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

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

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### The Harmonic Oscillator 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 Harmonic Oscillator II
- Diatomic Molecule
- Notion of Reduced Mass
- Harmonic Oscillator Potential & The Intermolecular Potential of a Vibrating Molecule
- The Schrӧdinger Equation for the 1-dimensional Quantum Mechanic Oscillator
- Quantized Values for the Energy Level
- Ground State & the Zero-Point Energy
- Vibrational Energy Levels
- Transition from One Energy Level to the Next
- Fundamental Vibrational Frequency for Diatomic Molecule
- Example: Calculate k

- Intro 0:00
- The Harmonic Oscillator II 0:08
- Diatomic Molecule
- Notion of Reduced Mass
- Harmonic Oscillator Potential & The Intermolecular Potential of a Vibrating Molecule
- The Schrӧdinger Equation for the 1-dimensional Quantum Mechanic Oscillator
- Quantized Values for the Energy Level
- Ground State & the Zero-Point Energy
- Vibrational Energy Levels
- Transition from One Energy Level to the Next
- Fundamental Vibrational Frequency for Diatomic Molecule
- Example: Calculate k

### Physical Chemistry Online Course

### Transcription: The Harmonic Oscillator 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 quantum mechanical harmonic oscillator.*0004

*Let us dive right on in.*0008

*A vibrating molecule can be modeled by the harmonic oscillator.*0011

*This is all that is happening, this is going back and forth like this.*0017

*A vibrating molecule can be modeled by the harmonic oscillator.*0028

*A diatomic molecule has two masses that are moving, not one fix end and the other one moving back and forth like that.*0053

*A diatomic molecule has two masses that oscillate and no fixed end.*0067

*In general, if you have like a really huge atom and a tiny atom, for all practical purposes, the huge atom is not going to move very much.*0088

*You can consider that fixed end and the tiny atoms itself would vibrate back and forth like that.*0098

*We cannot always guarantee you that so we need something a little bit more sophisticated than just a fixed wall and one mass moving.*0103

*The situation we have is this.*0111

*Something like that where you have mass one and mass two.*0116

*There are some modifications that we make actually to the equations of motion.*0122

*Instead of one equation of motion, we have two equations of motion, one for each mass.*0127

*We combine them and we come up with a new differential equation.*0131

*When I can to go through that process but will say this much.*0134

*We will not discuss the modifications due to the equations of motion but we end up with this.*0141

*But we end up with a differential equation that we end up with is μ D² X DT² + KX equal 0.*0169

*This is exactly the same equation that we came up with for the classical harmonic oscillator*0181

*with one fixed end of the wall and one body, one mass moving back and fourth.*0185

*The only difference is instead of the mass of the single ball moving back and forth, this is μ something called the reduced mass.*0191

*I will tell you what it is.*0198

*Where μ is called the reduced mass and is given by the μ is equal to 1/ M1 + 1/ M2.*0202

*It is also equal to M1 +,*0232

*I'm sorry to not μ, this is 1/ μ.*0235

*M1 + M2/ M1 M2.*0240

*If you like μ directly is M1 M2/ M1 + M2.*0245

*Basically, what we have done when we modify the equations of motion,*0253

*we have come up with this way of combining the masses into something called reduce mass.*0256

*By combining I do not just mean adding, it is this, this is the relationship.*0261

*The reduced mass is 1/ M = 1/ M1 + 1/ M2.*0265

*We have combined them and we have basically taken this two body problem and we have turned it into a one body problem.*0270

*This is a single equation, this is a single number.*0277

*This happens a lot in classical mechanics and classical physics.*0280

*In fact, we can do this, we can take a two body problem or three body problem and turn it into a one body problem.*0285

*It is actually quite extraordinary that we can do this.*0292

*It is all we have done.*0296

*Everything is the same, the only difference is this thing called the reduced mass which is a combination of the masses of the two objects.*0297

*In this case, the two atoms or maybe the two blocks of cement, whatever it is you have to be dealing with.*0305

*This is possible to do but you will end up with the same equation.*0310

*You end up with the same general solution, that is what makes this beautiful.*0314

*Let me go to blue actually.*0324

*This notion of reduced mass has allowed us to treat two body problem as a one body problem.*0328

*The same as before.*0357

*The equation, μ D² X DT² + KX is equal to 0, is exactly the same as before.*0362

