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 finish off our discussion of the particle in a box.
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In the next lesson, we are going to start on the example problems.
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Let us just jump right on in.
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In the last lesson, we discussed the particle in a box, the Schrӧdinger equation, and we did for a 1 dimensional box.
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Finding a particle on some interval.
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We are going to do the 2 dimensional box which is just a plane region and the 3 dimensional box
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which is really what you think about when you think about a box.
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Let us start with dimension 2.
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In dimension 2, the Schrӧdinger equation looks like this.
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It is going to end up being –H ̅²/ 2 M.
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I think it is always a good idea to keep writing these over and over and over again,
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until you just feel comfortable writing them until they are just another part of your day.
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D² ψ of DX² + D² ψ/ TY² = E × ψ.
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Where now X is going to be ≥ 0, ≤ A.
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Y is going to be ≥ 0 and ≤ B.
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The ψ is a function of two variables X and Y, not just some F of X.
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It is going to be some ψ which is a function of X and Y.
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When we do 3 dimensions, it is going to be XYZ.
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We are looking for some function that has all 2 or 3 variables in it.
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The boundary conditions are going to be as follows.
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Do not worry, I’m going to draw this often just a second.
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Our boundary conditions where the wave function goes to 0.
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We are actually containing it in this little square region.
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The wave does not exist outside of that.
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Those conditions mathematically are represented as follows.
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0 Y = 0 and ψ of X 0 = 0.
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This is our mathematical problem that we have to solve.
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This is their differential equation that we have to solve.
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Here are the constraints on the variables.
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Here are the constraints on the actual function itself on the boundaries namely on the outer edge.
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Here is what we are looking at now.
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I will do it over here.
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Let me do this in blue.
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We have our coordinate system.
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This is Y and this is X, I’m going to go ahead and put A over here.
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I will go ahead and put B over here.
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In general, A and B are going to be different but they can be the same, square box.
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We will go ahead and do this.
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In this region, we are going to contain the particle in that region.
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The particle is constrained to be in this 2 dimensional boxes.
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2 dimensional region, that is where the particle is going to be.
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These boundary conditions, all they are saying is that whenever the wave function for any one X
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is always 0, for any value of Y wave function is going to be 0.
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That is the same things here for all values of X is 0.
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All it is saying is it is going to be 0 on that boundaries.
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Let us go ahead and talk a little bit about this notation here, this partial derivative notation.
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Let me go ahead, I will stick to blue.
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If you are not familiar with this notation, this D ψ DX or D² ψ DX².
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If you are not familiar with that, these are called partial derivatives.
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Some of you may have seen them, for those of you have not, do not worry about it.
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I’m just going to take a couple of minutes to explain it, do an example.
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It is very simple, you actually know what to do.
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This is just a notational difference.
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They are called partial derivatives and what you are doing when you take a partial derivative of a function of 2 variables, 3 variables, or 4 variables,
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you differentiate with respect to the bottom variable.
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It is the same thing, it is like D ψ D.
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It is the same thing, you are taking a derivative but now we use a different notation because we have another variable.
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You differentiate with respect to the bottom variable treating the other variable as a constant.
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You are doing the same exact thing you have always done for the past several years with derivatives.
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Treating the other variable as constant.
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That is all this notation means.
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When we have ψ, we go ahead and we take the first derivative with respect to X.
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We take these derivative of that what we got again with respect to X.
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And then we take the derivative with respect to Y, the original function.
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We take the second derivative with respect to Y and then we add them together.
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That is what this means.
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Let us go ahead and do an example of this partial derivative just to see how it works and then we will move on with the Schrӧdinger equation.
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Our example is, let F of XY a function of 2 variables B X⁵ Y³.
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This is a very simple function.
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We want to find DF DX DF DY as well as D² of DX², D² of DY².
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And then find D² F DY DX, DX² DF DY.
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This one is just a derivative of this function with respect to X.
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This is the derivative with respect to Y holding the X constant.
