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

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

### Example Problems 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
- Review
- Wave Function
- Normalization Condition
- Observable in Classical Mechanics & Linear/Hermitian Operator in Quantum Mechanics
- Hermitian
- Eigenfunctions & Eigenvalue
- Normalized Wave Functions
- Average Value
- If Ψ is Written as a Linear Combination
- Commutator
- Example I: Normalize The Wave Function
- Example II: Probability of Finding of a Particle
- Example III: Orthogonal
- Example IV: Average Value of the Kinetic Energy Operator
- Example V: Evaluate These Commutators

- Intro 0:00
- Review 0:25
- Wave Function
- Normalization Condition
- Observable in Classical Mechanics & Linear/Hermitian Operator in Quantum Mechanics
- Hermitian
- Eigenfunctions & Eigenvalue
- Normalized Wave Functions
- Average Value
- If Ψ is Written as a Linear Combination
- Commutator
- Example I: Normalize The Wave Function 19:18
- Example II: Probability of Finding of a Particle 22:27
- Example III: Orthogonal 26:00
- Example IV: Average Value of the Kinetic Energy Operator 30:22
- Example V: Evaluate These Commutators 39:02

### Physical Chemistry Online Course

### Transcription: Example Problems II

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

*I apologize if I sat a little bit today, I’m just getting over a cold.*0004

*If I have some sniffles and things like that, I hope you will forgive me.*0007

*Today, we are going to continue on with our example problems.*0011

*We already did one set and then we talked a little bit more about the quantum mechanics, the formal hypotheses of the quantum mechanics.*0014

*Now, we are just going to do several lessons of problems.*0022

*Let us just jump right on in.*0024

*Before we start the example problems, I did want to go over just some of the high points.*0027

*Just recall some of the equations because there was a lot going on mathematically with quantum,*0032

*as there is with thermal, and anything else in physical chemistry.*0038

*Sometimes, you have to pull back and just make a listing of some of things that are important that we remember.*0041

*We solve the Schrӧdinger equation and we find this wave function ψ.*0049

*That is a wave function and it represents the particle that we are interested in a particular quantum mechanical system.*0057

*Instead of looking at the particle like a particle, we look at it like a wave.*0067

*What we do is we play with this wave function to extract information from it.*0070

*That is all that is actually happening in quantum mechanics.*0076

*The ψ conjugate × ψ, we said was the probability of finding the particle whose wave function is ψ*0080

*in a differential volume element called the DV at the point XYZ.*0112

*You have this wave function which is going to be a function of XYZ.*0131

*At some random XYZ, if you actually multiply, it is going to end up being the probability of*0135

*finding the particle in that little differential volume element.*0143

*Now we have the equation this ψ DV = 1.*0149

*Actually, I should say this is not the probability, the ψ* × ψ is the probability density.*0163

*But for all purposes, we can think of it as the probability.*0168

*The actual probability is the ψ × ψ* × the differential volume element so that you actually have the probability.*0171

*When we integrate all of the probabilities, we are going to get 1.*0178

*This is the normalization condition.*0182

*This was very important normalization condition.*0184

*Again, one of the frustrating things about quantum mechanics is wrapping your mind about around things conceptually.*0192

*But what is nice about it is, because it is so purely mathematical, even if you do not completely understand what is going on,*0198

*as long as you have a certain set of equations at your disposal,*0206

*You will at least get the right answer.*0210

*Eventually, if you become more comfortable and solve for problems, conceptually it will start to make more sense.*0212

*Now every observable in classical mechanics, corresponds to a linear hermitian operator in quantum mechanics.*0218

*If we observe a linear momentum in classical mechanics, we have a linear momentum operator in quantum.*0254

*If we observe angular momentum in classical mechanics,*0261

*for some particle moving in a circular path or curved path we have a angular momentum operator in quantum mechanics.*0266

*That operator, we apply it to wave function to give this information.*0273

*This is what I mean by we extract information from the wave function by operating on the wave function.*0276

*Doing something to it mathematically.*0282

*An operator applied to some wave function in a particular state, is equal to A sub N ψ sub N.*0287

