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 1 answerLast reply by: Professor HovasapianWed Nov 25, 2015 12:53 AMPost by Jupil Youn on November 18, 2015When we are talking about probability density, we should indicate the interval such as dx.  Here is my question:  what is the probability of finding a particle at a given point?  I know dx is zero in this case. But dx is also infinitesimally small, isn't it?  Even if probability density function is continuous at a give interval, why we can not calculate prob at a specific point? 3 answersLast reply by: Professor HovasapianMon Mar 2, 2015 6:43 PMPost by Frederic Hunt Hunt on February 7, 2015Hello, I noticed that you mentioned viewing the appendix for a more in depth explanation. I was wondering where exactly is the appendix? 1 answerLast reply by: Professor HovasapianWed Nov 19, 2014 5:34 AMPost by Joseph Szmulewicz on November 16, 2014Sorry, I went over the part where you set the boundary equal to 0 and determine that A is equal to 0, which eliminates the A part when you set the boundary to a. So, I figured it out myself. thanks 0 answersPost by Joseph Szmulewicz on November 16, 2014You lost me on the part where you discuss the boundary equal to a. Why is the Acospa crossed out immediately, leaving the Bsinpa part only? You don't explain this

### The Particle in a Box Part 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
• Free Particle in a Box 0:28
• Definition of a Free Particle in a Box
• Amplitude of the Matter Wave
• Intensity of the Wave
• Probability Density
• Probability that the Particle is Located Between x & dx
• Probability that the Particle will be Found Between o & a
• Wave Function & the Particle
• Boundary Conditions
• What Happened When There is No Constraint on the Particle
• Diagrams
• More on Probability Density
• The Correspondence Principle 46:45
• The Correspondence Principle
• Normalizing the Wave Function 47:46
• Normalizing the Wave Function
• Normalized Wave Function & Normalization Constant

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

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

Today, we are going to talk about the particle in a box.0004

The particle in a box is a reasonably simple Quantum Mechanical problem.0007

It is our first dealing with a quantum mechanical system.0012

We are going to solve for the wave equation.0016

We are going to investigate the energy levels and things like that.0018

It is very important.0021

What we do here is going to set the pattern for what we continue to do throughout the quantum mechanics.0023

Let us get started.0027

Let us recall the Schrӧdinger equation.0031

I will stick with black that is not a problem.0033

We have –H ̅²/ 2 M D² DX² + this potential energy × the wave function.0038

Actually, I’m not going to put the X of this wave function.0053

I will leave it like that = the energy × the wave function.0056

This is the Schrӧdinger equation, this is the partial differential equation that needs to be,0062

in this case it is an ordinary differential equation because it is just a single variable X.0067

In general, it was a partial differential equations that needs to be solved for ψ.0070

What we are looking for when we solve this equation is this wave function.0075

What is it that wave function represents the particle?0081

Instead of dealing with it as a particle, we are thinking about the particle as a wave and0084

this wave function which is a function of X represents how the particle behaves in any given circumstance.0090

Now the box in this particle in the box, what we are talking about is the following.0099

We are going to be dealing first with the particle in a box of 1 dimension.0105

The box is exactly what you think it is, just think of a box and if I drop a particle in there, it is going to be a 3 dimensional box.0109

Or a 2 dimensional box is just a plane and it could be square, rectangle, whatever.0115

A 1 dimensional box is just an interval.0120

The box here in 1 dimension, we are going to start with a 1 dimensional problem and we will go ahead and extend it to 2 and 3.0123

In 1 dimension, it is just an interval, that is it.0133

It is just an interval on a pure line.0137

Let us say from 0 to A, that is our box.0143

The particle is going to be basically found somewhere in here.0149

It can only be there.0152

It cannot be out here, it cannot be out here that is all these means.0154

We are going to study the free particle constrained to lie between 0 and A.0160

The free particle in a box.0169

The word free particle here, free particle means it experiences no potential energy.0172

It experiences no potential energy.0189

In other words, V in the Schrӧdinger equation is going to be 0, in this particular case.0192

