WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to talk about the particle in a box.
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The particle in a box is a reasonably simple Quantum Mechanical problem.
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It is our first dealing with a quantum mechanical system.
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We are going to solve for the wave equation.
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We are going to investigate the energy levels and things like that.
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It is very important.
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What we do here is going to set the pattern for what we continue to do throughout the quantum mechanics.
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Let us get started.
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Let us recall the Schrӧdinger equation.
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I will stick with black that is not a problem.
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We have –H ̅²/ 2 M D² DX² + this potential energy × the wave function.
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Actually, I’m not going to put the X of this wave function.
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I will leave it like that = the energy × the wave function.
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This is the Schrӧdinger equation, this is the partial differential equation that needs to be,
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in this case it is an ordinary differential equation because it is just a single variable X.
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In general, it was a partial differential equations that needs to be solved for ψ.
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What we are looking for when we solve this equation is this wave function.
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What is it that wave function represents the particle?
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Instead of dealing with it as a particle, we are thinking about the particle as a wave and
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this wave function which is a function of X represents how the particle behaves in any given circumstance.
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Now the box in this particle in the box, what we are talking about is the following.
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We are going to be dealing first with the particle in a box of 1 dimension.
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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.
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Or a 2 dimensional box is just a plane and it could be square, rectangle, whatever.
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A 1 dimensional box is just an interval.
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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.
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In 1 dimension, it is just an interval, that is it.
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It is just an interval on a pure line.
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Let us say from 0 to A, that is our box.
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The particle is going to be basically found somewhere in here.
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It can only be there.
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It cannot be out here, it cannot be out here that is all these means.
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We are going to study the free particle constrained to lie between 0 and A.
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The free particle in a box.
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The word free particle here, free particle means it experiences no potential energy.
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It experiences no potential energy.
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In other words, V in the Schrӧdinger equation is going to be 0, in this particular case.
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It simplifies our equation a little bit.
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It experiences no potential energy.
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Imagine just taking this particle, dropping it on this interval and saying where you are going to be.
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How fast you are going to be moving.
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Where is it go, things like that.
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What it is going to do.
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It can only do 1 of 2 things.
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It basically can go this way or it can go this way.
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The questions that we pose in quantum mechanics are which direction it is moving?
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How fast is it moving at any given moment, can it tell you where it is?
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Those are the questions that we want to ask.
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These are the questions that the wave function is hopefully going to answer for us.
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Free particle means it experiences no potential energy.
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Which means that V of X = 0.
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This equation actually ends up becoming the following.
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It ends up becoming a - H ̅²/ 2 M.
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The derivative of this squared, the second derivative of that = E × our wave function ψ.
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I’m going to go ahead and rearrange this and write it in a way that is more convenient for solving the differential equation.
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It is a way that you learn when you are taking the differential equation course.
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And it will make sense in just a minute.
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I’m going to rearrange this.
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Basically, we multiply by the 2 M, I divide by that, bring everything over to 1 side, and set everything equal to 0.
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It can look like this.
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It is going to be D² ψ DX² + 2 M E/ H ̅² × ψ = 0.
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And again, this wave function ψ, it is a function of X.
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It is just a normal function like anything else, sin X, cos X, log of X.
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That is all it is, that is what we are looking for.
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In algebra, we have an equation like 2X + 3 = 5.
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You are solving for X and you are trying to find a number.
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A differential equation is the same thing except that it kicked up a couple of levels.
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The variable that you are looking for is not a number, it is an actual function, that is all a differential equation is.
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It is just a fancy algebraic equation.
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In fact, there are techniques that actually reduce these straight to algebra.
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Do not get lost in the fancy mathematics here.
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It is just we are solving this, it is a little bit more complicated but we are just looking for some variable.
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Our variable happens to be a function.
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I’m going to leave off this X just to save some notation.
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Again, we have to specify X between A and 0.
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That is our constraints, they can only be between here and here.
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We are putting the constraints on it.
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The question is how can we interpret this ψ?
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How can we interpret that?
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We said that the wave function represents the amplitude of the matter wave in the previous lesson.
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In classical mechanics, when we square the amplitude of the wave, we get the intensity of that wave.
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The square of the amplitude which this is, represents the intensity of the wave.
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I’m going to do it this way.
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It is going to be ψ of X the complex conjugate × ψ.
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We are not just going to do ψ X × ψ of X × ψ of X.
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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.
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Anytime you have a given number that is complex, if you multiply the two conjugates together A + Bi A – Bi,
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You are going to end up getting a real quantity.
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We want a real quantity when we square this which is why this complex conjugate shows up.
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If ψ happens to actually end up being a real function like cos of X that it is not a problem.
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The square is just cos² of X because the conjugate of something real is the thing itself.
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The conjugate of 5 is 5, it is not a problem when we place that conjugate there, simply just in case ψ is complex.
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The square of the amplitude represents the intensity of the wave.
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We have this thing.
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Our problem is how the heck we are going to interpret intensity?
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What intensity mean when it comes to a particle?
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How do we interpret intensity?
