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

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

### Schrӧdinger Equation as an Eigenvalue Problem

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- 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

### Chemistry : Physical Chemistry

### 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?