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This one is taking the first derivative with respect to X and taking the second derivative
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again with respect to X and this is the same thing with respect to Y.
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This one here, it says take the derivative with respect to X and what you get then take the derivative with respect to Y, this time holding X constant.
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And this is the reverse of that.
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These are called mixed partial derivatives.
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Let us go ahead and do this.
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Let us do DF DX first.
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We have DF DX, we are going to differentiate this with respect to X.
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We are going to treat Y as a constant.
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Y³ will stay as Y³.
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This becomes 5X⁴ Y³.
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That is it, you are just holding the other variable constant while you differentiate with respect to one variable.
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Again, we can only do things one at a time.
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This is the symbol for it.
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Let us go ahead and do DF DY.
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We will go ahead and differentiate with respect to Y except we are going to hold X constant.
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The X⁵ stays the same.
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This is going to be 3X⁵ Y².
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Everything still applies, that you have learned from calculus.
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The chain rule, everything else is just holding,
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you just have to be a little extra careful holding something constant and remembering to actually carry it forward.
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That is it, like anything else in calculus.
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Just takes a little bit of extra vigilance and making sure because there is going to be a lot of things on the page.
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Let us go ahead and do the second derivatives.
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D² F DX², we have taken DF DX.
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Now we are going to take the derivative of that with respect to X holding Y constant.
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4 × 5 is 20 so we get 20 X³ Y³.
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Now we get D² F DY.
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Now, we are going to take the second derivative of the function with respect to Y so we have a DF DY.
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We differentiate that so we hold the X constants.
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2 × 3 =6, DX⁵ stays and the Y drops down to 1.
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I will go ahead and put the 1 there, that is not a problem.
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Let us go ahead and do the mix partials.
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D² F DY DX, the order on the bottom does matter.
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Here we are saying, we differentiate with respect to X first then differentiate with respect to Y.
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What this means is do DDY of DF DX that is what this means.
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This is why you get the D² F DY DX.
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I’m going to take the derivative with respect to Y of my DF DX.
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My DF DX is up here and this is 5 X⁴ Y³ but now I’m going to differentiate this function with respect to Y holding the X constant.
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It is going to be 15 X⁴ Y².
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Over here, we will go ahead and do D² F.
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This time we will do DX DY which means we will take the derivative of DF DY.
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We are going to do why first which we already did, now we are going to take the derivative of that with respect to X.
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3 × 5 is 15 X⁴ Y².
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Now notice something, DF DX DF DY is not the same.
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D² F DX² DX² of DY², it is exhausting saying all this, they are not the same.
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But when you take the derivative with respect to one variable then take the derivative with respect to the other.
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You get this and you get this.
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Notice that they are the same, this is not a coincidence.
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Mixed partials are going to be the same, provided the function is well behaved.
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You are always going to be dealing with well behaved functions.
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You have already run across a function that is not well behaved.
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In other words, continuous and continuous partial derivatives and things like that.
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This will always be the case.
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This is actually a very deep theorem in mathematics.
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This is not a coincidence.
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The equality of mixed partials is not a coincidence.
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For those of you that actually went through the thermodynamics portion of this particular course, I have discussed this in great detail.
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We actually used the property of the mixed partials being equal to simplify our equations.
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Equality of mixed partials is not a coincidence.
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Let us go ahead and get back to our particle in a box.
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Let me go back to black here.
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Once again, we have – H ̅²/ 2M × A² ψ DX² + D² ψ DY².
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I can derive these things over and over again simply to help me become comfortable with the equation.
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It is equal E × there we go.
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In operator notation, an Eigen value notation.
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Eigen function or Eigen value notation looks like this.
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-H ̅²/ 2M, I'm going to pull out the ψ aside.
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D² DX² + D²/ DY², the ψ out there = E × ψ.
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Remember, this is an operator and an operator distributes.
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It means do this, do this, do this to this function.
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And I do this to this function, it distributes like a normal algebraic binomial or trinomial or whatever it is.
