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

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

### Hydrogen Atom Example Problems I

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: Angular Momentum & Spherical Harmonics 0:20
- Example II: Pair-wise Orthogonal Legendre Polynomials 16:40
- Example III: General Normalization Condition for the Legendre Polynomials 25:06
- Example IV: Associated Legendre Functions 32:13

### Physical Chemistry Online Course

### Transcription: Hydrogen Atom Example Problems I

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

*We spent some time talking about the hydrogen atom and now we are going*0005

*to do the example problems for the hydrogen atoms.*0009

*There are going to be a lot of them.*0012

*No worries, we would spend plenty of time working in example problems because this is very important.*0015

*Let us jump right on in.*0021

*Example number 1, let S² equal to this expression.*0023

*This of course is the square of the angular momentum operator.*0028

*For L = 0 and L = 1, find this expression.*0033

*L² of S sub L super M θ φ, so these are the spherical harmonics.*0038

*We want you to find, use this operator to operate on the spherical harmonics for L = 0 and L = 1.*0044

*Let us see, let us go ahead and start.*0054

*Let me see, should I do black or blue, or red.*0056

*I will stick with black today, at least for this one.*0061

*For L = 0, we have that M is equal to 0.*0066

*You remember that, you know M is equal to, it goes all the way from -L all the way to +L, passing through 0.*0071

*In the case of L = 0, M just equal 0.*0088

*We have S0, 0 and it is equal to, let me go ahead actually use my θ and φ.*0092

*S0, 0 of θ φ is equal to 1/ 4 π ^½.*0100

*Of course, you will get all of these from your book.*0111

*There are list of the spherical harmonics are listed.*0116

*The Legendre polynomials are listed.*0118

*The Laguerre polynomials are listed.*0120

*All the full wave functions are listed.*0122

*And if for any reason you are not using a book, I will just go ahead and look them up online.*0124

*There they are listed. These are spherical harmonics.*0128

*This is our S0, 0, now we want to find L² S0, 0.*0132

*In this case, let us go ahead N equal to 0.*0139

*This one is an easy one because S0, 0 is not a function.*0143

*It is a function of θ and φ but it does not depend on θ and φ.*0152

*Notice, this is just some constant so θ and φ do not show up.*0157

*I had nothing to actually take the derivative of that.*0160

*When I take the derivative of a constant, it is just going to equal 0.*0162

*Because S0, 0 does not depend on θ or φ.*0170

*Let us go ahead and do for L = 1.*0181

*For L = 1, N is going to equal -1, 0 and + 1.*0186

*We have three of these to actually take care of.*0192

*Let us go ahead and do S1, 0 first.*0194

*Let us take care of this 0 and 1 first.*0197

*S1, 0 that is equal to 3/ 4 π to ½ cos θ.*0203

*-H ̅², we are going to use this operator and we are going to operate on this right here.*0217

*Let me go ahead and write it all out.*0224

*Again, as you know quantum mechanics is mostly, it is not particularly difficult,*0228

*it is just computationally intensive, notationally intensive.*0232

*That is the real issue and it can be intimidating.*0235

*Taking a look at an expression like this and realizing what is going on.*0239

*Nothing, it is just the second derivative here, first derivative, first derivative, that is all that is happening here.*0243

*No worries.*0249

*Let us see, sin θ DD θ + 1/ sin² θ D² D φ.*0253

*We are going to operate on the function 3/ 4π ^½ cos of θ.*0269

*Let us just take this one at a time.*0282

*We are going to go this on that and this on that.*0284

*Essentially, this distributes over that, this distributes over that.*0289

*I’m going to start here, I’m going to do this one first.*0292

*When we take DD θ of this cos θ, I’m going to go ahead and put the constant here.*0299

*I’m just going to deal with what the functions of θ and φ.*0307

*Stick with constants at the end.*0311

*DD θ of cos θ, this thing is going to end up being sin θ × - sin θ.*0313

*It is going to equal - sin² θ and then we are going to do this one, the DD θ of that.*0326

*We are going to end up getting -2 sin θ cos sin θ.*0338

*We are going to multiply that by 1/sin θ.*0348

*This is a nice way of actually doing it, in barrels with each operation.*0351

*We end up getting - 2 cos θ.*0357

*We end up getting that when we operate on that.*0363

*The second term, the second term does not matter.*0366

*The reason it does not matter is because this D² D φ of this is just 0.*0370

*There is no φ here, so when we take the derivative of it, it just goes to 0.*0376