It simplifies our equation a little bit.0197

It experiences no potential energy.0199

Imagine just taking this particle, dropping it on this interval and saying where you are going to be.0204

How fast you are going to be moving.0211

Where is it go, things like that.0213

What it is going to do.0214

It can only do 1 of 2 things.0215

It basically can go this way or it can go this way.0217

The questions that we pose in quantum mechanics are which direction it is moving?0221

How fast is it moving at any given moment, can it tell you where it is?0225

Those are the questions that we want to ask.0229

These are the questions that the wave function is hopefully going to answer for us.0231

Free particle means it experiences no potential energy.0236

Which means that V of X = 0.0239

This equation actually ends up becoming the following.0244

It ends up becoming a - H ̅²/ 2 M.0247

The derivative of this squared, the second derivative of that = E × our wave function ψ.0252

I’m going to go ahead and rearrange this and write it in a way that is more convenient for solving the differential equation.0264

It is a way that you learn when you are taking the differential equation course.0274

And it will make sense in just a minute.0277

I’m going to rearrange this.0279

Basically, we multiply by the 2 M, I divide by that, bring everything over to 1 side, and set everything equal to 0.0280

It can look like this.0287

It is going to be D² ψ DX² + 2 M E/ H ̅² × ψ = 0.0289

And again, this wave function ψ, it is a function of X.0305

It is just a normal function like anything else, sin X, cos X, log of X.0310

That is all it is, that is what we are looking for.0320

In algebra, we have an equation like 2X + 3 = 5.0322

You are solving for X and you are trying to find a number.0326

A differential equation is the same thing except that it kicked up a couple of levels.0328

The variable that you are looking for is not a number, it is an actual function, that is all a differential equation is.0331

It is just a fancy algebraic equation.0337

In fact, there are techniques that actually reduce these straight to algebra.0341

Do not get lost in the fancy mathematics here.0348

It is just we are solving this, it is a little bit more complicated but we are just looking for some variable.0350

Our variable happens to be a function.0356

I’m going to leave off this X just to save some notation.0358

Again, we have to specify X between A and 0.0362

That is our constraints, they can only be between here and here.0367

We are putting the constraints on it.0374

The question is how can we interpret this ψ?0377

How can we interpret that?0382

We said that the wave function represents the amplitude of the matter wave in the previous lesson.0383

In classical mechanics, when we square the amplitude of the wave, we get the intensity of that wave.0415

The square of the amplitude which this is, represents the intensity of the wave.0423

I’m going to do it this way.0452

It is going to be ψ of X the complex conjugate × ψ.0454

We are not just going to do ψ X × ψ of X × ψ of X.0459

Just in case this wave function happens to be a complex function, we need to take the complex conjugate × ψ, in order to get a real quantity.0465

Anytime you have a given number that is complex, if you multiply the two conjugates together A + Bi A – Bi,0475

You are going to end up getting a real quantity.0482

We want a real quantity when we square this which is why this complex conjugate shows up.0484

If ψ happens to actually end up being a real function like cos of X that it is not a problem.0489

The square is just cos² of X because the conjugate of something real is the thing itself.0495

The conjugate of 5 is 5, it is not a problem when we place that conjugate there, simply just in case ψ is complex.0501

The square of the amplitude represents the intensity of the wave.0513

We have this thing.0516

Our problem is how the heck we are going to interpret intensity?0518

What intensity mean when it comes to a particle?0520

How do we interpret intensity?0525

Here is how we do it.0527

Think of intensity as the extent to which a particle is actually present.0530

If a particle has a wave function and the square of that wave function which is the intensity is kind of low,0536

That means that chances are really low that you actually find a particle there.0545

If the intensity is really high, then, in other words if the square of the wave function is high then0552

that means chances are really good that you are going to find a particle there.0560

We are going to interpret intensity as a probability that a particular particle is at a given place.0565

We are going to express it as a probability, lower the intensity, lower the probability.0573

