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

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

### The Hydrogen Atom 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
- Associated Legendre Functions 0:07
- Associated Legendre Functions
- First Few Associated Legendre Functions
- s, p, & d Orbital
- The Normalization Condition
- Spherical Harmonics 20:03
- Equations We Have Found
- Wave Functions for the Angular Component & Rigid Rotator
- Spherical Harmonics Examples
- Angular Momentum 30:09
- Angular Momentum
- Square of the Angular Momentum
- Energies of the Rigid Rotator

### Physical Chemistry Online Course

### Transcription: The Hydrogen Atom II

*Hello, welcome back to www.educator.com.*0000

*Welcome back to Physical Chemistry.*0002

*Today, we are going to continue our discussion of the hydrogen atom.*0004

*Let us jump right on in.*0007

*Let me go ahead and take a blue here.*0009

*In the last lesson, we found the Legendre polynomials, those P sub L of X where M is equal to 0.*0012

*Remember, we were solving for that function of θ and we have found these Legendre polynomial*0044

*and we decided that was only for the value of N = 0.*0051

*Now, we are going to look at the polynomials 4M not equal to 0, + 1 -1, + 2 -2, and so on.*0058

*These are called the associated Legendre functions.*0079

*They are defined in terms of the Legendre polynomials that we saw, the P sub L of X.*0104

*They are defined terms of P sub L of X which are the actual Legendre polynomial.*0116

*Here is what they look like.*0130

*The symbol for P sub L absolute value of N of X, this is the symbolism.*0132

*It is equal to 1 - X² × V M derivative D absolute value of M DX, absolute value of M.*0141

*DDX will be the first derivative, that would be the first associated polynomial for M = 1.*0155

*If it was M was or equal to 2, it would be second derivative D² DX².*0164

*If M was = 3, it will be D3/ DX3, the third derivative, and so on.*0169

*That is what this M means, let me make this one little bit more clear, the P sub L of X.*0175

*All this is saying is that, for example, if you had L = 3, you would pick the P sub L of 3.*0186

*You would take the third meal, this would be P3.*0192

*Remember what we said, we said that L is going to equal 0, 1, 2, 3, and so on.*0201

*In the case of L = 3, we said that M takes on the value 0, + or -1, + or -2, and + or - all the way to L, so + or – 3.*0209

*In this case, we have 1, 2, 3, 4, 5, 6, 7, different values of M.*0221

*For those different values of M, we take various derivatives and we multiply by 1 - X², in order to get this function.*0226

*That is all we are doing here.*0234

*For every Legendre polynomial, we are going to end up with multiple associated Legendre functions.*0235

*That is was happening, when M does not equal 0.*0242

*Let me write out the specifications here for M a little bit more clearly.*0250

*This is for the value of L is going to be 0, 1, 2, and so on.*0259

*The value of M is going to equal 0, + or -1, + or – 2, all the way to + or – L, whatever that happens to be.*0267

*Because the leading term of P sub L of X because the leading term of the actually Legendre polynomial.*0279

*A polynomial has just a subscript.*0292

*The associated function has the subscript of the function comes from and it has a different value of M for the function itself.*0294

*Polynomial has just a subscript.*0303

*The associated Legendre function has both the sub and a superscript.*0305

*The subscript being L, basic polynomial that all of those associated functions come from.*0311

*Because the leading term of the polynomial P sub L of X is X ⁺L, for example the P sub 2 was X².*0318

*The P sub 3 had an X³, the P sub 4 had an X⁴, that is all of these means.*0330

*P sub L absolute value of M of X becomes 0.*0339

*0 is the absolute value of M is greater than L.*0348

*All that means is that, let us say we take L = 3, if L = 3 the leading term of the polynomial is going to be X³.*0358

*The coefficient does not matter, it is going to be some X³.*0369

*When I take the third derivative of that, the coefficients is going to be some constant.*0372

*If I take the fourth derivative, it is going to be the derivative of a constant which is going to be 0.*0380

*Whenever absolute value of M is bigger than L, these just go to 0.*0387

*It is just a property of the polynomial itself.*0394

*We decided to just throw that in there.*0397

*Let us go ahead and list the first few Legendre associated functions.*0400

*I know this is I want to take in, do not worry about that.*0421

*That is the nature of the beast, we have to run for the material.*0424

*All of this is going to make more sense when we do the problems, when we see over and over and over again.*0430

*It will start to make sense.*0436

*The nature of the beast is such that we have to present the material in such a way.*0439

*We have to present it in a rather large amount, before we can actually work problems that make sense.*0442

