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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### Schrӧdinger Equation as an Eigenvalue Problem

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
- Schrӧdinger Equation as an Eigenvalue Problem
- Operator: Multiplying the Original Function by Some Scalar
- Operator, Eigenfunction, & Eigenvalue
- Example: Eigenvalue Problem
- Schrӧdinger Equation as an Eigenvalue Problem
- Hamiltonian Operator
- Quantum Mechanical Operators
- Kinetic Energy Operator
- Potential Energy Operator
- Total Energy Operator
- Classical Point of View
- Linear Momentum Operator
- Example I

- Intro 0:00
- Schrӧdinger Equation as an Eigenvalue Problem 0:10
- Operator: Multiplying the Original Function by Some Scalar
- Operator, Eigenfunction, & Eigenvalue
- Example: Eigenvalue Problem
- Schrӧdinger Equation as an Eigenvalue Problem
- Hamiltonian Operator
- Quantum Mechanical Operators 16:46
- Kinetic Energy Operator
- Potential Energy Operator
- Total Energy Operator
- Classical Point of View
- Linear Momentum Operator
- Example I 26:01

### Physical Chemistry Online Course

### Transcription: Schrӧdinger Equation as an Eigenvalue Problem

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

*Today, we are going to talk about the Schrӧdinger equation as an Eigen value problem.*0004

*Let us get started, let us go ahead and work in black today.*0009

*Recall example 3 from the previous lesson.*0017

*We had an operator which was I believe C ̂ and we said that this operator was defined by -I × H ̅ DDX of something, whatever F was.*0023

*We said that F was equal to E ⁺I × n × X.*0039

*We operated on this and we found that this C ̂ of our particular F was equal to n × H e ⁺INX.*0045

*Now, notice that operating on F is the same as multiplying it by we just ended up, the original function was e ⁺INX.*0068

*We operated on function was nh × e ⁺INX.*0090

*We ended up just multiplying the original function by some scalar, by some number.*0095

*Positive or negative actually does not matter.*0099

*This is the general arrangement.*0101

*Operating on F is equal to some scalar × F.*0103

*The only thing this particular operator did was multiply F by some constant.*0124

*In other words, if my function is X², I may end up with 17X².*0149

*That is it, I just multiplied the X² by a factor 17, that is all.*0159

*Recall example 4, in example 4 we had that the operator D was actually equal to partial derivative of the function with respect to Z.*0162

*In that particular case, RF was equal to XY² and Z³.*0174

*In this particular case, the operation on F ended up equaling 3XY² Z².*0181

*Here operating on F did not just multiply it by some scalar factor, the D of F was 3X² Y² Z².*0192

*Make sure this is clearly to.*0207

*The original function was XY² Z³.*0209

*In this case, it did not do that.*0212

*When an operator operates on a function and gives back the original function multiplied by some scalar, this is profoundly important.*0220

*Here is what we are getting into some very deep and important mathematics.*0278

*The general expression is this, operating F is equal to some A × F.*0283

*I will actually go ahead and do here.*0292

*Let me go ahead and put the A of X = some constant × F of X.*0296

*Personally, I do not like the X, they tend to distract me.*0308

*I will just go ahead and do it this way.*0310

*A of F is equal to some constant A × F of some function that we are operating on.*0313

*Here the F of X or the F, it is called an Eigen function of the operator.*0321

*A is called an Eigen value for the Eigen function.*0345

*I actually called V, the Eigen value associated with that particular Eigen function.*0357

*Operating on F in a situation where when you operate on F, you just end up getting F back multiplied by some scalar.*0391

*The F of X is called the Eigen function of the operator.*0399

*A is called the Eigen value that is associated with that particular Eigen function, when this thing is satisfied.*0404

*Now, given a particular operator A ̂, finding F of X and its corresponding A is called an Eigen value problem.*0411

*You are going to be presented with some operator.*0452

*It is your task to find all of the functions and all of the values, the scalars that satisfy this equation.*0455

*That is what we are saying.*0464

*Which function when you operate using this particular operator that we give you,*0465

*gives you back the original function multiplied by a scalar, that is the problem.*0469