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This thing right here is very important.
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This is called the Laplace operator or just the Laplacian.
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It is profoundly important in science and mathematics.
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I promise it will be the last that you see it.
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It is profoundly important in science.
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We have a symbol for the Laplacian.
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Let me go back to black here.
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The symbol is called the Laplacian.
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It is going to be this upside down Δ².
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That is equal to this thing, this D²/ DX² + D²/ DY².
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Later, we have 3 variables for the particle in a 3 dimensional box, it is going to be D² DZ².
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The equation becomes –H ̅²/ 2N L² ψ = energy ψ.
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That is our equation.
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Once again, let us go ahead and do the boundary conditions.
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Our boundary conditions are ψ of 0 Y = 0 and ψ of X 0 = 0.
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Of course, we have X ≥ 0, ≤ A and Y ≥ 0, ≤ B.
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A and B are the limits here.
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When we solve this equation, we use the method called the separation of variables.
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I’m not going to go through it here, you can go ahead and see if you like, either in your book or in one of the appendices that I have here for this course.
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Separation of variables, this particular technique it requires us to assume that our function that we are looking for,
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this ψ of XY is actually equal to some of function of X × some function of Y and that is a product.
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It is not some mixed function.
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It is actually a separate function of X alone and a separate function of Y alone that are multiplied together.
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This particular technique requires that this be the case.
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I want to throw that in there.
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In other words, what we do is we find F of X and then we find G of Y.
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And we multiply them together that is it.
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Then multiply them together to form our ψ, our wave function which is now a function of X and Y.
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What is interesting is the following.
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The F of X and G of Y, they turn out to be solutions to the one dimensional case.
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The 1 dimensional particle in a box along the X axis or the 1 dimensional particle in a box along Y axis.
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We just happen to put them together to create the solution to the 2 dimensional particle in a box.
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It is extraordinary that we can do that.
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And I will talk about that a little bit more towards the end of this lesson.
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We find F of X, we find G of Y with this technique of separation of variables.
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We multiply them together to form our ψ of XY.
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F of X and G of Y turn out to be solutions to the 1 dimensional case.
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In other words, F of X is, from the last lesson we get some D sub X some constant × the sin of N sub X π A.
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Remember, these were the solution for the 1 dimensional particle in a box.
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I'm putting these little subscripts for the N, I can just call it N because there is going to be N for the X and there is going to be N for the Y.
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We are going to have two quantum numbers.
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It ends up being, if you remember 2/ A¹/2 × the sin of N sub X π/ A × X.
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F of Y ends up being some constant B sub Y sin of N sub Y × π/ BY.
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The constant is the same and the normalization constant, except it is going to be 2/ B.
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We are doing from 0 to B ^½, the sin of N sub Y π/ B × Y.
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Personally, I think the hardest part in quantum mechanics is not the quantum mechanics, it is not the concepts.
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It is not the math, it is writing the math down.
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There are so many things to write.
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Many indices, this star, that star, this subscript, that superscript.
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Keeping it all straight can be really difficult.
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We have the following.
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Our ψ which is now we have the subscript N sub X N sub Y is going to equal,
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You are multiplying F of X and F of Y.
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It is going to be B sub X B sub Y.
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Let us here.
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The sin of N sub X π/ A × X × the sin of N sub Y π/ B × Y.
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This B sub X and B sub Y, those are right here.
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That is the B sub X and that is the B sub Y.
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If you want you just put those in.
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It is just going to be 2/ A × B.
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The 2¹/2 and 2 ^½ they multiply it to form 2.
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The A ^½ and B ^½ ends up being AB ^½.
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I will write it over here.
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This BX BY, the coefficient, the normalization constant is just going to be 2/ AB ^½.
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I just left it in generic form right there.
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Now the energy, this is a solution to the particle in a box, this whole equation.
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And notice, it is a function of X and Y.
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It just happens to be two separate functions that we have multiplied together.
