue or red.0035

Let us go ahead and work in blue.0040

We have S1, 0 is equal to 3 / 4 π ^½ × cos of θ.0044

S1, 1 is equal to 3/ 8 π ^½ sin θ E ⁺I φ.0056

S1 -1 is equal to 3/8 π ^½ × sin θ E ⁻I φ.0069

We are going to be working in spherical coordinates.0084

It is very easy to forget these extra midterms that we have to put in the integrand.0089

We are working in spherical coordinates.0096

In other words, θ is going to run from 0 to π.0116

And φ is going to run from 0 to 2 π.0122

The integrals look like this.0129

The spherical coordinates has 3 variables, it has R, θ, and φ.0138

In this particular case, we are dealing just with a spherical harmonics so we are just concerned with θ and φ.0144

Since that is the case, we would be working with double integrals.0149

Later, when we start talking about ψ, ψ 211 and ψ 310, 0152

the entire hydrogen wave function that is what we are going to be working with R, θ, and φ.0157

That is when it is going to become a triple integral.0163

But for right now, it is going to be double integral, 2 variables.0165

The integrals actually look like this.0167

In general, we are going to be 0 to 2 π, 0 to π, and there is going to be some integrand whenever that happens to be.0171

Then, we are going to have the factors sin θ D θ D φ.0183

This extra thing has to be there.0189

Do not forget this factor, do not forget this sin θ factor when doing integrals of these types.0191

It is very easy to just start working with just D θ D φ to forget that we have 0204

to make a little bit of an adjustment because we did a change of variables,0207

when we went from Cartesian coordinates to spherical coordinates.0211

Let us go ahead and talk about this.0217

While I’m here, let me go ahead and mention it.0221

Later when we do start talking about R and θ and φ, the integral is going to look like this.0224

Let me write this out.0233

When we later include R, the general integral will look like this.0238

It is going to be the integral from 0 to R, integral from 0 to 2 π, integral from 0 to π, and there is going to be some integrand, 0261

whatever that happens to be depending on the functions we are dealing with.0281

Now the factor is going to be R² sin θ D θ D φ DR.0285

It does have to be in this order, in can be in any order you want depending on0295

where does that you are integrating the function that you are dealing with.0299

Just do not forget spherical coordinates, we have to have these factors.0301

Let go ahead and get started with the problem.0308

For S10, normalization looks like this.0311

Our standard normalization integral, we have seen it over and over again.0321

It is going to be the integral of S10 conjugate S10 and we want this which we want to equal 1.0325

We are trying to show that is normalized.0342

We are trying to show that S10 is normalized.0344

It means that when we take the S10 conjugate multiplied by S10 and integrate, that we should get 1.0347

It is equal to 1.0358

Let us go ahead and do S10 conjugate × S10.0360

It is actually going to just equal S10² because in this particular case, the S10 is a real function.0366

It does not have a complex part.0374

Therefore, it is just S10 × S10.0376

You already listed them before, what we are going to end up getting is 3/4 π × the cos² θ.0380

Again, because S10 is a real function, it is real.0390

There is no E ⁺I φ, E ⁻I φ, there is no I in there.0398

Our integral is going to end up looking like 0 to 2 π, 0 to π, 3/ 4 Π, that is this.0408

That is all we are doing, we are forming this integral and we are solving that integral.0419

That is all we are doing.0422

Cos² θ, that is our function.0426

Our factor is sin θ D θ 0 π θ, that is going to be the first integral we will do.0432

That is the integral, and the other integral is φ, we do that afterward.0438

We are going to take care of the inner integral first.0443

We are just going to deal with that one.0448

I'm going to pull the constant out, it is going to be 3/ 4 π the integral from 0 to π.0451

This is going to be cos² sin θ cos² θ sin θ D θ.0458

I’m doing one variable at a time.0465

I do not know if you remember these trigonometric integrals.0467

Again, you can just go ahead and have your software do it, it is not the end of the world.0470

But I thought it would be nice to actually do so by hand, just so we get to refresh our memories 0473

because we are actually going to have to be doing this on the quiz and the tests that you take.0478

You can go ahead and do use substitution here.0484

I’m going to set U equal to cos θ and I'm going to set Du is going to be - sin θ D θ.0486