*The P sub 0 is 0 of X is equal to 1.*0450

*Here L = 0 and M is going to be 0 but you can only go as high as L.*0456

*There is only one of those.*0464

*We have P sub 1 is 0 of X is going to equal cos θ.*0466

*I will do both, it is equal to X which is equal to cos of θ.*0478

*Because, again X is equal to cos of θ, X is not the coordinate.*0484

*It is the change of variable.*0488

*Here, L is 1 and M is 0.*0490

*P of 1, now N is going to go from 0 and is also do 1 and -1.*0497

*P1 1 is equal to 1 - X² ^ ½ and that is going to end up equaling sin θ.*0505

*I'm just putting all the values into the definition from the previous slide.*0518

*Let me write it now.*0533

*Let me do it in red.*0536

*There is also a P sub 1, M is 0, + or -1, + or -2, + or -3, all the way to + or – L.*0545

*In this case L =1.*0557

*We have done the one M = 0.*0561

*You have done the M, and we have done the one M =1.*0564

*There is also a 1 -1, we will go ahead and put in absolutes.*0568

*That ends up because this is an absolute value sign, the absolute value of -1 is just 1.*0574

*It actually ends up being the same as this.*0581

*For the L = 1, we actually have 3 associated Legendre functions.*0584

*We have this one and we have this one, and we have the one for 1 -1,*0589

*which is actually the same as this because absolute value of 1 is 1 so it ends up being the same as this function.*0594

*That is also 1 - X² ^½ which is equal to sin θ.*0601

*Let me go back to blue here.*0614

*P2 0 is going to equal ½ 3X² -1 = ½ 3 cos² θ -1.*0617

*We have P2 1 that is going to equal 3X × 1 - X² ^½.*0635

*It is going to be 3 cos θ sin θ, when we actually do the change of variable and wherever we see X we put in a cos θ.*0646

*There is also going to be a P2 2, write 2 0, 2 1, 2 2.*0658

*There is also going to be at 2 -1, there is going to be a 2 -2, they happen to be same as these.*0665

*It was a total of 5 of these, = 3 × 1 –X² = 3 sin² θ.*0671

*For L = 0, we have P0 0.*0690

*Let me go ahead and write this up here.*0698

*Notice again, X is not the variable.*0702

*Θ is the variable of interest.*0714

*We just had under change of variable X = cos of θ.*0722

*This is very important to remember.*0727

*Let us go back to blue.*0730

*For the L = 1, we have P1 of 0, we have P1 of 1, and we have P1 of -1.*0735

*The definition uses the absolute value sign so we can leave it like this.*0747

*Because this absolute value sign, the P1 -1 actually ends up being the same function as P1 1.*0752

*It is important to remember that there are 3 of them because L 0, 1, 2, and so on.*0759

*M takes some of value 0, + or -1, + or – 2, all the way to + or – L.*0768

*M depends on L.*0777

*For L = 2, we have P2 of 0, P2 of 1, we have P2 of -1, we have P2 of 2, and we have P2 of -2.*0781

*This should be starting to look familiar.*0800

*When L = 0, that is the S orbital, M = 0.*0822

*When L = 1, we call that the P orbital, when M = 0, + 1, -1.*0836

*When L = 2, that is the D orbital, M = 0, + 1, -1, + 2, -2, 1.*0848

*PX PY PZ, DXY DXZ, DXY DZ².*0867

*There are 5 D orbitals.*0880

*There are 3 P orbitals.*0883

*That is what is going on here.*0886

*The P orbital will end up being PX PY PZ.*0894

*When L = 2, that is the D orbital.*0913

*M takes on the value 0, + 1 -1, + 2 -2, we end up with D sub XY, D sub YZ, D sub XZ, D sub Z², D sub X² – Y².*0918

*That is what is going on here, this is where all this comes from.*0941

*The normalization condition is the integral from -1 to 1, P sub S absolute value of M conjugate.*0945

*P sub T absolute value of M DX is going to be S 0 to π P sub F of cos θ.*0974

*The value of M conjugate × P sub T absolute value of M cos θ.*0993

*This DX term is not just D θ, it is sin θ D θ.*1005

*It is that because we said X is equal to cos θ, this is the change of variable.*1012

*Therefore, DX is equal to sin θ D θ.*1022

*We do it in increments here.*1028

*DX D θ is equal to DX = sin θ D θ.*1032

*This normalization condition, this thing, when we actually solve it we end up with the following.*1049

*We end up with 2/ 2L + 1 × L + the absolute value of M!/ L - the absolute value of M!.*1057