Higher the intensity, higher the probability.0578

We have ψ conjugate × ψ is the probability density.0581

We do not need to concern ourselves too much with the probability density so much0596

because we are going to be integrating this thing and we would be concerned with probabilities.0602

That is what is going to be most important.0606

With the probability density, it is like any other density.0608

If there is a mass density that is a certain mass per volume, g/ ml.0611

The probability density is a certain probability per volume, or per length element.0618

In this particular case, we are sticking to 1 dimension.0626

If I take a little differential element DX like that, the product of the conjugate a ψ × ψ itself.0629

And that giving me the probability density.0642

It gives me the probability per unit of length or unit of area or unit of volume.0646

That is all it is, it is like any other density.0652

Our quantity is this, it is ψ and ψ × DX.0657

This right here, when I take the wave function conjugate × wave function,0666

I will just say the square of the wave function.0671

I will just say it that way.0673

The square of the function × some differential length element, this is the actual probability.0674

This is the probability that the particle is located between a given X and DX.0683

Let me redraw this thing right here.0705

If I have some value of X, if I take some differential length DX,0710

Now this is X + DX.0717

If I take the square of the wave function and multiply it by this length element,0720

I end up actually getting the probability that the particle is located here in that differential element.0724

If I integrate, all of the differential elements from 0 to A, I get the probability that the particle is somewhere between 0 and A.0729

That is the whole idea.0741

What is important is the square × the DX, that is the probability.0743

When I just take the square of the wave function, I get something called the probability density.0746

It is good to know but is not going to get in our way too much.0751

Let us go ahead and do the integration.0756

When I integrate from 0 to A, the square of the wave function.0759

This is basically I'm just adding all the probabilities.0769

The probability of the particles here or here.0775

All the DX is just like normal integration and mathematics.0780

When I add them all up, because probabilities are additive, I get the total probability that a particle is between 0 and A.0784

The probability that the particle will be found between 0 and A.0795

Since, I'm saying that the particle is going to be somewhere between 0 and A,0816

I can say for sure that somewhere between 0 and A, I’m going to find the particle.0821

I may not know where exactly it is, but I know it is going to be between 0 and A.0825

My probability is 100% it is equal to 1.0829

This thing is actually going to end up equaling 1.0832

We will see a little bit more of that in just a little bit.0836

Here, what is important is that the square of the wave function = the intensity.0838

We are going to interpret the intensity as the probability of finding the particle there.0843

The square of the wave function × some differential element, whether it is a length and area, or a volume,0850

it gives me the probability that the particle is located between or in that differential length element or the area element or volume element.0855

When I integrate that over my particular interval, I get the probability that the particle would be found between 0 and A.0864

Whenever I specify A to B, whatever my interval is.0872

This is what is important right here, profoundly important.0874

That is how we are going to interpret this wave function.0880

It is the square of it represents a probability density and the square × the differential element is the actual probability.0885

Let us concern ourselves with the ψ.0897

This wave function represents a particle, we know that.0900

What we have is this thing right here.0912

Between 0 and A, ψ of X represents a particle.0921

We are constraining you to lie between 0 and A so we know for sure that is not going to be to the left of 0, it is not going to be to the right of A.0927

Over here, our wave function is going to equal 0.0934

And over here, our wave function is going to equal 0.0939

Since we are differentiating not once but twice, the Schrӧdinger equation is a second derivative.0946

In order to take the derivative of this function, the function needs to be continuous.0953

Because it is 0 outside of the interval over here, 0 over here, and 0 over here, because the function is continuous,0958

Because we have to differentiate it, that means it has to be 0 actually at 0 and at A.0968

In other words, you are not going to have some wave function.0975

Let us say this is 0 and this is A, you are not going to have some wave function that goes like this.0978

It is all 0 from here, there is a discontinuity here.0983

That is not going to happen.0987

It needs to be 0 here because it is 0 pass those points.0988

Therefore, at 0 and A, the wave function ψ of 0 has to equal 0.0995

The ψ of A has to equal 0.1009

These things are called boundary conditions.1013

This is another set of constraints that we have to place on the particular problem in order for it to actually make sense.1024