Therefore, when I substitute these back into this here, I'm going to end up getting - 3/, 0497

Basically, what happens here is sin θ D θ is equal to – DU.0506

Sin θ D θ, I will put - DU cos², I put U².0512

It is going to be -3/ 4 π, you remember this right.0517

It should not be too long ago.0521

It was 0 to π U² DU.0522

I should change my limits of integration, I generally do not because I tend to just go back and put these functions back in when I solve it.0528

I would like to go ahead and put this into θ to get the lower limit and the π into θ to get the upper limit.0536

I just leave it as 0 to π because I’m going to go back to the θ in just a minute.0544

It is going to be equal -3/ 4 π × U³/ 3 0 to π and of course U is just cos θ.0551

It is going to be -3 / 4 π cos θ³ from 0 to π.0564

What we are going to end up here is getting -3/ 4 π × -1/ 3 -1/ 3, when you do that.0575

You are going to end up actually getting 2/ 4 π or 1/ 2 π, that is just the inner integral.0587

We are not in the outer integral yet.0598

1/2 π is that.0600

Now, we will go ahead and do the outer integral.0610

The outer integral, we have the integral from 0 to π of 1/ 2 π D φ.0620

And that is going to equal 1/ 2 π from 0 to 2 π of D φ which is going to equal 1/ 2 π × 2 π, and it is going to equal 1.0631

Yes, we did get the number 1 when we did the integral.0644

This particular function, this particular spherical harmonic is normalized.0648

Now for the rest, I will set the integral but I’m not going to go ahead and solve the integral.0653

You could do by hand, table, or just use your software.0660

For the rest, I will set up the integral but leave it to you to evaluate.0665

We want to do S11, for S11 the normalization condition is S11 conjugate × S11.0682

We want it to equal 1 when we do that, which we want equaling 1.0701

S11 is equal to 3/ 8 π ^½ sin θ E ⁺I φ.0715

Therefore, S11 conjugate is equal to, this does have something complex in it.0727

It does have a conjugate but is different than the original function.0736

S11 conjugate is equal to 3/ 8 π ^½ × sin θ E ^- I φ.0739

Therefore, S11 conjugate × S11 is equal to this × this.0752

We get 3/ 8 π, sin θ and sin θ is sin² θ, E ⁺I φ × E ⁻I φ is E⁰ which is equal to 1.0762

It is equal to this.0774

The integral of S11 conjugate S11 which is the normalization integral is equal to 0 to 2 π, 0 to π, 3/ 8 π sin² θ D θ.0779

When you solve this integral, do it by hand, you have to do the.0800

I'm sorry, D θ I forgot my D φ.0810

Just like I told you guys not to forget the factor, I forgot the factor.0821

This is the function, another factor is sin θ D θ D φ.0825

I think I should just go slow.0834

The first integral is that one, that is the inner and I will do the outer.0838

Or just have your software do it.0846

And when you do, do this, you are going to get 1.0848

It is also normalized, S11 is normalized.0852

We are confirming this.0856

We are just getting comfortable with the functions, manipulating the functions, 0857

dealing with the functions, writing them down, that is what we are doing.0862

Let me go back to red, I actually write red, it is nice.0867

For S1-1 normalization, that normalization is basic S1 -1 conjugate × S1 -1.0884

We want it to equal 1.0901

S1-1 is equal to 3/ 8 π ^½ sin of θ E ^- I φ.0904

Therefore, S1 -1 conjugate is equal to the conjugate of this is 3/ 8 π ^½ sin of θ E ⁺I φ.0918

Therefore, S1 -1 conjugate × S1 -1 is equally the same as before.0937

What we just did is equal to 3/8 π sin² θ.0945

Of course, the φ goes away.0955

The integral S1-1 conjugate S1 -1 is equal to 0 to 2 π, 0 to π, 3/8 π sin² θ, that is the function.0960

And the factor is sin θ D θ D φ, this is the integral that you would enter to your software.0977

And when you do so, you are going to get that is = 1.0985

That takes care of the normalization.0990

All of those are normalized, now we are going to deal with the pair wise orthogonality process.0995

