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

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

### 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
- The Rigid Rotator II
- 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

- 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

### Physical Chemistry Online Course

### 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 operator *0037

*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 axis *0497

*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 and *0711

*parcel of your mathematical education, your scientific literacy.*0718

*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 operator *0986

*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 and *1082

*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 operator *1112

*or actually the square of the anger momentum operator.*1119

*We see that the square of the angular momentum*1126

*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 molecule *1285

*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 out *1440

*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

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