The wave function has to be 0 at 0, it is 0 at A.1029

You are specifying what is happening at the boundary.1036

We just want to find out what is going on in between.1038

Our mathematical problem becomes something called a boundary value problem.1042

Here is what we are going to do.1046

We have to solve the equation + 2 ME/ H ̅² × ψ = 0,1048

Where X ≥ 0, ≤ A subject to the boundary conditions ψ of 0 = 0 and ψ of A = 0.1063

This is the mathematical problem that we have to solve.1076

We have to solve this thing subject to these constraints.1079

Particularly this, the boundary conditions.1083

Let us go ahead and do it.1087

As far as the solution of this going through this, you can take a look at the appendix1090

if you actually want to see how one goes through solving this particular differential equation.1094

Here I’m just going to go ahead and present the solutions.1098

And the solutions are really all that we need.1100

Only if you want the extra information, you are welcome to look at it.1102

Here, when we solve this we get the following.1107

We get the general solution is A × cos of PX + B × sin of PX,1111

Where P, I just put a P in here to make it a little easier instead of writing out everything.1126

Where P is actually equal to 2 ME¹/2, make it a radical sign if you want, / H ̅.1132

If you want, you can put this in here and here.1141

I just decided to call it P or you can call it whatever it is that you want.1144

We call the H ̅ is equal to H/ 2 π.1149

This is our general solution.1156

We found that the ψ, we found a way function.1158

Let us subject that wave function to our boundary conditions and see what A and B are going to be.1162

Now the boundary conditions.1169

This is normally how you handle all differential equations.1171

You solve the equation and then you take a look at whatever constraints that you have placed on it1173

to find the values of the individual constants.1178

Instead of the general solution, you try to find a specific solution or a specific set of solutions.1181

Now the boundary conditions.1190

Let us go ahead and deal with the first one.1199

I'm going to go to blue here.1200

Let us go ahead and deal with this one.1203

The ψ of 0 = A × I just put it into the equation, I see what I get,1206

A × cos of P × 0 because we are putting in for X, + B × sin of P × 0 = 0.1216

The sin 0 is 0, the cos of 0 is 1.1230

What we get here is A × 1 and this is 0, A × 1 = 0.1236

Therefore, A = 0.1243

We found what A is, it is equal to 0.1245

We will go ahead and this term just drops out.1248

We will go ahead and deal with the second boundary condition, that the ψ of A = 0.1252

Since we know that we are not dealing with this term anymore, we have B × the sin of P.1260

A, X is we are putting A in and that is also equal to 0.1269

There are two things that can happen, N can be 0 or sin of PA can be 0.1277

B equal to 0 is trivial so it does not give us anything.1281

We do not have to worry about that.1284

However, let us deal with the sin of PA equaling 0.1286

Which means that PA = the inverse sin of 0.1294

And what you end up getting here is PA = 0, you get π, you get 2π, you get 3π, 4π, and so on.1300

I’m going to do PA, and I’m going to write it as Nπ.1311

Where N is going to equal 1, 2, 3 and so on.1319

I’m not going to conclude the 0 because again it does not really give us anything.1324

Here is what we end up getting.1328

PA = Nπ.1330

N= 1, 2, 3, all of these make this boundary condition true.1333

Let us go to the next page here.1353

We said P was equal to 2 ME¹/2/ H ̅.1356

And we just said that PA = Nπ.1373

Let us put this in for P so what we end up getting is 2 ME¹/2/ H ̅.1377

I hope I’m not confusing my H and H ̅, that is a big problem here in Quantum Mechanics.1388

It is equal to = N π.1394

I’m going to rearrange this equation, when you rearrange and solve for E, the energy, you end up with the following.1396

