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 2 answersLast reply by: shashikanth sothukuThu Mar 26, 2015 12:07 AMPost by shashikanth sothuku on March 25, 2015hello prof,why do the laplacian op. has an exponent 2 instead of a caret? 1 answerLast reply by: Professor HovasapianFri Jan 2, 2015 9:02 AMPost by David Llewellyn on December 31, 2014In the one dimensional case you had the wave function = 0 at x<=0 and x>=a but in the two and three dimensional cases you only mentioned boundary conditions when x and y and z <=0 but not when they were >=a, >=b and >=c respectively. Is there a reason for this? 0 answersPost by Professor Hovasapian on September 27, 2014Hello Everyone.I hope the course is going well for all of you.I wanted to mention a slight error in this lesson: for the Particle in a 2-dimensional box, I wrote h^2/2m for the coefficient of the energy. It should be h^2/8m. My apologies.Raffi Hovasapian 2 answersLast reply by: xlr zMon Sep 29, 2014 5:17 PMPost by xlr z on September 27, 2014I think at 22:13 equation for energy it suppose to be (-h bar/2m)

### The Particle in a Box Part 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
• 2-Dimension 0:12
• Dimension 2
• Boundary Conditions
• Partial Derivatives
• Example I 6:08
• The Particle in a Box, cont. 11:28
• Operator Notation
• Symbol for the Laplacian
• The Equation Becomes…
• Boundary Conditions
• Separation of Variables
• Solution to the 1-dimensional Case
• Normalization Constant
• 3-Dimension 28:30
• Particle in a 3-dimensional Box
• In Del Notation
• The Solutions
• Expressing the State of the System for a Particle in a 3D Box
• Energy Level & Degeneracy

### Transcription: The Particle in a Box Part III

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

Today, we are going to finish off our discussion of the particle in a box.0004

In the next lesson, we are going to start on the example problems.0008

Let us just jump right on in.0011

In the last lesson, we discussed the particle in a box, the Schrӧdinger equation, and we did for a 1 dimensional box.0015

Finding a particle on some interval.0021

We are going to do the 2 dimensional box which is just a plane region and the 3 dimensional box0023

which is really what you think about when you think about a box.0029

In dimension 2, the Schrӧdinger equation looks like this.0039

It is going to end up being –H ̅²/ 2 M.0044

I think it is always a good idea to keep writing these over and over and over again,0050

until you just feel comfortable writing them until they are just another part of your day.0054

D² ψ of DX² + D² ψ/ TY² = E × ψ.0059

Where now X is going to be ≥ 0, ≤ A.0074

Y is going to be ≥ 0 and ≤ B.0078

The ψ is a function of two variables X and Y, not just some F of X.0089

It is going to be some ψ which is a function of X and Y.0097

When we do 3 dimensions, it is going to be XYZ.0107

We are looking for some function that has all 2 or 3 variables in it.0109

The boundary conditions are going to be as follows.0114

Do not worry, I’m going to draw this often just a second.0116

Our boundary conditions where the wave function goes to 0.0118

We are actually containing it in this little square region.0123

The wave does not exist outside of that.0126

Those conditions mathematically are represented as follows.0131

0 Y = 0 and ψ of X 0 = 0.0134

This is our mathematical problem that we have to solve.0143

This is their differential equation that we have to solve.0148

Here are the constraints on the variables.0150

Here are the constraints on the actual function itself on the boundaries namely on the outer edge.0152

Here is what we are looking at now.0161

I will do it over here.0164

Let me do this in blue.0166

We have our coordinate system.0172

This is Y and this is X, I’m going to go ahead and put A over here.0174

I will go ahead and put B over here.0179

In general, A and B are going to be different but they can be the same, square box.0182