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Here is how we do it.
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Think of intensity as the extent to which a particle is actually present.
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If a particle has a wave function and the square of that wave function which is the intensity is kind of low,
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That means that chances are really low that you actually find a particle there.
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If the intensity is really high, then, in other words if the square of the wave function is high then
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that means chances are really good that you are going to find a particle there.
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We are going to interpret intensity as a probability that a particular particle is at a given place.
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We are going to express it as a probability, lower the intensity, lower the probability.
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Higher the intensity, higher the probability.
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We have ψ conjugate × ψ is the probability density.
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We do not need to concern ourselves too much with the probability density so much
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because we are going to be integrating this thing and we would be concerned with probabilities.
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That is what is going to be most important.
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With the probability density, it is like any other density.
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If there is a mass density that is a certain mass per volume, g/ ml.
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The probability density is a certain probability per volume, or per length element.
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In this particular case, we are sticking to 1 dimension.
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If I take a little differential element DX like that, the product of the conjugate a ψ × ψ itself.
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And that giving me the probability density.
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It gives me the probability per unit of length or unit of area or unit of volume.
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That is all it is, it is like any other density.
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Our quantity is this, it is ψ and ψ × DX.
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This right here, when I take the wave function conjugate × wave function,
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I will just say the square of the wave function.
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I will just say it that way.
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The square of the function × some differential length element, this is the actual probability.
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This is the probability that the particle is located between a given X and DX.
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Let me redraw this thing right here.
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If I have some value of X, if I take some differential length DX,
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Now this is X + DX.
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If I take the square of the wave function and multiply it by this length element,
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I end up actually getting the probability that the particle is located here in that differential element.
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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.
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That is the whole idea.
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What is important is the square × the DX, that is the probability.
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When I just take the square of the wave function, I get something called the probability density.
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It is good to know but is not going to get in our way too much.
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Let us go ahead and do the integration.
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When I integrate from 0 to A, the square of the wave function.
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This is basically I'm just adding all the probabilities.
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The probability of the particles here or here.
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All the DX is just like normal integration and mathematics.
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When I add them all up, because probabilities are additive, I get the total probability that a particle is between 0 and A.
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The probability that the particle will be found between 0 and A.
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Since, I'm saying that the particle is going to be somewhere between 0 and A,
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I can say for sure that somewhere between 0 and A, I’m going to find the particle.
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I may not know where exactly it is, but I know it is going to be between 0 and A.
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My probability is 100% it is equal to 1.
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This thing is actually going to end up equaling 1.
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We will see a little bit more of that in just a little bit.
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Here, what is important is that the square of the wave function = the intensity.
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We are going to interpret the intensity as the probability of finding the particle there.
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The square of the wave function × some differential element, whether it is a length and area, or a volume,
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it gives me the probability that the particle is located between or in that differential length element or the area element or volume element.
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When I integrate that over my particular interval, I get the probability that the particle would be found between 0 and A.
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Whenever I specify A to B, whatever my interval is.
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This is what is important right here, profoundly important.
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That is how we are going to interpret this wave function.
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It is the square of it represents a probability density and the square × the differential element is the actual probability.
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Let us concern ourselves with the ψ.
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This wave function represents a particle, we know that.
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What we have is this thing right here.
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Between 0 and A, ψ of X represents a particle.
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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.
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Over here, our wave function is going to equal 0.
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And over here, our wave function is going to equal 0.
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Since we are differentiating not once but twice, the Schrӧdinger equation is a second derivative.
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In order to take the derivative of this function, the function needs to be continuous.
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Because it is 0 outside of the interval over here, 0 over here, and 0 over here, because the function is continuous,
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Because we have to differentiate it, that means it has to be 0 actually at 0 and at A.
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In other words, you are not going to have some wave function.
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Let us say this is 0 and this is A, you are not going to have some wave function that goes like this.
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It is all 0 from here, there is a discontinuity here.
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That is not going to happen.
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It needs to be 0 here because it is 0 pass those points.
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Therefore, at 0 and A, the wave function ψ of 0 has to equal 0.
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The ψ of A has to equal 0.
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These things are called boundary conditions.
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This is another set of constraints that we have to place on the particular problem in order for it to actually make sense.
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The wave function has to be 0 at 0, it is 0 at A.
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You are specifying what is happening at the boundary.
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We just want to find out what is going on in between.
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Our mathematical problem becomes something called a boundary value problem.
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Here is what we are going to do.
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We have to solve the equation + 2 ME/ H ̅² × ψ = 0,
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Where X ≥ 0, ≤ A subject to the boundary conditions ψ of 0 = 0 and ψ of A = 0.
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This is the mathematical problem that we have to solve.
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We have to solve this thing subject to these constraints.
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Particularly this, the boundary conditions.
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Let us go ahead and do it.
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As far as the solution of this going through this, you can take a look at the appendix
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if you actually want to see how one goes through solving this particular differential equation.
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Here I’m just going to go ahead and present the solutions.
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And the solutions are really all that we need.
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Only if you want the extra information, you are welcome to look at it.