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The energy N sub X N sub Y = A²/ 2M × N sub X²/ A² + N sub Y²/ B².
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These are your solutions right here.
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This is the solution to the Schrӧdinger equivalent for the problem for a particle in a 2 dimensional box.
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This is the wave equation, this being the normalization constant, and this is the energy for a different values of N.
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Let me go ahead and actually say N sub X takes on the values 1, 2, 3, and so on.
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N sub Y takes on the values 1, 2, 3, and so on.
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This N sub X and N sub Y, the N can vary independently.
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It is not just 11, 22, 33.
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It could be 111, 12, 13,14, 7, 8, 14, 16.
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They vary independently.
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There are all kinds of different energy levels.
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In a particle in a box, it actually have.
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The normalization constant, we found it just by multiplying the normalization constant for F of X and G of Y, that is fine.
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The normalization constant is found normally the same way as we did before.
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B sub X B sub Y is found the same way as before.
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Which was you have to take the integral over the entire region of C* × C.
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I will go ahead and put DA here and set it equal to 1.
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Again, that is all we did.
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In order to satisfy the normalization condition, this is the probability.
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When add up all the probabilities, the probability that the particles can be somewhere in the box has to equal 1.
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That is our normalization condition, it always = 1.
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We use this to solve for the BX BY.
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This is the equation that we actually have to solve in order to find our normalization constant.
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In order to normalize the wave function.
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You just take the function, multiply it by its conjugate, and you integrate.
00:24:01.000 --> 00:24:06.600
In this case, the conjugate happens to be the function itself because the functions are real functions, they are not complex.
00:24:06.600 --> 00:24:10.400
This is the process that you always go through.
00:24:10.400 --> 00:24:15.300
Here it is DA.
00:24:15.300 --> 00:24:19.100
I wrote DA because we are integrating over an area.
00:24:19.100 --> 00:24:22.200
We are going to integrate over X, then we are going to integrate over Y.
00:24:22.200 --> 00:24:23.800
We are integrating over an area.
00:24:23.800 --> 00:24:25.500
This is actually a double integral.
00:24:25.500 --> 00:24:30.800
You have to watch what dimension you are working in.
00:24:30.800 --> 00:24:51.700
Because we are integrating over an area, in other words we got this region here.
00:24:51.700 --> 00:24:58.300
We integrated in the X direction, and then we integrate in the Y direction, the double integral.
00:24:58.300 --> 00:25:01.000
Let me go back to black here.
00:25:01.000 --> 00:25:02.300
This integral is symbolic.
00:25:02.300 --> 00:25:04.800
This one right here, it is symbolic.
00:25:04.800 --> 00:25:06.700
It actually means this.
00:25:06.700 --> 00:25:19.600
In this particular case, a 2 dimensional case ψ* ψ DA= the integral from 0 to A of the integral from 0 to B of,
00:25:19.600 --> 00:25:22.300
when we take the function and multiply itself,
00:25:22.300 --> 00:25:45.100
we are going to end up with B sub X² B sub Y² ψ² N sub X π/ A × X and D sin² of N sub Y π/ B × Y DY DX.
00:25:45.100 --> 00:25:54.000
DY DX is the DA, it is a differential of area element.
00:25:54.000 --> 00:25:59.900
DX DY when you multiply them together you will get DA, the differential of area element.
00:25:59.900 --> 00:26:01.300
That is what this means.
00:26:01.300 --> 00:26:06.700
We have to solve this particular integral and do not worry about it, we will be doing these in the example problem.
00:26:06.700 --> 00:26:09.200
We will see.
00:26:09.200 --> 00:26:18.900
You can do it this way to solve the integral or you can just go ahead and take the coefficients for the 1 dimensional case and just multiply them together,
00:26:18.900 --> 00:26:21.300
which is exactly what we did.
00:26:21.300 --> 00:26:34.900
We had B sub X = 2/ A ^½ and we have the B sub Y = 2/ B ^½
00:26:34.900 --> 00:26:39.800
which means that the normalization constant BX BY is such as this × that.
