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

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

### Example Problems III

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
- Example I: Good Candidate for a Wave Function 0:08
- Example II: Variance of the Energy 7:00
- Example III: Evaluate the Angular Momentum Operators 15:00
- Example IV: Real Eigenvalues Imposes the Hermitian Property on Operators 28:44
- Example V: A Demonstration of Why the Eigenfunctions of Hermitian Operators are Orthogonal 35:33

### Physical Chemistry Online Course

### Transcription: Example Problems III

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

*Today we are going to continue on with our example problems.*0004

*Let us jump right on in.*0007

*Our first example is, is the function E ⁺to the power – X²/ 2 / the interval for - infinity to infinity is a good candidate for a wave function?*0010

*Recall that for a wave function to be viable, not any function can be a wave function, it has to satisfy certain properties.*0024

*In other words, it needs to be what we call well behaved.*0045

*The wave function to be viable must be well behaved.*0047

*Well behaved means, one, that the integral of ψ* ψ must converge.*0062

*By converge, we mean it must reach some finite number.*0073

*You must get actual number not infinity.*0076

*You must get 5, 7, 8, 6.2, π, 11 π, whatever it is, that is what we mean by converge.*0079

*It must converge to a number.*0085

*In other words, it must be finite.*0089

*I can get some value, it cannot be improper.*0092

*Two, that both the wave function and the derivative of the wave function must be finite.*0096

*The functions can go off into infinity over their particular interval of definition, over their domain.*0106

*Three, ψ and ψ prime must be continuous and must be smooth, no breaks in the graph.*0114

*Four, both the wave function ψ and the derivative of the wave function ψ prime, they must be single valued.*0133

*In other words, just a nice well defined function.*0145

*It have to be a function.*0148

*Let us go ahead and see if this particular function, E ⁻X²/ 2 over this interval, actually satisfies these properties.*0149

*Let us go ahead and talk about the first one.*0160

*We are going to have to integrate from -infinity to infinity.*0161

*We have to do this, ψ * × ψ.*0168

*In this particular case, ψ * is ψ because there is nothing complex in here.*0172

*There is no I involved.*0176

*It is E ⁻X²/ 2 × E ⁻X²/ 2 DX.*0178

*This is a symmetric interval and the function itself is an even function.*0187

*It is an even function, therefore, I can go ahead and write this as the integral from to infinity.*0195

*It is symmetric about the Y axis so it is not a problem.*0200

*I can just take half of it and just multiply by 2 to get the other half.*0206

*I do not have to do this -infinity to infinity thing.*0209

*E ⁻X² DX.*0213

*This definitely does converge.*0227

*Let us go ahead and write it out.*0231

*I will do this interval and I recommend you let your software do it for you.*0243

*It is going to be 2 × π/ 4 ^½.*0248

*The integral converges.*0260

*Again, we were able to do this thing because E ⁻X ^/ 2 is an even function.*0265

*Remember what an even function is.*0280

*It is where F of –X = F of X.*0282

*It is something that is going to be symmetric about the Y axis.*0288

*The same on the left it is on the right.*0293

*That takes care of one, that is good.*0299

*Now, we need to check to see whether it is finite.*0303

*Ψ is equal to E ⁻X²/ 2 and ψ prime = -X × E ⁻X²/ 2.*0309

*I hope you are checking my derivatives.*0337

*Yes, both of these are absolutely finite on their particular interval as X goes to + infinity or –infinity.*0339

*Here both are finite.*0346

*So far so good.*0352

*Now, we need to check continuity.*0354

*This is continuous and this is continuous, both are continuous.*0357

*There are no discontinuities, both are continuous, our domain of definition.*0362

*And last, both are single valued.*0371

*In other words, when you put 1X in there you will get 1Y value.*0378

*You are not going to get 2 different Y values.*0381

*Both are single valued.*0383

*Yes, this is a good candidate for a wave function.*0385

*This is not just any random function that can be a wave function.*0402

*It has to be well behaved.*0405

*It has to satisfy these properties.*0406

*I apologize for a little bit of sniffling and my voice being a little different.*0412