*This 1 / sin² θ does not matter because we are just multiplying by 0.*0380

*I’m going to go ahead and write that in blue here.*0385

*The second term of the L² operator does not matter because S1, 0 does not involve the φ.*0390

*We have L² of S1, 0 is equal to, I will go ahead and put all the constants back in.*0421

*-H² 3/ 4 π ^½ × - 2 cos θ.*0429

*That is what we got when we get the operation.*0439

*That is going to equal -2 H ̅² 3/4 π ^½ cos θ.*0442

*Notice this part, it is the same as that is the S1, 0.*0453

*Therefore, we have L ̅² of S1, 0 is equal to -2, the minus sign go away so we are left with.*0459

*I’m sorry this is a + here.*0474

*We have 2 H ̅ S1, this is the S1, 0 so I can just call it that.*0481

*LR² S1, 0 is equal to some constant × S1, 0.*0494

*Therefore, S1, 0 is an Eigen function of the square of the angular momentum operator.*0501

*L1 with Eigen value 2 H ̅, that is it.*0518

*We are just confirming something that we already know but it is nice to go through the process to do the mathematics.*0532

*We have taken care of S1, 0, let us go ahead and take care of S1, 1.*0540

*Let us do S1, 1, it is equal to 3 / 8 π to ½ and it is going to be sin θ E ⁺I φ.*0546

*This time, both θ and φ are involved.*0562

*We are going to operate, this is going to be –H ̅².*0566

*I hope you will forgive me if I keep repeating the operator over and over again.*0570

*To write down as much as possible and I’m not sure if that is bad.*0575

*I think that is probably pretty good.*0579

*Sin θ DD θ + 1/ sin² θ D² with respect to φ² this.*0582

*Of course, we are operating on for B/ 8 π ^½ sin θ E ⁺I φ.*0597

*We just have to do keep things straight, do our best to keep everything in line.*0607

*We are going to ignore the constant and I’m just going to do this one first.*0612

*This one is going to be DD θ of this, it is going to be the derivative of that is sin θ.*0617

*The derivative of that is cos θ so we are going to have sin θ × cos θ.*0628

*When we apply now this one, so we are going to apply the DD θ part.*0636

*We are going to get - sin² θ and I hope that I actually done my differentiation correctly.*0643

*- sin² θ + cos² θ and I'm going to multiply by 1/ sin θ.*0652

*That is going to end up giving me cos² θ.*0674

*Let us see if I can keep track of everything that is going on here.*0687

*We got cos² θ, let me drop it down here.*0700

*I'm going to end up getting this, I’m doing everything here.*0708

*Cos² θ - sin² θ/ sin θ.*0711

*Of course, we cannot forget about that.*0721

*E ⁺I φ that is going to be + sin θ or sin² θ I² E ⁺I φ.*0724

*That is why we operated this one on that.*0739

*It is going to end up being, cos² θ – sin² θ/ sin θ -1 ×,*0745

*this is going to be -1/ sin θ × E ⁺I φ.*0768

*I can go ahead and do on the next page here.*0781

*It is going to be -2 sin² θ, I’m just using basic identities.*0784

*Θ/ sin θ × E ⁺I φ which is equal to - 2 sin θ E ⁺I φ, there we go.*0792

*This is from our final answer after operating on that.*0808

*L ̅² of S1, 1 is equal to -H ̅² × 3/ 8 π ^½ × - 2 sin θ E ⁺I φ, which is equal to 2 H ̅ S1, 1.*0816

*S1, 1 is an Eigen function of L ̅².*0844

*The Eigen value is going to be 2 H ̅.*0855

*We see the pattern here and let us go ahead and do one more thing.*0861

*Now, we have to do the S1,-1.*0865

*That is going to equal, the constant is the same.*0869

*It is 3/ 8 π, this is listed in your books or on the net, sin θ.*0872

*Except, it is going to be E ⁻I φ.*0880

*When I go ahead and apply the L² operator to this function, I actually end up getting the same thing as I did before.*0884

*It is going to end up being L ̅² operating on S1, -1.*0893

*It is actually going to end up giving me 2 H ̅ S1, -1.*0899

*It is going to be the same as for the S1, 1.*0906

*S1, -1 is an Eigen function of the square of the angular momentum operator.*0914

*We already saw in the previous lesson, we already saw the general result in a previous lesson.*0926