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Here, when we solve this we get the following.
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We get the general solution is A × cos of PX + B × sin of PX,
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Where P, I just put a P in here to make it a little easier instead of writing out everything.
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Where P is actually equal to 2 ME¹/2, make it a radical sign if you want, / H ̅.
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If you want, you can put this in here and here.
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I just decided to call it P or you can call it whatever it is that you want.
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We call the H ̅ is equal to H/ 2 π.
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This is our general solution.
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We found that the ψ, we found a way function.
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Let us subject that wave function to our boundary conditions and see what A and B are going to be.
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Now the boundary conditions.
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This is normally how you handle all differential equations.
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You solve the equation and then you take a look at whatever constraints that you have placed on it
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to find the values of the individual constants.
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Instead of the general solution, you try to find a specific solution or a specific set of solutions.
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Now the boundary conditions.
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Let us go ahead and deal with the first one.
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I'm going to go to blue here.
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Let us go ahead and deal with this one.
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The ψ of 0 = A × I just put it into the equation, I see what I get,
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A × cos of P × 0 because we are putting in for X, + B × sin of P × 0 = 0.
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The sin 0 is 0, the cos of 0 is 1.
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What we get here is A × 1 and this is 0, A × 1 = 0.
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Therefore, A = 0.
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We found what A is, it is equal to 0.
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We will go ahead and this term just drops out.
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We will go ahead and deal with the second boundary condition, that the ψ of A = 0.
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Since we know that we are not dealing with this term anymore, we have B × the sin of P.
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A, X is we are putting A in and that is also equal to 0.
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There are two things that can happen, N can be 0 or sin of PA can be 0.
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B equal to 0 is trivial so it does not give us anything.
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We do not have to worry about that.
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However, let us deal with the sin of PA equaling 0.
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Which means that PA = the inverse sin of 0.
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And what you end up getting here is PA = 0, you get π, you get 2π, you get 3π, 4π, and so on.
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I’m going to do PA, and I’m going to write it as Nπ.
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Where N is going to equal 1, 2, 3 and so on.
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I’m not going to conclude the 0 because again it does not really give us anything.
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Here is what we end up getting.
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PA = Nπ.
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N= 1, 2, 3, all of these make this boundary condition true.
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Let us go to the next page here.
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We said P was equal to 2 ME¹/2/ H ̅.
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And we just said that PA = Nπ.
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Let us put this in for P so what we end up getting is 2 ME¹/2/ H ̅.
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I hope I’m not confusing my H and H ̅, that is a big problem here in Quantum Mechanics.
00:23:13.800 --> 00:23:16.000
It is equal to = N π.
00:23:16.000 --> 00:23:23.600
I’m going to rearrange this equation, when you rearrange and solve for E, the energy, you end up with the following.
00:23:23.600 --> 00:23:30.100
You end up with E for a given N because N is 1234.
00:23:30.100 --> 00:23:51.700
It is going to equal H ̅² N² I²/ A² 2M.
00:23:51.700 --> 00:24:07.400
Since, H ̅ = H/ 2 π which implies that H ̅² = H²/ 4 π²,
00:24:07.400 --> 00:24:22.800
An alternative version involving planks constant directly instead of H ̅ is going to be H² N²/ A² 8 M.
00:24:22.800 --> 00:24:25.000
And remember, M is the mass of the particle.
00:24:25.000 --> 00:24:34.100
What you have is, this is the energy of the particle, either you are this one or this one.
00:24:34.100 --> 00:24:35.800
It is totally up to you.
00:24:35.800 --> 00:24:38.800
Notice that the energy of the particle is quantize.
00:24:38.800 --> 00:24:42.800
It has very specific values depending on what N is.
00:24:42.800 --> 00:24:45.000
When N is 1, it has a certain energy.
00:24:45.000 --> 00:24:48.200
When N is 2, it has a certain energy.
00:24:48.200 --> 00:24:50.900
When N is 3, it has a certain energy.
00:24:50.900 --> 00:24:59.900
N is integral 12345, there is no 1.5, 1.6, 1.7, radical 14, things like that.
00:24:59.900 --> 00:25:02.800
This is what we mean by quantization.
00:25:02.800 --> 00:25:10.900
In other words, this particle in the box, the energy of the particle could only have specific values.
00:25:10.900 --> 00:25:13.100
There is no in between.
00:25:13.100 --> 00:25:18.300
It cannot take any value that it wants to. This is very different than classical behavior.
00:25:18.300 --> 00:25:23.500
Classical particle that have any energy value at all depending on what is going on.
00:25:23.500 --> 00:25:25.600
It does not matter.
00:25:25.600 --> 00:25:33.800
Here it is very specific and this energy is actually contingent on N, some quantum numbers, some integer.
00:25:33.800 --> 00:25:36.200
This N here is called a quantum number.
00:25:36.200 --> 00:25:40.500
This is a constant, that is a constant, A is whatever you happen to chose.
00:25:40.500 --> 00:25:43.800
You can change it but once you choose it, it is fixed amount.