You end up with E for a given N because N is 1234.1403

It is going to equal H ̅² N² I²/ A² 2M.1410

Since, H ̅ = H/ 2 π which implies that H ̅² = H²/ 4 π²,1432

An alternative version involving planks constant directly instead of H ̅ is going to be H² N²/ A² 8 M.1447

And remember, M is the mass of the particle.1463

What you have is, this is the energy of the particle, either you are this one or this one.1465

It is totally up to you.1474

Notice that the energy of the particle is quantize.1476

It has very specific values depending on what N is.1479

When N is 1, it has a certain energy.1483

When N is 2, it has a certain energy.1485

When N is 3, it has a certain energy.1488

N is integral 12345, there is no 1.5, 1.6, 1.7, radical 14, things like that.1491

This is what we mean by quantization.1500

In other words, this particle in the box, the energy of the particle could only have specific values.1503

There is no in between.1511

It cannot take any value that it wants to. This is very different than classical behavior.1513

Classical particle that have any energy value at all depending on what is going on.1518

It does not matter.1523

Here it is very specific and this energy is actually contingent on N, some quantum numbers, some integer.1525

This N here is called a quantum number.1534

This is a constant, that is a constant, A is whatever you happen to chose.1536

You can change it but once you choose it, it is fixed amount.1540

That is the length to M.1544

All of these are constants, this energy is a function of N some number that1547

shows up out of nowhere simply by virtue of the solution to the differential equation.1551

This is what is extraordinary about quantum mechanics.1555

This is what we mean by the quantum.1558

The energy is quantized.1560

N is called a quantum number.1563

Notice, this quantum number shows up naturally by virtue of our solution to the problem.1574

This is going to be a running theme in quantum mechanics.1580

There is going to be several quantum numbers that show up based on how we solve problem.1583

Let us see what we have got.1590

Well, PA = N π.1591

Therefore, P is actually equal to N π/ A.1598

Therefore, we can go ahead and write our ψ sub N of X depending on what N is.1605

It is equal to B × sin N π/ A × X and the energy for a given N = DH² N²/ A² 8M.1614

There we go.1635

You are probably saying to yourself what is B?1638

We will find out what B is in just a moment.1642

But this is the solution to our particular particle in a box.1646

For different numbers 1234567, the wave function is this thing, it represents the particle.1651

Any information that I want about the particle, I'm going to extract from that function.1659

If I want to know the energy for any given state 12345, I just plug it into here and I get the energy of the particle.1664

That is what is happening.1672

What would happen if we did not impose any constraints on the particle?1678

If we just said, here is a free particle, it is not experiencing any potential energy, tell me something about its energy?1684

What is happening?1691

If we did not place any restraints on the particle, in other words if we did not restrict the particle to lie between 0 and A,1694

Instead, the particle can be anywhere.1701

Mathematically it means that the ψ of 0, if we do not strain it to lie between 0 and A, the most boundary conditions that ψ of 0 = 0.1703

The ψ of A is equal 0, they vanish.1726

All you are left with is the solution to the mathematical equation.1728

The problem ends up just being the differential equation without the boundary value problems.1732

I end up without the boundary conditions.1738

Once we solve that, you end up getting the same solution.1741

It is the same differential equation.1745

You end up with something like ψ of X = A × cos of P of X + B × sin of P of X.1747

P is the same thing where P = √2 ME/ H ̅.1760

Now since there are no boundary conditions, I'm not constrained.1771

I do not end up having to find this A and this B.1779

I do not end up having to find this PA = something.1782

It turns out that P can actually be anything.1785

Because P can be anything, because P can be any number at all, before, we had P = N π/ A.1790

P depended on N.1804

There are no boundary conditions for a free particle completely.1806

That is not a particle in a box, it is free to move anywhere it want.1811

It is not dependent on N, P just equals this.1815

When I rearrange this for energy, the energy can actually take on any value at all.1819

And that is what is interesting here.1824

What you end up getting when you rearrange this, you end up with E = P² H ̅²/ 2 M.1825