*It turns out that the Schrӧdinger equation is just an Eigen value problem.*0474

*We will show it in just a minute.*0477

*For example 3, our C of F = A of F.*0481

*We have 3 things, we have the operator, we have the Eigen function, and we have the Eigen value.*0496

*In this particular case, the operator was n-I H ̅ DDX, that was the operator.*0500

*The Eigen function was E ⁺INX.*0510

*Any function E ⁺INX satisfies this relation.*0521

*Of course, the Eigen value.*0525

*You can write it as two words, I will write it as one word.*0531

*The Eigen value that was NH, that is the whole idea.*0535

*Operator, Eigen function, Eigen value, profoundly important.*0545

*Those of you who have taken linear algebra, chances are you already have seen this when we are talking about matrices.*0550

*This is this and that is that, that is what is going on here.*0558

*Let us go ahead and talk about what this has to do with the Schrӧdinger equation.*0566

*Let us look back at the Schrӧdinger equation and let us write it out like this.*0572

*We had - H ̅²/ 2M × D² DX² of our particular C.*0579

*I’m not going to go ahead and put the X there.*0593

*It is a function of X that we are looking for + this potential energy function × our Z or ψ if you like.*0594

*Equals E × Z this was Schrӧdinger equation, that is the function.*0607

*Notice something here, notice I have the Z here × something, the Z here × something.*0621

*I can factor out the Z and write this left hand side as an operator.*0628

*Here is what it looks like.*0634

*It ends up looking like this.*0636

*Let me go back to a black.*0638

*I’m going to factor out the, so it was going to be –H ̅²/ 2N D² DX² + V of X.*0642

*I’m going to write the Z out here.*0658

*Z equals E × C.*0660

*Notice, we know we can do this because operators they distribute the way that polynomials do.*0665

*Notice what we got.*0672

*Once again, here is my Z and here is Z, this as an operator.*0676

*This right here, this is just some number.*0687

*The energy of the system is just a constant.*0692

*If we call this operator, if we get a symbol, if you call this operator on the left, operator H ̂, we can rewrite this whole thing as H ̂.*0700

*Let me go back to black.*0727

*H ̂ of Z is equal to E × Z.*0730

*We just expressed the Schrӧdinger equation as an Eigen value problem.*0739

*If you given some wave function and if you operate on it with this thing, this we have not given a name to it yet, it is actually called a Hamiltonian operator.*0743

*If we operate on this wave function, we actually end up getting the wave function back multiplied by some constant.*0751

*The constant happens to be the total energy of the system.*0759

*This is profoundly beautiful.*0761

*We expressed the Schrӧdinger equation as an Eigen value problem.*0772

*That is exactly what it is, as an Eigen value problem.*0782

*We have presented with a particular Schrӧdinger equation for a particular system.*0787

*Our task is to find the function Z, the wave function and the associated Eigen values that happened to coincide with these operators.*0792

*The operator is the same, it is the Hamiltonian operator.*0802

*We are going to take this operator and see if we can find a function and the Eigen values that are associated with it.*0805

*For each function, there is some energy for the system.*0811

*When the system is in this particular energy state, the particle is behaving this way according to the wave function, that is all we are doing.*0816

*The rest of it is just math, it really is.*0824

*When we have solved the equation, we will not only have found a wave function and Z of X but also the total energy of the particle.*0829

*The total energy of the particle in the state is represented by the wave function.*0877

*In the state that is represented by the particular wave function the Z of X.*0890

*Once again, the operator is called the Hamiltonian.*0906

*HR is called the Hamiltonian operator.*0910

*Let us go ahead.*0918

*Our Hamiltonian operator happens to equal - H ̅²/ 2 × the mass, the second derivative +*0928

*this potential energy function that is the Hamiltonian operator right there.*0941

*Notice, if we have - H ̅²/ 2M D² DX² + V of X × some C is equal to the energy × the Z.*0946

*Notice, this is potential energy, this part right here.*0970

*This is the total energy is equal to the potential energy + the kinetic energy.*0981

*Therefore, this makes the kinetic energy operator.*0991

*I will use E kinetic.*0998

*Operators are quantum mechanical.*1008

*The operators, this and this, or this thing together, are quantum mechanical.*1022

*The energy is classical mechanical, that is the relationship.*1030

*There is an association.*1036

*Classical quantities, things like energy, momentum, angular momentum, position, things like that, are represented in quantum mechanics,*1039