Let us take a look and deal with that.1001

Let us see if I should start in a new page or not.1005

That is fine, I will just go ahead and continue here.1009

For pair wise orthogonality.1014

The integral of the integrals that we would be looking at, the orthogonality condition1028

is going to be the the integral of S sub LM conjugate × S sub L primer M prime,1038

these are different because now we are taking 2 functions that are actually different.1047

We want this, which we want equal to 0 that will confirm that they are orthogonal.1053

Let us see what we have got for S10 and S11.1068

S10 is equal to 3/ 4 π ^½ cos of θ and S11 is equal to 3/ 8 π ^½ sin θ E ⁺I φ.1086

Therefore, the integral of S10 conjugate S11 is going to equal the integral from 0 to 2 π.1107

The integral from 0 to π of this function × the conjugate of this function × this function.1125

The conjugate of this function is the same because this is real.1134

There is no I part to it.1138

It is going to be 3/4 π ^½ cos θ × 3/8 π ^½ sin θ E ⁺I φ.1140

And it is going to be sin θ D θ D φ.1172

We are going to get of 3/ 4 π √ 2, when we take care of the constants.1180

It is going to be 0 to 2 π, 0 to π.1189

We are going to have cos θ.1194

I’m going to put this sin θ and that sin θ together, sin² θ E ⁺I φ D θ D φ.1201

Here is something we can do which is really nice.1213

Now that we have a function which is a function of both θ and φ, we can actually separate the functions out.1217

We can write the integral like this, it is very convenient.1223

3/4 π √ 2, 0 to 2 π.1227

The φ, we can just take this function E ⁺I φ D φ.1232

Then, we can take the θ portion, cos θ sin² θ D θ.1240

We can separate this out and do it this way because again these are individual functions of 1 variable 1251

and we are integrating 1 variable at a time.1256

You do have a lot of integrals that actually end up looking like this.1261

With this E ⁺I φ, E ⁺2I φ, E ⁺3I φ, things like that.1264

For integrals that look like this, this is a good thing to know.1269

For integrals that look like this, it is good to know the following.1275

It is good to know the following.1292

We are just making our life easier because the integral, you have seen over and over again.1297

We do not want to keep evaluating them and good to know the following.1300

The integral from 0 to 2 π of E ^+ or - I φ is actually always going to be equal to 0.1308

In fact, it is true for any multiple of the I φ.1327

In fact, the integral from 0 to 2 π of E 6+ or – IN φ D φ is equal to 0.1333

I forgot my D φ which I often forget.1347

This integral is automatically equal to 0.1350

It saves me from having to actually solve the rest of the integral, very nice.1352

This integral right here, because this is true, this part is 0.1359

0 × whatever it is does not matter, it is going to equal 0.1365

Therefore, this integral, 3/ 4 π √ 2 integral from 0 to 2 π E ⁺I φ D φ 0 to π cos θ sin² θ D θ is equal to 0, 1368

which means that they are orthogonal.1398

Which is what we wanted, very nice and convenient property.1400

Now we do S10 and S1 -1.1405

The integral for that is going to be the integral of S10 conjugate S1 -1, we want it to equal 0.1417

This integral S10 conjugate S1 -1, the integral is going to turn out to be 3/4 π √ 2 the integral from 0 to 2 π,1425

this time E ^- I φ D φ, because this S1 -1 that was that particular function 0 to π.1456

Again, you are going to get cos θ sin² θ D θ, that is equal to 0.1467

Therefore, the whole integral is equal to 0.1478

Again, S10 and S1 -1 are orthogonal, very nice.1480

Let us go ahead and see the last one we got.1490

I think we got one more.1493

You have S11 and S1 -1.1494

For S11 and S1 -1, our integral is going to be S11 conjugate, S1 -1, we want the integral to be equal to 0.1499

S11 conjugate is going to equal 3/ 8 π ^½ sin θ E ⁻I φ.1514

And S1-1 is equal to 3/ 8 π ^½ sin θ E ⁺I φ.1529

When we multiply those 2 together, we are going to get the integral of S11 conjugate S1 -1 is going to equal 3/ 8 π,1547

the integral 0 to 2 π, the integral 0 to π, it is going to be sin² θ E ⁻I 2 φ sin θ D θ D φ, which we can separate.1567

