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### The Rigid Rotator 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
• The Rigid Rotator II 0:08
• Cartesian Coordinates
• Spherical Coordinates
• r
• θ
• φ
• Moving a Distance 'r'
• Moving a Distance 'r' in the Spherical Coordinates
• For a Rigid Rotator, r is Constant
• Hamiltonian Operator
• Square of the Angular Momentum Operator
• Orientation of the Rotation in Space
• Wave Functions for the Rigid Rotator
• The Schrӧdinger Equation for the Quantum Mechanic Rigid Rotator
• Energy Levels for the Rigid Rotator

### Transcription: The Rigid Rotator II

Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.0000

Today, we are going to continue our discussion of the rigid rotator.0004

Let us jump right on in.0007

In the last lesson, we said that because there is no potential energy term V of X,0010

we said that the hamiltonian operator is just equal to the kinetic energy operator0037

which is equal to -H ̅² / twice the reduced mass × the del² operator.0042

We know or we have seen that the del² is equal to D² DX² + D² DY² + D² DZ²,0051

as an operator in Cartesian coordinates.0071

For an object rotating in space, Cartesian coordinates are not exactly the most,0076

We can certainly use it, it just that one thing is rotating in a circle, polar coordinates,0095

spherical coordinates, tend to be a more natural choice.0101

They actually make the math easier which is why we actually have these various coordinate systems.0104

Certain problems just by the virtue of their particular symmetry lend themselves to a particular coordinate system.0109

However, for an object rotating about a fixed center in any spatial orientation, and when I say spatial orientation,0116

I mean it can rotate like that or like this or like this.0151

The molecule itself can be oriented in any way and can rotate however it wants.0159

Rotating in any special orientation, spherical coordinates are more natural choice.0165

This is going to take a couple of minutes to review, spherical coordinates and what they are.0182

I’m not going to say too much about them.0185

A point in 2 space requires 2 coordinates to represent.0195

A point in 3 space requires 3 coordinates.0198

A point in N space requires N coordinates.0201

Whatever the dimension of the space is, you need at least that many numbers to represent that point.0204

A point in 3 space requires 3 coordinates to describe its position.0209

We know in Cartesian coordinates we use X, Y, and Z.0236

In spherical coordinates, which is the 3 dimensional version of polar coordinates,0243

the spherical coordinates we are going to use R θ and ϕ.0248

Let us talk about what R θ and ϕ actually are.0255

I will draw on the next page, that is not a problem.0260

Let us go ahead and draw ourselves a little coordinates system here.0267

A 3 dimensional coordinates system is something like this.0270

We have our X axis, we have our Y axis, and we have our Z axis.0275

I’m going to pick some random point in the first octet and this is going to be the vector.0279

This is going to be our point R θ and ϕ, in terms of R θ and ϕ.0287

Let me go ahead and draw a couple of things.0299

We dropped a perpendicular projection down to the XY plane.0301

That is going to give me this thing right here from there and that is it.0308

This is the point.0315

Here is where the R θ ϕ comes in.0316

R is the length of the actual vector.0318

It is a length from the origin to the point X, Y, Z.0320

In this case, it is represented by R θ and ϕ.0325

R θ is going to be this angle right here.0330

That is angle of θ.0337

That is the angle that makes, the R makes with the +Z axis.0340

Φ is this angle right here.0346

It is angle that it makes from the +X axis.0351

R is going to be greater than 0, greater than or equal to 0.0358

Θ is going to run from 0 π and ϕ is going to be greater than or equal to 0 and less that or equal to 2 π.0363

Those are the possible values.0374

R is the length of the vector.0376

Do not worry if this picture is not making complete sense.0385

Right now, I’m going to explain what it means.0387

It is the length of the vector’s θ.0389

It is the angle that a vector makes with the +Z axis.0392

Θ runs from 0 to π.0418

Φ is the angle at the projection of R onto the XY plane makes with the +X axis.0421

We need a reference point, so we have something that we start with.0456

Let me go this way.0460

This is the Z axis right, if I move along the Z axis at distance R and this is the Z axis, this is going to be the X axis.0461

Why you, yourself, from your perspective you do not see it.0472

If I move along the Z axis and I pull away an angle θ from the Z axis and then from here, if I end up rotating this way.0476

Now R is here and I’m going to rotate it this way.0485

This third angle rotation that is the ϕ.0489

R θ and then swing out the ϕ.0492

If you move a distance R from the origin along the positive Z axis0497

and then run θ from 0 all the way to π,0529

then run ϕ from 0 all the way to 2 π, you are going to end up sweeping out the sphere of radius R.0539