*I’m just getting over a bit of cold.*0418

*Example number 2, look at the solutions to the problem of a free particle in a 2 dimensional box,*0424

*These are the solutions to the Eigen value problem, is the Schrodinger equation expressed as an Eigen value problem.*0431

*H is the hamiltonian operator.*0439

*It is just a shorthand notation and E is the energy Eigen value.*0440

*It shows that whenever the system is in one of its energy Eigen states,*0446

*In other words whenever the system is one of the wave functions,*0449

*One of the Eigen functions, that the variance of the energy is equal to 0.*0453

*The standard deviation is equal to 0.*0459

*In other words, that every time you measure the energy, you are going to get the same value.*0462

*You are going to get some of that particular value and you will never get anything else.*0468

*In other words, there is no going to be deviation.*0472

*You are going to get the same point every time.*0474

*55555555 a million × and the average of that is 5.*0478

*The deviation is 0 because there is no other number except 5.*0484

*That is what this means.*0488

*Let us go ahead and find out what ψ is, look up the solutions.*0492

*For a 2 dimensional box, ψ N sub X N sub Y, we have 2 quantum numbers,*0497

*Is going to equal 2/ AB¹/2 × the sin of B sub X π/ A × X × the sin of N sub Y π/ B × Y.*0503

*That is the wave function.*0523

*We want to show that this is equal to 0.*0529

*This is equal to the average value of E² – the average value of E².*0534

*Whenever that system is in one of its energy Eigen states, that means this is satisfied.*0550

*Its Eigen value is satisfied.*0557

*When the system is in one of its energy Eigen states, remember the Hamiltonian operator is the operator for the total energy of the system.*0568

*The E and H, when we talk about the total energy of the system, we are talking about the Hamiltonian operator.*0586

*It is one of its energy Eigen states, that means that the H of ψ N sub X N sub Y is equal to the energy N sub X N sub Y of ψ N sub X N sub Y.*0595

*We are just applying this, we are just working symbolically and formally.*0611

*N² ψ, just apply it twice.*0617

*N sub X N sub Y= E², N sub X N sub Y ψ, N sub X N sub Y.*0629

*It is just an application of this.*0639

*This is true.*0642

*The expectation value of the energy is going to equal the double integral.*0646

*Double integral because we are talking about 2 variables.*0653

*We have X and Y of ψ * N sub X N sub Y × the energy Hamiltonian operator are the same.*0654

*Ψ N sub X N sub Y.*0671

*This is the definition of the average value.*0674

*This is a double integral because it is a two variable problem.*0683

*This is going to equal the double integral.*0691

*This is this, this part this part is this part, I’m just going to substitute in.*0698

*It is going to be ψ* N sub X N sub Y × E N sub X N sub Y ψ N sub X N sub Y.*0709

*This is the definition, this thing is just this thing and I just substituted this in for here to get this.*0727

*E now is a scalar, it comes out of the integral.*0740

*This equals E N sub X N sub Y × the double integral of ψ*.*0747

*N sub X N sub Y × ψ N sub X N sub Y which is equal to E, because this integral =1.*0757

*This is a normalized wave function and this is the normalization condition.*0767

*Because the integral equals 1.*0773

*D² that is equal to the integral of ψ* N sub X N sub Y × the hamiltonian² ψ N sub X N sub Y.*0786

*I just substitute in, equals the double integral of ψ sub * N sub X N sub Y E N sub X N sub Y ψ N sub X N sub Y.*0800

*This comes out because it is a scalar.*0815

*This is², N sub X N sub Y × the double integral.*0820

*We are back to N sub X N sub Y ψ N sub X N sub Y is equal to E² N sub X N sub Y.*0828