It is going to be 3/ 8 π × the integral from 0 to 2 π E ⁻I 2 φ D φ.1589

The integral from 0 to φ of sin³ θ D θ.1601

This integral is equal to 0, therefore this is equal to 0.1609

Therefore, we have shown that they are orthogonal.1612

And of course, it is generally true.1617

All of the spherical harmonics are pair wise orthogonal.1619

Let us see what we have got, lot of extra pages here.1626

In the previous exercise, we use the fact that 0 to 2 π E ⁺IN φ D φ equal 0 for any integer N.1633

We want to actually have you show explicitly that this is true.1641

Show that this is true rather than just using it.1644

It should not be a problem.1647

E ⁺IN φ, we are going to use the formula that we now.1651

We are going to separate it into its real and complex parts.1657

We are going to factor the cos θ and sin θ.1661

This is cos of N φ + I × the sin of N φ.1664

Therefore, the integral from 0 to 2 π of E ⁺IM φ D φ is going to equal1673

the integral from 0 to 2 π of cos N φ D φ + I × the integral from 0 to 2 π of sin N φ D φ.1682

We just separate it out so we can actually solve the normal integrals.1702

This is going to equal 1/ N × sin N φ from 0 to 2 π + I ×.1706

It is going to be actually ±I because when we integrate the sin, it is going to be a negative cos.1720

Cos N φ 0 to 2 π and you are going to get 1/ N 0 -0 + - I/ N, sorry I forgot the N here.1727

This is going to be 1 -1 0, there you go.1742

We just show explicitly that this integral is always equal to 0 and this integral will come up a lot.1749

You can just, on a test and on a quiz, it is equal to 0.1755

You do not have to go ahead and actually evaluate the θ portion of the integral.1759

Let us see what we have got next.1767

In spherical coordinates, the angular momentum operator for the X direction of the angular momentum, 1771

the X component of the angular momentum operator is this expression right here.1779

This is in spherical coordinates and we have already seen previously in Cartesian coordinates, 1787

we are not going to go through the exercise of partial differentiation to change the variables or to actually turn it into this,1792

but this is what it looks like when you expressed in spherical coordinates.1801

Using the spherical harmonics for L = 1, in other words S10 S11 S1 -1, 1805

show that the average value of the angular momentum in the X direction is 0.1814

Using the spherical harmonic S11, for L = 1 N = 1, 0 and -1, we deal with S10 S11 and S1-1.1828

For these 3, we want to show that the average value of the X component of the angular momentum is going to be 0.1855

Let us take a look, for S10 the average value integral looks like this.1865

It is going to be S10 conjugate and you actually put the operator in between and then you operate on S10.1878

You operate on S10 first and then you multiply on the left by S10 and you integrate that.1887

We want to see if this is actually going to equal 0.1897

Let us go ahead and take care of this part first.1901

Let us go ahead and actually operate, its form L sub X of S10, let us operate on it using this operator.1904

This is going to equal - I H ̅ - sin φ DD θ - cot θ cos φ DD φ.1914

We are going to be operating on S10 or S10 is 3/ 4 π ^½ cos θ.1937

This is going to equal ,we are going to distribute and operate on this.1950

In this particular case DD φ, there is no φ in this S10 spherical harmonic.1957

Therefore, this term just goes to 0 so this is the only one that matters.1962

I’m going to go ahead and pull these out, the constants - I H ̅ × 3/ 4 π ^½ × - sin of φ × DD θ of cos θ, which is -sin θ.1966

When I put this, I do this to this.1989

DD φ of this is just 0 so it is going to be -0.1997

This is just going to equal - I H ̅ × 3/ 4 π ^½ × O.2001

This is - × - × -, one of the – stays, this becomes sin φ sin θ.2014

That takes care of just this part.2029

Now, we are going to multiply on the left, multiply what we got on the left by S10 conjugate.2032

S10 conjugate × what it is that we just got which was L sub X of S10, that is going to be 3/ 4 π ^½ cos θ × – I H ̅ 3/4 π ^½ sin φ sin θ.2048