We will go ahead and do this.0188

In this region, we are going to contain the particle in that region.0192

The particle is constrained to be in this 2 dimensional boxes.0204

2 dimensional region, that is where the particle is going to be.0220

These boundary conditions, all they are saying is that whenever the wave function for any one X0223

is always 0, for any value of Y wave function is going to be 0.0229

That is the same things here for all values of X is 0.0237

All it is saying is it is going to be 0 on that boundaries.0243

Let me go ahead, I will stick to blue.0254

If you are not familiar with this notation, this D ψ DX or D² ψ DX².0257

If you are not familiar with that, these are called partial derivatives.0267

Some of you may have seen them, for those of you have not, do not worry about it.0273

I’m just going to take a couple of minutes to explain it, do an example.0277

It is very simple, you actually know what to do.0279

This is just a notational difference.0282

They are called partial derivatives and what you are doing when you take a partial derivative of a function of 2 variables, 3 variables, or 4 variables,0284

you differentiate with respect to the bottom variable.0292

It is the same thing, it is like D ψ D.0308

It is the same thing, you are taking a derivative but now we use a different notation because we have another variable.0310

You differentiate with respect to the bottom variable treating the other variable as a constant.0316

You are doing the same exact thing you have always done for the past several years with derivatives.0323

Treating the other variable as constant.0327

That is all this notation means.0342

When we have ψ, we go ahead and we take the first derivative with respect to X.0344

We take these derivative of that what we got again with respect to X.0349

And then we take the derivative with respect to Y, the original function.0353

We take the second derivative with respect to Y and then we add them together.0356

That is what this means.0360

Let us go ahead and do an example of this partial derivative just to see how it works and then we will move on with the Schrӧdinger equation.0361

Our example is, let F of XY a function of 2 variables B X⁵ Y³.0371

This is a very simple function.0375

We want to find DF DX DF DY as well as D² of DX², D² of DY².0377

And then find D² F DY DX, DX² DF DY.0385

This one is just a derivative of this function with respect to X.0389

This is the derivative with respect to Y holding the X constant.0393

This one is taking the first derivative with respect to X and taking the second derivative0397

again with respect to X and this is the same thing with respect to Y.0402

This one here, it says take the derivative with respect to X and what you get then take the derivative with respect to Y, this time holding X constant.0406

And this is the reverse of that.0414

These are called mixed partial derivatives.0415

Let us go ahead and do this.0418

Let us do DF DX first.0421

We have DF DX, we are going to differentiate this with respect to X.0425

We are going to treat Y as a constant.0433

Y³ will stay as Y³.0435

This becomes 5X⁴ Y³.0437

That is it, you are just holding the other variable constant while you differentiate with respect to one variable.0444

Again, we can only do things one at a time.0449

This is the symbol for it.0451

Let us go ahead and do DF DY.0454

We will go ahead and differentiate with respect to Y except we are going to hold X constant.0456

The X⁵ stays the same.0464

This is going to be 3X⁵ Y².0466

Everything still applies, that you have learned from calculus.0472

The chain rule, everything else is just holding,0475

you just have to be a little extra careful holding something constant and remembering to actually carry it forward.0479

That is it, like anything else in calculus.0484

Just takes a little bit of extra vigilance and making sure because there is going to be a lot of things on the page.0486

Let us go ahead and do the second derivatives.0493

D² F DX², we have taken DF DX.0495

Now we are going to take the derivative of that with respect to X holding Y constant.0499

4 × 5 is 20 so we get 20 X³ Y³.0505

Now we get D² F DY.0512

Now, we are going to take the second derivative of the function with respect to Y so we have a DF DY.0514

We differentiate that so we hold the X constants.0521

2 × 3 =6, DX⁵ stays and the Y drops down to 1.0524

I will go ahead and put the 1 there, that is not a problem.0531

Let us go ahead and do the mix partials.0534

D² F DY DX, the order on the bottom does matter.0538

Here we are saying, we differentiate with respect to X first then differentiate with respect to Y.0543

What this means is do DDY of DF DX that is what this means.0548

This is why you get the D² F DY DX.0557

I’m going to take the derivative with respect to Y of my DF DX.0562

My DF DX is up here and this is 5 X⁴ Y³ but now I’m going to differentiate this function with respect to Y holding the X constant.0566