*Our σ² of E which is equal to the expectation value of E² - the expectation value of E that is² is equal to E² N sub X N sub Y - E N sub X N sub Y² is equal to 0.*0841

*We did what we were supposed to do just by manipulating the basic definition of the Eigen value problem.*0871

*That and that were the case of the hamiltonian operator and hamiltonian operator of ψ is equal to the energy × ψ.*0877

*The Schrodinger equation and we just use this and the definition of the expectation value.*0888

*Evaluate L sub X L sub Y.*0902

*Very simple statement in the problem.*0908

*Let us see what we can do.*0911

*This is the angular momentum operator in the X direction.*0915

*Angular momentum in the Y direction.*0919

*The X component of the angular momentum, the Y component of the angular momentum, the commutator of those two.*0922

*Let us go ahead and write out what these are.*0932

*Let me go back to blue here.*0934

*You know what, let me go back to black, I think.*0936

*I have L sub X = - I H ̅ Y DDZ - Z DDY and the Y component of the angular momentum operator equals - I H ̅ Z DDX - X DDZ.*0940

*Trying to keep all of these symbolism straight is the hardest part of quantum mechanics, I promise you.*0976

*I really do not like these brackets.*0986

*I do not like physically actually drawing them out so I hope you will forgive me, I tend to use a slightly different notation.*0990

*I just tend to write comma, L sub X, L sub Y, commutators are not a big deal.*0994

*You can use whatever symbolism you want as long as you know what is being said.*1002

*I’m just making the brackets for some odd reason just that bothers me.*1006

*That is equal to L sub X L sub Y of F – L sub Y L sub X of F.*1010

*Basic definition of commutator.*1023

*Let us go ahead and do this first one.*1027

*I’m going to blue, let us go ahead and do this first one.*1029

*L sub X L sub Y of F = - IH Y DDZ - Z DDY of - I H ̅ Z DF DX - X DF DZ.*1033

*We have to multiply this out.*1069

*Let me go ahead and take care of the constants first.*1075

*This and this, - IH and - IH ends up becoming a -H ̅².*1078

*An operator, you can treat it just like a binomial or a polynomial.*1087

*You have a binomial operator.*1092

*A binomial operator is when I do this × this, this × that, this × this, this × that.*1093

*We are just multiplying, we are operating so we have to be careful here.*1099

*This one here, let us do this first.*1102

*This and this, this is the DDZ differential operator.*1106

*Here we have Z × a function of XYZ so this is going to be a product rule.*1112

*This one is going to end up being , let us see here.*1118

*YZ is going to be this Y and then the derivative of this is going to be this × the derivative of that.*1124

*It is going to be D² F DZ DX + that × the derivative of this.*1144

*The Y and Z part, we just multiply that.*1158

*Y the derivative of this thing is this time the derivative of that which is why I get the Y × Z.*1161

*D² of DZ DX.*1168

*It is going to be the derivative of this.*1171

*This × the derivative of this which is 1.*1173

*We are going to get Y DF DX.*1176

*I hope that made sense because this is Z and this is a function of XY and Z.*1180

*This is the product rule so when I apply this differential operator to this term, I have to do the product rule.*1187

*The rest are easy.*1195

*This × this, we will do this × this.*1197

*This is going to be - XY D² F DZ².*1204

*We will do this × this, not a problem these are just straight because there is no function of that variable.*1217

*I'm not taking the derivative with respect to a variable of, you know something that includes that variable like this one.*1228

*They are the normal product rules.*1234

*This we get - Z² D² F DY DX +,*1236

*I will do it this right here.*1248

*It is going to be XZ D² F DY DZ.*1250

*Let us go ahead and do L of Y, L of X of F.*1263

*That is going to be - I H ̅ Z DDX - X DDZ that is the operator of LY, × - I H ̅ of Y DF DX - X DF DY.*1271