00:25:43.800 --> 00:25:46.800
That is the length to M.
00:25:46.800 --> 00:25:51.400
All of these are constants, this energy is a function of N some number that
00:25:51.400 --> 00:25:54.900
shows up out of nowhere simply by virtue of the solution to the differential equation.
00:25:54.900 --> 00:25:57.900
This is what is extraordinary about quantum mechanics.
00:25:57.900 --> 00:25:59.800
This is what we mean by the quantum.
00:25:59.800 --> 00:26:02.700
The energy is quantized.
00:26:02.700 --> 00:26:13.800
N is called a quantum number.
00:26:13.800 --> 00:26:19.800
Notice, this quantum number shows up naturally by virtue of our solution to the problem.
00:26:19.800 --> 00:26:23.300
This is going to be a running theme in quantum mechanics.
00:26:23.300 --> 00:26:29.800
There is going to be several quantum numbers that show up based on how we solve problem.
00:26:29.800 --> 00:26:31.400
Let us see what we have got.
00:26:31.400 --> 00:26:38.200
Well, PA = N π.
00:26:38.200 --> 00:26:45.200
Therefore, P is actually equal to N π/ A.
00:26:45.200 --> 00:26:54.400
Therefore, we can go ahead and write our ψ sub N of X depending on what N is.
00:26:54.400 --> 00:27:15.000
It is equal to B × sin N π/ A × X and the energy for a given N = DH² N²/ A² 8M.
00:27:15.000 --> 00:27:17.700
There we go.
00:27:17.700 --> 00:27:21.200
You are probably saying to yourself what is B?
00:27:21.200 --> 00:27:22.300
We have not found out what B is, you are right.
00:27:22.300 --> 00:27:26.100
We will find out what B is in just a moment.
00:27:26.100 --> 00:27:31.200
But this is the solution to our particular particle in a box.
00:27:31.200 --> 00:27:39.300
For different numbers 1234567, the wave function is this thing, it represents the particle.
00:27:39.300 --> 00:27:44.200
Any information that I want about the particle, I'm going to extract from that function.
00:27:44.200 --> 00:27:52.300
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.
00:27:52.300 --> 00:27:58.300
That is what is happening.
00:27:58.300 --> 00:28:03.700
What would happen if we did not impose any constraints on the particle?
00:28:03.700 --> 00:28:11.000
If we just said, here is a free particle, it is not experiencing any potential energy, tell me something about its energy?
00:28:11.000 --> 00:28:14.000
What is happening?
00:28:14.000 --> 00:28:21.000
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,
00:28:21.000 --> 00:28:23.200
Instead, the particle can be anywhere.
00:28:23.200 --> 00:28:45.900
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.
00:28:45.900 --> 00:28:48.300
The ψ of A is equal 0, they vanish.
00:28:48.300 --> 00:28:52.400
All you are left with is the solution to the mathematical equation.
00:28:52.400 --> 00:28:58.300
The problem ends up just being the differential equation without the boundary value problems.
00:28:58.300 --> 00:29:01.000
I end up without the boundary conditions.
00:29:01.000 --> 00:29:05.600
Once we solve that, you end up getting the same solution.
00:29:05.600 --> 00:29:07.500
It is the same differential equation.
00:29:07.500 --> 00:29:20.500
You end up with something like ψ of X = A × cos of P of X + B × sin of P of X.
00:29:20.500 --> 00:29:30.800
P is the same thing where P = √2 ME/ H ̅.
00:29:30.800 --> 00:29:38.900
Now since there are no boundary conditions, I'm not constrained.
00:29:38.900 --> 00:29:42.000
I do not end up having to find this A and this B.
00:29:42.000 --> 00:29:45.500
I do not end up having to find this PA = something.
00:29:45.500 --> 00:29:50.100
It turns out that P can actually be anything.
00:29:50.100 --> 00:30:04.400
Because P can be anything, because P can be any number at all, before, we had P = N π/ A.
00:30:04.400 --> 00:30:06.600
P depended on N.
00:30:06.600 --> 00:30:11.500
There are no boundary conditions for a free particle completely.
00:30:11.500 --> 00:30:15.500
That is not a particle in a box, it is free to move anywhere it want.
00:30:15.500 --> 00:30:19.000
It is not dependent on N, P just equals this.
00:30:19.000 --> 00:30:23.900
When I rearrange this for energy, the energy can actually take on any value at all.
00:30:23.900 --> 00:30:25.300
And that is what is interesting here.
00:30:25.300 --> 00:30:36.500
What you end up getting when you rearrange this, you end up with E = P² H ̅²/ 2 M.
00:30:36.500 --> 00:30:39.700
I hope that the mathematics is properly here.
00:30:39.700 --> 00:30:44.000
But the idea is that P is no longer contingent on N.
00:30:44.000 --> 00:30:45.700
It can be anything.
00:30:45.700 --> 00:30:47.600
Therefore, the energy can be anything.
00:30:47.600 --> 00:30:56.800
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.
00:30:56.800 --> 00:31:00.000
In other words, it is not quantized.