It is going to be 15 X⁴ Y².0574

Over here, we will go ahead and do D² F.0582

This time we will do DX DY which means we will take the derivative of DF DY.0586

We are going to do why first which we already did, now we are going to take the derivative of that with respect to X.0594

3 × 5 is 15 X⁴ Y².0600

Now notice something, DF DX DF DY is not the same.0608

D² F DX² DX² of DY², it is exhausting saying all this, they are not the same.0613

But when you take the derivative with respect to one variable then take the derivative with respect to the other.0619

You get this and you get this.0626

Notice that they are the same, this is not a coincidence.0629

Mixed partials are going to be the same, provided the function is well behaved.0631

You are always going to be dealing with well behaved functions.0636

You have already run across a function that is not well behaved.0639

In other words, continuous and continuous partial derivatives and things like that.0642

This will always be the case.0645

This is actually a very deep theorem in mathematics.0647

This is not a coincidence.0651

The equality of mixed partials is not a coincidence.0659

For those of you that actually went through the thermodynamics portion of this particular course, I have discussed this in great detail.0671

We actually used the property of the mixed partials being equal to simplify our equations.0677

Equality of mixed partials is not a coincidence.0683

Let us go ahead and get back to our particle in a box.0689

Let me go back to black here.0694

Once again, we have – H ̅²/ 2M × A² ψ DX² + D² ψ DY².0696

I can derive these things over and over again simply to help me become comfortable with the equation.0712

It is equal E × there we go.0720

In operator notation, an Eigen value notation.0724

Eigen function or Eigen value notation looks like this.0735

-H ̅²/ 2M, I'm going to pull out the ψ aside.0739

D² DX² + D²/ DY², the ψ out there = E × ψ.0745

Remember, this is an operator and an operator distributes.0759

It means do this, do this, do this to this function.0762

And I do this to this function, it distributes like a normal algebraic binomial or trinomial or whatever it is.0766

This thing right here is very important.0773

This is called the Laplace operator or just the Laplacian.0777

It is profoundly important in science and mathematics.0811

I promise it will be the last that you see it.0818

It is profoundly important in science.0821

We have a symbol for the Laplacian.0832

Let me go back to black here.0834

The symbol is called the Laplacian.0838

It is going to be this upside down Δ².0853

That is equal to this thing, this D²/ DX² + D²/ DY².0858

Later, we have 3 variables for the particle in a 3 dimensional box, it is going to be D² DZ².0865

The equation becomes –H ̅²/ 2N L² ψ = energy ψ.0873

That is our equation.0892

Once again, let us go ahead and do the boundary conditions.0894

Our boundary conditions are ψ of 0 Y = 0 and ψ of X 0 = 0.0906

Of course, we have X ≥ 0, ≤ A and Y ≥ 0, ≤ B.0918

A and B are the limits here.0926

When we solve this equation, we use the method called the separation of variables.0930

I’m not going to go through it here, you can go ahead and see if you like, either in your book or in one of the appendices that I have here for this course.0938

Separation of variables, this particular technique it requires us to assume that our function that we are looking for,0946

this ψ of XY is actually equal to some of function of X × some function of Y and that is a product.0965

It is not some mixed function.0977

It is actually a separate function of X alone and a separate function of Y alone that are multiplied together.0979

This particular technique requires that this be the case.0985

I want to throw that in there.0988

In other words, what we do is we find F of X and then we find G of Y.0991

And we multiply them together that is it.1007

Then multiply them together to form our ψ, our wave function which is now a function of X and Y.1011

What is interesting is the following.1030

The F of X and G of Y, they turn out to be solutions to the one dimensional case.1032

The 1 dimensional particle in a box along the X axis or the 1 dimensional particle in a box along Y axis.1041

We just happen to put them together to create the solution to the 2 dimensional particle in a box.1046

It is extraordinary that we can do that.1053

And I will talk about that a little bit more towards the end of this lesson.1056

We find F of X, we find G of Y with this technique of separation of variables.1059

We multiply them together to form our ψ of XY.1063

F of X and G of Y turn out to be solutions to the 1 dimensional case.1068

In other words, F of X is, from the last lesson we get some D sub X some constant × the sin of N sub X π A.1091

