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 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

### Physical Chemistry Online Course

### 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 on*0295

*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 condition*1028

*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 equal*1673

*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 projection*3518

*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 because*3570

*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

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