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The Hydrogen Atom 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
  • The Hydrogen Atom III 0:34
    • Angular Momentum is a Vector Quantity
    • The Operators Corresponding to the Three Components of Angular Momentum Operator: In Cartesian Coordinates
    • The Operators Corresponding to the Three Components of Angular Momentum Operator: In Spherical Coordinates
    • Z Component of the Angular Momentum Operator & the Spherical Harmonic
    • Magnitude of the Angular Momentum Vector
    • Classical Interpretation of Angular Momentum
    • Projection of the Angular Momentum Vector onto the xy-plane

Transcription: The Hydrogen Atom III

Hello and welcome back to www.educator.com.0000

Welcome back to Physical Chemistry.0002

Today, we are going to continue our discussion of the hydrogen atom.0004

Let us jump right on in.0007

Make sure we are on this page.0009

I’m going to go to blue.0011

In the last lesson, we finished off by a little discussion of the angular momentum then 0015

we talked about how the angular component of the hydrogen wave function the S of θ φ 0020

is actually an Eigen function of the square of the angular momentum operator.0025

We would like to continue along that track today.0031

Angular momentum is a vector quantity, in other words there is some angular momentum vector and 0035

it is going to have three components.0059

It is going to have an X component, Y component, and a Z component.0060

It is going to be L sub X in the I direction, X direction + L sub Y in the J direction, the Y direction and L sub Z in the K direction.0064

This is just the standard unit vector representation of the angular moment vector.0077

By the operators corresponding to the three components, the LX, LY, and LZ, we saw them before.0084

The operator is corresponding to the three components of L.0092

The three components of the angular momentum operator, we are talking about R.0115

We had L sub X is going to equal, I will write out everything.0122

It is Y PZ – Z PY.0129

The PY and PZ are the linear momentum components.0137

It = - I H ̅ Y DDZ - Z DDY.0142

I’m just recalling what these are.0155

The Y component is going to end up being Y P sub X - X P sub Z - I H ̅ Z DDX - X DDZ.0157

These are the operators, I’m going to list them out.0179

Of course, we have the final one which is L sub Z and that is going to equal the X sub Y.0182

It is going to be X P sub Y – Y P sub X.0187

It is going to be - I H ̅ X DDY - Y DDX.0194

I’m sorry, is my order correct?0203

These are Cartesian coordinate representations, we are working at spherical coordinates.0206

Let us go ahead and see what these look like in spherical coordinates.0213

Let me do this in red and hopefully you do not have to actually go through 0218

the process of converting it because it is a very tedious exercise.0230

It is part of your exercises in your book, I know it is but hopefully you do not have to do it yourself.0235

In spherical coordinates, it looks like this.0241

The X component of the angular momentum is going to equal - I H ̅ that is going to be – sin φ.0243

Let me sure I will get all of this correct here.0254

DD θ, I’m telling you try to keep track of all the symbols is the real challenge with all the quantum mechanics, - the cot of θ.0260

The cos of φ DD φ and the Y component is going to equal - I H ̅ is going to be cos of φ DD θ - cot of θ 0272

and this is going to be the sin φ and it should be DD φ.0294

We have L sub Z, it is going to look very different.0301

This one looks really simple.0304

It is going to be -I H ̅ DD φ.0306

This looks very different from the Y and the X.0311

Let us go ahead and apply this last one, the Z component of the angular momentum.0315

Let us apply it to our angular component of the rigid rotator function of S of θ φ.0320

Let me go back to blue.0330

Let us apply L of Z to the S sub LM of θ φ.0337

Let us see what we get.0347

I will write this one very carefully when I go through it.0351

L sub Z of S LM, I wonder if I should use my θ and φ.0354

It is going to equal, we have the normalization component LN and then we have the L sub Z of S sub L sub M, 0368

that is going to equal the associated Legendre function P sub L absolute of M of cos θ × E ⁺IM φ.0379

That is going to equal N LM P sub L absolute M of cos θ × –I H ̅ DD φ applied to E ⁺IM φ.0394

The L sub Z is you are taking the partial derivative with respect to φ.0422

There is no θ involved so I can just go ahead and skip this all together.0428

It is like a constant, I just apply it to the function that has a φ in it which is the exponential portion of it.0433

We end up with N LM P sub L absolute of M of cos of θ × IM × – I H ̅ × E ⁺IM φ.0439

