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The Hydrogen Atom I

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 I 1:31
    • Review of the Rigid Rotator
    • Hydrogen Atom & the Coulomb Potential
    • Using the Spherical Coordinates
    • Applying This Last Expression to Equation 1
    • Angular Component & Radial Component
    • Angular Equation
    • Solution for F(φ)
    • Determine The Normalization Constant
    • Differential Equation for T(a)
    • Legendre Equation
    • Legendre Polynomials
    • The Legendre Polynomials are Mutually Orthogonal
    • Limits
    • Coefficients

Transcription: The Hydrogen Atom I

Hello, and welcome to, and welcome back to Physical Chemistry.0000

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

We are ready to put everything together and we are ready to talk about the orbitals,0008

the energy levels of the hydrogen atom.0012

One little bit of warning before we start here.0014

This is going to consist of multiple lessons and is going to be very heavily mathematical,0018

at least in the sense of all the symbols floating around on the page.0023

Ultimately what is important is, we want to be able to find the solutions for ψ,0026

the wave function for the electron that is moving around the proton in the hydrogen atom.0034

That is what is most important.0039

It is easy to lose our way, I'm going to present it in the normal fashion.0041

I’m going to present most of the mathematics, not all of it.0047

It is really important that you separate, do not get intimidated and just alienated from all the symbols on the page.0050

I will try to keep it clear as to what is important and what is just cultural information.0060

What is just part of your scientific literacy.0066

I could just present the solutions and say this is what it is and jump one to some problems0068

but we also want you to understand where does that these things come from because that is where real understanding takes place.0072

Ultimately, you are only responsible for the solutions0078

but we want to show you where the solutions come from and why they take the form it takes.0081

Just a little bit of warning, do not be intimidated, it is just symbols.0085

Let us get started.0089

In discussing the rigid rotator, we held R, the radius of the turn constant.0092

We allowed θ and φ to vary.0120

Something rotating this way, this way, is basically just something spinning around some fixed center.0130

You had the θ which is this way and the φ which is this way.0138

Those are the things that vary so it gave you all of the points on a sphere.0141

We are going to allow R to actually change.0146

We are going to allow not just move it about a sphere, we are going to allow it to be close.0149

We are going to allow it to be far, so all of space is going to be considered now.0153

We allow R to vary also.0163

For the hydrogen atom, we take as our model a fixed proton of the center,0171

a fixed proton at the origin so that will be our 0 position or origin position.0197

An electron of mass M sub E interacting with the proton via the Coulombic potential.0205

Basically, this is just a fancy way of saying we have a positive charge and a negative charge and0231

they are going to interact with each other based on the fact that positive and negative charge attract each other.0234

That is it, that is all we are saying.0240

The Coulomb’s, the potential is the potential energy is 1 charge of positive charge and negative charge,0243

the distance between them being R.0250

The potential energy is going to equal – E² / 4 π ε sub 0 R.0253

E here, it is the charge on 1 proton, the charge an electron in Coulomb’s.0264

It is going to be 1.602 × 10⁻¹⁹ C.0271

This ε sub 0, that is a permittivity of free space and its value is 8.854 × 10⁻¹² C² / J-m.0278

It is called the permittivity of free space.0299

R is the distance between the proton and the electron.0315

Basically, you just have this proton, center, and you already know it is already electron is moving around all over.0329

Not necessarily in circular orbits but now it is moving around this way, it is getting close, it is getting far, it can be anywhere.0342

Let us see what we have got.0351

Basically, we dealt with this rigid rotator which is a perfect sphere and all we have really done is allow the electron to move in and out.0354

Get close to the proton, get far from the proton.0362

We still want to work in spherical coordinates because the symmetry of the situation naturally lends itself to spherical coordinates.0365

That is why these various coordinates systems exist0374

because certain problems lend themselves to a particular coordinate system so the math actually becomes easier.0377

In spherical coordinates, let me go ahead and erase this.0384

Let me see, should I go ahead and put on this page.0391

A spherical symmetry of the model compels us to use spherical coordinates.0394

We are going to end up with something like this.0426

The Schrӧdinger equation is going to be H of ψ.0428

Ψ is a function of 3 variables, R, θ, and φ.0431

R, θ, and φ is equal to E × ψ of R, θ, and φ.0436

The Eigen function, Eigen value representation of that Schrӧdinger equation.0447

H is going to be, the Hamiltonian operator here is going to be – H ̅² / 2 × the mass of the electron Del² - the potential energy.0451