Remember, these were the solution for the 1 dimensional particle in a box.1105

I'm putting these little subscripts for the N, I can just call it N because there is going to be N for the X and there is going to be N for the Y.1108

We are going to have two quantum numbers.1115

It ends up being, if you remember 2/ A¹/2 × the sin of N sub X π/ A × X.1119

F of Y ends up being some constant B sub Y sin of N sub Y × π/ BY.1131

The constant is the same and the normalization constant, except it is going to be 2/ B.1143

We are doing from 0 to B ^½, the sin of N sub Y π/ B × Y.1149

Personally, I think the hardest part in quantum mechanics is not the quantum mechanics, it is not the concepts.1162

It is not the math, it is writing the math down.1166

There are so many things to write.1170

Many indices, this star, that star, this subscript, that superscript.1172

Keeping it all straight can be really difficult.1177

We have the following.1182

Our ψ which is now we have the subscript N sub X N sub Y is going to equal,1184

You are multiplying F of X and F of Y.1196

It is going to be B sub X B sub Y.1198

Let us here.1202

The sin of N sub X π/ A × X × the sin of N sub Y π/ B × Y.1209

This B sub X and B sub Y, those are right here.1227

That is the B sub X and that is the B sub Y.1231

If you want you just put those in.1233

It is just going to be 2/ A × B.1235

The 2¹/2 and 2 ^½ they multiply it to form 2.1238

The A ^½ and B ^½ ends up being AB ^½.1241

I will write it over here.1248

This BX BY, the coefficient, the normalization constant is just going to be 2/ AB ^½.1251

I just left it in generic form right there.1261

Now the energy, this is a solution to the particle in a box, this whole equation.1264

And notice, it is a function of X and Y.1271

It just happens to be two separate functions that we have multiplied together.1275

The energy N sub X N sub Y = A²/ 2M × N sub X²/ A² + N sub Y²/ B².1280

These are your solutions right here.1301

This is the solution to the Schrӧdinger equivalent for the problem for a particle in a 2 dimensional box.1304

This is the wave equation, this being the normalization constant, and this is the energy for a different values of N.1310

Let me go ahead and actually say N sub X takes on the values 1, 2, 3, and so on.1317

N sub Y takes on the values 1, 2, 3, and so on.1325

This N sub X and N sub Y, the N can vary independently.1331

It is not just 11, 22, 33.1334

It could be 111, 12, 13,14, 7, 8, 14, 16.1337

They vary independently.1341

There are all kinds of different energy levels.1344

In a particle in a box, it actually have.1346

The normalization constant, we found it just by multiplying the normalization constant for F of X and G of Y, that is fine.1355

The normalization constant is found normally the same way as we did before.1367

B sub X B sub Y is found the same way as before.1373

Which was you have to take the integral over the entire region of C* × C.1384

I will go ahead and put DA here and set it equal to 1.1402

Again, that is all we did.1404

In order to satisfy the normalization condition, this is the probability.1407

When add up all the probabilities, the probability that the particles can be somewhere in the box has to equal 1.1413

That is our normalization condition, it always = 1.1420

We use this to solve for the BX BY.1422

This is the equation that we actually have to solve in order to find our normalization constant.1427

In order to normalize the wave function.1433

You just take the function, multiply it by its conjugate, and you integrate.1436

In this case, the conjugate happens to be the function itself because the functions are real functions, they are not complex.1441

This is the process that you always go through.1446

Here it is DA.1450

I wrote DA because we are integrating over an area.1455

We are going to integrate over X, then we are going to integrate over Y.1459

We are integrating over an area.1462

This is actually a double integral.1464

You have to watch what dimension you are working in.1465

Because we are integrating over an area, in other words we got this region here.1471

We integrated in the X direction, and then we integrate in the Y direction, the double integral.1492

