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Lecture Comments (6)

1 answer

Last reply by: Professor Hovasapian
Sat Apr 23, 2016 6:56 PM

Post by Tram T on April 20 at 10:16:08 PM

Hello prof. Hovasapian,

-Since Harmonic Oscilators has Potential Energy V(x), what is the external force applied on the oscillating diatomic molecule?

-Is it because the IR photon has more energy than microwave which cause the external force to pull atom and make it vibrate?

1 answer

Last reply by: Anthony Salamanca
Sat Apr 9, 2016 2:28 AM

Post by Anthony Salamanca on April 9 at 02:25:27 AM

HOW DID U GO FROM r1 to r and back to r1

1 answer

Last reply by: Professor Hovasapian
Mon Feb 16, 2015 3:04 PM

Post by Ahmed Thabet on February 14, 2015

I would like about determining LS coupling? Finding Ms & Ml & degeneracies?  Can you give some examples?
Or give the link if its here?

The Rigid Rotator 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
  • Possible Confusion from the Previous Discussion 0:07
    • Possible Confusion from the Previous Discussion
  • Rotation of a Single Mass Around a Fixed Center 8:17
    • Rotation of a Single Mass Around a Fixed Center
    • Angular Velocity
    • Rotational Inertia
    • Rotational Frequency
    • Kinetic Energy for a Linear System
    • Kinetic Energy for a Rotational System
  • Rotating Diatomic Molecule 19:40
    • Rotating Diatomic Molecule: Part 1
    • Rotating Diatomic Molecule: Part 2
    • Rotating Diatomic Molecule: Part 3
  • Hamiltonian of the Rigid Rotor 36:48
    • Hamiltonian of the Rigid Rotor

Transcription: The Rigid Rotator I

Hello and welcome back to and welcome back to Physical Chemistry.0000

Today, we are going to talk about the rigid rotator.0004

Let us just go ahead and get started.0007

Now, before I actually start discussing the rigid rotator, I want to go back and0010

actually clear up a possible fuzzy point from our previous discussions.0014

Let us go ahead and write that down.0020

Before I discuss the rigid rotator, I would like to go back and clear up the possible confusion, 0028

there may not be any confusion but just in case, I like to make sure that it is pretty clear.0051

A possible confusion from the previous discussion being the harmonic oscillator.0056

Let me go ahead and draw out this parabolic thing for the harmonic oscillator, the energy levels.0069

Just put a couple of them here.0084

We have got R = 1, R = 2, R = 3, R = 4, R being the quantum number for the harmonic oscillator and 0087

these being the particular energy levels.0096

I’m going to have a 0 point energy so let us go ahead and this is going to be R=0, 1, 2, and 3.0100

Some students are a little fuzzy about what this parabola is and 0111

why are we using this parabola to represent these different energy levels, about what this parabola is.0132

They sometimes think the motion of the particle itself, of the mass of the particle or the mass is somehow parabolic.0144

And that is not true.0168

It is really easy to lose your way in what it is that is going on in the big picture.0169

There is so much math going on, there are so much derivation of equations, manipulation of equations, 0174

that oftentimes it is very simple to lose your way in terms of remembering what are we trying to do and 0180

what is it that we are actually trying to elucidate, what are we looking for.0186

Let us talk about what is going on.0191

They think that the motion of the particle is parabolic.0193

The answer is no, the motion of the particle is not parabolic.0195

Let me go ahead draw and a line here.0199

This is a 0 point, the equilibrium position of the harmonic oscillator.0203

You remember the harmonic oscillator is just a particle that moves back and forth like this.0206

Let us go ahead and put A and A, this is A1, A1, or – A1.0211

For the first energy, the amplitude is that far.0220

We put a little bit more energy into the system, the amplitude is going to increase.0225

We know that, it will just go a little bit farther .0229

A3 and – A3, they correspond to these energy levels.0232

The more energy you put in, the bigger the amplitude.0238

That is what is actually happening here.0242

The motion of the particle is harmonic, it just goes back and forth like this.0246

As it deviates from the equilibrium position, what this parabola describes 0249

is the amount of potential energy that is being stored in the system as making its way out to its farthest point.0255

As it goes here, the potential energy goes to here.0262

As it goes here, the potential energy goes to here.0267

As it comes back and goes to 0, the potential energy falls and all of it becomes kinetic.0272

