Physical Chemistry The Particle in a Box Part II

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

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

Let us jump right on in.0008

We have from the previous lesson, for a free particle in a one dimensional box, we have the following.0010

I’m going to stop using the letter ψ because I personally do not like the letters.0017

I’m going to use a different letter for the wave function.0024

I’m going to use W.0026

For a free particle and a one dimensional box, the particles just moving back and forth.0034

It can only move one dimension between 0 and A.0053

We have the wave function W sub N of X = 2/ A¹/2 × the sin of N π / A × X.0059

The energy for that particular state of N was equal to H² N²/ A² × 8M.0074

Let me make M a little bit clear here.0086

We are going to say X is between 0 and A, and N of course takes on integer values 123 and so on.0094

This was the solution to our particle in a box problem.0105

Recall from the lesson on the mathematical interlude on probability and statistics, we have the following.0108

We had the average value of X was equal to the integral of X × the probability of X DX.0117

We have the average value of X² that was something that we call the second moment.0130

That was going to be integral of X² × the probability of X DX.0135

And we have something called the variance.0143

The variance which we symbolized as that, that was equal to the integral of the real value of X - the average value of X² × the probability of X DX.0145

And again, just treat these mathematically.0166

A lot of what we are doing in quantum mechanics.0168

If we do not entirely understand what is happening, it is okay.0170

Just treat it mathematically, become accustomed to the mechanics and eventually the understanding will emerge as you do more and more problems.0173

As you and your teacher and your friends and colleagues discuss things more.0181

Just deal mathematically if all of this does not entirely make sense.0187

That was also equal to the averages value of the second moment which is X² – the square of the average value.0191

There are 2 ways, you can find this and find that, and define the variance that way.0199

Or you can just go ahead and integrate this function.0204

In all these cases, this piece of X , piece of X ,piece of X, the PX DX including the DX that is the probability.0209

I will write where P of X DX is the probability of a given state.0224

For a particle in a box, our probability, our PX DX = W* × W × DX.0239

W* W DX = this × itself DX.0262

I will go ahead and just write it, that is not a problem.0275

2/ A¹/2 × sin N π AX × 2/ A¹/2 sin of N π / AX DX.0281

This thing is our probability.0303

Again, 0 less than or equal to X, less than or equal to A and it is equal to 0 otherwise outside of the interval.0308

Let us go ahead and write that here just in case we forgot.0317

The left of 0 to the right of A, the probability is 0.0320

In other words, the particle will never be found there.0323

That is all that means.0326

We are going to use this probability to put it into these equations to calculate some average values and some standard deviations and variances.0328

We use this W* W DX to calculate some average values and standard deviations.0346

And if you remember, the standard deviation is the square root of the variance.0373

The variance is the σ² X.0385

Let us go ahead and calculate the average value of X.0393

The average value of X, in other words the average value of the position of the particle.0397

That is what X represents.0401

X is just where the particle is.0403

I take a measurement, I take another measurement, each measurement that I take, the particles are going to be somewhere.0409

It is just going to be somewhere.0415

Sometimes it is here, sometimes it is there.0417

We take 10 measurements, 50 measurements, 100, 10,000, 100,000, a 1,000,000 measurements.0421

Once I have those million measurements, I want to find the average value.0426

It is going to equal this.0432

It is going to equal by definition the integral of X PX DX, that is the definition of the average value.0434

In this particular case, it is going to be the integral from 0 to A of X W* W DX.0442

X × the probability, X × in this particular case the probability of a particle in a box is this.0452

And we integrate it from 0 to A.0458

What we end up having is the integral from 0 to A of 2/ A.0461

I'm sorry X × 2/ A sin² N π/ A X DX.0469

If you look at this particular integral, let me pull the constant out.0483

2/ A, 0 to A of X sin² N π A/ X DX.0491

If you look up this integral in a table, you will find it.0501

Or have your software do it, you are going to end up with the following.0506

It turns out that the average value of X is going to equal 2/ A × when you do this integral here, you are going to end up with A²/ 4 A/ 2.0511