This is going to equal -3 IH ̅/ 4 π × cos θ sin φ sin θ. 2085

The integral of S10 conjugate L sub X operating on S10, the average value integral 2105

is actually going to equal -3 I H ̅/ 4 π the integral from 0 to 2 π.2119

Let me go ahead and separate these integrals out.2132

I think I’m going to write the whole thing first.2135

0 to π this thing, cos θ sin θ, sin θ that is the function and the factor, sin θ D θ D φ.2138

One thing at a time, I’m going to go ahead and separate these out.2157

It is going to be -3 I H ̅/ 4 π the integral from 0 to 2 π sin of φ D φ × the integral from 0 to π cos θ sin² θ cos sin D θ.2160

The integral from 0 to 2 π of sin φ D φ, let us just deal with this right here.2201

Recall the graph of sin φ, the graph looks like this.2210

From 0 to 2 Π of the sin function goes like this, here is 2 π.2227

The integral from 0 to 2 π of the sin function gives us that area and that area.2234

This is going to end up being positive, this is going to end up being negative.2242

The integral / this is going to equal 0.2245

The integral of the sin of φ sin of θ sin of X does not matter.2248

The sin φ from 0 to 2 π is equal to 0.2253

Let me write this out.2258

The integral from 0 to 2 π sin φ D φ is the shaded region.2261

This integral is equal to 0 because above the X axis is positive and below the X axis is negative.2272

The magnitudes of the area, if they ask for the total area,2286

you just take the absolute value of this particular portion and just get twice that of the integral itself is equal to 0.2291

Therefore, this is equal to 0 which means that the average value of L sub X which was this integral, is equal to 0.2300

For S10, we have the average value is equal to 0.2329

Let us go ahead and take a look at S11.2335

For S11, S11 is equal to 3/ 8 π ^½ sin θ E ⁺I φ.2344

Therefore, the LX of S11 is equal to -I H ̅ - sin φ DD θ – cot θ cos of φ DD φ.2364

Of course, we are operating on this function 3/ 8 π ^½ sin θ E ⁺I φ.2393

If you operate on this, you operate on this, and you are going to get some long expression.2406

Let us see what this actually looks like here.2411

We have – IH ̅ × 3/ 8 π ^½.2416

Hopefully, I got everything right here.2429

This is - sin φ the derivative of this is cos θ × cos θ.2432

And of course, I have to include that.2450

The product function × E ⁺I φ – cot of θ, the cos of φ × DD φ of this which is sin θ × IE ⁺I φ.2453

I will just apply this to that.2481

It is going to equal – I H ̅ 3/ 8 π ^½ × - sin of φ cos of θ E ⁺I φ – cot θ/ sin θ, this is cos φ, 2485

this is sin θ × IE ⁺I φ, sin θ and sin θ cancel and I'm left with - I H ̅ 3/ 8 π ^½ ×, - and – is +.2518

Negative and negative this becomes positive.2554

We get sin φ cos θ E ⁺I φ + I cos θ cos φ E ⁺I φ.2558

We have to multiply this just LX of S11.2581

We have to take S11 conjugate multiply on the left by S11 conjugate L sub X of S11.2586

That is going to equal 3/8 π ^½ sin θ E ⁻I φ because it is S11 conjugate × this thing which is,2602

This goes away, we have actually taken care of a positive.2623

I H ̅ 3/8 π ^½.2627

This is crazy, this is actually crazy.2634

Sin φ cos θ E ⁺I φ + I cos φ cos of θ E ⁺I φ.2638

This is really something.2662

Let us see what we have got when we multiply everything out.2664

We have got the 3, I , H, = 3 I H ̅/ 8 π × sin θ cos θ sin φ E ⁺I φ.2669

E ⁻I φ and E ⁺I φ that goes away, + I × sin θ cos θ cos φ.2704

This is our final, the integral of S11 conjugate L sub X S11 is equal,2723

Let me go ahead and write it over here.2737

It is going to be I H ̅ × 3/ 8 π × the integral 0 to π.2740

0 to π of everything that we have just wrote, which is going to be sin θ cos θ sin of φ × sin θ cos θ cos φ × sin θ D θ D φ.2751

Let the software do this for you.2785

It all = 3 I H ̅/ 8 π × 0 to 2 π.2788

We are going to separate this φ and θ.2806

Sin φ D φ × the integral from 0 to π of sin² θ cos θ D θ + I × the integral from 0 to 2 π.2809