If it goes the opposite way, either here or here, the potential energy of the system rises to here or to here.0277

You remember that the potential energy function of this was ½ KX².0284

If I want the potential energy of the harmonic oscillator, it is ½ KX².0292

This is the parabola ½ KX² is the parabola.0296

It is describing how much potential energy this system is experiencing as it deviates at distance X from that.0300

That is why we are using this parabola to describe it and these of course are the energy levels of higher energy levels.0310

You have bigger amplitudes and the potential energy goes off into infinity.0316

Do you remember when we are talking about the particle in a box?0320

The particle in a box, it was just something like this.0323

This particle is moving back and forth, there is no potential energy in there.0326

But the minute it hits the wall, the potential energy shoots up to infinity.0330

In other words, it cannot escape that box.0334

Here, the potential energy rises parabolically, it does not just boom rise up to infinity that way.0338

It rises parabolically.0345

As you get farther from the center, there is a potential energy that is building up in the system.0347

That is what the parabola represents.0355

The motion itself is harmonic.0357

It is just back and forth represented by a sin or cos curve.0359

This just represents the potential energy function.0363

I hope that clears up a little bit.0367

Let me actually write that down so we have it.0369

The motion of a particle is harmonic which is just a fancy word for back and forth, repeating, periodic, whatever you want to call it.0372

The further the mass goes from equilibrium position, which is this position right there, the parabola describes its potential energy.0395

It is a potential energy that is important in all of this.0433

It is the potential energy function that actually changes the Schrӧdinger equation.0437

Recall that V of X = ½ KX².0444

As the particle or the mass deviates from equilibrium position at distance X its potential energy increases parabolically.0454

Increases as ½ KX².0487

I hope that clears a little bit up, if there was confusion.0494

If not, no worries.0496

Let us start talking about the rigid rotator.0499

Now I diatomic molecule oscillates along this bond.0503

We already discussed the harmonic oscillator but that is not the only thing that it does.0507

A diatomic molecule also rotates in space, rotates this way.0510

It is a rotating molecule also, so we have rotational motion.0518

We can model that rotational motion by something called the rigid rotator.0522

A molecule not only oscillates along its bond, it also rotates about its center of mass.0531

Let us say you have a big ball here and a little ball here, the center of the mass is not going to be in the center.0565

It is going to be a little closer to over here.0571

It can actually end up rotating about that thing like that .0572

That is all, nothing strange.0576

Let us start by looking at the rotation of a single mass about a fixed center and then we will talk about 2 masses, the actual diatomic molecule.0582

Let us start by looking at the rotation of a single mass around a fixed center.0595

We have a center like this and we have maybe a mass like that and it is moving a circle.0624

That is it, we have a nice, fixed circle that way and the radius of the rotation we will just go ahead and call it R.0633

I’m going to start with the basic equation for circular motion and that is going to be S= R θ.0641

S is the distance that the particle actually makes.0650

It is the distance along the circle.0654

R is the radius and θ is the angle and radians.0657

If it travels 90° or π/2, let us say something like this.0660

From here to here, S would be this distance and θ would be this fact that there.0665

Basic fundamental definition of radian measure.0671

S= R θ, let us go ahead and differentiate this with respect to θ.0677

We have DS D θ = DD θ of R θ.0682

R is a constant so it comes out.0689

I’m getting a little confuse here.0699

We are going to differentiate this with respect to time, my apologies.0705

S is a distance and θ is an angle.0708

We are going to do DS DT = DDT of R θ.0712

The R is constant so it comes out and what we are left with is R D θ DT.0721

DS DT is the time derivative of a distance which is the velocity.0727

What we get is velocity = R and the time derivative with respect to angle rotation gives me the angular velocity RO.0732

Velocity = R × the angular velocity.0741

The linear velocity of the particle is equal to R × the angular velocity or the angular velocity is equal to the linear velocity ÷ the radius.0746

Here, ω is the angular velocity and we have seen it before.0754

Its unit is radians per second.0764

Let us go ahead and go.0772

We know that the equation for the kinetic energy of a particle is ½ its mass × velocity².0774

Let us go ahead and actually put this for V into here.0785

We are going to get ½ mass × R ω² = ½ M R² ω².0793

This thing right here, the ½ M R², we refer to that as I.0806

It is called the moment inertia or more commonly known as it is called the rotational inertia of the system.0812