*L ̅² of SLM is equal to H ̅² × L × L + 1 × SLM.*0951

*Here is the Eigen value and the spherical harmonic is an Eigen function of the square of the angular momentum operator.*0965

*Of course, the magnitude of the angular momentum vector is going to equal H ̅ × L × L + 1.*0974

*That is the general result.*0989

*Let us go ahead and go to example problem 2.*0999

*For the Legendre polynomials, P1 of X, P2 of X, and P3 of X, they are orthogonal.*1003

*We look them up either online or in our books, these are listed at least for the first few, for the first 4 and 5.*1012

*P1 of X is just equal to X, P2 of X is equal to ½ 3X² -1 and P3 of X is equal to ½ 5X³ -3X.*1021

*Recall that X is equal to cos Θ, a little change of variable.*1046

*And that θ ran from 0 to π.*1056

*Cos of 0 is equal to 1, cos π is equal to -1.*1063

*Therefore, the value of X actually goes from -1 all the way to 1.*1072

*This is change of variable, when we put X in here, we actually get the actual function of θ.*1080

*When we do the change of variable and we call the cos θ X, just to make it look a little simpler, that is all.*1086

*Let us see what we can do for P1 and P2.*1093

*We have the P1 P2, P1 P3, and P2 P3.*1103

*We are going to check that they are orthogonal.*1107

*In other words, what we are going to check is the following.*1110

*We want to check to see that this integral P1 P2, we want that to equal 0, that is what we want.*1111

*-1 to 1, P1 is X, X conjugate is just X.*1128

*It is a real number so it is just going to be X × ½ 3X² -1 DX.*1133

*You can solve the integral directly if you want, it is not a problem.*1147

*Just go ahead and multiply out, integrate it, just like you would in any other function.*1155

*You can do this one by hand very quickly.*1159

*Integral directly or just like the Hermite polynomials of the harmonic oscillator.*1161

*We will try to make our lives a little bit easier if we can by dealing with integrals*1192

*that we are going to see over and over again.*1196

*Instead of just actually integrating over and over again.*1198

*P sub L, the Legendre polynomial sub L is even when L is even.*1205

*It is an even function.*1212

*It is even when L is even and it is odd when L is odd.*1216

*This integral -1 to 1 of this thing X × ½ 3X² -1 DX, it is actually equivalent to the integral for -1 to 1, an odd function.*1232

*X is an odd function × an even function, this ½ 3X² -1 is actually an even function.*1248

*We know from previous work that an odd function × an even function is an odd function.*1257

*The whole thing when we multiply that out, it is actually going to be an odd function.*1266

*The integral of an odd function / a symmetric interval, symmetric about 0 -1, -π, something like that is going to equal 0.*1271

*The integral of an odd function/ a symmetric interval, in this case from -1 to 1.*1289

*Symmetric interval is 0.*1307

*The integral from -1 to 1 of P1* P2 is equal to 0.*1315

*P1 and P2 are orthogonal.*1323

*Let us do P1 and P3.*1328

*For P1 and P3, we are going to have the integral from -1 to 1.*1335

*P1 conjugate P3, that is going to equal the integral from -1 to 1 of X × ½.*1348

*This is going to be 5 X³ -3 X DX.*1357

*X is an odd function and this one also happens to be an odd function.*1371

*An odd × an odd function is actually equal to an even function.*1375

*This one we have to integrate directly.*1384

*We integrate directly and it is not a problem at all.*1391

*This is going to equal 1/2 integral from -1 to 1.*1397

*Let us go ahead and pull the ½ out, distribute the X.*1402

*We have 5 X⁴ -3 X² DX = ½.*1407

*Of course, this one is going to be X⁵ – X³ from -1 to 1 = ½ is going to be 1 -1 - -1.*1419

*--1 that is going to be the hard part to keep track of all of these.*1437

*It is going to end up being equal to 00, it is going to end up being 1/2 is 0 = 0.*1441

*These are orthogonal.*1454

*For our last one, for P2 and P3.*1460

*P2 is an even function, P3 is an odd function.*1467

*We know that they are even × an odd is going to equal an odd function.*1473

*We know that the integral / symmetric interval of an odd function is equal to 0.*1478

*These are also orthogonal.*1488

*There we go, and of course this is true for all the Legendre Polynomials.*1491

*They are all going to be pair wise orthogonal.*1496

*Example 3, for the Legendre polynomials P2 and P3 show that they satisfy the general normalization condition*1509

*for the Legendre polynomials which is the integral of its² DX is equal to 2/ 2L + 1.*1519