It turns out that the average value of the position of the particle is A/ 2.0524

What is that mean?0530

Our interval is 0 to A.0531

If I take a million measurements on average, sometimes it is going to be here, sometimes here.0534

If I average it out over many, it is going to be A or 2.0543

In other words, on average I'm going to find it right in the middle.0547

That is all what average is.0551

It is a mean value.0554

On average, I'm going to find the particle right there, that is all this is saying.0557

This make sense, I mean the particles can sometimes be here, sometimes here,0563

Over a bunch of measurements, it is going to average out to right down in the middle.0572

If you flip a coin and get heads, if you flip a coin you get tails.0577

If you keep flipping 100, 200, 300, or 1000 flips, you are going to end up getting just as many heads as you get tails on average.0579

Let us go ahead and find the average value of X² which is something called the second moment.0591

The average value of X² that is equal to the integral of X² × PX DX, that is the definition.0597

It is going to equal the integral from 0 to A of X² × the wave function × itself DX, that is going to equal,0608

I’m going to pull the constant out, the integral from 0 to A of X² sin² N π/ A DX.0620

When you solve this integral, you are going to end up with A²/ 3 - A²/ 2 N² π².0631

The variance is equal to the average value of X² - the average value of X².0654

When I do that, I get A²/ 3 - A²/ 2 N² π² - A/ 2²,0667

Because the average value of X was A/ 2.0681

It was A/ 2².0684

That is fine, I will just do it.0693

It is going to be A²/3 – A²/ 2 N² π² – A²/ 4.0697

I find myself a common denominator with the 4 and 3.0714

Let us go ahead and do this A²/ 12.0718

4A² – 3A²/ 12.0722

A²/ 12 – A²/ 2 N² π².0724

This is our σ² X.0733

When I take the square root of that, I get the actual standard deviation S sub X.0736

I get σ sub X is going to equal A²/ 12 - A²/ 2 N² π² all raise to the ½.0744

I found the average value of X.0762

I found the average value of X².0766

And I use these two to find this one.0768

That is that right there.0772

What happens when I want to calculate the average value of the energy or the average value of the momentum?0780

Now, we want to calculate the average value of the energy or the average value of the momentum.0789

I recall that things like energy and momentum they are represented by operators.0801

In this particular case, they are represented by differential operators so it creates a little bit of a problem how do we actually do that?0807

Recall that energy and momentum are represented in quantum mechanics by differential operators.0814

Remember, the energy operator which was the Hamiltonian operator which was – H ̅²/ 2 MD² DX² + VX that was the operator.0849

Of course, we had the momentum operator which was -I H ̅ DDX.0868

Now the question is how do I find the average value of the momentum?0879

When the momentum is represented by this differential operator, on what function do I actually operate?0886

We have a wave function, that is not a problem.0893

We have our wave function, the 2/ A ⁺square root.0895

√2/ A × the sin N π/ AX.0900

We have the wave function and we know that if we want to extract some information,0905

like something about momentum, we operate on that function.0908

We want to find the average value so are we operating on the complex conjugate W?0913

Are we are going to operate on W*?0919

Are we going to operate on W? Are we going to operate on W* × W?0923

What is it that we do?0928

The question is, on which function does the operator operate?0930

Is it the conjugate?0946

Is it the function itself?0948

Is it the probability density? Is it the square of the wave function?0952

Which one is it?0957

Let us go ahead and see if we can find out.0959

Let us go ahead and recall how our operator version, our Eigen value problem.0960

I have got the Hamiltonian operator operating on the wave function WN.0974

It is going to equal the energy, the wave function.0980

This was our Eigen value problem.0984

This was the Schrӧdinger equation expressed.0986

This is the Schrӧdinger equation in operator form or Eigen function, Eigen value form.0988

WN is the Eigen function, E sub N is the Eigen value.1014

Here is what I’m going to do.1019

I’m going to fiddle around with this a little bit.1020

I’m going to multiply on the left by the conjugate of the wave function and I’m going to integrate.1023

I'm going to get the following.1028

I'm going to get the integral of W* HW = the integral of W* E sub NW.1030

I can pull the Z sub N, it is just a number.1051

It is a scalar so I can pull it out.1053

That is equal to E sub N × the integral of W* W.1056

The wave function is normalized so the integral of the square of the wave function, this is just going to end up being 1.1062