This is called the rotational inertia of the system and it plays the same role that mass does for a particle moving in a straight line.0824

We just call that inertia, linear inertia, that is the mass.0839

This I is the same thing for bodies that rotate.0843

The rotational inertia and we also refer to as the moment of inertia of the system.0848

So far so good.0863

I, and then we have O².0868

The kinetic energy = the rotational inertia × the angular velocity².0870

That is an equation for the kinetic energy of a rotating body.0883

I want to freshly do it on this page.0892

That is fine, I can go ahead and do it on this page.0894

We said that this is in radians per second.0897

I'm going to go ahead and multiply it by one cycle.0902

It means one complete cycle, one complete circle, that is 2 π radians.0905

It is 2 π radians, radians cancels with radians and we are left with ω / 2 π cycles per second which is the same as the Hz.0911

ν which is equal to the angular velocity ÷ two π is the rotational frequency.0927

Ω is the angular velocity.0936

ν which is the ω ÷ two π gives me the rotational frequency in cycles per second.0941

We are just setting up some variables here, that is all.0956

This right here is the rotational inertia, I’m going to keep the ½ separate.0982

We will do it that way so the kinetic energy is equal to ½ I ω².0988

We are ready to move on.1000

For a linear system, we have the kinetic energy is equal to ½ the mass × the velocity² = M² V² / 2M.1001

Just do the mathematical manipulation.1026

= MV² / 2M.1030

What we have is the kinetic energy is equal to the linear momentum ÷ twice the mass.1034

This is just kinetic energy in terms of momentum.1042

That is all, nothing strange about it.1047

Momentum is mass × velocity so the kinetic energy is the square of the linear momentum ÷ twice the mass.1057

Now for a rotational system, we have the kinetic energy = ½ the rotational inertial × the angular velocity².1063

Again, notice the correspondents ½, ½ mass rotational inertia, linear velocity, angular velocity.1082

That is going to equal I² O² / 2 I which is equal to IO² / 2 I.1089

Therefore, for the rotation we have something analogous, the kinetic energy is equal to L² 1103

which is the angular momentum of the rotational system ÷ twice the rotational inertia.1112

Again, the rotational inertia was the M R², the mass × the R.1120

I is equal to M R².1125

Here where L is the angular momentum.1129

Notice the correspondence between linear system, and rotational systems, the angular momentum of the system.1136

Nice and basic discussion of simple rotational motion, a single mass rotating about a fixed center.1151

I can talk about its linear velocity and momentum.1158

I can talk about its angular velocity and its angular momentum.1162

I can talk about its rotational inertia.1165

These are all the equations that represent it.1167

I can talk about its rotational frequency and things like that.1170

Let us go ahead and talk about a rotating molecule.1175

We have a rotating diatomic molecule.1182

Let us go ahead and set up the system here.1195

We have one of the atoms here, let us make it slightly bigger.1197

Let us go ahead and make this one over here a little bit smaller.1202

Let us go ahead and call this mass one over here.1208

Let us call this mass two and the rotation is going to be in this direction.1211

Let us go ahead and pick a counterclockwise rotation.1217

And its center of mass is going to be let us say somewhere around right there.1222

Let us go ahead and call the distance from the center of mass to mass one, we will call that R1.1227

And from the center of mass to mass 2, we will go ahead and we will call that R2.1232

We will call a total distance between the masses R.1238

This total distance is going to be R.1243

R is going to equal R1 + R2.1245

That is our basic setup of a rotating diatomic molecule.1251

Atom, atom, it is going to rotate about its center of mass.1254

If they happen to be the same mass, the same atom, then it is going to rotate right down in the middle.1257

That is all, basic physics.1262

What is interesting here is the angular rotation is actually going to be the same for both.1266

I will go to the next page.1280

Let me draw my molecule here, the center of mass, R1, mass 1, R2, mass 2 and this is going to be R.1282

Now the velocity 1 is equal to R1 ω 1.1301

We also have the velocity of mass 2,the linear velocity is going to equal R2 × ω 2.1317