This is the origin of the name spherical coordinates.0567

Let us see what this actually this looks.0570

I'm guessing that you guys are familiar with this already but again for those of you that are not,0572

we definitely want to make sure that you understand this.0576

Let me draw this out.0579

What I'm going to draw out here is I'm going to draw just the Z axis and the X axis.0582

We said from the origin we want to move at distance R up to Z axis.0591

Let us come up here.0596

Now the next thing we are going to do is we are going to swing this R, we are going to take θ from 0 to π.0598

We are going to swing this 180°.0605

I’m going to take this and I'm going to swing it around this way, a full circle 2 π.0611

When I do that, when I take this and I sweep it around,0620

Remember, rotations in calculus class, I'm going to end up actually sweeping out a sphere.0624

What I end up getting is something that looks like this.0629

I’m going to start from R, the origin then I go up the Z axis, that is my length R.0634

I’m going to end up swinging this down.0643

I already used up so much space, it does not have to be so big you will still understand it.0648

I'm going to end up swinging it up R and I’m going to swing it down this way.0655

I will draw a sphere and make it a lot better.0666

This is the center along the Z axis.0678

I swing it down this way, this right there is my θ.0683

My θ goes from 0 all the way down to π.0688

I’m going to take this and I'm going to swing around full circle.0693

It is going to go all the way around that is my ϕ.0697

I’m going to end up sweeping out a sphere and that is where this comes from.0702

We do not want to say any more about that.0708

Hopefully that is pretty clear.0709

We are going through a lot of these mathematical stuff because it is part and0711

You need to see these techniques.0722

You do not have to know it, you do not have to do it yourself, you do not have to do the derivations.0724

They are so many partial differential equations but you want to be able to see it.0728

It is important, we do not want to leave you with a feeling that they just dropped out of the sky.0731

At least it would make things possible for you, it is easier to wrap your mind around it.0736

That is why we are doing this.0740

We do not want to get caught up in the derivation.0743

We want to be able to follow derivation.0745

We want to spend most of our time thinking about the final equations that we get.0747

The energy functions and wave functions, but we do need to go through the derivation.0751

It is part of your education.0755

In spherical coordinates, our del² is actually equal to this.0757

It looks like a very complicated expression and it is, but we can simplify it.0765

It is going to be 1/ R² DDR.0770

I, myself, do not memorize this.0777

That is fine, I will go ahead and move θ and ϕ here anytime we have some partial derivative.0790

Out here as a subscript, that means holding this constant while we take the partial derivative.0794

It is implicit, it is part of the notation.0800

I will just go put it this one time and not going to keep repeating it.0803

+1/ R² sin θ × DD θ of sin θ DD θ.0808

This is going to be R and ϕ + our final term which is 1/ R² sin² θ and D², this time D ϕ².0821

This is going to be RN θ.0834

For the rigid rotator, this is where the R, θ and ϕ, all three of these things varying.0838

R changes length, θ changes length, ϕ changes length.0848

It could be anywhere in space.0854

For the rigid rotator, R is fixed, R is constant.0856

R is the length in between the two masses.0865

R is constant so del² ends up actually simplifying a little bit.0873

We end up with 1/ R² sin θ DD θ sin θ DD θ + 1/ R² sin² θ D² D ϕ².0880

I know it does not look well altogether but it is, believe me.0902

We had that the rotational inertia is equal to the reduced mass × the R².0912

Let me go ahead and solve for the reduced mass.0922

M= I / R².0924

The hamiltonian operator is equal to – H ̅² / twice the reduce mass × Del².0929

I'm going to put this M here and I end up with.0939

I get –H ̅² / 2 I / R² del² which is equal to - H ̅² all / 2 I.0945

-H ̅² R² / del².0964

Previously, we had that 1/ R² 1/ R².0971

R² is on the numerator here so for the del², the R² on top cancels the R² on the bottom.0975

The R² in the numerator cancels the R² terms in the denominator our del² operator0986

leaving the hamiltonian equaling –H ̅² all / two I × 1/ sin θ DD θ sin θ DD θ + 1/ sin² θ × D² with respect to ϕ.1016

That ends up being our hamiltonian operator.1052

We said that the hamiltonian operator is also equal to the kinetic energy operator.1059

The kinetic energy operator expressed in terms of angular momentum is L²/ 2I.1064

This L² as an operator/ 2I happens to equal the Hamiltonian.1073

I’m going to go ahead and call, everything that is in this bracket, I'm just going to go ahead and1082

call it Z just to make my life a little bit easier.1089

That is equal to - H ̅² / 2 I × Z which implies that L bar² is equal to -H ̅² × Z.1092

