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Lecture Comments (6)

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

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

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 polynomial1135

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 operator2207

÷ 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