*The general solution is the same, everything is the same.*0391

*Everything that we did in the previous lesson applies here.*0396

*The general solution is the same.*0400

*We have X of T equals C1 × the cos of ω T + C2 × the sin of ω T.*0407

*Where ω is the angular velocity.*0419

*Ω is K/ M, the only difference is ω is K.*0422

*This time instead of /M it is /μ.*0428

*Everything else is the same, nothing is changed.*0430

*Wherever we see mass before, we are replacing it with a reduced mass.*0433

*Let us take a look, we are talking about molecules.*0440

*We are talking about atoms and atoms that we come together back and fourth, they oscillate like this.*0445

*Let us look at how good an approximation of the harmonic oscillator potential is to the actual intra molecular potential.*0454

*This intermolecular should be inter atomic potential but that is fine.*0503

*I will just leave this as intermolecular.*0513

*They are potential of the vibrating molecule.*0520

*These are one of those terms that are stuck.*0524

*We are talking about potential that exist between the two atoms of the molecule.*0527

*Intermolecular means between two molecules.*0533

*Basically, you have something that can look like this.*0539

*I will go ahead and draw something like that.*0542

*In this axis, we are going to have the energy.*0545

*On this axis, we are going to have the distance that one atom is to the other atom.*0550

*I will go ahead and mark S of 0 here.*0557

*And this is just S, this is the distance that the two atoms are away from each other.*0560

*We are end up getting something like this.*0564

*Let me go ahead and put a minimum mark here.*0567

*I will go ahead in red.*0580

*I will go ahead and draw something like that, sort of magnified.*0582

*This right here, we will go back to blue.*0599

*This curve right here is the potential energy, that is the ½ KX².*0601

*That is the parabola, that is the potential energy of our harmonic oscillator.*0608

*As we pull something further apart the potential energy increases.*0614

*They want to be pulled back together.*0619

*As we squeeze them together, the potential energy increases.*0622

*They want to push themselves apart.*0625

*There are someplace where the equilibrium position is actually perfect where the energy is minimized.*0626

*That is this point right here.*0632

*This is the potential energy, this is the actual potential energy curve of*0634

*what happens when you take two atoms and you squish from together.*0647

*Over here, there are an infinite distance apart from each other.*0651

*As you start bring them together, bring them together,*0654

*as we are bringing the atoms from an infinite distance apart from each other, the closer and closer and closer.*0657

*The attraction that they feel toward each other is actually going to drop the potential energy.*0662

*The potential energy is going to drop.*0667

*It is going to hit a point where they are as close as they are going to be to each other where energy is minimized.*0670

*If you push them any closer together, the energy rises again.*0675

*This point right here, this little minimum that is the actual bond length.*0680

*That is the length at which the atoms are comfortable.*0685

*They are as close as they are going to be to each other.*0689

*It pull them apart, the potential energy gets bigger.*0691

*You push them together, the potential energy gets bigger.*0694

*This is an approximation based on the harmonic oscillator.*0698

*Again, harmonic oscillator you pull them apart, the potential energy increases down to 0.*0702

*It push them away from equilibrium position goes to 0.*0707

*For very small displacements which is pretty much what happens when a molecule vibrates and a molecule is not going like this.*0713

*There are not huge displacements away from the equilibrium position.*0721

*They are very small and for very small displacements this parabolic approximation to the actual potential energy curve is very good.*0724

*In fact, you can see them overlap beautifully and that is what we are doing.*0736

*We are a approximating this inter nuclear potential.*0741

*It is not intermolecular, it is inter nuclear potential.*0746

*Sorry, earlier in the day I was thinking about intermolecular forces and I think I’m getting them mixed up.*0759

*The inter nuclear potential, this is what actually happens when you take atoms and bring them closer together.*0764

*This pair of parabola is an approximation, this is the harmonic oscillator.*0770

*For very small displacements, they are right on top of each other.*0774

*It was a perfect match.*0778

*It is only when we get it to huge oscillations that they start to deviate from each other*0779

*or our parabolic model and the potential energy does not fit with the data anymore.*0787

*But it is never going to be that way.*0793

*For normal vibration of frequencies the displacement is very small.*0795

*We are good, let us go ahead and write this down.*0802

*For small vibrations about the equilibrium position it is an excellent approximation.*0808

*Of course that place which is the minimum, that distance, that distance they are from each other, that is the bond length.*0846

*It makes sense.*0853

*Now the Schroeder equation and we want talk about a quantum mechanical harmonic oscillator.*0856

*The Schrӧdinger equation for the 1 dimensional quantum mechanical harmonic oscillator looks like this.*0865