*Again, you notice I use an F, I did not just work with the operators.*1302

*For this one, I included the F.*1308

*For this part here, I include the F.*1311

*I have working on some function, I do not want to lose my way.*1315

*This is going to equal to – H ̅².*1318

*It is going to be this in this, it is going to be a YZ D² X.*1329

*Let me see if I got this here.*1350

*DF DZ this is XYZ.*1354

*We have to be careful here.*1364

*Notice, the LX operator, X is going to be YZ ZY.*1366

*This is the Y operator which is going to be ZX XZ.*1374

*There we go, now we are good.*1379

*We have YZ D² F DX DZ.*1381

*I will do this one, this is going to be –Z² D² F DX DY.*1390

*Let me have this one which is going to be –XY D² F DZ².*1401

*I have Z and Z, I’m going to get two terms.*1414

*This is going to be product rule, again.*1418

*It is going to be + XZ D² F DZ DY.*1420

*In other words, this × the derivative of this and it is going to be + X.*1429

*This × the derivative of this DF DY.*1440

*Everything actually ends up canceling out.*1451

*YZ D² of DZ DX YZ D² F DX DZ.*1454

*Remember, mixed partials are equal so that cancels with that.*1459

*That stays and that stays.*1469

*XY D² of DZ² XY D² of DZ².*1471

*This - - becomes +, so that cancels.*1477

*Z² D² of DY DX D² of DX DY cancels.*1483

*X of Z D² of DY DZ D² of DZ DY cancels.*1490

*What I’m left with is the following.*1498

*I’m left with LX LY F – LY LX of F = -H ̅² × Y DF DX – X DF DY.*1501

*I'm going to go ahead and switch this around.*1535

*I’m going to go ahead and write this as H ̅², flip that.*1537

*Basically pull out a -1 from here I get X DF DY - Y DF DX.*1543

*I’m going to write this as - I H ̅ × - I H ̅.*1557

*- and – is a + here, X D FDY - Y DF DX.*1567

*If you notice this thing right here, this part right here happens to be L of Z.*1582

*We get I H ̅ L of Z of F and we can drop the F.*1594

*Our final operator notation I H ̅ L of Z.*1605

*I will just go ahead and use the XLY, it equals that.*1616

*Notice that this does not equal to 0 operator.*1624

*It is not possible.*1633

*These operators the LX and LY do not commute.*1635

*It is not possible to measure any two components of the angular momentum simultaneously to an arbitrary degree of precision.*1639

*In other words, this is the same as the linear momentum position operator.*1684

*If you measure one really well, you have to lose the measurement of the other.*1689

*You cannot measure both to any degree of accuracy or precision that you want.*1693

*In the case of the angular momentum, any 2 components of the angular momentum cannot be measured simultaneously to any degree of precision.*1697

*The same holds for LX LZ and LY LZ.*1704

*LX LY and LY LZ, LX LZ, you are always going to get something like this.*1709

*A little messy but nice.*1717

*This is the stuff that you have to do.*1720

*You just have to do it through it.*1721

*Example 4, this is going to be a not so much of a problem, this is going to be a demonstration.*1728

*It is a demonstration of why the necessity of a real Eigen values imposes the hermitian property on operators.*1733

*Remember what we said about when we take our measurements in quantum mechanics,*1740

*What we are measuring, we need the measurements to be real numbers.*1743

*In other words, we need the Eigen values to be real numbers.*1748

*Operators and Eigen functions can be complex.*1751

*They can be real or they can be complex.*1753

*There has to be a certain property of the operator that guarantees that the Eigen value is always real because what we measure has to be a real number.*1757

*It cannot be a complex, we cannot measure complex numbers.*1766

*We need a real number.*1768

*Here we are going to demonstrate why this is the case, before we just threw it out there and say that it is the case.*1771

*We are going to demonstrate why the necessity of having a real Eigen value forces the operator to be hermitian.*1776