*But operators that is the whole idea.*1052

*Operators are quantum mechanical.*1057

*Scalars like the energy are classical mechanical.*1060

*When there is some quantity in classical mechanics like momentum, position, energy, angular momentum, whatever it is,*1076

*In quantum mechanics those are represented by the kinetic energy operator.*1083

*The position operator, the momentum operator, they are represented by operators.*1087

*The reason they are is because all the information is contained about a particular system of quantum mechanical system is contained in the wave function.*1092

*In order to extract information from that function, we have to operate on it.*1103

*If I want to find the angular momentum of a particle, I'm going to take the angular momentum operator of the wave function.*1108

*If I want to find the position, I'm going to take the position operator of that function*1114

*and it is going to give me some certain information that I can do something with.*1118

*Operators are quantum mechanical.*1126

*Scalars are classical mechanical.*1128

*Such associations are the very heart of quantum mechanics.*1131

*Now given what is above,*1136

*Let me do this on the next page, that is fine I can do it here.*1140

*Given what is above, in another words this thing that we just did.*1144

*Let us define the kinetic energy operator that is equal to -HR²/ 2N DDX².*1151

*This is the kinetic energy operator.*1171

*If I have a wave function and I want to know what the kinetic energy is at a given moment,*1178

*I will go ahead and apply the kinetic energy operator to that wave function and it tells me something.*1182

*When I say apply it to that wave function, I will be more specific about that when we actually talk about*1188

*how we are going to extract information that we can actually measure and see from the wave function.*1194

*But that is what we are doing.*1200

*We defined V which is just V of X which means multiply the function by this function V of X.*1203

*This is the potential energy operator.*1219

*In other words, if I want to know what the potential energy of a particular particle is,*1222

*of an electron in this particular system whatever happens to be, I take the wave function and multiply it by the potential energy function.*1228

*This is the potential energy operator.*1239

*In this particular case, the operation is just multiply by.*1241

*It is fine, let me go ahead and in the next page.*1250

*The Hamiltonian operator that is equal to the sum of the kinetic and potential energy operators, that is equal to.*1252

*I think I can go ahead and put on the next page.*1260

*Let us do this.*1270

*H ̅ that is equal to –H ̅²/ 2M D² DX² + VX this is the total energy operator.*1273

*Or just the energy operator.*1295

*These are operators.*1304

*Let us go ahead and fiddle with this a little bit.*1307

*From the classical point of view, classically the kinetic energy of a particle you know was equal to ½ MV² that is also equal to this.*1311

*If you take ½ MV² it is going to be equal to the momentum squared divided by twice the mass.*1323

*We will go ahead and just start playing with this formally, mathematically.*1333

*Formally just means we are working with this symbolically.*1338

*We are going to define the kinetic energy operator is equal to this momentum operator squared/ 2M.*1340

*Let us go ahead and multiply here, the momentum operator squared is going to equal 2M ×*1356

*the kinetic energy operator and it is going to equal 2M × the kinetic energy operator which happens to be - H ̅²/ 2M D² DX².*1365

*The 2M cancels and we get ourselves a square of the linear momentum operator which is going to equal - H²² DX².*1380

*We are able to derive another operator.*1396

*These exponent here on the operator, it is nothing more than sequential operation.*1403

*Here is operator² is equivalent to doing P again.*1408

*If you saw the linear momentum operator cubed, it will just be P again.*1414

*The squared is just a symbol for sequential operation.*1424

*This does not mean this is symbolic for the operation.*1428

*It does not mean take the linear momentum operator, whatever you get square it.*1431

*That is not what it means.*1436

*When you see an operator squared, it means operate sequentially.*1440

*Let us go ahead and break this down.*1444

*Squared is equal to -H² D² DX².*1448

*Let us go ahead and separate this out.*1458

*- I H DDX in other words we are going to factor this out and - I H DDX.*1468

*This linear momentum operator squared actually factors out.*1475

*If I multiply these two, if I operate and operate, I end up getting this thing.*1479

*Therefore, we can go ahead and define the straight linear momentum operator in the X direction which is one of these – I H ̅ DDX.*1484

*This is the linear momentum operator,*1497

*if I have some quantum mechanical system and if I have another wave function for it.*1505