00:31:00.000 --> 00:31:06.200
Quantization, this whole idea of quantum mechanics, the quantum property of a particle,
00:31:06.200 --> 00:31:11.700
only appears by virtue of the constraints that we place upon our particular system.
00:31:11.700 --> 00:31:18.400
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.
00:31:18.400 --> 00:31:21.100
Everything is fine, everything behaves normally.
00:31:21.100 --> 00:31:23.300
The energy is not quantized at all.
00:31:23.300 --> 00:31:26.000
It can take on any value just like a normal classical particle.
00:31:26.000 --> 00:31:28.200
That is pretty extraordinary.
00:31:28.200 --> 00:31:34.100
Quantization appears only when we begin to place constraints on a given system.
00:31:34.100 --> 00:31:59.500
What this means, in the case of the particle in a box, the constraints that we put on it, this particle wave,
00:31:59.500 --> 00:32:10.600
what we are saying is that it has to fit inside the box.
00:32:10.600 --> 00:32:12.300
It is what quantization means.
00:32:12.300 --> 00:32:19.800
That means there are only certain waves that will satisfy this fit property that I draw in just a second.
00:32:19.800 --> 00:32:22.500
If I do not place any constraints then it can be any wave that all.
00:32:22.500 --> 00:32:24.400
It doe not really matter, that is right there.
00:32:24.400 --> 00:32:26.300
It can be any wave at all.
00:32:26.300 --> 00:32:31.500
When I place constraints on it, it can only be specific waves, waves that fit into that box.
00:32:31.500 --> 00:32:36.300
And because it can be only specific waves, those waves can have only specific energy values.
00:32:36.300 --> 00:32:38.000
That is what is happening.
00:32:38.000 --> 00:32:40.400
Quantization is an emergent property.
00:32:40.400 --> 00:32:46.100
It is something that comes about by placing constraints on the system.
00:32:46.100 --> 00:33:00.500
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.
00:33:00.500 --> 00:33:03.200
Drawing wise, it means this.
00:33:03.200 --> 00:33:10.500
If this is 0 and A, that is one possibility for the wave.
00:33:10.500 --> 00:33:16.000
0 and A, that is another possibility for the wave.
00:33:16.000 --> 00:33:23.600
Notice, I have to be the wave that has to begin and end there.
00:33:23.600 --> 00:33:33.000
This is 0 and this is A, begins and ends there.
00:33:33.000 --> 00:33:39.300
What you are not going to see is something like this.
00:33:39.300 --> 00:33:41.400
This is 0 and this is A.
00:33:41.400 --> 00:33:43.300
It is not going to be up here.
00:33:43.300 --> 00:33:45.200
It is not going to be down here.
00:33:45.200 --> 00:33:52.800
It is not just some random wave, very specific waves with very specific energies, the constraint.
00:33:52.800 --> 00:33:59.300
Quantum behavior emerges as a result to the constraints that we place it on the system.
00:33:59.300 --> 00:34:10.700
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.
00:34:10.700 --> 00:34:14.700
Let us go ahead and draw this formally here.
00:34:14.700 --> 00:34:28.500
Let us go ahead and this is going to be N and I'm going to start at 1, 2, 3, and 4.
00:34:28.500 --> 00:34:39.900
I have this one, this is 0 to A, 0 to A.
00:34:39.900 --> 00:34:43.700
This is 0 and this is A.
00:34:43.700 --> 00:34:52.400
This is going to be our ψ₁, our ψ₂, our ψ₃, our ψ₄, and so on.
00:34:52.400 --> 00:34:54.300
I will just do the first four.
00:34:54.300 --> 00:34:59.900
Over here, I’m going to go ahead and here I’m going to draw the wave function.
00:34:59.900 --> 00:35:07.500
Here I’m going to draw the probability density, the square of the wave function.
00:35:07.500 --> 00:35:17.800
It is that.
00:35:17.800 --> 00:35:39.200
This is going to be ψ₁ conjugate, ψ₁, ψ₂ conjugate, ψ₂, ψ₃ conjugate, ψ₃ and ψ₄ conjugate, ψ₄.
00:35:39.200 --> 00:35:48.600
That is our first wave when we have the B sin N π/ X.
00:35:48.600 --> 00:35:50.500
Our ψ₁.
00:35:50.500 --> 00:36:00.000
Let us try this again.
00:36:00.000 --> 00:36:09.700
Our ψ sub N of X we said was equal to B sin N π/ A × X.
00:36:09.700 --> 00:36:14.300
If N = 1, we end up with B sin π/ A × X.
00:36:14.300 --> 00:36:32.700
We have B sin 2 π/ A × X, it is going to be this one.
00:36:32.700 --> 00:36:43.300
Here and here, we have that.
00:36:43.300 --> 00:36:48.400
1 node, 2 node, 3 node, and so on.
00:36:48.400 --> 00:36:59.200
We have this and this.
00:36:59.200 --> 00:37:01.400
Begins and ends at A.
00:37:01.400 --> 00:37:03.600
This is the wave function.