When I actually differentiate DD φ with respect to this, this IM comes here.0461

This is going to equal, I'm going to bring this so I and I is -1, negative and negative is positive.0467

I’m going to bring this H and this M out front so I get H ̅ × M × N sub LM P sub L absolute M cos θ × E ⁺IM φ.0475

This is equal to, this part right here, this is just S LM of θ φ.0495

That is just S, it is the same thing.0512

Now, I have an operator, something operating on a function gives me the function back × a constant.0515

That is equal to H ̅ M × S sub LM of θ φ.0524

Our spherical harmonics, S sub L M of θ φ are Eigen functions of the Z component angular momentum operator with Eigen value H ̅ M. 0535

Let us go back to blue.0574

In other words, when something is an Eigen function, when some function is an Eigen function of an operator.0577

In other words, when we measure values for the Z component of the angular momentum we get integral multiples of H ̅.0582

H ̅ is a basic unit.0615

Notice H ̅ is a basic unit of measuring angular momentum for a quantum mechanical system.0622

H ̅ is a basic unit of measure for angular momentum in the quantum mechanical system.0628

Spherical harmonics are Eigen functions of the square of the angular momentum operator L².0655

They are also Eigen functions of the Z component of the angular momentum operator.0665

The spherical harmonics S sub LM are not Eigen functions of the X and Y components of the angular momentum operators.0672

They are not Eigen functions of L sub Y and L sub X.0704

The spherical harmonics of sub LM are Eigen functions of L² and L sub Z.0718

Because the S sub LM are Eigen functions of both L² and LZ, these two operators commute.0742

These two operators, they commute.0771

Remember, when we said when we talked about operators that commute.0779

For two operators that do commute, their commutator = 0.0785

It think we can apply the operators on either order.0789

When that is the case, that means that we can measure each one to any degree of precision that we want.0792

When operators do not commute, like position and linear momentum, 0800

that is when Heisenberg uncertainty principle comes into play.0806

It says that if you more accurately measure one, the less accurately you know the other.0808

There is a certain compromise that I have to come to.0814

But for operators that do commute, I can measure them both to any degree of accuracy that I want.0817

In this case, I can measure the square of the angular momentum or the angular momentum and 0822

I can measure the Z component of the angular momentum to any degree of accuracy I want0828

but I cannot say anything about the X and Y components.0833

I can tell you in which direction the X and Y components are.0838

I can tell you in which direction the actual angular momentum vector is.0841

I can tell you the direction of the Z component but I cannot tell you where X and Y are pointed.0844

That is what this is saying.0849

Because they are Eigen functions of both L² and L sub Z, these two operators commute.0851

This means that we can measure values for L² and LZ simultaneously to an arbitrary degree of precision or accuracy.0861

I’m a little loose with those words but it is not going to hurt us here.0895

I was going to actually prove this as an exercise but it will be the extra mathematics here that is not altogether necessary.0908

We may instead be doing it in the part of the problem sets, if I decide to do that.0916

For now, I think we will actually save ourselves a little bit of time and little bit of mathematical grief 0920

and we will just take it on fate if this is the case.0925

The angular momentum operator and the Z component of the angular momentum operator, they commute.0929

Let us go ahead and move forward here.0943

We already know that L takes on the values 0, 1, 2, . . . And M 0947

which is dependent on L takes on the values of 0, + or -1, + or -2, all the way to + or –L.0962

Those are the highest values.0970

For every value of L, there are 2L + 1 values of M.0975

There are 2L + 1 values of M for each value of L.0982

Let us recall what we have here.0999

We have L ̅² of S L sub M.1002

I going to skip this and say of φ.1009

We said that was equal to H ̅² × L × L + 1 × S LM.1010

This just is a repetition of the fact that the spherical harmonics are Eigen functions of the angular momentum square operator.1018

The square of the angular momentum operator.1026

This is exhausting.1029

We also have that, they are Eigen functions of the Z component so L sub Z applied to S LM is that is equal to, we found H ̅ M S LM.1032

If we choose the state L = 1, we are going to end up with the following.1049

We are going to get L ̅² S1 M is equal to H ̅² × 1 × 1 + 1 S1 M.1067

We are going to end up with L sub Z S1 M is equal to H ̅ M S1 M.1086

Let me erase this and let me we just put and here.1103

For M = 0, + 1 -1.1108

L is 1 so M is going to be 0 + 1 - 1.1114

Over here, because this is the case when we measure the L², 1119

we are going to end up with these values because these are Eigen values for the Eigen function.1126