*It is an Eigen value problem.*0296

*Remember, we can express the Schrӧdinger equation as an Eigen value problem.*0300

*Again, we are just going over some highlights of what is that we covered so that we have them*0308

*in a one quick place before we start the example problems.*0311

*That the ψ sub N or called the Eigen functions of the operator A.*0315

*The A sub N are called Eigen values of A corresponding to the Eigen function, corresponding to the ψ sub N Eigen function.*0329

*When a particular function is in a given state, let us say ψ sub 3, it is in that Eigen state for the operator.*0357

*We speak of Eigen states, we speak of Eigen functions, we speak of Eigen values.*0366

*Let us talk about what hermitian means.*0372

*Hermitian also has a mathematical definition.*0379

*Hermitian operator means it has to satisfy this.*0385

*F* AG= the integral of GA* F*.*0394

*If you have 2 wave functions F and G, if you do the left integral and if you do the right integral, those equal each other,*0405

*then this operator is something that we call hermitian.*0419

*If it is hermitian, if it satisfies this property.*0423

*The hermitian operator implies that the Eigen values are real numbers.*0427

*It is very important and I will actually do a lot to demonstrate why this hermitian property implies reality*0434

*and implies orthogonality in some of the example problems.*0442

*One of the first things that hermitian implies is the fact that the Eigen values are real.*0447

*The other thing, that this hermitian property of the operator implies, double arrow for application.*0451

*It implies that the Eigen functions are actually orthogonal.*0457

*The integral of ψsub N conjugate × ψ sub P is equal to 0 for N not equal to P.*0463

*If I have one Eigen function ψ sub 1 and I have another Eigen function ψ sub 15,*0472

*If the operator is hermitian, the operator that gave rise to the Eigen functions, the Eigen functions are going to be orthogonal.*0478

*That is analogous to two vectors being perpendicular.*0484

*Two vectors are orthogonal when their dot product is equal to 0.*0487

*Two Eigen functions are orthogonal when their integral of their product is equal to 0.*0491

*It is completely analogous, that is all that is happening here.*0497

*When measuring an observable in quantum mechanics, we only get the Eigen value of*0505

*the operator corresponding to the observable when the wave function is an Eigen function of the operator.*0538

*In other words, when a quantum mechanical system happens to be in a state that is represented*0572

*by a wave function that happens to be an Eigen function of the operator,*0581

*then what we observe when we take a measurement is going to be one of the Eigen values.*0585

*When the quantum mechanical system is in a state that is represented by the Eigen function of*0592

*the operator of interest then what we observe is going to be one of the Eigen values of the operator.*0599

*If the wave function ψ is equal to ψ 1 ψ 1 + ψ 2 ψ 2 + so so, is written.*0609

*Ψ is the wave function of the quantum mechanical system.*0631

*If this wave function happens to be written as a linear combination also called the super position.*0634

*I do not like the word super position but that is fine.*0650

*It is written as a linear combination of Eigen functions of the operator of interest, whatever operator we happen to be dealing with.*0653

*Then what we observe are the Eigen values A sub 1, A sub 2, A sub 3, and so on.*0680

*Let me go to the next page.*0702

*With probabilities C sub 1² C sub 2² C sub 3² and so on.*0704

*This is for normalized wave functions.*0725

*For the most part, all of our wave functions are going to be normalized.*0734

*If they are not normalized, we are going to normalize them.*0738

*That is not a problem.*0739

*Basically, what we are saying is if we have some wave function ψ of a quantum mechanical system and*0740

*let us say it is represented by 1/2 I × ψ sub 1 – 1/5 ψ sub 3 + 2/7 ψ sub 14.*0745

*Let us say it is represented as a linear combination of Eigen functions of the operator of interest.*0767

*Then what I'm going to observe are the Eigen values A sub 1, A sub 3, A sub 14, every time I make a measurement,*0775

*I’m going to see one of these gets one of these three numbers.*0783

*The extent to which I get one number over the other is going to be square of that, the square of that, the square of that.*0787