Let me go back to black here.1498

This integral is symbolic.1501

This one right here, it is symbolic.1502

It actually means this.1505

In this particular case, a 2 dimensional case ψ* ψ DA= the integral from 0 to A of the integral from 0 to B of,1507

when we take the function and multiply itself,1519

we are going to end up with B sub X² B sub Y² ψ² N sub X π/ A × X and D sin² of N sub Y π/ B × Y DY DX.1522

DY DX is the DA, it is a differential of area element.1545

DX DY when you multiply them together you will get DA, the differential of area element.1554

That is what this means.1560

We have to solve this particular integral and do not worry about it, we will be doing these in the example problem.1561

We will see.1567

You can do it this way to solve the integral or you can just go ahead and take the coefficients for the 1 dimensional case and just multiply them together,1569

which is exactly what we did.1579

We had B sub X = 2/ A ^½ and we have the B sub Y = 2/ B ^½1581

which means that the normalization constant BX BY is such as this × that.1595

It is equal to 2/ AB ^½.1600

That is it, nice and simple.1605

Let us go ahead and say a little bit more here.1611

This ψ N sub X N sub Y which is a function of X and Y is the product of ψ, for the 1 dimensional case sub X and ψ NY.1620

For the 1 dimensional case in the Y variable.1651

Let me write it out.1667

Our wave function for 2 dimensions is the product of the wave functions for 1 dimensional case.1672

The energy for the 2 dimensions is the sum, here we have the product,1679

the energy is the sum of the energies for the individual cases, for the 1 dimensional cases.1684

N sub X N sub Y.1693

This is actually very important result and I will be talking about it again a little bit more formally towards the end.1698

Let us go ahead and move on to the 3 dimensional case.1707

I will go ahead and go back to black here.1712

We have dimension 3.1714

A particle in a 3 dimensional box, now our equation is going to be –H ̅²/ 2 M × the Laplacian operator for 3 variables.1723

DX² D²/ DY² + the second derivative with respect to ψ of our wave function that we are looking for,1746

It is going to equal that × ψ.1756

This time we have X ≥ 0, ≤ A, Y ≥ 0 and ≤ B, Z ≥ 0 and ≤ C.1760

We are constraining at a box which is A × B × C.1769

A long, B wide, C high.1777

However, you want to look at it.1778

The ψ is a function of 3 variables.1781

Ψ is now a function of 3 variables X, Y, and Z.1788

Ψ is ψ of X, ψ of Y, ψ of Z.1800

It is going to be exactly what you think it is.1804

We are just going to end up forming the product of the 3 functions.1807

It looks like this1811

Let us go ahead and draw this one out.1813

The coordinates is in here.1816

This time what we have is the following.1821

This is going to be our X coordinate, and this is going to be our Y coordinate, this is going to be our Z coordinate.1825

This is the standard 2 dimensional representation of a 3 dimensional object.1829

The right hand coordinate system.1835

The X is this way, the Y is this way, and the Z is this way.1837

A is over here and B is over here and let us go ahead and put C like right there.1841

Here is what we have.1852

Here is our box, let me go ahead and do the box in blue.1854

I’m going to go across and I will go down.1879

Our particle in a 3 dimensional box.1887

The particle is somewhere there.1890

That is all we are doing.1892

We have our equation which is this one.1895

We have the constraints on the variables.1898

Our boundary conditions are going to be as follows.1901

Our boundary conditions are ψ of 0 YZ = 0 ψ of X 0 Z = 0 and the ψ of X Y 0 = 0.1908

All these boundary condition say is that once you actually hit the walls of this box, the wave function goes to 0.1927

The wave function is 0 at that point and beyond.1936

That is all we are doing.1941

In Del notation, it is nice to see it.1944

-H ̅²/ 2N Del² ψ = energy × ψ.1952

Again the Del² is just this Laplacian operator, that is all it is.1965

Our ψ again, we do it with the method of separation of variables.1983

Our wave function of XYZ is going to equal the product of F of X × G of Y × H of Z.1987

The F of X = 2/ A ^½ that is just the wave function for the particle in a 1 dimensional box in the X direction.2000

