For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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## Transcription

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### The Particle in a Box Part II

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
- Free Particle in a Box
- Calculating Average Values & Standard Deviations
- Average Value for the Position of a Particle
- Standard Deviations for the Position of a Particle
- Recall: Energy & Momentum are Represented by Operators
- Recall: Schrӧdinger Equation in Operator Form
- Average Value of a Physical Quantity that is Associated with an Operator
- Average Momentum of a Free Particle in a Box
- The Uncertainty Principle

- Intro 0:00
- Free Particle in a Box 0:08
- Free Particle in a 1-dimensional Box
- For a Particle in a Box
- Calculating Average Values & Standard Deviations 5:42
- Average Value for the Position of a Particle
- Standard Deviations for the Position of a Particle
- Recall: Energy & Momentum are Represented by Operators
- Recall: Schrӧdinger Equation in Operator Form
- Average Value of a Physical Quantity that is Associated with an Operator
- Average Momentum of a Free Particle in a Box
- The Uncertainty Principle 24:42
- Finding the Standard Deviation of the Momentum
- Expression for the Uncertainty Principle
- Summary of the Uncertainty Principle

### Physical Chemistry Online Course

### Transcription: 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 and*1079

*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 that*1858

*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 smaller*2305

*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 uncertainty*2423

*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

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