*Those are the probabilities.*0796

*1/5² is going to be 1/25.*0798

*1/ 25 of the time, at every 25 measurements, one of those measurements I'm going to get an A3.*0802

*That is all this is saying.*0809

*That is all this represents.*0814

*This probably will not play a bigger role in what we do.*0816

*These are one of the hypotheses that we discussed.*0818

*Let us go ahead and say a little bit more.*0823

*After many measurements, the average value also called the expectation value.*0828

*The average value is symbolized like that and it is going to be the integral of ψ sub * the operator and ψ.*0845

*And this is for normalized.*0860

*We will go ahead and put the one for un normalized.*0863

*This definition right here, it applies when the ψ is written as a linear combination or not.*0868

*If it is if this thing, then this thing goes in here and here.*0873

*The definition is universal.*0877

*The average value of a particular observable is this.*0879

*The general definition for an un normalized wave function, it is just good to see it.*0884

*We have the average value of A is going to equal the integral of ψ sub *.*0894

*These are just integrals, all you are doing is literally plugging the functions in.*0899

*Operating on this, multiplying it by to ψ conjugate.*0905

*Putting it in the integrand and integrating it with respect to the variables.*0908

*If it is a one dimensional system, it is a single integral.*0911

*If it is a 2 dimensional system, it is a double integral.*0914

*If it is a 3 dimensional system, XYZ, it is a triple integral.*0916

*You have your software to do the integral for you.*0918

*The integral of ψ A ψ ÷ the integral of the normalization condition, this thing.*0924

*Remember, when it is normalize, this thing is equal to 1 which is why it is equal this, just the numerator.*0934

*This is the definition for an un normalized wave function.*0938

*If ψ is a linear combination is written as a linear combination.*0946

*In other words, ψ = C1 C1 + C2 C2 +… ,*0960

*Then the average value is really simple.*0972

*It is actually equal to C 1² × A1, the Eigen value + C 2² × the Eigen value + …,*0978

*It is equal to the sum I, C sub I² A sub I.*0992

*There is another way of actually finding it when it is written as a linear combination.*1001

*The final thing you want to review is something called the commutator.*1005

*We have operator AB, the symbol this means this is called the commutator of the 2 operators.*1010

*And it is defined as AB – BA.*1019

*You apply AB to the function then you apply BA to the function and you subtract one from the other.*1024

*This is called the commutator.*1030

*And we also have sigma A² = A² - A², there is that one.*1036

*The uncertainty in the measurement, the variance, if you take the square root of that you get the standard deviation.*1049

*And the sigma of B² is equal to squared.*1056

*Of course, the final relation which is the general expression for the Heisenberg uncertainty principle is the following.*1067

*The sigma of A, sigma of B is greater than or equal to ½ the absolute value of the integral of ψ sub *.*1074

*The commutator of AB applied to ψ.*1088

*That is the general expression for the uncertainty principle and it is based on this commutator.*1094

*If you do AB of the function of the BA of the function.*1101

*If you subtract one from the other you get 0 and those operators commute.*1105

*If they commute then you can measure any of those 2 things to an arbitrary degree of precision.*1110

*If they do not commute like for example the position of the momentum,*1119

*the position of the momentum operator do not commute.*1122

*Based on the original thing that we saw, the original version of the Heisenberg uncertainty principle that we saw,*1126

*we know that we cannot measure the momentum and*1132

*the position of a particle to an arbitrary degree of an accuracy or precision simultaneously.*1137

*We have to sacrifice one for the other and we have to find the balance.*1143

*Whatever it is that we happen to want depending on the situation.*1148

*With that, let us go ahead and start some example problems.*1153

*I do not know it that helped or not but that was nice to see.*1156

*Let ψ sub θ = E ⁺I θ for θ greater than or equal to 0 and less than or equal to 2 π.*1160

*We want to normalize this wave function.*1167

*Quite nice and easy.*1168

*Normalize the wave function.*1171

*Let me go ahead and do this in blue, just to change the color a little bit.*1172

*Normalized means we have some constant that we have multiply the wave function by, to make the normalization condition satisfied.*1176