*Let us go ahead and do that.*1787

*Should I work in red or black?*1789

*Let us go ahead and work with red.*1794

*We have A ψ = λ ψ, where λ is a real number.*1797

*It is in a real number system.*1809

*This is the general Eigen value problem.*1811

*The operator × ψ = ψ × some constant.*1814

*You are not getting ψ back when you operate on it but you end up getting it back multiplied by some real value.*1819

*What I'm going to do is I'm going to go ahead and take this and multiply on the left by ψ* both sides.*1828

*Whatever you do to both sides, you retain the equality.*1837

*Ψ * A ψ = ψ * λ ψ.*1841

*All of them have taken the basic definition and I have done this to it.*1849

*I’m going to integrate both sides.*1853

*The integral of that side is equal to the integral of this side.*1857

*This integral over here on the right, this is a real number so I could pull it out of the integral side.*1862

*This is going to be λ × the integral of ψ * ψ.*1869

*Ψ * ψ is the normalization condition and it equals 1.*1874

*This equal to λ × 1 which equals λ.*1877

*That is it, that is the first part.*1883

*Let us take the complex conjugate of this thing, start with that and see what we can do.*1886

*Let us start again with A ψ = λ × ψ.*1897

*Let us go ahead and take the complex conjugate.*1903

*This is going to be this conjugate, ψ conjugate = λ conjugate ψ conjugate.*1905

*We are going to manipulate this.*1914

*I’m going to multiply on the left by ψ.*1915

*I have ψ, I’m just manipulating.*1919

*All I’m doing is a mathematical manipulation and see where it takes me.*1922

*This is ho we did it.*1926

*Ψ λ * ψ *.*1929

*I’m going to integrate both sides.*1938

*When I integrate the side, I integrate this side.*1941

*Λ * is just a scalar so it comes out as λ * × ψ ψ *.*1947

*This is just the normalization condition so this is equal to λ *,but Λ is a real number.*1957

*Therefore, the conjugate of a real number is the number itself.*1968

*This equals λ.*1972

*Both integrals equal λ.*1976

*Let me go to blue.*1978

*This integral equals λ, this integral equals λ.*1980

*If they are both equal λ, then they are both equal to each other.*1994

*Therefore, I have the integral of ψ* A of ψ equals the integral of ψ A *ψ * which is the definition of the hermitian property.*2002

*In other words, we arrived at this λ being real forces these integrals to be equal.*2035

*We use this final step in our derivation as the definition of hermitian property.*2043

*In other words, their operator has to satisfy this condition.*2048

*That integral has to equal this integral, that is the definition.*2053

*The necessity of then being real forces to operator to have this property.*2056

*This property we call hermitian.*2060

*There you go.*2063

*If some operator satisfies the hermitian property regardless of whether*2067

*The operator or the wave function is complex or not, the Eigen values λ sub N are always real.*2092

*This is very profound on so many levels.*2116

*Those of you who are going to continue on with some higher mathematics or go further with quantum mechanics,*2120

*It is quite extraordinary what is going on here mathematically.*2124

*Let us see something else.*2132

*We said that the hermitian property guarantees that the Eigen values are real.*2136

*We say why, that is the case.*2139

*We also said that it guarantees that the Eigen functions are orthogonal.*2141

*In other words, the integral is equal to 0 but they are perpendicular to each other in some sense.*2145

*In this problem, it is going to be a demonstration of why the Eigen functions of hermitian operators are orthogonal.*2151

*Let us go ahead and start with A ψ sub M = λ sub M ψ sub M.*2159

*Let us start with A of ψ sub M= λ sub M ψ sub M.*2176

*I’m going to go ahead and do the same thing that I did before.*2187

*We are going to multiply on the left by the conjugate and take the integral.*2190

*I got ψ conjugate AN, ψ sub N = ψ N conjugate × λ sub N × ψ of N.*2195

*I’m going to integrate those two.*2213

*I’m going to call this integral 1 and I’m going to call this integral 1 prime.*2217