This is just 1 + or -2, we are going to end up with L² as a value for the angular momentum.1132

Now, it is going to equal to H ̅² which means that the magnitude of the angular momentum is actually going to be H ̅ × √ 2.1142

That is the magnitude of the angular momentum that I measure.1155

Over on this side, I have L sub Z.1159

When I measure these, these are the Eigen values so those are the values that I get.1164

L sub Z is going to equal H ̅ M.1168

Well, M is equal to 0 + 1 -1.1173

Therefore, LZ is going to take on the values –H ̅ 0 and +H ̅.1177

I have to put the positive there.1185

My angular momentum for the state L = 1, my angular momentum itself is going to have a value of √ 2 × H ̅.1191

The Z component of the angular momentum is going to have the value of H or 0 or –H ̅.1200

H ̅ is 0 or – H ̅.1207

The absolute value of L is the magnitude of the angular momentum vector.1212

In other words, this electron is flying around this proton.1233

It was moving a circular orbit.1237

It is going in and out, there is a circular component to it.1238

Anything that moves in a curve around a fix center is going to have an angular momentum.1245

This actually gives me the magnitude of the angular momentum of electron for wherever it is.1249

That is what is happening here, that is what we are doing.1257

L is the magnitude of the angular moment vector.1259

Notice that the magnitude of L is not equal to the magnitude of LZ.1262

The magnitude of L is H ̅ × √ 2.1275

The magnitude of LZ is just H ̅, they are not the same.1279

The angular momentum vector and the Z component of the angular momentum vector, they do not point to the same direction.1287

Otherwise, they would have the same value.1306

They do not point to the same direction.1309

We have the magnitude to the angular momentum vector.1319

We have the magnitude and the direction of the Z component of the angular momentum, 1322

it is moving in the Z direction and has certain magnitude, in this case H ̅.1328

In order to specify the direction of the angular momentum vector L itself, 1332

we need to know which direction X and Y are pointed at also.1339

We do not know that, all we know is the Z component.1343

We have the magnitude of the total angular momentum but we do not know its direction.1347

In order to know the direction of L, we must also be able to specify LX and LY.1354

We need to know which direction they are in to but this is not possible.1387

Let me erase these.1400

We know the directions, the direction of X and the direction of Y is the direction of Y, 1403

we do not know their values but this is not possible.1407

This commutes with this, so those commute.1417

The square of the angular momentum operator commutes with LZ.1426

It will commute with LY.1432

It will commute with LX.1433

One of them, it does not commute with all of them simultaneously.1437

LZ or LY, in other words there is nothing particular special about Z.1444

It just depends on how we fit for coordinate system.1449

The Z ended up being the one that ended up being the easiest to deal with.1452

The truth is, L² commutes with LZ or it commutes with LY, or it commutes with LX.1456

But the issue is the LX, LY, and LZ, do not commute among each other.1469

What that means is if I can specify the Z component of the angular momentum, I cannot say anything about the X or the Y.1484

If I specify the Y component of the angular momentum, I cannot say anything precisely about the Z or the X.1499

That is what is going on here, I cannot specify all of them simultaneously.1507

I have to choose one of them but only one, that is what this means by commute among each other.1511

We get the following picture.1520

The picture that we end up getting looks like this.1523

For a particular point in space, this is going to be the origin.1541

Here is what is going to happen, this is the Z axis, this is the X axis, and this is the Y axis.1556

I’m going to go ahead and draw my angular momentum vector.1562

Let me go ahead and draw this, let me make a circle here.1572

Let me work in red.1584

This vector right here, that is the Z component.1589

This is L sub Z that is equal to H ̅.1598

This right here, this is equal to H ̅ √ 2.1603

This is equal to the actual magnitude of the angular momentum.1607

This right here is the magnitude of the angular momentum.1615

We know the Z component, we know the actual angular momentum, the problem is we do not know X or Y.1621

We do not know how much and what direction.1631

What ends up happening is this is the angular momentum vector, this has to be this thing right here.1634

We know the Z component straight up, the problem is we do not know about the X or the Y.1642