00:37:03.600 --> 00:37:06.000
Notice, they are just normal standing waves.
00:37:06.000 --> 00:37:10.900
You have a string that you are holding at one end, a string that you are holding at the other end,
00:37:10.900 --> 00:37:16.800
You pluck that string, it is going to vibrate in different frequencies.
00:37:16.800 --> 00:37:24.100
1 frequency, 1 frequency, another frequency and another frequency, they represent the different and the N you have different energy.
00:37:24.100 --> 00:37:26.300
There is certain energy.
00:37:26.300 --> 00:37:32.500
That certain energy you get from the equation that we saw.
00:37:32.500 --> 00:37:40.400
When I square these wave functions, I get the probability density or I can just think of it as the probability at this point.
00:37:40.400 --> 00:37:46.000
When I square this, this one looks like this.
00:37:46.000 --> 00:37:53.400
What this means is that the particles more likely to be found here towards the center than it is to be found here.
00:37:53.400 --> 00:37:56.600
Notice, the probability, the intensity of the wave is lower here.
00:37:56.600 --> 00:38:00.700
There is less of a chance that I’m going to find a particle here or here.
00:38:00.700 --> 00:38:06.600
When a particular particle was in this state, it is in the 1 state, the probability I’m going to find it here.
00:38:06.600 --> 00:38:16.900
This one ends up being, when I square it, it is going to look like this.
00:38:16.900 --> 00:38:19.600
The probability of finding the particle at the center is 0.
00:38:19.600 --> 00:38:23.100
More than likely that I’m going to find the particle here or here.
00:38:23.100 --> 00:38:33.400
That is what is going on here.
00:38:33.400 --> 00:38:42.600
And so on, and so forth.
00:38:42.600 --> 00:38:49.000
These high points are the greatest probability of finding the particles.
00:38:49.000 --> 00:38:58.000
If there is our particle happens to be in the state 3, the probability of finding the particle here or here or here.
00:38:58.000 --> 00:39:02.300
There is very little probability that I will find it here or here, and so forth.
00:39:02.300 --> 00:39:04.700
Notice, here mostly it is concentrated in the center.
00:39:04.700 --> 00:39:06.600
Here it is off to the sides.
00:39:06.600 --> 00:39:08.500
Here it is a little bit more distributed.
00:39:08.500 --> 00:39:10.700
Here it is a little bit more distributed evenly.
00:39:10.700 --> 00:39:19.400
As N gets bigger and bigger, the distribution of the particle actually becomes a little bit more uniform.
00:39:19.400 --> 00:39:20.800
That is what is happening here.
00:39:20.800 --> 00:39:27.300
This is just a pictorial representation of the wave function and the probability density.
00:39:27.300 --> 00:39:30.000
In other words, where you are going to find the particle.
00:39:30.000 --> 00:39:33.600
The high points have the highest probability of finding the particle.
00:39:33.600 --> 00:39:39.800
As you get lower and lower, there is a lower probability of finding the particle there.
00:39:39.800 --> 00:39:44.900
That is all that is going on here.
00:39:44.900 --> 00:39:57.900
Since this thing is actually a real function, in this particular case, we do not need the conjugate
00:39:57.900 --> 00:40:00.300
but we will go ahead and use it because that is the symbolism.
00:40:00.300 --> 00:40:08.700
This × that = ψ is this.
00:40:08.700 --> 00:40:15.800
Ψ conjugate is also that because this is a real function, it is not a complex function.
00:40:15.800 --> 00:40:23.600
The product actually = B² sin² N π/ A × X.
00:40:23.600 --> 00:40:35.200
That is it, we are just multiplying this by itself because it is a real quantity so the conjugate is just ψ.
00:40:35.200 --> 00:40:45.600
Since, ψ of X is real, my conjugate happens to equal ψ.
00:40:45.600 --> 00:40:52.600
When I do that, I get that.
00:40:52.600 --> 00:40:54.800
Let us say a little bit more here.
00:40:54.800 --> 00:41:05.300
The probability density, the second graph.
00:41:05.300 --> 00:41:23.700
The probability density for N = 1, it shows that a particle is most likely to be found near the center.
00:41:23.700 --> 00:41:34.000
It is most likely to be found near the center of the interval.
00:41:34.000 --> 00:42:02.600
For N = 2 it is more likely to be found near A/ 4 or 3A/ 4.
00:42:02.600 --> 00:42:05.100
Those are high points.
00:42:05.100 --> 00:42:42.900
As N increases, the particles are more likely to be found more uniformly distributed across the interval.
00:42:42.900 --> 00:42:46.400
In other words, here is what happens.
00:42:46.400 --> 00:42:55.300
As N gets bigger and bigger, let us say for something like N = 30.
00:42:55.300 --> 00:43:04.500
For something like N= 30, you do have something looks like this.
00:43:04.500 --> 00:43:15.600
We have this and this is 0, this is A, you are going to have something like.
00:43:15.600 --> 00:43:17.500
The distribution becomes broader.
00:43:17.500 --> 00:43:23.200
In other words, now you are very likely to find it almost anywhere where there is a high point.