*Therefore, all of that is going to end up equaling 1.*1083

*The integral of this thing is equal to this thing.*1089

*This normalization condition is equal to 1.*1093

*Therefore, when we actually solve this we end up with the following normalization constant.*1096

*The normalization constant LM is equal to 20 + 1/ 2 × the absolute value of L – that !/ L - the absolute value of M.*1105

*I hope you got them keeping all this straight, this is absolutely crazy, ½.*1128

*That is the normalization constant for the particular associated Legendre polynomial*1135

*P sub L absolute value of M, LM each one has a differential normalization constant based on this.*1145

*It looks very complicated, it is not.*1154

*L and M are really going to be just 1, 2, 3, that is about it 0, 1, 2, 3.*1156

*Let us make sure we know where it is that we are.*1168

*We started all this by looking for the angular component of the hydrogen wave function.*1170

*We were looking for S of θ of φ which is equal to some T of θ and some F of φ.*1189

*We have found this then we found that.*1205

*We put them together now, right.*1210

*We found F of φ, we found T of θ, now we are going to multiply them because S of θ φ is equal to T of θ × F of φ.*1221

*So now we are going to write out the entire wave function symbolized.*1231

*The angular component L M of θ φ is going to equal 2L + 1.*1236

*It is going to look a lot more complicated than actually is in practice.*1251

*× L - the absolute value of M!/ the absolute value of L + the absolute value of M! ½.*1256

*The takes care of the normalization constant for that.*1274

*Now, we are going to multiply by the normalization constant for the F of φ which is 1/ 2 π ^½ × P sub L absolute value of M cos θ × E ⁺IM φ.*1277

*Where, L takes on the values 0, 1, 2, and M takes on the values 0, + or -1, + or – 2, all the way to + or – L.*1300

*This is what we wanted, we wanted the angular component of the hydrogen atom wave function is this.*1315

*That is the normalized wave function for the angular component of the hydrogen atomic orbital.*1328

*This is what it looks like.*1339

*Let us go ahead and combine the two normalization constants.*1342

*Make it a little bit easier to look at here.*1345

*We have S sub L M of θ of φ is going to equal 2 O + 1, when we combine the two constants*1348

*it is going to end up being that, / 4 π × L - absolute of M!/ L + absolute of M! ^½ P sub LM cos θ E ⁺IM φ.*1361

*Where L is 0, 1, 1…*1410

*M = 0, + or -1, + or – 2, all the way to + or – L.*1415

*These are the wave functions for the angular component of*1424

*the hydrogen atom wave function ψ of R θ φ, which is R of R, S of θ φ.*1452

*This is just the part right there.*1473

*They are also the wave functions for the quantum mechanical rigid rotator that we promised earlier, when we are just dealing with the energies.*1477

*These normalized wave functions are here.*1510

*The normalized wave functions a formal orthogonal set.*1517

*They are mutually orthogonal functions form and orthonormal set.*1523

*Let me go back to blue here.*1538

*These functions are called the spherical harmonics.*1541

*The first few spherical harmonics, let us list them out.*1561

*The first few spherical harmonics S0 of 0 is equal to 1/ 4 π ^½, it is a constant.*1568

*We are just putting for every value of L with the value of M and we list them out.*1590

*We just put in the equation.*1596

*The equation looks big but L and M are such tiny numbers but everything ends up condensing becoming really easy to deal with.*1598

*S1 of 0 is equal to 3/ 4 π ^½ × the cos of θ.*1606

*S sub 1 1 is equal to 3/ 8 π ^½ sin θ E ⁺I φ.*1621

*S1 -1, 1 0, 1 1, 1 -1, it = 3/ 8 π ^½ sin θ E ⁻I φ, because now N is -1.*1641

*Remember, it is E ⁺IM φ.*1663

*We are going to have E ⁻I φ, that is always going to be like that for the + or -1, + 2 -2.*1666

*Let us do, where are we next?*1675

*We are at S20 that is going to equal 5/ 16 π ^½ × 3 + sin² θ -1.*1677

*Let me write then, I can continue on this page, not a problem.*1698

*We will do S21 is going to equal 15/ 8 π.*1702

*It looks like I forgot about my notes, it has a ½ there.*1717

*We have sin of θ cos θ E ⁺I φ.*1721

*Let us come up here and do S2 -1 is equal 15/8 π.*1729

*Everything else is absolutely the same.*1737

*Sin θ cos θ E ⁻I φ -1 + 1.*1741

*Let us do S22, it = 15/ 32 π, just running for the numbers.*1753

*L0 and M0, L1 M0 1 -1, L2 M0 1 -1 2 – 2.*1764

*That is all that we are doing.*1774

*32 π ^½ this one is going to be sin² θ E ⁺I φ.*1777

*S2 -2 = 15/ 32 π ^½ sin² θ.*1793

*What do you think?*1802

*E ⁻I2 φ, that is all that is happening here.*1804

*Let us talk a little bit about angular momentum.*1813

*When we talk about the rigid rotator, when we discussed the rigid rotator,*1817

*we found that the square of the angular momentum operator*1832

*was actually equal to – H ̅² × 1/ sin θ DD θ sin θ DD θ + 1/ sin² θ D² D φ².*1843