But the angular velocities are the same, they are rotating at the same rate.1330

Our O1 is equal to ω 2.1336

Velocity 1 = R1 ω and velocity 2 = R2 ω.1344

We just call it ω, it is an ω 1 and ω 2.1351

I have got the kinetic energy = ½.1356

The kinetic energy of the systems is just the kinetic energy of the first mass + the kinetic energy 1363

of the second mass 1/2 of M1 V1² + ½ M2 V2², that is going to equal ½ M1 R1² ω².1367

I hope can keep all of my variables straight here.1392

I hope that you will be vigilant and make sure I do not make too many mistakes.1395

½ M2, we have R2² and ω² so we can rearrange this.1399

It becomes ½ M1 R1² + M2 R2².1408

I’m going to pull out the ω² right there and this is going to equal ½.1424

I'm going to call this thing ,the M1 R1² + M2 R2² I ω².1432

Where I is equal to the M1 R1² + M2 R2².1440

This is the rotational inertia of the two body system.1457

I just manipulated some equations from the same thing.1462

I started with the kinetic energy and I work my way and I wrote it, I expressed it in terms of something × the angular velocity².1466

That something is what we call the rotational inertia.1474

The rotational inertia of this 2 body system is this right here.1478

It is the rotational inertia of the 2 body system.1483

Let us see what are we going to do now.1495

We count what we have.1501

We have the kinetic energy = ½ the rotational inertia × the angular velocity.1504

And we have this equation right here.1512

Sorry, I tend to rewrite things several times.1514

M1 R1² + M2 R2².1518

When we look up here, we also have an equation for the center of mass.1527

We know where the center of mass is.1531

The center of mass is right here and there is a relationship that exists for the center of mass from classical physics.1533

It is going to be this mass × its distance from the center of mass and that is going to equal this mass × its distance from the center of mass.1539

We have another relationship here.1549

We have the center of mass relationship which is going to be M2 R2 = M1 R1.1553

We also have the relation that R is equal to R1 + R2.1570

The total R is equal to the sum of the radii.1576

These are the relationships that we have and I want to play with this a little bit.1579

I’m going to start with this one.1582

I’m going to solve for R1 and R2.1585

R1 is equal to R - R2 and R2 is equal to R - R1.1590

I’m going to start fiddling with these equations and see what it is that I can come up with.1599

I have the center of mass relation M1 R1 = M2 R2.1605

I'm going to substitute one of those other equations that I just created from R.1614

It is going to be M1 and we said that R1 is equal to R - R2 is equal to M2 R2.1621

I get M1 R- M1 R2 = M2 R2.1632

I'm going to move this / here and then factor out.1641

I get M1 R1 is equal to M1 + M2 × R2.1645

I’m just going to solve for R2.1655

I get R2 is equal to M1 R1 / M1 + R1.1657

Interesting, let us do the other one.1666

For all practical purposes, I do not really need to go through all these derivations for you.1672

Basically, what I can do is I can just present the Schrӧdinger equation for the rigid rotator 1676

and I can present the wave functions and the energy functions for you.1682

It is important to actually go through this.1688

The stuff that we are going through, please know you do not necessarily have to actually know this or1689

even recreate it unless your teacher asks you to.1696

That is just not going to happen.1701

This is just part and parcel of your basic scientific literacy.1702

It is important to see these derivations to at least have seen what a derivation is, 1706

what kind of mathematical manipulation is involved in this.1711

Again, this is a part of your scientific training.1715

You need to see these techniques over and over again, that is why we are going through these.1719

You do not have to know this, as long as you can follow it, 1725

that is all that matters and it is all that you going to be expected to do, I promise.1727

Let us do the other one.1732

We started again with the M1 R1= M2 R2.1734

This time we are going to go M1 R1 = M × R – R1.1745

We are going to solve this one for R1.1753

We get M1 R1 = M2 R – M2 R1.1756

I’m going to bring this over here and factor.1762

Again, I'm going to get M1 + M2.1764

This time it is going to be × R1 = M2 R.1769

When I solve for R1, I end up with.1774

It is going to be M2 R ÷ M1 + M2.1780

I made a little mistake here.1787

We always have M, R, and things floating around. 1789

Let me take a second and make sure we got everything straight. 1793

M1 R1/ M1 + M2, M2 R ÷ M1 + M2.1796

So far so good.1805

We said that I, the rotational inertia, is equal to M1 R1² + M2 R2².1808

I’m going to take these values that I calculated.1820

This is R, I have to make sure that all of these things are correct here.1823

I do not want to make any crazy mistakes.1839

So far so good.1840

We are going to plug in R2 and R1 into the equation for the rotational inertia.1841