*It is going to be –H ̅²/ 2 and again we use the reduced mass.*0888

*D² ψ/ DX² + the potential energy × ψ equals the energy × ψ.*0896

*We know what the potential energy is, it is ½ KX².*0905

*When we put B equals ½ KX².*0909

*When we put that into here, rearrange this equation, we get D² ψ DX² + 2 μ/ H ̅² × energy - ½ KX² × ψ equals 0.*0914

*This is the equation that we end up solving.*0939

*This is the Schrӧdinger equation for the harmonic oscillator.*0941

*When we solve this, when we solve this equation which is actually not an easy thing to do*0944

*because the coefficients are no longer constant.*0954

*We will start with the energies first.*0959

*We get the following quantized values for the energy, that right there.*0963

*For the energy E of the system, we get that E is equal.*0984

*Let me do this in blue.*0997

*The energy of the system was equal to H ̅ × K/ μ ^½ × R + ½,*1000

*Where R is equal to 0, 1, 2, and so on.*1013

*R is another quantum number.*1019

*It is another quantum number so are 0, 1, 2, it can only take on those values.*1026

*The energy of the system is quantized, it cannot be any energy.*1032

*It is going to be one energy and then it is going to be another energy.*1036

*It does not make a nice smooth transition jumps.*1041

*It is quantized.*1044

*K/ M ^½ is just ω.*1047

*H ω R + ½ those are the energies for the quantum mechanical harmonic oscillator.*1051

*Since frequency equals ω/ 2 π and E is also equal to planks constant × frequency × R + ½.*1061

*Again, ω and K/ μ ^½ power and frequency equals ω/ 2 π,*1084

*Which is going to equal 1/ 2 π K μ ^½.*1107

*In most books you will see that the energy is equal to H ̅ ω × something that looks like a μ + ½.*1119

*Μ= 0, 1, 2, and so on.*1142

*These V are modified μ, what ever it is that they call this thing in the books.*1148

*Let me go ahead and write it out.*1157

*This and this, look too much alike in a book.*1161

*I reserve that symbol for frequency and R for the quantum number.*1175

*When we plot the energies, we end up doing something like this.*1201

*We have our parabolic.*1205

*This is R = 0, R =1, R=2, R=3.*1212

*Again, as we are getting further and further away, our potential is increasing.*1224

*This one, our energy sub 0, our 0 state energy is equal to ½ H ̅ ω.*1230

*Our energy for level 1 is equal to 3/2 H ̅ ω.*1239

*Our energy 2, is equal to 5/2 H ̅ ω.*1247

*Our energy 3, and so on, equals 7/2 H ̅ ω.*1252

*Again, we are just plugging them into these values.*1257

*Notice that the successive energy levels ½ H ̅ W, 3/2 H ̅ W, 5/2 H ̅ W, 7/2 H ̅ W.*1264

*They are the same, the jump is the same.*1275

*It is H ̅ ω.*1278

*Notice that the successive energy levels have equal spacing.*1284

*That equal spacing is H ̅ ω.*1308

*Notice, specially that the ground state energy is E0 ½ H ̅ ω.*1313

*It is not equal to 0.*1338

*This is very different from the classical harmonic oscillator.*1342

*The classical harmonic oscillator, when the mass is sitting at its equilibrium position, it is not moving, it is not stretched or compressed.*1345

*There is no potential energy, there is no kinetic energy, it is 0.*1351

*The ground state of the classical harmonic oscillator is 0.*1354

*The ground state of the quantum mechanical harmonic oscillator is not 0.*1357

*There is always some vibration going on, that is what is happening here.*1361

*The ground state, the one with a quantum number R=0 is called the 0 point energy.*1367

*The 0 point energy does not mean it is 0.*1389

*Sometimes it will be, sometimes it will not be.*1393

*In the case of a quantum mechanical harmonic oscillator, it is not.*1397

*The fact that it is not 0 it actually comes from the uncertainty principle.*1404

*It does not come from as a result of the uncertainty principle.*1418

*It is a result of the uncertainty principle and we will show you how.*1422

*The energy of the system is equal to its kinetic energy + its potential energy.*1436

*The kinetic energy can be written as P²/ 2 × the mass.*1442

*½ mass × velocity² is the same as the momentum² / twice the mass + ½ KX².*1448

*Let me go to red here.*1459

*We have momentum and we have position.*1463

*In order for the energy to be 0, I have to be able to make that position arbitrarily 0 and the momentum arbitrarily 0.*1466