*Normalized is ψ of θ is equal to some normalization constant × the function.*1188

*The normalization condition is this.*1200

*It is that equal to 1.*1202

*We need to solve this integral and find N, the normalization constant.*1206

*That is what we do.*1211

*If we take the integral of ψ sub *.*1214

*In this particular case, ψ*= E ⁻I θ because it is a conjugate and ψ is equal to E ⁺I θ.*1222

*We do not have to watch out for it.*1235

*Sometimes the conjugate is not the same as the real number.*1236

*This become N × E ⁻I θ × ψ which is NE ⁺I θ.*1241

*It is going to be E θ and we are going to set it equal to 1.*1252

*We are going to get N² × the integral of E ⁻I θ × E ⁺I θ E θ.*1256

*This is going to equal 1.*1265

*We are going to get N² of E ⁺I E ⁻I θ × E ⁺I θ is E⁰ which is 1.*1267

*It is going to be D θ.*1275

*We are integrating from 0 to 2 π.*1276

*D θ is equal to 1.*1280

*This is going to be N² × 2 π is equal to 1.*1287

*N² is equal to 1/ 2 π which implies that N is equal to 1/ 2 π ^½, or if you like 1/ √2 π if you prefer older notation.*1296

*I should do it down here.*1319

*Ψ sub θ of the normalize wave function is equal to 1/, 2 π ^½ E ⁺I θ.*1322

*That is your normalize wave function.*1336

*You want to normalize a wave function, apply the normalization condition.*1339

*Give me that extra page here.*1347

*There is a little one missing here.*1349

*The wave function in example 1 is that a particle moving in a circle, what is the probability that the particle will be found between π/ 6 and π/ 3?*1354

*The probability density we said is ψ * ψ which is also equal to the modulus of that.*1365

*This was equal to the probability density.*1377

*Ψ is equal to 1/ radical 2 π × E ⁺I θ.*1383

*Ψ* is equal to 1/ radical 2 π × E ⁻I θ.*1394

*So far so good, let us go ahead and find the probability density.*1402

*We will just multiply these 2 together.*1406

*Ψ* × ψ is going to equal 1/ 2 π × E ⁺I θ × E ⁻I θ which is going to equal 1/ 2 π.*1409

*The probability is equal to the probability density × the differential element.*1425

*D θ in this case because we are working with θ.*1436

*Therefore, our probability is going to equal 1/ 2 π which is equal to this part, D θ.*1439

*Now, we want to find the total probability of finding it within a particular region and we said π/ 6 and π/ 3.*1452

*We are going to integrate from π/ 6 to π/ 3.*1458

*Therefore, the probability of finding the particle when θ is between π/ 6 and π/ 3 is equal to the integral π/ 6 to π/ 3 of the probability.*1462

*I actually prefer to write it differently.*1487

*I prefer my differential element to be separate.*1489

*I do not like to write it on top.*1492

*This is going to equal 1/ 2 π × θ as it goes from π/ 6.*1495

*2 π/ 3 which is equal to 1/ 2 π × π/ 3 - π/ 6, which is going to equal 1/ 2 π.*1509

*Π/ 3 – π/ 6, 2 π/ 6 – π/ 6 is π/ 6.*1525

*The π cancels, leaving you with the probability of 1/ 12.*1531

*The probability density is ψ* ψ.*1537

*The probability ψ* ψ D θ.*1539

*If you want the probability between two certain points, in this case two certain angles, use integrate from the point to the other point.*1543

*We will see later that the general wave equation for a particle moving in a circle is ψ sub θ = E ⁺I × M sub L θ.*1563

*Where M sub L is a quantum number like the N in the equation for the particle in a box.*1573

*Just another quantum number for a circular motion.*1577

*Shows that ψ sub 2 and ψ sub 3 are orthogonal.*1580

*In order to show orthogonality, we need to show the following.*1588

*We need to show that the integral of ψ* of 2, ψ of 3 is equal to 0.*1592

*We need to show that they are perpendicular.*1607

*We need to show that they are orthogonal.*1609

*Orthogonal was the general definition.*1610

*We need to show that the integral of their product is equal to 0.*1613

*Let us go ahead and do it.*1617

*The integral of ψ* to ψ 3, it does not matter which order you do it.*1621

*You can do ψ* ψ 3, it really does not matter.*1628

*That is going to equal the integral from 0 to 2 π, that is our space from 0 to 2 π.*1632