I hope that the mathematics is properly here.1836

But the idea is that P is no longer contingent on N.1840

It can be anything.1844

Therefore, the energy can be anything.1846

Now the energy of a completely free particle that is not constrained to lie in a particular region, it can take on any energy value at all.1847

In other words, it is not quantized.1857

Quantization, this whole idea of quantum mechanics, the quantum property of a particle,1860

only appears by virtue of the constraints that we place upon our particular system.1866

When we remove those constraints, it allow a particle just do what it wants to do, whenever it wants to do and however it wants to do it.1872

Everything is fine, everything behaves normally.1878

The energy is not quantized at all.1881

It can take on any value just like a normal classical particle.1883

That is pretty extraordinary.1886

Quantization appears only when we begin to place constraints on a given system.1888

What this means, in the case of the particle in a box, the constraints that we put on it, this particle wave,1894

what we are saying is that it has to fit inside the box.1919

It is what quantization means.1930

That means there are only certain waves that will satisfy this fit property that I draw in just a second.1932

If I do not place any constraints then it can be any wave that all.1940

It doe not really matter, that is right there.1942

It can be any wave at all.1944

When I place constraints on it, it can only be specific waves, waves that fit into that box.1946

And because it can be only specific waves, those waves can have only specific energy values.1951

That is what is happening.1956

Quantization is an emergent property.1958

It is something that comes about by placing constraints on the system.1960

Let us fit into a box, inside the box, such that as we said our ψ of 0 = 0 and our ψ of A equal 0.1966

Drawing wise, it means this.1980

If this is 0 and A, that is one possibility for the wave.1983

0 and A, that is another possibility for the wave.1990

Notice, I have to be the wave that has to begin and end there.1996

This is 0 and this is A, begins and ends there.2003

What you are not going to see is something like this.2013

This is 0 and this is A.2019

It is not going to be up here.2021

It is not going to be down here.2023

It is not just some random wave, very specific waves with very specific energies, the constraint.2025

Quantum behavior emerges as a result to the constraints that we place it on the system.2033

The more constraints we place upon the system, the more restricted we are in the particular values that the energy of the particle can be.2039

Let us go ahead and draw this formally here.2051

Let us go ahead and this is going to be N and I'm going to start at 1, 2, 3, and 4.2055

I have this one, this is 0 to A, 0 to A.2068

This is 0 and this is A.2080

This is going to be our ψ sub 1, our ψ sub 2, our ψ sub 3, our ψ sub 4, and so on.2084

I will just do the first four.2092

Over here, I’m going to go ahead and here I’m going to draw the wave function.2094

Here I’m going to draw the probability density, the square of the wave function.2100

It is that.2107

This is going to be ψ sub 1 conjugate, ψ sub 1, ψ sub 2 conjugate, ψ sub 2, ψ sub 3 conjugate, ψ sub 3 and ψ sub 4 conjugate, ψ sub 4.2118

That is our first wave when we have the B sin N π/ X.2139

Our ψ sub 1.2148

Let us try this again.2150

Our ψ sub N of X we said was equal to B sin N π/ A × X.2160

If N = 1, we end up with B sin π/ A × X.2170

We have B sin 2 π/ A × X, it is going to be this one.2174

Here and here, we have that.2193

1 node, 2 node, 3 node, and so on.2203

We have this and this.2208

Begins and ends at A.2219

This is the wave function.2221

Notice, they are just normal standing waves.2223

You have a string that you are holding at one end, a string that you are holding at the other end,2226

You pluck that string, it is going to vibrate in different frequencies.2231

1 frequency, 1 frequency, another frequency and another frequency, they represent the different and the N you have different energy.2237

There is certain energy.2244

That certain energy you get from the equation that we saw.2246

When I square these wave functions, I get the probability density or I can just think of it as the probability at this point.2252

When I square this, this one looks like this.2260

What this means is that the particles more likely to be found here towards the center than it is to be found here.2266

Notice, the probability, the intensity of the wave is lower here.2273

There is less of a chance that I’m going to find a particle here or here.2276

When a particular particle was in this state, it is in the 1 state, the probability I’m going to find it here.2281