*And we know from previous work that we cannot do that.*1474

*These two operators do not commute.*1476

*You cannot specify to an arbitrary degree of precision or accuracy both the position and the momentum simultaneously.*1479

*As you make one better, the other one gets worse.*1488

*But you cannot arbitrarily make them both.*1491

*You can bring the error of both to 0, which would make the 0.*1494

*As a result of that, because the energy is kinetic + potential, you have the momentum and*1498

*the position showing up in the same expression, you are never going to get something which is 0.*1503

*That is where it comes from.*1509

*Let us go back to black here.*1520

*If we take the quantum mechanical harmonic oscillator as the model for an oscillating diatomic molecule or a vibrating diatomic molecule,*1526

*then the vibration energy levels of the molecule are given by what we said before E sub R = H ̅ × ω × R + ½.*1561

*It is giving me the different energy values of the different vibrational states of this vibrating molecule.*1590

*R goes from 0, 1, 2, and so on.*1597

*Molecule can transition from one energy level to the next if it absorbs or emits radiation energy of frequency μ.*1604

*It is vibrating, its molecule is vibrating.*1649

*If I hit it with some radiation, the frequency of the radiation matches the energy change, the H ̅ ω,*1651

*the system is going to start vibrating at the next level up.*1660

*Or if it is vibrating to the next level up, it releases enough energy that happens to match certain frequency which is given by,*1664

*if the energy of that particular frequency happens to be H ̅ ω, then it will go from a higher energy state to the next lower energy state down.*1673

*It is just making transitions between energy states.*1681

*A molecule can transition from one energy level to the next, if it absorbs or emits radiation of frequency μ.*1685

*The change in energy is equal to planks constant × μ.*1692

*The radiation of a given frequency has energy equal to planks constant × that frequency.*1698

*Later in the course, we will demonstrate that the quantum mechanical harmonic oscillator*1704

*only allows transitions between successive energy levels, that is the δ R =+ or -1.*1727

*In other words, if I'm at the energy level 1 and if I want to get to the energy 5, I cannot just go directly from 1 to 5.*1755

*I have to go to 2, to 3, to 4, to 5.*1762

*In can only jump go up or down in individual stages.*1765

*Successive energy levels, I cannot make a huge leap like an electronic transition or something.*1769

*In this particular case, I have to pass through the successive stages.*1773

*This right here, this DR is equal to + or -1.*1777

*In other words, R 12345 54321 is called the selection rule.*1782

*Let us write here it is called the selection rule.*1789

*+ or - 1 is called the selection rule and you are going to see quite of them.*1793

*The change in energy from one state to another is the energy of R + 1 - the energy of R, the one stage above,*1805

*- the one stage that you are coming from.*1814

*Let us do it this way, H ̅ × K of μ ^½ × R + 1 + ½ - R × K/ μ ^½ × R + ½.*1819

*This is going to equal H ̅ × K/ μ ^½ × R + 3/2 - R - ½.*1841

*R and R, you end up with H ̅ K/ μ ^½ 3/2 - 1/2 is 2/2 = 1.*1856

*It equals that, and that equals H ̅ ω.*1874

*The change in energy, the transition from one energy level to the next one, either up or down is going to be H ̅ ω.*1881

*The change in energy is equal to planks constant × the frequency.*1892

*You see that the change in energy between successive energy levels of the quantum mechanical harmonic oscillator is H ̅ ω or H ̅ × K/ μ ^½.*1896

*Let us set them equal to each other.*1911

*We have H μ is equal to, this H ̅ is equal to H/ 2 π.*1915

*I’m just going to write it that way.*1924

*It is equal to H/ 2 π × K/ μ ^½.*1925

*The H's cancel and you are left with μ is equal to 1/ 2 π × K/ μ ^½.*1935

*In order for it to actually make the transition from one energy level to the other,*1949

*the quantum mechanical harmonic oscillator it has to absorb or emit energy of frequency that is given by this.*1953

*Nothing more than the spring constant and the actual reduced mass.*1960

*We are going to express this in terms of something called a wave number,*1965

*which is something that is what you actually the scales that you see in spectroscopic data are actually expressed in wave numbers,*1968

*not necessarily in frequencies, more often than not.*1976

*Anything with a little tilde sign over it is a wave number.*1981

*It is nothing more than the actual thing divided by the speed of light.*1986

*In this particular case, it is going to be 1/ 2 π × the speed of light K / μ ^½.*1990