*We are talking about circular motion.*1638

*Ψ sub 2 is equal to, that is the 2 and 3, that is the NL.*1642

*We have 1/ radical 2 π × E ^- I 2 θ × 1/ radical 2 π.*1650

*All I’m doing is just putting in the equation, plugging them into the equations that I have developed already.*1663

*That is the nice thing about quantum mechanics.*1668

*There is a lot going on but at least it is reasonably handle able because you have the equations.*1671

*In fact, you just plugged them in.*1679

*As far as the integration is concerned, sometimes you are going to have something that you can integrate really easily like these.*1681

*Sometimes you are going to have to use your software, not a big deal.*1686

*If you have long integration problems, please do not do the integration yourself.*1689

*If you want to use tables, that is fine.*1693

*I think it is nice but at this level you want to concentrate more on what is going on underneath.*1694

*You want to leave the mechanics to machines.*1699

*Let the machines do it for us.*1701

*That is what they are for.*1703

*We have E ⁺I × 3 θ D θ.*1707

*This is the integral that we have to solve.*1711

*It turns out to be really nice integral.*1714

*We have 1/ 2 π, let us pull that out.*1717

*0 to 2 π E ^- I 2 θ or 2 I θ × E³ I θ.*1721

*Just add them up and you are going to end up with E ⁺I θ D θ.*1729

*That is going to equal 1/ 2 π × when I integrate this, I'm going to get 1/ IE ⁺I θ.*1735

*I'm going to take it from 0 to 2 π.*1746

*I will do all of this in one page.*1751

*= 1/ 2 π I × E² π I – E⁰ which is equal to 1/ 2 π I.*1756

*Remember, E² π is cos of 2 π + I × sin of 2 π.*1776

*The Euler’s relation, cos of 2 π + I × the sin of 2 π -1 is equal to 1/ 2 π I × cos of 2 π is 1 + sin of 2 π 0 -1 = 0.*1782

*They are orthogonal, nice and simple.*1811

*Let us see what we got here.*1820

*Let ψ be a wave function for a particle in a 1 dimensional box.*1824

*Calculate the expectation value, average value of the kinetic energy operator for this function.*1829

*The average value of the kinetic energy operator is equal to the integral of ψ* × the operator and apply to ψ.*1837

*That is the integral that we have to solve.*1848

*We just plug everything in.*1850

*The particular ψ, I had written here.*1855

*Sometimes the problem will give you the equation.*1861

*Sometimes it will not give you the equation.*1862

*You have to be able to go to the tables or places in your book where you are going to find the equations you need.*1864

*Much of the work that you actually do will knowing where to get the information you need.*1870

*You do not necessarily have to keep the information in your head, you just have to know where to get it.*1874

*If we recall or if we can look it up, the equation for a particle in a 1 dimensional box is equal to 2/ A¹/2 × sin of N Π/ A × X.*1879

*The length of the box is from 0 to A.*1895

*That is the equation that we want to work with, that is ψ.*1899

*In this particular case, this is a real.*1903

*Ψ* is equal to ψ so we can go ahead and write that down.*1905

*Ψ* is equal to ψ, it is not a problem.*1910

*The kinetic energy operator, let us go ahead and write down what that is.*1916

*The kinetic energy operator is –H ̅/ 2 M D² DX².*1919

*We are going to apply that.*1928

*We are going to do this part first.*1930

*We are going to apply the kinetic energy operator to ψ.*1931

*K apply to ψ is equal to –H ̅.*1936

*I would recommend you actually write everything out during the entire course.*1941

*If you want to get in the habit of writing everything out, do not do anything in your head.*1948