We end up with that.1076

I found the energy simply by multiplying on the left by operating on the function and1079

then multiplying on left by the complex conjugate, and then integrating over the particular interval.1087

That ends up giving me my energy.1093

This is extraordinary.1095

Find the average value of a physical quantity like energy or like momentum that is associated with a quantum mechanical operator...1099

Whatever, we have to find the physical quantity that is associated with an operator.1163

If I have the momentum operator and if I want to find the average momentum, I take the wave function,1167

I operate on the wave function, I multiply it on the left by the conjugate of the wave function, and then I integrate over the entire interval.1175

That gives me the average value.1183

This is the definition of finding the average value of a physical quantity that is associated with an operator.1186

Here, L is the operator and L is the average value of that operator.1194

Is the average value, I should say of the quantity for the particle in the state described by W sub N.1215

Let us go ahead and calculate the average momentum of the particle in a box.1243

We have the average momentum of the particle in a box.1251

That is equal to the integral of W conjugate × the momentum operator W DX.1254

That is going to equal the integral of 2/ A ^½.1265

You literally just put everything in.1274

It looks really complicated, but it is not.1276

Sin of N π/ A × X.1281

This momentum operator you have – I H ̅² DDX.1287

We are going to operate on 2/ A ^½ sin N π/ A × X.1295

And you are going to integrate all of that from 0 to A.1309

Here, let us pull some things out.1321

I pull this out, I pull this out, I can pull this out, and when I differentiate this,1327

I’m not going to go through all the steps, here is what I end up with.1333

-I H ̅ that takes care of that.1338

-I H this is not squared, this is a momentum operator.1347

-IH and then this and this, gives me 2/ A.1350

All I’m left with is, take the derivative of this function, that is what we are doing.1361

You are going to apply this operator to this function and then multiply it by that.1366

When I take the derivative of sin of N π A/ N π/ A of X, I end up with N π/ A × cos of N π A/ X.1371

That constant also comes out N π/ A × integral from 0 to A of the sin of N π/ AX × the cos of N π/ A × X × DX.1386

This integral = 0.1410

Therefore, my average momentum is equal to 0.1415

Again, this makes sense and here is why.1419

If I have 0 to A, this is an average value.1424

This is that if I take 10,000 measurements, there going to be times that the particle is moving in this direction.1427

There are going to be times that the particle is moving in that direction.1432

This direction, that direction, that direction.1435

When I average it out, these directions are going to cancel out.1442

The average momentum of the particle is going to end up being 0, that is what this means.1447

The average value that you get from taking thousands and thousands of measurements.1453

Not even thousands, maybe just hundreds of measurements, maybe 50.1458

On average, this is what is going to happen and it makes sense physically.1462

In a way of looking at is your equally likely to find a particle moving to the left as it is moving to the right.1471

On average, it is not moving at all.1477

Let us go ahead and talk about something called the uncertainty principle.1484

This is very important.1487

It is fine, I will stick with blue.1491

The uncertainty principle or the Heisenberg uncertainty principle.1497

We have calculated the average value of X and we also found the σ of X.1509

We found the standard deviation and we also found the average momentum.1522

Let us go ahead and let us find.1529

We found the average value of X and the standard deviation of X.1538

We found the average momentum, the average value of P.1541

Now, let us find the standard deviation of the momentum.1543

Let us find σ sub P.1546

OK so we know that σ² of P that is going to equal the average value of the P² - the average value for P quantity².1550

I need to find this value now and subtract in order to find this, and take the square root of it, in order to get that right there.1564

The first thing we need to do is find the second moment of the momentum.1572

The average value of P² that is going to equal, what you got is the integral of the conjugate, the operator² that.1577

Remember, an operator² is the same as just doing the operator and doing it again.1595

That just means do it twice, that is all the squared means, do it twice.1602

What we are going to have is the following.1610

This is equal to.1614

Let me write the whole thing.1616

0 to A, 2/ A × sin N π/ A × X × -I H ̅ DDX, that is one operation.1619

We have – I H ̅ DDX is the second operation and we are operating on that function which is sin of N π/ A × X.1636