A × the sin of N sub X π/ A × X.2015

If we have G of Y = 2/ B ^½ × the sin of N sub Y × π not A.2023

Be really careful, I always do that.2039

It is like I get stuck on one value A and I write A for everything.2045

This is N sub Y × π/ B.2048

The variable is Y and of course our H of Z is going to = 2/ Z ^½ × the sin of N sub Z × π/ A × Z.2053

Again, N sub X and N sub Y and N sub Z, they all range from 1, 2, 3, and so forth.2069

They are very independently.2084

You have 3 quantum numbers for a particle in a 3 dimensional box.2086

Now, our wave function N sub X N sub Y N sub Z, it ends up very independently,2106

= 8/ ABC ^½.2114

N sub X π/ A × X × sin of N sub Y × π/ B × Y × sin of N sub Z × π/ ψ × Z.2129

This is crazy, these functions get really big and yes we are going to be integrating with them.2148

We are going to be differentiating them.2154

It does tend to get notational intensive and again personally I think it is the most difficult part of quantum mechanics.2156

It is just keeping all of the notations straight.2163

The energy of a particle in a box, first state N sub X N sub Y N sub Z.2168

This is going to equal planks constant²/ 8M.2177

N sub X²/ A² + N sub Y²/ B² + N sub Z²/ C².2184

The energies are just the some of the energies in the individual directions.2203

A particle in a 1 dimensional box in the X, 1 dimensional box in the Z, 1 dimensional box in the Z.2208

Put them together, you form the product of the functions, you get the overall wave equation for a particle in a 3 dimensional box.2213

You add up the energies to get the energy of the particle in a 3 dimensional box.2219

Once again, the N sub I they vary independently.2225

Do not get stuck on the idea that it all has to be 111, 222, 333.2232

It does not.2236

Let us talk about something interesting that happens.2240

Something happens when the sides of the box are equal.2244

A can be anything, B can be anything, C can be anything.2248

When they are equal, here is what happens.2250

The energy N sub X, let us specify what we mean by this sides being equal.2254

We mean when A = B = C, when you have a cubed.2262

Here is what happens to the energy.2270

The energy of N sub X, N sub Y, N sub Z, that is equal to H²/ 8M.2271

Now, it is going to be N sub X²/ A² + N sub Y²/ A²,2279

Because B = A so I can just put in A².2287

They are all the same length, + the N sub Z²/ A².2292

The A² is a constant so you can just pull it out.2299

The energy = H²/ 8 M A² × N sub X² + N sub Y² + N sub Z².2302

This is just a constant.2316

What is important is this thing.2318

The N depending on what N is, the energy of that level is going to be different.2321

If N is 1, if N sub X is 1, N sub Y is 1, N sub Z is 1.2326

You are going to end up with 1 + 1 + 1 it is going to be 3 A²/ 8 MA² that is the energy of the 111 level.2331

Now, for a particle in a 3 dimensional box.2341

The state of the system is expressed by the N sub X N sub Y N sub Z.2345

Let me say that again.2351

For a particle in a 3 dimensional box, the state of the system is expressed by these quantum numbers.2358

It is expressed by N sub X, N sub Y, and N sub Z.2376

Let us go ahead and actually work out some energy for some different states.2382

When we have N sub X = 1, N sub Y = 1, and N sub Z = 1, this is going to be the energy of the 111 state,2388

That is going to equal H² / 8 MA² 1² + 1² + 1².2399

I’m just putting it into the equation.2409

I end up with 3 H²/ 8 MA², that is the energy of the 111 state of a particle in a 3 dimensional box.2412

If I said I have a particle in this 3 dimensional box, the sides are length A.2425

What is the energy of the 111 state?2433

That particular wave function, what is the energy of that particle?2441

There it is, planks constant², multiply by 3, divide by 8, divide by the mass of whatever the particle is,2444

electron, proton, whatever and then divide by the square of the sum of the side length.2450

