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

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

### 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
- Free Particle in a Box
- 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
- Normalizing the Wave Function

- 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

### Physical Chemistry Online Course

### 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 and*0084

*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 then*0552

*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 much*0596

*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 appendix*1090

*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 it*1173

*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 that*1547

*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 have not found out what B is, you are right.*1641

*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 conjugate*2385

*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

*The distribution becomes broader.*2595

*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 courses*2618

*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 that*2933

*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

*That is what is nice about this.*3223

*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

1 answer

Last reply by: Professor Hovasapian

Wed Nov 25, 2015 12:53 AM

Post by Jupil Youn on November 18, 2015

When 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 answers

Last reply by: Professor Hovasapian

Mon Mar 2, 2015 6:43 PM

Post by Frederic Hunt Hunt on February 7, 2015

Hello, I noticed that you mentioned viewing the appendix for a more in depth explanation. I was wondering where exactly is the appendix?

1 answer

Last reply by: Professor Hovasapian

Wed Nov 19, 2014 5:34 AM

Post by Joseph Szmulewicz on November 16, 2014

Sorry, 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 answers

Post by Joseph Szmulewicz on November 16, 2014

You 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