We are integrating from 0 to A.1653

This is what we are integrating.1657

This means take the derivative of this and then take the derivative of it again.1659

That is a constant, that is a constant, I forgot the 2A over here.1668

It is the hardest part of quantum mechanics, just keeping all of that straight.1693

It is not that it is conceptually difficult.1697

Let us go back to red.1700

This is a constant, that is a constant.1702

When you pull all of that out, you end up with the following.1707

You will end up with - H ̅² × 2/ A the integral from 0 to A.1712

I’m just going to go ahead.1724

That is fine, I will write it all out.1728

Sin of N π/ A × X.1730

Now we have D² DX² of the sin N π/ A × X DX.1734

What you will end up with is, we have 2 H ̅² N² π²/ A × A² × the integral from 0 to A of sin² N π/ A X DX.1746

The derivative of sin is cos, the derivative of cos is negative sin.1772

The negative and negative cancel to give me a positive.1778

The derivative of sin of this thing, that constant comes out once, the constant comes out twice.1780

You are going to get N² π²/ A².1786

Here is the N² π²/ A².1788

We have H ̅ 2/ A, H ̅ 2/ A.1791

That is all where this comes from.1794

Now when I do this, I’m going to get the following.1796

I'm going to get 2 H ̅² N² π²/ A × A² × A / 2.1801

In other words, when I solve this integral I'm going to end up with A/ 2.1815

To again, just use the table, use mathematical software whatever it is that you need to do in order to integrate it.1822

A cancels with A, 2 cancels with 2, and what I am left with is the average value of the momentum² is going to equal H ̅² N² π²/ A².1828

There we have that.1853

We know that the average value of the momentum = 0 and we know that1858

the average value of the second moment of the momentum = H ̅² N² π²/ A².1864

Now, the variance = this² - that².1875

This is just 0 so this goes to 0.1885

I'm left with P = H ̅² N² π²/ A²,1887

Which implies that the σ of the P = H ̅ N π/ A.1900

There we have it.1908

We have σ P, we have the σ X, let us see what we can do.1912

Let us go back to blue here.1920

Both σ P standard deviation and σ P² variance, are measures of the extent of deviation from the mean value.1926

That is what they represent.1963

You have a certain set of data.1965

That certain set of data has an average value.1966

The standard deviation is a numerical measure of the extent to which all of the data as a whole deviate from that mean value.1968

The standard deviation of the mean value.1981

Therefore, we can interpret σ P as a measure of the uncertainty involved in the measurement.1985

You have a set of data, that set of data has an average value.2032

If I take any particular measurement that I have made and if I subtract from it the average value, if I take the absolute value,2039

Basically, it is the difference between any one measurement and the mean value of all the measurements,2049

there is going to be some sort of a gap there.2055

That gap is what the variance is.2057

It is a numerical measure of the actual deviation from the mean value.2060

For example, if I had a bunch of values and a mean value happens to be 5 and if I take some random data point 5.3.2068

That 5.3 and the 5, there is a difference of 0.3.2076

There is some sort of an error if you will, in that measurement.2081

The standard deviation, it is a measure of the extent to which any given measurement actually deviates from the average value.2086

We are going to interpret it as the uncertainty in any given measurement.2094

That is what we are going to do.2099

Let us go ahead and take the variance of our position σ sub X.2102

We said that was equal to A²/ 12 - A²/ 2 N² π².2121

I’m going to write this in a way that makes it a little bit more convenient.2132

I’m going to write this is A²/ 4 π² N² × N² π²/ 3 – 2.2134

I also have the variance of the momentum which is H ̅² N² π²/ A².2147

Notice, as far as the variance with a measure of the standard deviation,2157

measure of the uncertainty and as far as the position is concerned, everything else here is a constant.2164

It is a function of A but it is A in the numerator.2173

For the uncertainty and the momentum, A is in the denominator.2178

This is important.2184

Watch what happens here.2186

Let me go ahead and write it actually on the next page again.2189

Σ² of X = A²/ 4 π² N² × N² π²/ 3 – 2.2195

And then I have over here, I have the variance of the momentum which is equal to H ̅² N² π²/ A².2208