00:26:39.800 --> 00:26:45.500
It is equal to 2/ AB ^½.
00:26:45.500 --> 00:26:50.900
That is it, nice and simple.
00:26:50.900 --> 00:26:59.800
Let us go ahead and say a little bit more here.
00:26:59.800 --> 00:27:31.500
This ψ N sub X N sub Y which is a function of X and Y is the product of ψ, for the 1 dimensional case sub X and ψ NY.
00:27:31.500 --> 00:27:47.000
For the 1 dimensional case in the Y variable.
00:27:47.000 --> 00:27:52.200
Let me write it out.
00:27:52.200 --> 00:27:58.900
Our wave function for 2 dimensions is the product of the wave functions for 1 dimensional case.
00:27:58.900 --> 00:28:03.800
The energy for the 2 dimensions is the sum, here we have the product,
00:28:03.800 --> 00:28:13.600
the energy is the sum of the energies for the individual cases, for the 1 dimensional cases.
00:28:13.600 --> 00:28:17.700
N sub X N sub Y.
00:28:17.700 --> 00:28:27.100
This is actually very important result and I will be talking about it again a little bit more formally towards the end.
00:28:27.100 --> 00:28:32.000
Let us go ahead and move on to the 3 dimensional case.
00:28:32.000 --> 00:28:34.400
I will go ahead and go back to black here.
00:28:34.400 --> 00:28:43.400
We have dimension 3.
00:28:43.400 --> 00:29:05.800
A particle in a 3 dimensional box, now our equation is going to be –H ̅²/ 2 M × the Laplacian operator for 3 variables.
00:29:05.800 --> 00:29:16.400
DX² D²/ DY² + the second derivative with respect to ψ of our wave function that we are looking for,
00:29:16.400 --> 00:29:20.400
It is going to equal that × ψ.
00:29:20.400 --> 00:29:29.600
This time we have X ≥ 0, ≤ A, Y ≥ 0 and ≤ B, Z ≥ 0 and ≤ C.
00:29:29.600 --> 00:29:36.900
We are constraining at a box which is A × B × C.
00:29:36.900 --> 00:29:38.600
A long, B wide, C high.
00:29:38.600 --> 00:29:40.800
However, you want to look at it.
00:29:40.800 --> 00:29:47.800
The ψ is a function of 3 variables.
00:29:47.800 --> 00:30:00.200
Ψ is now a function of 3 variables X, Y, and Z.
00:30:00.200 --> 00:30:04.500
Ψ is ψ of X, ψ of Y, ψ of Z.
00:30:04.500 --> 00:30:06.700
It is going to be exactly what you think it is.
00:30:06.700 --> 00:30:11.500
We are just going to end up forming the product of the 3 functions.
00:30:11.500 --> 00:30:13.100
It looks like this
00:30:13.100 --> 00:30:16.600
Let us go ahead and draw this one out.
00:30:16.600 --> 00:30:20.900
The coordinates is in here.
00:30:20.900 --> 00:30:24.700
This time what we have is the following.
00:30:24.700 --> 00:30:29.400
This is going to be our X coordinate, and this is going to be our Y coordinate, this is going to be our Z coordinate.
00:30:29.400 --> 00:30:35.600
This is the standard 2 dimensional representation of a 3 dimensional object.
00:30:35.600 --> 00:30:37.300
The right hand coordinate system.
00:30:37.300 --> 00:30:41.400
The X is this way, the Y is this way, and the Z is this way.
00:30:41.400 --> 00:30:52.200
A is over here and B is over here and let us go ahead and put C like right there.
00:30:52.200 --> 00:30:53.800
Here is what we have.
00:30:53.800 --> 00:31:18.700
Here is our box, let me go ahead and do the box in blue.
00:31:18.700 --> 00:31:27.600
I’m going to go across and I will go down.
00:31:27.600 --> 00:31:30.600
Our particle in a 3 dimensional box.