This one ends up being, when I square it, it is going to look like this.2286

The probability of finding the particle at the center is 0.2297

More than likely that I’m going to find the particle here or here.2299

That is what is going on here.2303

And so on, and so forth.2313

These high points are the greatest probability of finding the particles.2322

If there is our particle happens to be in the state 3, the probability of finding the particle here or here or here.2329

There is very little probability that I will find it here or here, and so forth.2338

Notice, here mostly it is concentrated in the center.2342

Here it is off to the sides.2345

Here it is a little bit more distributed.2346

Here it is a little bit more distributed evenly.2348

As N gets bigger and bigger, the distribution of the particle actually becomes a little bit more uniform.2351

That is what is happening here.2359

This is just a pictorial representation of the wave function and the probability density.2361

In other words, where you are going to find the particle.2367

The high points have the highest probability of finding the particle.2370

As you get lower and lower, there is a lower probability of finding the particle there.2373

That is all that is going on here.2380

Since this thing is actually a real function, in this particular case, we do not need the conjugate2385

but we will go ahead and use it because that is the symbolism.2398

This × that = ψ is this.2400

Ψ conjugate is also that because this is a real function, it is not a complex function.2409

The product actually = B² sin² N π/ A × X.2416

That is it, we are just multiplying this by itself because it is a real quantity so the conjugate is just ψ.2423

Since, ψ of X is real, my conjugate happens to equal ψ.2435

When I do that, I get that.2445

Let us say a little bit more here.2452

The probability density, the second graph.2455

The probability density for N = 1, it shows that a particle is most likely to be found near the center.2465

It is most likely to be found near the center of the interval.2484

For N = 2 it is more likely to be found near A/ 4 or 3A/ 4.2494

Those are high points.2522

As N increases, the particles are more likely to be found more uniformly distributed across the interval.2525

In other words, here is what happens.2563

As N gets bigger and bigger, let us say for something like N = 30.2566

For something like N= 30, you do have something looks like this.2575

We have this and this is 0, this is A, you are going to have something like.2584

In other words, now you are very likely to find it almost anywhere where there is a high point.2597

Now you are not restricted.2603

As N increases, that particle is more likely to be found more uniformly distributed across the interval.2606

This is how a normal classical particle would behave.2616

In other words, when we treat a particle the way we have treated it in the physics courses2618

that you have taken as a particle, another wave, the particle can be anywhere.2622

It has no preference.2627

It can be absolutely anywhere.2631

When we treat it like a wave, now it has a preference.2633

It has a greater probability of being here or here or here.2637

As N increases, these high points tend to also increase.2642

Now, you end up finding it more and more places.2647

As N increases, it starts to behave like a classical particle.2651

That is what is going on.2658

As N increases, the probability density or the probability becomes more uniform which is how a classical particle behaves.2661

In other words, it shows no preference for where it is in the interval.2704

It is just as likely to be here, as it is here.2717

When we are looking at the N = 1, it is something that looks like this.2732

Not like that, it is a little bit worse.2741

We have something that looks like that.2746

Here the particle shows a preference.2750

The probability density, the square of the wave function when we graph it,2753

it shows that the particles more likely to be here somewhere in this area, that is not so likely to be here.2756

That is really interesting.2763

Under quantum behavior, it actually has a preference for where it is going to be.2766

As N increases, you get more and more.2772

It can be here, here, or here.2775

Now, it spends more time everywhere, that is the idea.2777

It is spending more time everywhere.2782

As N increases, the distribution becomes more uniform.2784

Here, this is not a uniform distribution.2788

Basically, it is going to be like somewhere in this region.2790

It is going to avoid this area and this area.2793

As N increases, it starts to show the behavior of a classical particle.2796

This illustrates something called the correspondence principle.2802

We will see this again, this correspondence principle.2806

Quantum mechanical results, they approach classical mechanics as quantum numbers get bigger.2818

As the quantum numbers get bigger, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40.2852