Let us do some more.2459

This time let us do N sub X = 2 and N sub Y = 1 and N sub Z = 1.2461

This is the energy of the 211 state.2470

That is going to equal A²/ 8 M, A² this time it is going to be 2² + 1² + 1².2473

N sub X, N sub Y, N sub Z.2486

You will end up with 6 H²/ 8 MA².2489

Let us do another level.2501

For N sub X = 1, this time N sub Y = 2, N sub Z = 1.2503

This is not the same state 211, this is 121, this is an entirely different wave function.2511

An entirely different wave, an entirely different set of probabilities, an entirely different of state of the system.2517

We know the energy, the energy of the 121 state = H²/ 8 MA².2524

This time it is going to be 1² + 2² + 1².2533

Again, we end up with 6 H²/ 8 MA².2538

It ends up being the same energy as the 211 state.2543

You can imagine what we are going to do next.2549

This time we are going to do X = 1 and Y = 1 and this time we are going to take the N sub Z = 2.2551

This is going to be the 112 state.2557

Again, a completely different state 112.2562

Sorry to elaborate the points here, I think it is actually good to go through it like this.2570

+1² + 2²,2575

Once again we end up with 6 H²/ 8 MA².2578

For 3 different states, the 211, 121, and 112, the energies are the same.2585

For 3 different states, the energies are the same, this is what we call degeneracy.2599

Let me go ahead and do this in red.2615

We say this particular energy level, this energy level in this particular case, the 6 H², this particular energy level.2622

Let us say the 6 H²/ 8 MA², whatever the energy happens to be.2637

That level is 3 fold degenerate or has degeneracy of order 3.2642

In other words, there are 3 states that have that energy.2651

That should be the case.2657

A particle in one state is given by a particular wave function should have a different energy.2660

All of a sudden, simply by making the side of the box equal, whether it is a perfect square or a perfect cubed,2665

all of a sudden you have these particles that are completely different states, completely different wave functions,2673

if they have the same energy that is what we call degeneracy.2680

We say that this energy level is 3 fold degenerate.2683

That is the level has 3 states that have that particular energy.2693

Let me go ahead and go back to blue here.2726

I will just write in general, degeneracy is the number of different states having the same energy.2729

That is it that all degeneracy is.2747

It is a very important concept.2753

Degeneracy emerges when the system becomes symmetric.2759

Remove the symmetry and you actually end up removing the degeneracy.2762

Once again we have some rectangular box A, B, C are different.2767

You are going to get different values.2772

With them, you actually end up making the size of the box equal, you actually introduced degeneracy into the system.2776

You actually solve by introducing symmetry into the system, the cube is perfectly symmetric.2783

The square is perfectly symmetric.2788

You actually introduced degeneracy.2790

That was actually a fundamental principle of quantum mechanics.2792

It says that degeneracy is up here, they emerge, they show up when you introduce symmetry into the system.2795

If you remove the symmetry, you end up removing the degeneracy.2803

Degeneracy is not necessarily like an open door and a closed door.2807

It is not like yes it symmetrical, not it is not symmetrical.2811

There are actually degrees of symmetry and we are going to see some of that when we actually did example problems.2814

But again, degeneracy emerges as a result of introducing symmetry.2820

Take away the symmetry, you reduce the degeneracy.2826

You take away the degeneracy.2829

This is actually a very profound quantum mechanical principle.2830

Not just quantum mechanical.2833

The idea of symmetry is what permeates nature, permeates all of mathematics and physics.2835

I’m going to go ahead off with what I said earlier, when I said I would discuss more formally the idea of the particle in a 3 dimensional box2842

being a product of the functions of the 1 dimensional box and the energies being the sum.2850

Let me go back to black here.2857

We said that the ψ of N sub X N sub Y N sub Z = ψ of N sub X × ψ of N sub Y × ψ of N sub Z,2860

and that the energy of N sub X N sub Y N sub Z state is just the energy of the N sub X state +2880

the energy of N sub Y state + the energy of the N sub Z state.2888

I think I’m going to leave this for another time, I apologize.2906

I think I can go ahead and close out this lesson right here and I will discuss this formality a little bit later.2910

Perhaps, I will actually do some problems.2915

That should not be a problem at all.2917

Thank you so much for joining us here at www.educator.com.2919

We will see you next time, bye.2921