The potential energy was the Coulombic potential here, - E² / 4 π ε NR.0466

We know what del² looks like in Cartesian coordinates.0483

A spherical coordinates, it looks like this.0492

Let me go ahead and rewrite this.0495

H R is equal to - R² / 2 ME Del² -E².0497

Where the del² operator in spherical coordinates, when we actually make0522

the change of variable from cartesian to spherical, it looks like this.0527

It looks really complicated.0531

Do not let it scare you, it is just symbols.0533

We just want you to see it.0534

It is in your books, we want you to see it.0536

We want you to at least recognize it, that is all we ask.0538

Whatever R², DDR, R², DDR, so this is the del² operator in a spherical coordinate.0542

I'm sure that I’m forgetting something here, R sin θ DD θ sin θ DD θ +,0561

let me actually write it here, so I’m not squeezing it anywhere, + 1/ R² sin θ × E² D φ².0578

We see we have R, we have θ, and we have the φ.0588

We got the Laplacian operator in spherical coordinates.0593

That is all it is.0598

If we put this last expression into the equation 1, the original equation that we had0601

which was this ψ of R θ and φ is equal to energy × ψ R θ and φ.0607

In other words, if we apply this to this, we end up with something like this.0620

We apply this last expression, meaning this thing right, the whole thing,0633

that where this del square is equal to as, we apply this expression to, let us call this equation 1 and multiply by 2 M sub ER².0642

We get this thing.0671

Let us see if we can, that is getting crazy here.0676

We get –H ̅² DDR² D of ψ DR –H ̅² 1/ sin θ × DD θ sin θ D of ψ D θ + 1/ sin².0680

Here are the mistakes, I think there is an R² there and an R² there.0726

1/ sin² θ D² ψ.0731

Once again, I’m going to keep saying it this is not something that you have to know.0740

We just want you to see it because I think it is actually need to see it.0744

It is in your books but more than likely you are just sorting orders in your books, like the rest of us in school.0748

-2 M sub E R² E/ 4 π E sub 0 R + the energy × ψ = 0.0754

This is the equation that we end up solving.0776

It is really nice to see it because I think it is one intellectual achievement back0779

but we have managed to come up with this and would actually managed to solve it and be very successful in our solutions.0783

It is actually quite extraordinary.0790

It is really nice to know that you are in a position to appreciate this, see it, and understand what it means.0792

It is not necessarily the mechanics of the actual solution of it so you really should be very proud of where it is that you are.0798

This ψ, we said is actually a function of 3 variables.0810

It is a function of R, θ, and φ, so in order to solve this, we are going to separate variables.0815

We are going to write this ψ R θ and φ, we are going to write it in terms of some function just of R and some function of θ and φ.0823

The wave function for the hydrogen atom, we are going to separate them into two functions, a product of two functions.0842

One of them is just a function of the radius.0849

One of them is just a function of the angles θ and φ.0851

This one right here, this is called the radial component of the hydrogen wave function.0858

This is called the angular component.0871

We are going to deal these one at a time.0876

Since we have separated variables and written ψ as a product of two functions, we recovered two separate differential equations.0887

We are going to have a radial equation and we are going to have an angular equation.0922

We are going to deal with the angular one first.0931

We will deal with the angular equation first.0945

Let us go ahead and start working in blue.0955

When we do this, we are going to take these angular differential equation that we get0958

and we are going to subject it to further manipulation.0964

I’m not going to go through all of the manipulations but when we do that,0966

I will just say with further manipulation which is nothing more than simplifying it so we are going to deal with it better.0971

With further manipulation, we get sin of θ DD θ sin of θ DS D θ.0978

S is the function of θ and φ, the angular component that consist of the two variables.0995

Θ + D² S D φ² + β sin² θ × S is equal to 0.1003

This is for the angular component, the S θ of φ, this is the differential equation that we have solved in order to find S.1020

We are looking for S.1032

Notice that this equation is the same as the Schrӧdinger equation for the rigid rotator.1034

It is the same as the Schrӧdinger equation for the rigid rotator.1058

The angular portion of the hydrogen atom wave function is the same as the rigid rotator wave functions.1071

Let us see here, let us go ahead and do this.1116

We are looking for this S of θ and φ.1121

That is what we are looking for, the angular components.1128

We are dealing with this so we subject this to another separation of variables.1131

We are going to take this S of θ and φ because we are dealing only with that and deal with it by separating variables.1138

We are going to express this as a product of two functions, 1 just of θ, 1 just of φ.1145