The particle starts to behave more like a classical particle.2861

Let us talk about something very important called normalizing the wave function.2864

Let me go back to black for this one and now we are going to go ahead and figure out what that B is.2873

Normalizing the wave function.2881

We said that our ψ sub N of X = B × sin N π/ A × X.2893

This is our wave function for a particle in a box.2902

We are going to take the ψ conjugate × ψ × DX = B² sin² N π AX × DX.2908

This is the probability that the particle will be found in the interval DX, wherever happen to take that DX.2924

We are restricting the particle to lie between 0 and A, the probability that2933

a particle is going to be found somewhere between 0 and A is 100% it is equal to 1.2943

Therefore, when I integrate this probability density over the entire interval from 0 all the way to A, I get the total probability.2949

The probability that I'm going to find the particle somewhere between 0 and A.2971

Again, the square of the wave function × DX is the probability that I'm going to find the particle in that little interval,2976

that little differential interval DX.2983

When I integrate over the entire interval from 0 to A, I get the probability that the particles can be found between 0 and A.2986

I know I’m going to find it somewhere.2993

Therefore, this is equal to 1.2996

Let us go ahead and solve this.3002

I’m going to pull the B² out.3004

When I go ahead and I solve this integral, you can do it either with mathematical software that you have,3019

your mathematical or maple, or mathcad, whatever is that you happen to be using.3024

You can solve this by looking in a table of integrals online.3029

You have table of integrals in the back of your first year calculus text.3033

You have table of integrals in the CRZ handbook.3037

I'm not going to bother with the actual integration all that much.3039

I’m going to be concerned with setting up the integral.3043

Mostly, I will just use software or tables to do the integrals.3046

I'm not going to go through the process.3050

When we solve this integral right here, we actually end up getting this.3052

It is going to be B² × A/ 2 = 1.3065

Therefore, we end up with.3079

I will stay on the same page.3081

I get B² = 2/ A.3083

Therefore, B = 2/ A ^½ or √2/ A.3090

We went ahead and we found B by using this property of probabilities that we know.3100

Now, we can write the individual wave function ψ sub N, they are equal to 2/ A¹/2 × sin of N π/ A × X.3107

X ≤ A, ≥ 0.3127

Energy sub N = A² N² / A² 8M.3130

This is the final solution to our particle in a 1 dimensional box.3141

A wave function that satisfies the integral when I have a particular wave function,3147

If I multiply it by a complex conjugate and I integrated it over the particular interval,3155

If I end up getting 1, that way function is said to be normalized.3173

A wave function that satisfies this relation is said to be normalized.3179

What we did was use this normalization condition to actually find B which was what we call this B²/ A ^½.3189

In this particular case, it is called a normalization constant.3199

If a function is the solution to a differential equation, any constant × that function is also a solution to the differential equations,3201

Because the Hamiltonian operator is linear, we are dealing with linear operators.3209

If F of X is a solution to a function then K × F of X.3215

Any constant × F of X is also a solution to that differential equation.3220

We can always adjust the constant which is what we did to make this happen.3225

If I take this function and if I multiply it by its conjugate which is just multiplying it by itself,3232

And if I integrate it from 0 to A, I’m going to get 1.3239

I have normalized this wave function.3243

We used this condition to find B which is called the normalization constant.3249

This is very important, normalization constant and a normalized wave function.3280

I will close it out with the following.3292

Because this DX is the probability of finding the particle between X and DX,3294

the integral from any X sub 1 to any X sub 2 within a particular interval of ψ conjugate × ψ × DX,3322

It gives the probability of finding the particle within that interval.3334

Again, this thing is the probability of finding it within that particular DX.3339

You can integrate from anywhere.3344

You do not have to integrate from 0 to A, that gives you the probability of finding it over the whole interval.3346

You can take a piece of the interval, you just change your upper and lower limits of integration.3351

This gives the probability of finding the particle between X1 and X2.3355

Thank you so much for joining us here at www.educator.com.3379

We will see you next time, bye.3381