Here is A is in the numerator, here A is in the denominator.2219

Here is what happens.2223

As A increases, the σ sub X also increases.2232

Σ sub X², I just took the square root of this, it also increases.2240

The σ sub P, as A increases the σ sub P decreases.2248

As A decreases, the uncertainty in the position decreases but the uncertainty on the momentum increases.2258

Basically, if I have some interval from 0 to A, if I now make a bigger.2269

In other words, if I give more room for the particle to be my uncertainty in where the particle is, goes up.2274

Now, it is very delocalize.2284

It could be anywhere from 0 to A.2287

It is a huge area but mathematically, as A gets bigger, the uncertainty and the momentum drops.2289

Now, I can be very certain about what the momentum is.2295

If I make A smaller, I’m actually localizing the particle.2298

I'm saying the particle is there.2302

If I’m making A smaller and smaller, my uncertainty in where the particle actually is become smaller2305

but the problem is as A gets smaller, this whole quantity gets bigger.2312

The momentum of the particle now I can say anything about the momentum.2318

This is the relationship and it is based on the mathematics like that.2322

Let us go ahead and take the product of the two.2328

Σ X σ sub P = this is going to be A/ 2 π N × N² π²/ 3 – 2 ^½ × H ̅ N π/ A.2331

When I multiply the two, the A cancels with the A.2360

N cancels the N, the π cancels the π.2368

And I'm left with the following.2374

The σ X σ P = H ̅/ 2 × N² π²/ 3 - 2 all to the ½.2376

This right here is greater than H ̅/ 2.2395

The reason is because this term right here, because N² π²/ 3 -2¹/2 is always greater than 1.2402

Because it is greater than 1, this is always to be going to be greater than this.2416

We have it.2420

Any uncertainty in the measurement of the position, multiplied by the uncertainty2423

in the measurement of the momentum is always going to be greater than H ̅/ 2.2428

In other words, if I become more certain of the position, I become less certain of the momentum.2433

As I become less certain of the position, I become more certain of the momentum.2445

Maximizing and minimizing the relationship between them is this.2451

This is an expression of the uncertainty principle.2455

When it comes to position an momentum, I can only maximize.2457

If I maximize one, I minimize the other.2463

If I minimize one, I maximize the other.2465

There is a point, I have to come to some sort of compromise.2467

I have to decide what is important to me.2470

Do I want to know more about the position?2472

Do I want to know more about the momentum?2473

Or do I want to know a little bit about both?2475

This expresses the relationship between the uncertainties in these measurements.2477

Again, let me write final page here.2485

As I increase the space over which the particle can move, the uncertainty in where the particle is rises.2490

But the uncertainty of the particle’s momentum drops, vice versa.2546

As I decrease the space over which particle can roam.2569

In other words, as I can find the particle more and more, as I can find the particle to a smaller region,2573

I have a better idea of where the particle is.2599

In other words, my uncertainty of my particle’s position drops but I have a better idea of where the particle is.2611

I have a worse idea of the particle’s momentum.2634

The uncertainty in the position and the uncertainty of the momentum are inversely related.2651

Once again, the uncertainty in the particle’s position × uncertainty in the particle’s momentum is going to be greater than H/2.2656

This is one of the statements of the uncertainty principle.2666

In the last couple of lessons we have been just been going over material and presenting theory, we have not done any problems.2673

I want you to know that the problem sets are going to be in several lessons to come.2681

I’m going to be doing them all at once.2687

The nature of the material was such that with quantum mechanics, it is true that you can present a little bit of the topic and do a problem.2689

I think it is better to just go ahead and present a certain amount of theory and then go back and then do a whole bunch of problems,2697

Because I’m given a chance to actually review the material as we are doing the problems.2707

If you are wondering where the problems are, do not worry we are going to be doing it and absolute ton of them and a variety of them.2712

Do not worry about that.2718

Thank you so much for joining us here at www.educator.com.2721

We will see you next time, bye.2722