*That is it, just take the frequency and divide by ψ.*2000

*That is all you are doing.*2002

*Where this μ tilde is the wave number.*2007

*It is called the wave number in units of inverse centimeter.*2016

*Since successive energy states are separated by the same energy which is H ̅ ω δ E is the same for every transition.*2027

*Δ E is the same for every transition so when we actually irradiate something or when we measure the radiation that actually being radiated.*2069

*We measure the radiation that actually gives off when it is dropping back down to is ground state.*2074

*It is all the same energy level from 5 to 4, 4 to 3, 3 to 2, 2 to 1.*2080

*All we measure is one value.*2088

*The spectrum for this only gives us one line at one frequency or one particular wave number.*2090

*Spectrum predicted by μ or μ tilde consists of a single line.*2098

*We see the spectrum predicted by that, predicts that it should be represented by a single line.*2122

*The prediction is good and fits well with the actual data that we correct.*2134

*The prediction is good and this single line is called the fundamental vibration of frequency.*2146

*For diatomic molecules, this line appears at around 10⁻³ inverse cm.*2178

*Or somewhere in the range of 10⁻⁴ 10⁻⁵ Hz, if you want to express it in terms of frequency.*2209

*This falls into the infrared range.*2225

*Now we can use IR spectra data and this thing.*2231

*We can use that to actually find force constant defined values of K for individual diatomic molecules.*2251

*How stiff is the bond.*2265

*That is what we are doing, we are finding K.*2268

*How stiff is the bond when it vibrates, is the atom going a lot or it is going a little really fast, what is it?*2270

*Let us go ahead and do an example.*2279

*Let me go ahead and do this in blue.*2283

*Our example is going to be, the IR spectrum of H 35 CO has a single line at 2886 inverse cm.*2288

*Calculate K, the force constant, for this molecule.*2327

*Let us go ahead and do it.*2335

*Let us see, we start with our basic equation which is μ tilde = 1/ 2 π C × K / μ ½.*2338

*We are going to get, 2 π C that² is equal to K/ μ.*2364

*Therefore, our force constant is going to be μ × 2 π C that².*2378

*We just have to work out all the values.*2389

*Let me write it again, K = μ × 2 π C μ tilde².*2395

*Let us go ahead and calculate μ.*2403

*Μ we said it is going to be M1 M2/ M1 + M2.*2408

*Hydrogen is 1 atomic mass unit or dealing with HCL 35, the chlorine is 35.*2420

*The atomic mass units/ 1 + 35 atomic mass units × 1.661 × 10⁻²⁷ kg/ atomic mass unit.*2431

*And we end up with μ = 1.615 × 10⁻²⁷ kg.*2453

*Therefore, K is equal to 1.615 × 10⁻²⁷ kg × 2 π × 3.0 C, which is 3.0 × 10⁸.*2464

*This is going to be m/ S × 2886 inverse cm.*2495

*It is going to be ×, I’m dealing with meters and cm.*2508

*I have to go 100 cm/ 1m and all of that is going to be².*2511

*This whole thing, that is going to be².*2524

*When I actually do this, I get K is equal to 478 kg/ s².*2529

*A Newton is a kg m/ s².*2541

*Therefore, a N/ m is equal to kg m/s² m.*2549

*The m cancels leaving me kg/ s² which is exactly the unit that we got.*2557

*We got N/m.*2563

*K =478 N/ m that is the force constant of the bond between Hydrogen and Chlorine 35.*2566

*Thank you so much for joining us here at www.educator.com.*2582

*We will see you next time, bye.*2583

1 answer

Last reply by: Professor Hovasapian

Tue Apr 12, 2016 3:42 PM

Post by Tammy T on April 12, 2016

Dear Prof. Hovasapian,

-Watching from 14-20 minute mark, the Energy of Harmonic Oscillator or the Energy of vibrating diatomic molecule E=hv(r+1/2) is the Potential aka. the Potential Energy of the vibrating molecule? Does Kinetic Energy counted in this Energy of Harmonic Oscillator?

-Is "r" in Energy E formula the Energy level? It is mentioned that it is 'a quantum number'. Is r the n quantum number?

Thank you!

2 answers

Last reply by: Professor Hovasapian

Wed Dec 30, 2015 1:18 AM

Post by bohdan schatschneider on December 29, 2015

Dr. Hovasapian, why no mathematical explanation regarding Taylor series expansion and the selection of the quadratic term for the potential? I came here looking for why the linear term in the expansion vanishes...