*There is too much going on.*1951

*I do not do anything in my head.*1952

*I write everything out.*1954

*D² DX² of ψ which is 2/ A.*1956

*Do not let the notation intimidate you.*1964

*Most of it is just constants that go away.*1966

*Sin N π/ A × X.*1970

*Like I said, most of it is just constant.*1977

*When I take the derivative of the sin N π of A twice, the derivative of sin is cos.*1979

*The derivative of cos is –sin.*1984

*The - and – go away and I'm left with a +.*1988

*Let me write everything out here.*1992

*We are going to get the H ̅²/ 2 M.*2004

*We are going to pull this one out 2/ A¹/2.*2011

*Again, we have the sin when we differentiate twice but because of this N π/ A × X,*2016

*that is going to come out twice and it is going to be N² π²/ A².*2022

*And you are going to get sin of N π A/X.*2029

*This is just basic differential from first year calculus.*2033

*Nothing going on here.*2035

*This is the K of ψ, the ψ* × K ψ is going to equal 2/ A¹/2 sin of N π/ A × X × H ̅² N² π²/ 2 MA² × 2/ A¹/2.*2037

*I’m just putting things together.*2070

*Sin of N π/ A × X and that is going to equal 2/ A.*2073

*Let me write everything.*2089

*2/ A ^½ and 2/ A¹/2, I’m going to do it like this.*2092

*It is going to be 2 on top, there is going to be A × A ^½, that is A on the bottom.*2096

*A and A² becomes A³.*2103

*We get H ̅² N² π²/ 2MA³ and we get sin² N π/ A × X.*2105

*The 2 and 2 cancel.*2121

*Now, we need to integrate this thing so we are going to have.*2124

*I hope I have not forgotten any of my symbols here.*2130

*H ̅², I should have an N², I should have a π², I should have an M and I should have an A³ ×*2132

*the integral from 0 to A of sin² N π/ A × X.*2141

*This is going to equal H ̅² N² π²/ MA³.*2150

*This is going to be, when I look this up in a table or in this particular case I will use the table entry.*2159

*You can have the software do it for you.*2164

*This integral is going to end up being A/ 2.*2169

*I will go ahead and write it out.*2172

*-A × sin of N π/ A × X/ 4 N π from 0 to A.*2174

*And it is going to equal H ̅² N² π²/ MA³ × A/ 2.*2186

*This is A/ 2, A cancels one of these and turns it into A² and we are left with H ̅² N² π²/ 2 MA².*2195

*That is correct, yes.*2211

*That was what we wanted.*2213

*Let me see, do I have an extra page here?*2215

*Yes, I do.*2216

*The expectation value of the kinetic energy operator.*2219

*When I measure the kinetic energy, this is what I'm going to get.*2223

*Let us do another approach to this problem.*2231

*We are going to do that to the next page.*2233

*Another approach to this problem.*2235

*It was nice to revisit momentum every so often because momentum and*2238

*angular momentum are huge in quantum mechanics, in all physics actually.*2244

*Another approach to the problem.*2250

*We know that K is equal to P²/ 2 M.*2258

*That is just another way of writing the kinetic energy, ½ mass × velocity²*2263

*is actually equal to the mass × the velocity which is the momentum²/ 2M.*2266

*The average value of K is equal to the average value of P²/ 2.*2276

*2M is just a constant so it ends up being the average value of P²/ 2M.*2282

*From our previous lesson, we have already calculated this PM.*2288

*It was H ̅² M² π²/ A².*2294

*We have H ̅² M² π²/ A² / 2M.*2306

*Just put this over the 2 M.*2316

*We will put the 2M down here and we get the same answer as before.*2318

*You can do it with the definition of expectation value or you can do it with something else based on something that you already done.*2323

*This is a really great relation to remember.*2331

*Kinetic energy is the momentum² or twice the mass.*2333

*Where are we now?*2340

*Evaluate the commutator of P sub X P sub Y and the commutator X² P of X.*2344

*Let us go ahead and do this first one.*2350

*When we evaluate these commutator relations, use a generic function F.*2353

*Just use F, do not try to do these symbolically without a function.*2358

*At least until you become very comfortable with this.*2363

*I, myself is not comfortable with it.*2366

*I like to put my function in there because I know I’m operating on a function and the end just drop the function*2368