I will call this T of θ and F of φ.1153

We are just making our life easier, this is what we do.1159

We start with 3 variables, we separated into 1 and 2 variables.1161

We deal with the 1 and 2 variables, we separate that into 1 and 1, that is what we are doing here.1165

The solutions for F of φ are F of φ is equal to this normalization constant A sub M E ⁺IM φ,1173

where M is another quantum number and it is equal to 0 + or -1, + or -2, and so on.1204

This is the normalization constant.1216

Let us determine what the normalization constant is so we can actually write a full function for this.1228

Let us determine the A sub M.1238

The normalization condition, we know that the normalization condition is that1243

the integral of the function conjugate × the function has to equal 1.1256

In this particular case, φ itself is greater than or equal to 0 and less than or equal to 2 π.1265

Those are our limits of integration, 0 to 2 π.1271

We have the integral from 0 to 2 π of S * sub M × FM = 1.1276

This is going to be the integral from 0 to 2 π.1287

We are going to take F sub M*, we are going to star this, we are going to conjugate this.1294

It is going to be A sub M, when we conjugate, we conjugate everything.1303

It is going to be the constant, conjugate of the constant E and – IM φ1312

because the conjugate of the IM φ is – IM φ × the function itself F sub M which is A sub M E ⁺IM φ D φ.1318

This is this, that ends up becoming this and this, ends up becoming the absolute value of A sub M².1338

The integral from 0 to 2 π, this and this may cancel out, - IN φ + IN φ is 0, which is equal to 1.1352

All you are left with is D φ, that is equal to 1.1362

Therefore, what we have is absolute value of A sub M² × 2 π.1368

This integral from 0 to 2 π of D φ is equal to 2 π is equal to 1.1375

That implies that A sub M is equal to 1 / 2 π M =0 + or -1, + or -2, and so on.1382

Our final function F of φ, this thing right here, what it looks like is the following.1402

F of φ is equal to 1/ 2 π, I will write it as, the exponent of ½ E ⁺IM φ.1409

Again, subject to these values for M.1421

This is the φ component of the angular component of the hydrogen wave function, fully normalized.1425

This is definitely one of the equations, one of the functions that we are interested in.1436

We are making good progress here.1445

We will see what we have got.1448

We had S of θ and φ.1456

When we separate variables for this, we end up with two differential equations.1462

We did mention the differential equation for the φ component, we just gave you the solutions.1467

For the other part, the T of θ.1477

This one I’m going to write down the differential equation.1486

We go to blue here.1496

We had this S of θ and φ, we separated into a function just of θ, and the function just of φ.1499

Let me go back to red.1508

We just found the F of φ, now we want to find the T of θ so we can multiply them together to get our full angular solution.1508

The differential equation for T of θ is the following.1521

It is going to be sin of θ T of θ DD θ sin of θ DT D θ + this β sin² θ = M².1530

The solution to this differential equation involves something called the change of variable.1556

We are going to change the variable and we are going to actually solve this equation.1561

The solutions will get our act going to be polynomials.1567

The variable inside those polynomials is actually going to be a cos θ.1570

It is actually is going to be a function of θ.1574

Solving this involves using a change of variable.1582

You do not have to know this, you do not have to reproduce this.1596

If you can follow it, that is great.1602

If not, this is just part of your general scientific literacy.1604

The change of variable we are going to use is going to be X = cos θ.1607

Θ is running from 0 to π so T of θ becomes a polynomial in X comes which will designate as P of X.1615

Under this change of variable, what ends up happening to the differential equation is the following.1642

I will go ahead and stay with red.1648

Under this change of variable, the differential equation becomes the following.1651

It becomes 1 -X² D² ψ DX² -2 X DP DX + β – M²/ 1 - X² × P is equal to 0.1665

Again, M is equal to 0 + or -1, + or -2, and so on.1693

This is a very important equation in physics, it is called the Legendre equation.1702

This is called the Legendre equation.1710

It shows up a lot.1720

Let me go ahead and go back to blue here.1729

What is the equation we solve?1731

When we solve this equation, in order for these polynomials, the solutions P of X to stay finite, we need it to stay finite.1739

This β, it must equal λ × L.1755

In order for the solutions to stay finite, this constant β has to equal this L × L + 1.1768

The value of L is another quantum number.1778

It is actually equal to 0, 1, 2, and so on but with a restriction that the absolute value of M has to be less than or equal to L.1781