00:43:23.200 --> 00:43:26.400
Now you are not restricted.
00:43:26.400 --> 00:43:35.900
As N increases, that particle is more likely to be found more uniformly distributed across the interval.
00:43:35.900 --> 00:43:38.200
This is how a normal classical particle would behave.
00:43:38.200 --> 00:43:42.600
In other words, when we treat a particle the way we have treated it in the physics courses
00:43:42.600 --> 00:43:47.400
that you have taken as a particle, another wave, the particle can be anywhere.
00:43:47.400 --> 00:43:50.700
It has no preference.
00:43:50.700 --> 00:43:53.100
It can be absolutely anywhere.
00:43:53.100 --> 00:43:56.900
When we treat it like a wave, now it has a preference.
00:43:56.900 --> 00:44:02.300
It has a greater probability of being here or here or here.
00:44:02.300 --> 00:44:06.700
As N increases, these high points tend to also increase.
00:44:06.700 --> 00:44:11.000
Now, you end up finding it more and more places.
00:44:11.000 --> 00:44:18.600
As N increases, it starts to behave like a classical particle.
00:44:18.600 --> 00:44:21.000
That is what is going on.
00:44:21.000 --> 00:45:04.500
As N increases, the probability density or the probability becomes more uniform which is how a classical particle behaves.
00:45:04.500 --> 00:45:17.500
In other words, it shows no preference for where it is in the interval.
00:45:17.500 --> 00:45:32.400
It is just as likely to be here, as it is here.
00:45:32.400 --> 00:45:40.800
When we are looking at the N = 1, it is something that looks like this.
00:45:40.800 --> 00:45:46.000
Not like that, it is a little bit worse.
00:45:46.000 --> 00:45:50.300
We have something that looks like that.
00:45:50.300 --> 00:45:52.800
Here the particle shows a preference.
00:45:52.800 --> 00:45:56.300
The probability density, the square of the wave function when we graph it,
00:45:56.300 --> 00:46:03.000
it shows that the particles more likely to be here somewhere in this area, that is not so likely to be here.
00:46:03.000 --> 00:46:05.800
That is really interesting.
00:46:05.800 --> 00:46:12.000
Under quantum behavior, it actually has a preference for where it is going to be.
00:46:12.000 --> 00:46:15.000
As N increases, you get more and more.
00:46:15.000 --> 00:46:17.400
It can be here, here, or here.
00:46:17.400 --> 00:46:21.700
Now, it spends more time everywhere, that is the idea.
00:46:21.700 --> 00:46:23.900
It is spending more time everywhere.
00:46:23.900 --> 00:46:27.700
As N increases, the distribution becomes more uniform.
00:46:27.700 --> 00:46:30.100
Here, this is not a uniform distribution.
00:46:30.100 --> 00:46:33.100
Basically, it is going to be like somewhere in this region.
00:46:33.100 --> 00:46:35.800
It is going to avoid this area and this area.
00:46:35.800 --> 00:46:42.000
As N increases, it starts to show the behavior of a classical particle.
00:46:42.000 --> 00:46:46.400
This illustrates something called the correspondence principle.
00:46:46.400 --> 00:46:57.700
We will see this again, this correspondence principle.
00:46:57.700 --> 00:47:32.600
Quantum mechanical results, they approach classical mechanics as quantum numbers get bigger.
00:47:32.600 --> 00:47:41.300
As the quantum numbers get bigger, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40.
00:47:41.300 --> 00:47:44.200
The particle starts to behave more like a classical particle.
00:47:44.200 --> 00:47:53.200
Let us talk about something very important called normalizing the wave function.
00:47:53.200 --> 00:48:01.000
Let me go back to black for this one and now we are going to go ahead and figure out what that B is.
00:48:01.000 --> 00:48:12.900
Normalizing the wave function.
00:48:12.900 --> 00:48:22.400
We said that our ψ sub N of X = B × sin N π/ A × X.
00:48:22.400 --> 00:48:28.100
This is our wave function for a particle in a box.
00:48:28.100 --> 00:48:43.700
We are going to take the ψ conjugate × ψ × DX = B² sin² N π AX × DX.
00:48:43.700 --> 00:48:53.500
This is the probability that the particle will be found in the interval DX, wherever happen to take that DX.
00:48:53.500 --> 00:49:03.500
We are restricting the particle to lie between 0 and A, the probability that
00:49:03.500 --> 00:49:09.200
a particle is going to be found somewhere between 0 and A is 100% it is equal to 1.
00:49:09.200 --> 00:49:31.600
Therefore, when I integrate this probability density over the entire interval from 0 all the way to A, I get the total probability.
00:49:31.600 --> 00:49:35.900
The probability that I'm going to find the particle somewhere between 0 and A.
00:49:35.900 --> 00:49:43.500
Again, the square of the wave function × DX is the probability that I'm going to find the particle in that little interval,
00:49:43.500 --> 00:49:45.700
that little differential interval DX.
00:49:45.700 --> 00:49:53.500
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.