*If I want to know something about a linear momentum, whatever is that I want to know, I operate on the wave function with this operator and it gave me some information.*1510

*And again, what I mean by operating on it is not just going to be a direct operation.*1521

*We will see what it is, we are still going to play with this mathematically but is essentially what I’m doing.*1526

*I will leave that for the time being.*1534

*Let us see, now we have our linear momentum operator.*1538

*We have our Hamiltonian which is the total energy operator.*1542

*We have our kinetic energy operator.*1545

*We have a potential energy operator.*1548

*We have a majority of things that we need to actually get started.*1549

*Let us go ahead and finish off with an example here.*1553

*We will close this lesson out.*1557

*One more, here we go.*1561

*Let P = -I H ̅ DDX and in this particular case our functions are going to be sin X.*1564

*We want to show that the operator squared of F does not equal the operation on F².*1571

*In the last lesson we show that operators do not commute.*1581

*We want to also make sure that you understand the difference between squaring operator and taking something and getting a function and squaring that.*1585

*They are not the same thing.*1592

*Let us go ahead and do this.*1594

*The operator squared of F is equal to,*1597

*Again, we set it as a sequential operation on F.*1601

*It is going to equal this one, that is the far left one and it is going to be – I H ̅ DDX of sin X.*1606

*We end up with this being - I H ̅, the derivative of sin X is cos of X.*1617

*I think I’m going to it up here.*1629

*We have the - I H ̅ of the –I H ̅ of cos of X and we end up with - -, we end up with I² H ̅².*1635

*I’m sorry I forgot my derivative operator.*1658

*Let me go back and erase this, it is – I H DDX of - i H ̅ cos X.*1662

*When I take the derivative of cos X, I get - sin X - -is + I and I is I²,*1674

*H ̅ and H ̅ is H ̅².*1685

*I² is -1.*1691

*Let us skip a few steps.*1693

*- and - is going to be positive, I and I is going to be I².*1695

*I end up I² is -1 and – and - is going to cancel and become a positive.*1702

*We are just left with an I² H ̅² and sin X.*1705

*Let us go ahead and do P of F and we will square that.*1716

*The P of F is going to be – I H ̅ DDX sin X².*1721

*It is going to be – I H ̅ cos X² is going to equal -1 × -1 is positive.*1736

*I × I is going to be I².*1752

*H ̅² cos² X is equal to – H ̅² cos² X.*1756

*Clearly, this and this do not equal each other.*1767

*In general, some operator raised to some exponent and then applied to F absolutely*1772

*does not equal that operator applied to F and then raised to the exponent.*1783

*This is a symbolic representation of how many times you are going to operate in sequence.*1789

*This squared right here is an actual mathematical operation.*1793

*These are not the same thing.*1796

*We have to be very careful about our symbolism, about our mathematics.*1797

*Profoundly important.*1805

*Just take your time and work slowly.*1806

*Now that we have introduce this notion of an operator, there are a lot of symbols floating around.*1809

*In any case, we will go ahead and leave it that.*1818

*Thank you so much for joining us here at www.educator.com.*1821

*We will see you next time for a continue discussion of operators in Quantum Mechanics, bye.*1823

1 answer

Last reply by: Professor Hovasapian

Mon Feb 23, 2015 6:43 PM

Post by Carly Sisk on February 23, 2015

Hey Professor Raffi!

Just to clarify what Anhtuan was saying; the final answer should be positive (hbar^2 * sinx), correct?

1 answer

Last reply by: Professor Hovasapian

Wed Jan 28, 2015 12:32 PM

Post by Anhtuan Tran on January 28, 2015

Hi Professor Hovasapian,

On your last example, when you were doing P^2f, your second last step was i^2 * hbar^2 (-sinx). So the answer should be hbar^2 * sinx, because you already took into account the minus sign from sinx and minus sign from i^2. Therefore, there should be no i^2 left at the end.

Could you please double check it? Thank you.

2 answers

Last reply by: Professor Hovasapian

Sat Sep 27, 2014 6:27 AM

Post by Raj Singh on September 26, 2014

Where did the i in (-iHbar d/dx) came from when deriving linear momentum operator? additionally if you multiply 2 -i wouldnt you get positive i or positive 1?