There is a connection, M is actually dependent on L.1798

We write, we are just going to put this into β.1805

We write 1 - X² D² P DX² -2X DP DX + L × L + 1 - M²/ 1 - X² P = 0.1814

Where, L is going to take on the value 0, 1, 2, and so on.1845

M is going to take on the values + or -1, + or -2, all the way to + or – L.1853

The absolute value is less than or equal to L, which means that M is going to take on the values when we write them out explicitly 0, 1, -1, 2, -2, 3, -3.1861

If L is 4, then is going to be 4, -4 all the way up to that value.1872

Now, we have two quantum numbers that are going to arise here.1876

There are different sets of solutions for the various values of L.1882

Let us actually start writing this so we can read it.1907

For the various values of M, we are going to look first, let us first look at M = 0.1911

When M is equal to 0, the solutions which we symbolize as1931

P sub L of X are called appropriately Legendre polynomials because they are coming from Legendre equation.1941

The Legendre polynomials are very important.1956

I’m going to go ahead and list the first few Legendre polynomials.1967

We are taking M = 0.1970

The first few Legendre polynomials, P sub 0 of X is equal to 1.1986

P sub 1 of X now M is equal to 0, this subscript right here, that is the L value.2000

P sub 1 of X is going to equal X.2006

P sub 2 of X is going to equal ½ × 3 X² – 1.2012

P sub 3 of X is going to equal ½ 5X³ -3X.2022

We will go ahead and stop with P sub 4, that is going to equal 1/8 35 X⁴ - 30 X² + 3.2032

It is very important to remember.2048

Recall that X is not a variable here, they are the change of variable.2054

X is actually equal to cos of θ.2061

If I do P2 of X that is actually P2 of cos θ.2071

Wherever X is, I'm just going to stick in to cos θ.2080

In the case of the P2, I'm going to have ½ 3 cos² θ -1.2083

Θ is still the variable that we are concerned with.2100

Remember ψ is a function of R, θ, and φ.2103

All we have done is change the variable in order for us to actually solve this differential equation2107

and make it a little bit easier to deal with, that is all we have done here.2111

X is not the variable, XYZ is not a coordinate.2114

It is just a variable that we are using a hold the place of cos θ, the change of variable.2117

The function is really this thing right here because again we need a function of θ2123

but we are expressing it in terms of these polynomials.2128

This is very important to remember.2131

I will remind you of that several times because it is easy to lose ones way.2132

We all do, I do all the time.2138

Let us go back to blue here.2140

We said that L takes all the values from 0, 1, 2, 3, 4, and so on.2145

A couple of properties here of these polynomials.2150

When L is odd, the Legendre polynomial P of X is an odd function.2153

When L is even, the Legendre polynomial is an even function.2169

It is also true that the integral from -1 to 1 of P sub Q of X, P sub R of X, is equal to 0.2181

In other words, these Legendre polynomials are actually orthogonal.2204

They are mutually orthogonal.2208

That is the Legendre polynomials are mutually orthogonal.2212

If you are wondering where I get this -1 to 1, let me go to the next page and tell you.2230

The limits here are -1 to 1 because again X is equal to cos of θ.2237

Θ is running from 0 all the way to π.2255

X is cos θ so X goes from cos of 0 to cos of π.2261

Cos of 0 is 1 to -1.2277

It is θ is the variable that we are interested, θ runs from 0 to π.2281

When we take the cos of 0 like this one, we take the cos of π that gives us -1.2285

Our limit of integration, when we are integrating the polynomial is going to be -1 to 1 or 1 to -1.2290

The order does not matter, you are just going to switch signs.2297

When we are integrating with respect to θ, that is when we put 0 to π.2301

That is why -1 to 1, that is where the limits come from.2304

The last property, the coefficients in front of the polynomials, the ½, the 1/8, and so on,2310

the coefficients in front of the P sub L of X are chosen so that when we do the P sub L of 1 it always ends up equaling 1.2320

It is just for convenience, we have chosen it that way.2341

For example, P sub 2 of X is equal to ½ 3 X² -1.2344

P sub 2 of 1 is equal to ½ 3 × 1² - 1 is equal to ½.2359

3 × 3 – 2, we actually end up getting a value of 2.2371

We stick a ½ in front of it, in order for us to end up with a final value of 1.2377

This is the actual polynomial, we have adjusted it by putting the coefficient in front of it to make sure that P of 1 is equal to 1.2382

That is what gives us our final form of Legendre polynomials.2391

We will go ahead and stop this lesson here.2396

Thank you so much for joining us here at

We will see you next time for a continuation.2399