We ended up with finding an expression for the angular momentum operator1112

or actually the square of the anger momentum operator.1119

We see that the square of the angular momentum1126

is a naturally occurring operator for quantum mechanical rotating systems.1150

Anything that rotates, the angular momentum is the most important aspect of that system.1173

Let us go ahead and continue on here.1183

Let us go back to blue.1186

Since R is fixed, the orientation of the rotation in space depends only on the two other variables, θ and ϕ.1191

It does not depend on R, R is fixed.1220

With the rigid rotator, R does not change.1223

It is only θ and ϕ change so you end up with a function of two variables only.1226

Since R is fixed, the orientation of the rotation is space depends on θ and ϕ.1230

The wave functions for the rigid rotator are functions of θ and ϕ only.1242

I’m going to call them S for spherical.1266

S, θ, and ϕ.1271

The wave function of a rigid rotator S is a function of two variables θ and ϕ.1274

The Schrӧdinger equation for the quantum mechanical rigid rotator which is our model for a rotating diatomic molecule1285

is the hamiltonian of S which is a function of θ and ϕ is equal to the energy × the function S θ and ϕ.1312

This is the equation that we have to solve, where the hamiltonian is –H ̅² / 2I × 1/ sin θ DD θ of sin θ.1324

DD θ is an operator, +1/ sin² θ D² D ϕ² × S of θ and ϕ = E × the function S which is the function of θ and ϕ.1345

I’m going to end up multiplying this whole thing by sin ⁺θ and I’m going to end up rearranging it.1371

I'm going to get the following.1379

I'm going to get sin θ DD θ sin θ DD θ + D² D ϕ² of S + sin² θ × 2 I E/ H ̅² × S = 0.1382

I’m going to actually perform the operation by distributing this / the S and multiplying out1440

so we are going to get sin θ DD θ × sin θ DSD θ + I’m going to do this one.1450

D² of S D ϕ² + this is going to be M sin² θ × S =0.1467

I have where M is actually equal to 2 I E/ H ̅² to make my life simpler by putting everything all into one.1485

For right now, I’m going to discuss the energies of the rigid rotator.1512

I’m going to save the discussion of the wave functions.1517

The S I’m going to wait until we talk about the hydrogen atom so we can talk about the wave function.1519

It is going to end up working a little more, I do not want to complicate things too much right now.1529

Right now, let us worry about the energy functions and later we will talk about the actual wave function.1532

Let me go back to blue here.1544

When we solve this equation, the M must obey the following constraint.1556

The constraint is M has to equal J × J + 1, where J =0, 1, 2, and so on, the whole numbers.1580

M= 2I E/ H ̅², that has to equal to J × J + 1.1601

Let me make this J a little bit more clear so it do not look like a 6.1613

J + 1, when I rearrange to solve for E, which is what we always do we end up with the following relation E sub J.1617

J is the level of energy = H ̅²/ 2 I × J × J + 1 or again J =0, 1, 2, and so on.1629

These are the energies of the rigid rotator.1648

Anytime I notice J is in our quantum number, the 0 level of energy, the energy is 0.1652

The first level of energy is going to be 1 × 1 + 1=2.1660

It is going to be 2 H ̅² / 2 I.1664

It is going to be H ̅² / I.1667

I put in this 2, the 3, these are the different energy levels of the rotating diatomic molecule for rigid rotator.1671

Notice, the quantize energy levels cannot just take on any values at all.1686

There are very specific values that it can take, the quantize energy levels.1694

Now each energy level J has a degeneracy.1702

Degeneracy D sub J of 2J + 1.1723

All that means is that when I have a particular energy level, let us say the second energy level,1731

when the particle is in the wave functions for that particular state, for the J = 2 state, there are going to be 2 × 2 + 1.1743

There are going to be 5 wave functions that have that same energy level.1751

Remember what degeneracy was.1755

Degeneracy was if you have a degeneracy of 10 that means there are 10 wave functions that actually share that same energy level.1757

For a particular energy level J, that level has 2 J + 1 degeneracy.1767

It has that many wave functions that share that same energy level, that is all that means.1775

These are the two important relations so far.1783

Again, we are going to save our discussion of the wave functions which is our S of θ and ϕ for our future lesson.1787

We are going to talk about that when we discuss the hydrogen atom and the Schrӧdinger equation for the hydrogen atom.1819

We will go ahead and stop it here for now.1827

Thank you so much for joining us here at www.educator.com.1829

We will see you next time, bye.1831