*This is the general normalization condition for the Legendre polynomials.*1530

*This is how we get R, if we needed to, to get our particular constants that make the integral equal to 1.*1542

*We just need to put in a P2 and put in a P3 to show that it actually satisfies this.*1552

*Let us see what we have got.*1560

*P2 is equal to ½ 3X² -1.*1563

*In this particular case, L is equal to 2.*1572

*When we do the integral, we should get 2/ 2 × 2 + 1.*1577

*L is 2 so we should get 2/5.*1584

*That is our target, let us go ahead and see if that is true.*1591

*P sub 2² that is going to give us ¼ × 9 X² , we should do our algebra correctly, -6X² + 1.*1597

*We have ¼ of the integral from -1 to 1 of 9X⁴ -6 X² + 1 DX.*1618

*That is going to equal ¼ 9X⁵/ 5 – 2 X³ + X from -1 to 1.*1633

*This is going to equal ¼, that is what it is going to give you for actually going through all the tedium of these integrations.*1649

*I would not do to all the time but at least for now, I think it is fine.*1657

*We are going to have 9/5 -2 + 1, a – and --9/5 + 2 -1 = 1/4 4/5 –and - 4/5 is going to equal 2/5, when you do the arithmetic.*1661

*Yes, it satisfied that relation for P2.*1689

*Let us go ahead and see if it actually works for P3.*1694

*P3 of X is equal to ½ 5X³ -3X, in this particular case L is equal to 3 so we should get the 2/2 × 3 + 1.*1699

*We should get our final answer of 2/7 when we do the integral.*1721

*P3² is going to equal ¼ and is going to be 25 X⁶ -30 X⁴ + 9 X².*1727

*We have a 1/4 of integral from -1 to 1 of this thing which is 25 X⁶ -30 X⁴ + 9 X² DX.*1744

*This is going to end up equaling, let me go ahead and put is here.*1758

*The ¼ and it is going to be 25 X⁷/ 7 -6 X⁵ + 3X³, all the way from -1 to 1.*1764

*This is going to equal ¼ 25/7 -6 + 3 -8 -25/ 7 + 6 -3.*1780

*It is going to be ¼ × 4/7 -and -4/7.*1795

*When you do the arithmetic, it is going to equal 2/7.*1804

*Yes, it also worked.*1808

*That is your general formula.*1810

*Let us see what we have got next.*1816

*These are the Legendre polynomials, add them aside.*1822

*Your book is going to list the first 5 or 6 Legendre polynomials.*1830

*But I'm guessing that your book probably does not have a general formula for generating the Legendre polynomials.*1834

*They are actually several them available, I'm going to go ahead and give you the one that is the most easiest.*1841

*As inside, there are several formulas for the nth degree Legendre polynomial P sub N of X.*1846

*The easiest is probably Rodriguez’s formula.*1876

*It is the nth degree Legendre polynomial is equal to 1/ N! 2 ⁺nth power.*1894

*The nth derivative, let me make my L a little bit clearer here.*1904

*The nth derivative of X² -1 ⁺N.*1912

*If you actually want to generate a particular polynomial, let us say the 15th degree Legendre polynomial, this is the formula that you would use.*1918

*Let us see what we have got.*1931

*Example number 4, the general formula for generating the associated Legendre function.*1933

*This is not the Legendre polynomial, this is the associated Legendre function.*1940

*It actually has 2 indices, a subscript and a superscript is this right here.*1946

*Use this relation to generate P2, 0, P2, 1, and P2, 2.*1956

*Express in terms of θ not X.*1961

*Recall the X is equal to cos θ, when you find the function of X, you are just going to put sin θ for that and simplify if you can.*1964

*I will use the basic trigonometric identities that you remember from pre calculus.*1972

*Let us see what we have here.*1977

*P2 of 0, we are just going to put them into this.*1979

*That is it, that is all we are going to do.*1988

*It = 1 - X², this is L the subscript, the superscript is M.*1990

*This is L = 2, N = 0, so we put it up here and it is going to be 0/ 2 D0 DX is 0 of P2.*2001

*In this particular case, I can go to notational state here.*2028

*L is the associated Legendre function.*2035

*This should be M, there we go.*2040

*In this case D00 P2, this just is 1 and this 0 or the derivative is just P2.*2053

*It = ½ 3X² -1.*2069

*It is LL P2.*2080

*X = cos θ, therefore, our P2 0 as a function of θ is equal to ½ × 3 cos² θ – 1.*2086