What ends up happening is anything is possible, any direction is possible.1649

This angular momentum vector actually ends up recessing, going in a circle around the Z axis 1653

and actually ends up sweeping out of cone.1661

It is actually going to sweep out this cone, that is what is happening.1670

We do not know about X and Y, all we know is Z.1671

And because we do not, anything is possible in the X or Y direction.1675

We end up getting this angular momentum vector, the actual L sweeping out this cone.1679

I hope that makes sense.1684

Let us go ahead and write this down.1687

Because we cannot specify LX or LY, this is the magnitude, the vector L, this thing right here, it sweeps out the cone.1689

This is a classical interpretation of what it is that actually going on.1720

We are just interpreting it in terms of what it is we know from classical mechanics.1726

The angular momentum be a vector value thing.1731

This is a classical interpretation because we need to make sense of it somehow for all purposes.1737

We need to know what is it that we are looking at.1748

With respect to something that we know in classical mechanics angular momentum, we can go ahead and talk about what is happening.1752

We can go ahead and assign values to different vectors, that is what we are doing.1759

This is a classical interpretation of what is happening.1764

The angular momentum vector processes around the Z axis to sweep out a cone.1776

LX and LY do not have values that can be specified precisely because they cannot commute with LZ.1794

Because they do not commute, there is no way of actually specifying the LX and LY,1807

or LX and LY precisely because of the uncertainty principle.1813

LX do not have values that can be specified precisely but they have the average values.1817

We can always find average values.1841

If we cannot specify precisely what value is, we can always get an average value.1843

Let us stop for a second and talk about what we mean by precise and average value.1852

We said that these spherical harmonics are Eigen functions of the square of the angular momentum operator 1857

and they are also Eigen functions of the Z component of the angular momentum operator.1865

That means any time that we measure the Z component, we are going to end up getting the same value over and over and over again.1871

It is either H + or H-, they are very specific.1877

When we take measurements of the LX and LY, we get a whole bunch of different values all over the place.1881

That is what it means for them not to be Eigen functions, they are going to be different values.1887

They are not going to be specified as this or this precisely.1892

There is going to be lots of deviation.1896

There is going to be a standard deviation for the X component and Y component.1898

On average, when I take all of those values that 200, 500, 1000 values that measure, 1903

I averaged them out, I can find an average value for it.1910

But they have average values.1914

The average values are as follows.1916

The average value for the X component of the angular momentum is going to be 0.1918

And the average value for Y component is also going to be 0.1922

The reason is this is the following.1928

I will draw it out in just a minute.1930

In the picture above, we do not know in which direction the LX and LY are going.1932

Because we do not know which direction they are going, they can go in any direction.1946

Because they can go in any direction, this way and that way, on average everything cancels out to 0.1949

We do not know in which direction the LX and LY point because we know that 1956

the unit vectors has to point to the X and the Y direction but they are going to be certain magnitudes.1981

Because of that, the vector can be this way or this way or this way or this way.1984

I will draw it out in just a minute here.1991

The picture that we get is the following.1993

If I take this thing, if I turn around and look at it from the top, looking down along the Z axis on the XY plane, here is a picture that I get.1996

This is the X axis and this is the Y axis, the Z axis is right here, it is coming straight up.2012

If I'm looking down on this, the vector this is the Z axis.2018

This angular momentum vector is sweeping out a cone.2024

I'm looking at it from up above.2027

From your perspective, what is happening is this thing.2030

It is actually sweeping a circle.2034

Looking at it from above, the vector actually sweeps out a circle as it swings this way,2036

the projection of the precessing motion of the angular momentum vector. 2042

This vector right here looking at it from above, the projection of a precise motion of that along to the XY plane is a circle.2060

Now, I cannot specify the X and Y components which means that the X and Y vector.2076

Let me write what we have.2087

L we said is equal to LX I + LY J + LZ K.2092

I’m able to specify this, I can specify this and this so that means it can be this vector or this vector.2101

It could be any of the vectors I do not know.2109

Because it can be any of them, if I take a bunch of measurements,2112

I'm actually going to end up with every single measurement possible.2115

On average, it is going to end up being 0 because for every one of these, I'm going to have one of these.2119

For every one of these, I'm going to have one of these.2125

It is all going to average out to 0, that is why the average values.2127

At any given instant, L sub XI + L sub YJ, let me make sure I got B, any vector along the circle on average,2133

you are going to end up with 0 LY = 0.2166

I hope that makes sense.2174

Thank you for joining us here at www.educator.com.2175

We will see you next time, bye. 2177