*This is the operator for the square of the angular momentum.*1867

*I do not want to keep repeating myself over and over again so I’m just going to call of that Z.*1871

*We had for S of θ φ, we had -1/ S × 1/ sin θ DD θ sin θ DS D θ + 1/ sin² θ D² S D φ²,*1888

*all of that was equal to θ which is equal to L × L + 1.*1946

*We said θ is equal to L × L + 1.*1951

*If we multiply both sides by H ̅² S of θ φ, we end up with – H ̅² Z is equal to H ̅² × S × L × L + 1.*1954

*We have L ̅² S of θ φ is equal to H ̅² × L × L × L + 1 × S.*2003

*I just rearrange this thing.*2030

*Notice A of ψ = λ of ψ, Eigen value, Eigen function kind of thing.*2038

*We have L ̅² of S is equal to H ̅² × L × L + 1 × S.*2046

*Here is the operator, here is the function.*2057

*It is equal to this scalar × the function which means that the S sub L of M of θ φ, they are Eigen functions.*2060

*In other words, when we apply the angular momentum operator to S, when we apply it to S we end up getting this thing.*2081

*We end up getting some constant × S back.*2092

*This is the condition for Eigen value Eigen function.*2095

*The angular component, the wave functions for the rigid rotator happens to be Eigen functions of DL² operator.*2099

*Let me rewrite this so that we have it on one page.*2123

*L ̅² of S, I will get rid of θ and φ, is equal to H ̅² × L × L + 1 × S.*2127

*The square of the angular momentum, when I actually take measurements*2145

*of the angular momentum because this is an Eigen function of the operator.*2154

*Because when we operate what we actually get the function back × the constant,*2161

*that means when we take measurements, those are the only values that we actually get.*2164

*Square of the angular momentum can only have the values² = H ̅² × L × L + 1.*2168

*Or again L = 0, 1, 2, and so on.*2188

*Those are the only possible values for the square of the angular momentum.*2192

*If I measure the angular momentum, the square of the value that I end up getting*2195

*is going to be one of these values depending on what L is.*2199

*It was the only values that can take.*2202

*The angular momentum is quantized.*2204

*Recall that the Hamiltonian operator is equal to the square of the angular momentum operator*2207

*÷ twice the rotational inertia for the rigid rotator.*2217

*If I apply this, I’m going to do H ̅ of S LM, it is equal to L ̅²/2I S LM is equal to H ̅² × L × L + 1/ 2I S LM.*2227

*This thing should look familiar.*2266

*All I have done is, I already had this L² of S = something, = this thing.*2268

*H is just L²/ 2I.*2275

*I took this thing and just divide it by 2I.*2280

*They should be familiar.*2284

*These are just the energies of the quantum mechanical rigid rotator.*2289

*Remember, the Hamiltonian is just the energy function.*2295

*It is the energy operator, the total energy operator.*2299

*They should look familiar.*2302

*They are the energies of the rigid rotator.*2306

*Back when we discuss the rigid rotator, we used J E sub J = H ̅²/ 2I J × J + 1.*2315

*Now, we are replacing J with this L that actually shows up.*2325

*That is all that is going on here.*2330

*Thank you so much for joining us here at www.educator.com.*2332

*We will see you next time for a continuation of the hydrogen atom, bye.*2335

3 answers

Last reply by: Professor Hovasapian

Thu Mar 26, 2015 5:26 PM

Post by matt kruk on March 24, 2015

hi professor so how does the legendre polynomial correlate to the legendre equation. do you multiply it through the equation. i'm trying to understand it mathematically and physically. Thanks so much

1 answer

Last reply by: Professor Hovasapian

Sun Nov 2, 2014 3:22 AM

Post by xlr z on November 1, 2014

Is the associated legendre function = (1-x^2)^1/2 or is it = (1-x^2). it seems for l=2 m=2 you are using (1-x^2)^1/2 and for l=2 m=1 you are using (1-x^2). so is there a square root?

thank you