This is a cos φ D φ, make sure that you understand how was I separated this.2830

There is this one and there is the I part.2837

I separated two integrals, the integral of this and the double integral of that.2842

Within E I have separated out the φ and the θ, that is what I have done.2850

The integral from 0 to π of sin² θ cos θ D θ.2856

Once again, the integral of 0 to π of sin of φ, this is equal to 0, this is equal to 0.2872

Our integral is equal to 0.2881

Once again, the average value for S11 which is this integral, this whole integral = 0.2884

For S1 -1, I’m not going to go through the process.2895

S1 -1 you actually end up with the same thing that you get for S11.2899

For S1 -1, the integral ends up the same as for the S11.2904

The average value of S sub X for S1 -1 also = 0.2924

It is true in general.2938

We just had checked 3 of the spherical harmonics.2947

It is true in general that for all L, the average value of the X component of the angular momentum is equal to 0.2949

The average value of the Y component of angular momentum is equal to 0.2967

Let us not forget what the average value is.2973

If I take a particular measurement, I'm going to get whatever I happen to measure for the angular momentum.2975

On average, if I take 100 measurement, a 1000 measurement, a 1,000,000 measurements, 2981

on average I’m going to get many different values that they are all going to average to 0.2986

That means for every one that I get to the left, I will get the same to the right.2991

For every one that I get going up, I get when going down.2997

For every one that I get going this way, I get when going that way.2999

On average, you can end up canceling, that is what is going on here.3003

What this means is that, remember we know what L² is.3008

We know the L², it is equal to H ̅² L × L + 1.3026

We know the magnitude of the angular momentum. 3038

We also know this one, we also know the Z component of the angular momentum.3043

In other words, the projection of the angular momentum vector on the Z access is specified by M.3049

The magnitude is actually specified by L.3054

What we do not know is X and Y.3057

We do not know them specifically.3060

This is an application of the Heisenberg uncertainty principle.3062

This operator, they commute but they do not commute with the X and Y.3067

X and Y are uncertain which is why on average, we are going to end up being equal to 0.3074

What this means is that we can specify L² and LZ.3080

That we can do but we cannot specify the X and Y component of the angular momentum.3096

We know how long the angular momentum is.3121

We also know the X component of it, the problem is we do not know what direction 3123

the angular momentum vector is at a given moment.3129

We can specify the X and Y, we can only tell you its projection along the Z axis.3133

We know how long it is but we do not know which direction it is pointing.3139

It can be pointing anywhere but we do know how long it is and how long the Z value is.3142

We will do a little bit more just in a moment.3150

This problem is actually sufficiently important to further discussion.3155

We did discuss in a previous lesson but I actually did go through it again here.3160

This problem is sufficiently important.3169

In fact, very important.3182

Angular momentum is everything in quantum mechanics.3184

This problem is sufficiently important to discuss it in context.3187

We know the magnitude of the angular momentum vector is equal to H ̅ × √ L × L + 1.3203

When I have L, I can tell you how long the angular momentum vector is.3214

It is just a function of L.3220

I also know the Z component of the angular momentum.3222

That is H ̅ M, that is specified by M.3226

I know what N is, because M is just -L to + L.3233

The quantum numbers L and M K.3238

The magnitude of the angular momentum.3249

The magnitude of L can never equal LZ.3254

This is a function of L and this is a function of M.3264

They are never going to equal each other because they never equal each other, they do not lie on the same direction.3266

Therefore, the actual angular momentum vector itself can never lie along the Z axis.3274

Let us see what we have got.3292

Let me actually go over here.3296

The average value of L sub X is equal to 0.3303

We just demonstrate it with the first few spherical harmonics but we know that is true in general.3307

The average value of L sub Y is equal to 0.3311

In other words, the X and Y components on average for the angular momentum are equal to 0.3317

These two equaling 0 mean that when we measure L sub X and L sub Y,3325

which are the components of the angular momentum along the X axis and along the Y axis.3340

Respectively, the values will average to 0.3345

We will get a bunch of values but the average is going to be 0.3358

This means that each and every value is represented.3365

For every positive value we get in the X direction, you have at least one negative in the X direction 3369

because they have to cancel out to become 0.3375

Let us to take a look at what this actually means graphically and I think it will make sense.3380