*In other words, we are generating the full function, the associated Legendre function from the Legendre polynomial.*2117

*The one with just the subscript is the Legendre polynomial.*2125

*The one with subscript and superscript is the associated Legendre function.*2128

*Let us go ahead and do P2, 1.*2135

*P2, 1 that is going to equal 1 - X² ^½ D1 D X1 of P sub L P sub 2 which is ½ 3 X² -1.*2138

*That is going to equal 1 -X² ^½.*2161

*When I take the derivative of this, it is going to equal 6 X/ 2.*2168

*It is just going to equal 3 X so I end up getting 3 X × 1 -X²¹/2.*2174

*I have X equal cos θ, this is my function associated Legendre function in X.*2190

*X = cos θ, remember we want to express it in terms of θ.*2197

*We are going to have 3 × cos θ × 1 - cos² θ ^½.*2201

*You remember basic trigonometry identity 1 – cos² θ is equal to sin² θ.*2214

*We have 3 cos θ time sin² θ ^½.*2219

*We end up with 3 cos θ sin θ, there you go.*2225

*This is the associated Legendre function P2, 1 expressed as a function of θ.*2233

*Note, L = 2 which mean that N is equal to -2 -1, 0, 1, and 2.*2250

*We have taken care of this one, the 0 one.*2265

*I will be taking care of the 1.*2268

*The formula actually has the absolute value of M.*2272

*There is also a P2 absolute value of -1 but the absolute value of -1 is 1.*2277

*It is actually the same as P2, 1 so there is another one.*2294

*There is a P2 absolute value of -1 which happens to be the same as this.*2298

*We are going to end up with three of them but we actually have 5 of them.*2303

*One for each value of M because M runs from -L all the way to +L, in increments of 1 passing through 0.*2306

*There is also at which is the same as P2, 1 because the absolute value of -1 is equal to 1.*2318

*Let us go ahead and do P2, 2 here.*2343

*We have P2, 2, it is going to be the formula 1 - X²²/ 2.*2348

*We have D2² DX² of ½ 3X² -1.*2362

*We are going to differentiate this twice.*2373

*Let us go ahead and just do this part.*2376

*We have ½ 3 X² -1.*2382

*Now, we are going to differentiate it the first time and we would end up with 3X.*2387

*We take the second derivative, we are going to end up with 3.*2393

*Our P2, 2 is going to equal 1 - X²¹ × 3.*2400

*It is going to equal three × 1 - cos² θ.*2412

*1 – cos² is sin² θ.*2418

*We have three sin² θ.*2420

*This is our P2, 2 expressed as a function of θ.*2425

*And we also have a P2 absolute value of -2, that is going to be the same.*2431

*It is going to be 3 sin² θ.*2437

*Recall what these are, this is going to be the important part here.*2444

*Recall what these associated Legendre functions are, in case you have forgotten.*2451

*They are the T θ portion of the spherical harmonics.*2480

*The angular portion of the wave function for the hydrogen atom.*2497

*It is also the wave function of the rigid rotator.*2505

*They are the T θ portion of the spherical harmonics.*2511

*Let me go ahead and do one more page here.*2518

*Basically, what we have is we have this ψ which is a function of R θ and φ.*2520

*As a function of three variables, spherical coordinates, the radius changes, the φ changes, the θ changes.*2527

*That is how we get the wave function for the electron of the hydrogen atom.*2533

*We separated that into a radial portion R N sub L which is just a function of R and S sub LM which was a function of θ and φ.*2539

*These are the spherical harmonic, the radial function, the angular function.*2553

*The angular function we call the spherical harmonics.*2557

*When we are solving this equation, this is a function of 2 variables θ and φ.*2559

*We broke out that up into 2.*2563

*S L sub M of θ φ.*2566

*We brought that up into a function T of θ and a function F of φ.*2570

*This right here, this T of θ, those are the associated Legendre functions.*2579

*A lots of functions flowing around now.*2592

*It is very easy to get them all confused.*2595

*Hermite polynomials, Legendre polynomials, Legendre functions, Laguerre polynomials,*2599

*radial functions, having keep them all straight, this is what we did.*2606

*We broken it up into 2 and we broke this up into further down.*2610

*Right now, we are dealing with the associated Legendre functions.*2613

*These are the associated Legendre functions.*2617

*Thank you so much for joining us here at www.educator.com.*2623

*We will see you next time for a continuation of example problems.*2625

*Take care, bye.*2628

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