00:49:53.500 --> 00:49:55.700
I know I’m going to find it somewhere.
00:49:55.700 --> 00:50:01.900
Therefore, this is equal to 1.
00:50:01.900 --> 00:50:04.100
Let us go ahead and solve this.
00:50:04.100 --> 00:50:19.200
I’m going to pull the B² out.
00:50:19.200 --> 00:50:24.600
When I go ahead and I solve this integral, you can do it either with mathematical software that you have,
00:50:24.600 --> 00:50:29.000
your mathematical or maple, or mathcad, whatever is that you happen to be using.
00:50:29.000 --> 00:50:33.000
You can solve this by looking in a table of integrals online.
00:50:33.000 --> 00:50:37.100
You have table of integrals in the back of your first year calculus text.
00:50:37.100 --> 00:50:39.000
You have table of integrals in the CRZ handbook.
00:50:39.000 --> 00:50:43.300
I'm not going to bother with the actual integration all that much.
00:50:43.300 --> 00:50:45.700
I’m going to be concerned with setting up the integral.
00:50:45.700 --> 00:50:50.300
Mostly, I will just use software or tables to do the integrals.
00:50:50.300 --> 00:50:52.500
I'm not going to go through the process.
00:50:52.500 --> 00:51:04.900
When we solve this integral right here, we actually end up getting this.
00:51:04.900 --> 00:51:19.000
It is going to be B² × A/ 2 = 1.
00:51:19.000 --> 00:51:21.200
Therefore, we end up with.
00:51:21.200 --> 00:51:23.300
I will stay on the same page.
00:51:23.300 --> 00:51:29.800
I get B² = 2/ A.
00:51:29.800 --> 00:51:40.200
Therefore, B = 2/ A ^½ or √2/ A.
00:51:40.200 --> 00:51:46.900
We went ahead and we found B by using this property of probabilities that we know.
00:51:46.900 --> 00:52:07.000
Now, we can write the individual wave function ψ sub N, they are equal to 2/ A¹/2 × sin of N π/ A × X.
00:52:07.000 --> 00:52:10.300
X ≤ A, ≥ 0.
00:52:10.300 --> 00:52:21.100
Energy sub N = A² N² / A² 8M.
00:52:21.100 --> 00:52:27.100
This is the final solution to our particle in a 1 dimensional box.
00:52:27.100 --> 00:52:35.500
A wave function that satisfies the integral when I have a particular wave function,
00:52:35.500 --> 00:52:53.300
If I multiply it by a complex conjugate and I integrated it over the particular interval,
00:52:53.300 --> 00:52:58.900
If I end up getting 1, that way function is said to be normalized.
00:52:58.900 --> 00:53:09.500
A wave function that satisfies this relation is said to be normalized.
00:53:09.500 --> 00:53:18.700
What we did was use this normalization condition to actually find B which was what we call this B²/ A ^½.
00:53:18.700 --> 00:53:21.100
In this particular case, it is called a normalization constant.
00:53:21.100 --> 00:53:29.000
If a function is the solution to a differential equation, any constant × that function is also a solution to the differential equations,
00:53:29.000 --> 00:53:35.200
Because the Hamiltonian operator is linear, we are dealing with linear operators.
00:53:35.200 --> 00:53:39.800
If F of X is a solution to a function then K × F of X.
00:53:39.800 --> 00:53:43.600
Any constant × F of X is also a solution to that differential equation.
00:53:43.600 --> 00:53:45.500
That is what is nice about this.
00:53:45.500 --> 00:53:51.700
We can always adjust the constant which is what we did to make this happen.
00:53:51.700 --> 00:53:59.000
If I take this function and if I multiply it by its conjugate which is just multiplying it by itself,
00:53:59.000 --> 00:54:03.600
And if I integrate it from 0 to A, I’m going to get 1.
00:54:03.600 --> 00:54:09.000
I have normalized this wave function.
00:54:09.000 --> 00:54:39.900
We used this condition to find B which is called the normalization constant.
00:54:39.900 --> 00:54:51.800
This is very important, normalization constant and a normalized wave function.
00:54:51.800 --> 00:54:53.700
I will close it out with the following.
00:54:53.700 --> 00:55:21.800
Because this DX is the probability of finding the particle between X and DX,
00:55:21.800 --> 00:55:34.200
the integral from any X₁ to any X₂ within a particular interval of ψ conjugate × ψ × DX,
00:55:34.200 --> 00:55:39.300
It gives the probability of finding the particle within that interval.
00:55:39.300 --> 00:55:43.900
Again, this thing is the probability of finding it within that particular DX.
00:55:43.900 --> 00:55:45.800
You can integrate from anywhere.
00:55:45.800 --> 00:55:51.500
You do not have to integrate from 0 to A, that gives you the probability of finding it over the whole interval.
00:55:51.500 --> 00:55:55.400
You can take a piece of the interval, you just change your upper and lower limits of integration.
00:55:55.400 --> 00:56:19.100
This gives the probability of finding the particle between X1 and X2.
00:56:19.100 --> 00:56:21.000
Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.