Graphically, we mean this.3384

Let us see here.3393

Here is what is going on.3400

We have a coordinate system here, the Z axis is vertical.3400

In this particular picture, they have the Y axis lying this way.3404

They have the X axis going this way.3409

This is a 3 dimensional version.3411

We can specify the length of the angular momentum vector.3413

In other words, we can specify this vector right here.3417

That is the angular momentum vector, we know how long it is.3423

We also know its projection on the Z axis.3426

In other words, if I shine a light this way towards the vector from this side, this is going to be a shadow.3431

That is what we call the projection.3439

The projection of a vector on the Z axis is the Z component of that vector.3440

Again, we are talking about a vector in 3 space.3445

L it has an X component in the I direction, it has a Y component in the J direction, Y direction, 3448

and it has a Z component in the Z direction which is specified by the unit vector K.3459

If we use this engineering notation I, J, K.3464

We know this and we know the magnitude of this.3467

That is this, we know how long it is.3472

We also know how long this is.3474

We know that value.3477

The problem is we cannot specify simultaneously all of them.3482

In other words, we do not know its projection along the Y axis.3487

We actually do not know this length and we do not know its projection along the X axis, 3490

we do not know that it is a length.3496

The Heisenberg uncertainty principle, the L² operator and LZ operator they commute 3499

but that does not commute with all of simultaneously.3506

The LZ and the LY do not commute.3510

The LZ and LX do not commute.3512

This commutes with one of them, one at a time.3514

We can specify how long the angular momentum vector is and we can tell you the projection3518

along the Z axis but we cannot tell you where, how far along.3524

This vector, we know that it is pointing this way.3529

But notice the projection down on to Y or down on to the X axis, we do not know how long these are.3532

Therefore, they can be any value at all.3541

This one could be one of this, this can be 5 but what the average value tells us is that for every one.3544

In other words, if I project this down onto a circle, it can be this way, this way, this way.3551

It can be in any direction along the circle in the XY plane, which is why you have this angular momentum vector.3564

I know the length, I know along Z but it is could actually sweep out at a cone because3570

I do not know where it is a long the X and Y directions.3575

I do not know, it can be anywhere in the X and Y directions which is why it sweeps out a circle in the XY plane.3579

And depending on where it is, you are going to get different values.3586

For L1, it is going to be this one.3589

For L2, it is going to be this one.3591

For L3, it is going to be this one.3593

It is going to sweep out a cone, that is what these columns are.3596

Recall the angular momentum cones because the vector itself, we do not know which direction is pointing.3599

We only know how long it is, that is all we can say.3606

We know its projection along the Z axis but we do not know in which direction it is going, so it can be in any direction.3609

Another picture that may or may not help, I do not know.3617

It is going to be the following.3619

This is a physical interpretation of the orbital angular momentum in quantum mechanics.3621

This is a way of thinking about it.3625

We know that in angular momentum vector, it happens when something spins.3626

Something spins this way, it has an angular momentum in that direction.3630

If it spins this way, its angular momentum is in this direction.3634

We can think of electrons spinning like that in orbit.3637

And it is going to give rise to this orbital angular momentum vector.3642

That is this vector right here.3647

We know how long it is, we also know its projection along the Z axis.3649

That is specified by M.3653

However, we do not know what its projection is along the Y or the X.3656

It can be anywhere along those.3661

It can be anywhere, which is why it actually sweeps out a cone.3663

That is what is happening.3669

This angular momentum vector sweeps out that way and because it can be this way or this way,3670

when we take measurements, for every measurement we get in one direction,3678

we are going to find a measurement of the opposite direction.3682

If we get a measurement this way, we are going to get a measurement this way.3685

When I add all of these up, they average to 0, that is what this means.3688

That is with this cone means.3692

Sometimes the angular momentum in the Y direction is going to be here, sometimes here, sometimes there.3695

That is what is going on, I hope this makes sense.3702

In any case, let us go ahead and leave it that.3707

Thank you so much for joining us here at www.educator.com.3711

We will see you next time for a continuation of more examples.3713

Take care, bye.3717