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Lecture Comments (7)

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?

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 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

Transcription: Schrӧdinger Equation as an Eigenvalue Problem

Hello, welcome back to 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

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