Physical Chemistry 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 going0005

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 integrals1192

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 condition1509

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