*and you are left with your operator symbol.*2372

*You write that down here.*2378

*When doing these, use a generic F and by generic F I mean just the symbol F.*2379

*Do not use just the operators until you become much more proficient and familiar with operators.*2401

*This P sub X P sub Y, the most exhausting part of quantum mechanics is writing everything down.*2414

*This is the symbolism, this is just so tedious.*2421

*Applied to some generic function F.*2425

*That is equal to P sub X P sub Y of F – P sub Y P sub X of F.*2428

*We know what we are doing here.*2440

*P sub X - IH DDX that is the P sub X operator and the P sub Y.*2443

*This is P sub Y applied to F, then do P of X applied what you got.*2454

*We are working from right to left.*2460

*Remember, sequential operators.*2462

*This is going to be - I H ̅ DF DY.*2464

*Notice, I put the F in there so operate on a function.*2471

*- I H ̅ DDY - I H ̅ DF DX.*2475

*Here we get – H ̅² D² F DX DY - H ̅² D ⁺F DY DX is actually = 0.*2488

*And the reason it is equal 0 because for all of the functions that you are going to be dealing with, use mixed partial derivatives.*2521

*And again, we saw this in thermodynamics.*2544

*Mixed partial derivatives, by mix partial we mean the partial with respect to X first then the partial with*2546

*the respect to Y is equal to the partial with respect to Y first and then the partial with respect to X.*2553

*The order in which you operate, the order in which you take the derivative, it does not matter for all well behaved functions.*2559

*By well behaved, it just means to satisfy certain continuity conditions.*2566

*For our purposes, you will never run across a function that does not satisfy this.*2570

*We will always be dealing with functions that satisfy this property.*2575

*This and this, even though the orders are different, they are actually equal.*2578

*- + you end up with 0.*2583

*Mixed partial derivatives are equal.*2586

*In other words, the D ⁺2F DX DY is absolutely equal to D² F DY DX.*2591

*That is a fundamental theorem in multivariable calculus.*2601

*The order of differentiation does not matter.*2604

*Let us try our next commutator.*2609

*We want to do the X² PX was going to equal X².*2612

*If you remember the X operator, the position operator just means multiply by X and the PX operator is - I H ̅.*2623

*Again, we are going to do DF DX -, now we are going to switch them.*2633

*We are going to do PX X² - IH DDX.*2639

*We are going to do X² F.*2649

*It is this × this - this × this and this × this order.*2654

*Be very careful here.*2662

*This is going to be - I H ̅ X² DF DX, I just change the order here.*2665

*Nothing strange happening.*2676

*And then this one is going to be +.*2677

*Notice, now I have an X² F.*2683

*Let me go ahead and write this up.*2690

*This is , X² PX - PX X².*2694

*We have X² F, this is a function × a function and differentiating that.*2702

*I have to use the product rule so it is going to be this × the derivative of that + that × the derivative of this.*2706

*It is going to be, the negative cancels so I get + I H ̅ this × the derivative of that is going to be X² DF DX*2713

*+ that × the derivative of this + I H ̅ 2 XF – I X² DF DX + IHX² DF DX.*2725

*These go away, I'm left with I H ̅ 2 X F.*2741

*I know I can go ahead and drop that F in terms of we know it is not equal 0.*2750

*What is happening now, I can go ahead and drop the F part and just deal with the operator part.*2755

*It is equal I H ̅ 2 X which definitely does not equal the 0 operator.*2765

*This is the operator, this is our answer.*2774

*The a commutator of this is equal to that.*2783

*We include F in order to keep track of our differentiation properly.*2788

*If we did not include the F, we would not have F here, we would not have the F here.*2793

*It might cause some confusion as far as where is the product rule.*2797

*That is why we are putting it in there.*2802

*It is very important to put it in there until you become very accustomed to operators.*2803

*I, myself, do not, I use F.*2808

*That is it, thank you so much for joining us here at www.educator.com.*2811

*We will see you next time for a continuation of example problems.*2814

*Take good care, bye.*2817

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