*Over here, I’m going to get the conjugate first.*2225

*I’m going to take A conjugate ψ sub M conjugate = λ sub M conjugate ψ sub M conjugate.*2229

*I’m going to multiply both sides on the left.*2242

*I’m sorry this is going to be ψ sub M not ψ sub N.*2250

*Ψ sub M and over here I’m going to multiply on the left by ψ sub N.*2255

*I got ψ sub N × A conjugate ψ sub M conjugate = ψ sub N × λ sub M conjugate ψ sub M conjugate.*2264

*Let me see if I got this right.*2286

*I’m going to call this integral 2 and I’m going to call this integral 2 prime.*2290

*I’m going to take 1 prime, this integral – 2 prime.*2298

*I’m going to go to red.*2306

*I got 1 prime – 2 prime = the integral ψ sub M conjugate × λ sub N ψ sub N – 2 prime which is.*2310

*I forgot to once I get to this point I’m going to integrate both sides.*2327

*-2 prime which is ψ sub N λ sub M conjugate ψ sub M conjugate.*2335

*That is going to equal, we are going to pull the λ out.*2350

*Λ N – λ M conjugate × the integral of ψ sub M conjugate ψ sub N.*2354

*Ψ sub N ψ sub N conjugate is the same thing.*2365

*I’m just going to write it as ψ sub M conjugate ψ sub N.*2369

*That is going to be that one.*2379

*I’m going to take 1-2, that integral.*2382

*1-2 is going to equal the integral of ψ sub M conjugate A ψ sub N – the integral of ψ sub N A conjugate ψ sub M conjugate.*2386

*This integral is going to equal 0.*2415

*The reason it is equal to 0 is because the operator we said was hermitian.*2418

*This is the definition of hermitian.*2424

*This thing and this thing, they are hermitian.*2426

*They are the same integral.*2430

*Therefore, this integral is equal to this integral.*2431

*This - that is equal to 0.*2436

*This is the case because A is hermitian.*2438

*This is the definition of hermitian ψ sub M * A ψ sub N is ψ sub N A * ψ sub M*.*2446

*So far so good.*2464

*I have got this thing and I got this thing.*2466

*The 1 prime -2 prime is equal to this.*2474

*1 -2 is equal to 0.*2477

*1 is 1 prime, 2 is equal to 2 prime.*2482

*1 prime -2 prime is equal to 1 -2.*2492

*Therefore, my 1 prime -2 prime which we said was λ sub N – λ sub M conjugate ×*2498

*The integral of ψ sub M conjugate ψ sub N is equal to 1 - 2 which we said is equal to 0.*2509

*I have something × something is equal to 0.*2521

*That means this is 0 or this is 0.*2529

*If M does not equal to N, then this λ sub N – λ sub M conjugate do not equal each other.*2533

*They are different Eigen values.*2551

*A different Eigen values are not going to equal each other.*2552

*These are not going to be equal to each other.*2555

*If they are not equal to each other, this thing is not 0, which means that this has to be 0.*2558

*Let me say it again.*2565

*If M does not equal N, then λ sub N and λ sub M do not equal each other.*2565

*Which means that λ sub N – λ sub M does not equal 0.*2573

*This equation =0 so that means this one has to equal 0, this term.*2577

*Therefore, the integral of ψ sub M conjugate × ψ sub N does equal 0 which we take as the definition orthogonality.*2582

*We already use the definition and now the justification for why it works, for why it is, what it is.*2603

*Which we take as the definition of orthogonality.*2611

*A hermitian operator, all the operators in quantum mechanics are linear.*2625

*They are hermitian, they guarantee that the Eigen values are going to be real and*2629

*They guarantee that the Eigen functions are going to be orthogonal with respect to each other.*2632

*The integral is going to equal 0.*2638

*It is actually quite extraordinary.*2642

*With that, we will go ahead and end it here.*2646

*Thank you so much for joining us here at www.educator.com.*2648

*We will see you next time, bye.*2651

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