We are going to end up with M1 R1² R1.1847

I’m going to square this.1855

It is going to be M2² R² / M1 + M2², that × that, + M2.1858

For R2, I’m going to plug this value in.1875

It is going to become M1² R² ÷ M1 + M2².1878

I’m going to go ahead and pull out the R².1901

I’m going to express this as M1 M2.1904

M2, I’m going to separate that out, / M1 + M2.1915

Again, this is just some mathematical manipulation here.1922

+ M2 × M1 × M1.1926

I’m expressing the M1² / M1 + M2².1931

And all of that I'm going to multiply by R².1943

I'm going to express this in a slightly different way.1949

This is going to be M1 M2 ÷ M1 + M2 × M2/ M1 + M2 + M1 M2/ M1 1955

+ M2 × M1/ M1 + M2, all of this × R².1982

This is I, right?1999

Well guess what, this right here M1 M2/ M1 + M2.2001

M1 M2/ M1 + M2 that is equal to μ, the reduced mass.2009

I put this and this together and I get M2 + M1/ M1 + M2 × R², this is just 1.2025

I have the rotational inertia = μ × R².2042

Where again, μ is the reduced mass.2050

We saw this already when we discussed the harmonic oscillator.2058

We ended up taking this 2 body problem and reducing it to a 1 body problem.2063

Where now, we are no longer talking about the mass of 1 or mass of 2.2067

We actually combined it in this thing called the reduced mass.2071

Μ = M1 × M2/ M1 + M2.2075

We combine the two masses and now we can treat this as a single particle of mass μ rotating about a fixed center.2087

This is what we have done.2098

Let me go back to black here.2102

Where μ is the reduced mass and R = R1 + R2.2105

It is just the distance between those two masses.2120

What we have done, we have again turned a 2 body problem into a 1 body problem.2123

In other words, we can treat 2 masses rotating about their center of mass as a single mass,2150

which we call μ rotating in a circle of radius R.2180

Where R is given by this and μ is given by that.2196

Let us go ahead and move over to the next page.2207

Earlier, we found that the angular momentum is equal to the rotational inertia × the angular velocity.2209

Which is just like the linear momentum is analogous to linear momentum equaling mass × linear velocity.2228

Angular momentum = rotational inertia × angular velocity2235

We also found that the rotational inertia is the μ × R².2241

We also found that kinetic energy is equal to angular momentum² / twice the rotational inertia.2249

This is very important relationship for rotating systems.2263

The angular momentum is given by rotational inertia × angular velocity.2266

The rotational inertia is actually given by the reduced mass × the R², the distance between the two masses.2270

Again, we treat this as if it is a single mass.2277

A reduced mass rotating about a fixed center.2281

The kinetic energy of this rotational system is equal to the square of the angular momentum ÷ twice the rotational inertia.2283

Again, the rotational inertia is given by this μ R².2291

There is no potential term here.2296

There is no potential energy term, notice.2300

For some diatomic molecule that is rotating in free space, in any orientation actually does not matter, 2310

there is no potential energy term for this.2317

There is no potential energy term because there are no external forces on the masses.2319

If we want to, we can certainly put these molecules in a magnetic field and create some external forces.2337

But as it is, it is just a rotating molecule, there no external forces.2342

The orientation of the rigid rotator.2355

When we say rigid rotator we are just talking about the thing that is actually turning,2366

the diatomic molecule or whatever it is that we happen to be discussing.2369

The rigid rotator which is just a molecule.2373

The orientation of the rigid rotator does not affect the energy because there is no external force.2379

It does not matter how it is rotating, it does not change the energy of the rotating system.2392

Since there is no potential energy term, the hamiltonian of the rigid rotator is just the kinetic energy operator.2400

It is just K, the kinetic energy operator.2428

The kinetic energy operator which is going to equal –H ̅/ 2M but now it is twice the reduced mass and δ².2435

This is just the initial discussion.2461

In the next lesson we will continue on and talk about the Schrӧdinger equation.2463

Thank you so much for joining us here at

We will see you next time, bye.2469