00:31:30.600 --> 00:31:32.500
The particle is somewhere there.
00:31:32.500 --> 00:31:34.700
That is all we are doing.
00:31:34.700 --> 00:31:38.200
We have our equation which is this one.
00:31:38.200 --> 00:31:40.900
We have the constraints on the variables.
00:31:40.900 --> 00:31:47.700
Our boundary conditions are going to be as follows.
00:31:47.700 --> 00:32:07.400
Our boundary conditions are ψ of 0 YZ = 0 ψ of X 0 Z = 0 and the ψ of X Y 0 = 0.
00:32:07.400 --> 00:32:16.300
All these boundary condition say is that once you actually hit the walls of this box, the wave function goes to 0.
00:32:16.300 --> 00:32:21.500
The wave function is 0 at that point and beyond.
00:32:21.500 --> 00:32:24.100
That is all we are doing.
00:32:24.100 --> 00:32:32.200
In Del notation, it is nice to see it.
00:32:32.200 --> 00:32:44.900
-H ̅²/ 2N Del² ψ = energy × ψ.
00:32:44.900 --> 00:33:03.300
Again the Del² is just this Laplacian operator, that is all it is.
00:33:03.300 --> 00:33:07.100
Our ψ again, we do it with the method of separation of variables.
00:33:07.100 --> 00:33:19.800
Our wave function of XYZ is going to equal the product of F of X × G of Y × H of Z.
00:33:19.800 --> 00:33:35.200
The F of X = 2/ A ^½ that is just the wave function for the particle in a 1 dimensional box in the X direction.
00:33:35.200 --> 00:33:43.600
A × the sin of N sub X π/ A × X.
00:33:43.600 --> 00:33:59.300
If we have G of Y = 2/ B ^½ × the sin of N sub Y × π not A.
00:33:59.300 --> 00:34:04.700
Be really careful, I always do that.
00:34:04.700 --> 00:34:08.500
It is like I get stuck on one value A and I write A for everything.
00:34:08.500 --> 00:34:12.900
This is N sub Y × π/ B.
00:34:12.900 --> 00:34:29.100
The variable is Y and of course our H of Z is going to = 2/ Z ^½ × the sin of N sub Z × π/ A × Z.
00:34:29.100 --> 00:34:44.200
Again, N sub X and N sub Y and N sub Z, they all range from 1, 2, 3, and so forth.
00:34:44.200 --> 00:34:46.400
They are very independently.
00:34:46.400 --> 00:35:06.100
You have 3 quantum numbers for a particle in a 3 dimensional box.
00:35:06.100 --> 00:35:14.500
Now, our wave function N sub X N sub Y N sub Z, it ends up very independently,
00:35:14.500 --> 00:35:29.400
= 8/ ABC ^½.
00:35:29.400 --> 00:35:48.100
N sub X π/ A × X × sin of N sub Y × π/ B × Y × sin of N sub Z × π/ ψ × Z.
00:35:48.100 --> 00:35:54.300
This is crazy, these functions get really big and yes we are going to be integrating with them.
00:35:54.300 --> 00:35:56.400
We are going to be differentiating them.
00:35:56.400 --> 00:36:02.800
It does tend to get notational intensive and again personally I think it is the most difficult part of quantum mechanics.
00:36:02.800 --> 00:36:07.700
It is just keeping all of the notations straight.
00:36:07.700 --> 00:36:17.000
The energy of a particle in a box, first state N sub X N sub Y N sub Z.
00:36:17.000 --> 00:36:23.800
This is going to equal planks constant²/ 8M.
00:36:23.800 --> 00:36:42.700
N sub X²/ A² + N sub Y²/ B² + N sub Z²/ C².
00:36:42.700 --> 00:36:48.200
The energies are just the some of the energies in the individual directions.
00:36:48.200 --> 00:36:53.300
A particle in a 1 dimensional box in the X, 1 dimensional box in the Z, 1 dimensional box in the Z.
00:36:53.300 --> 00:36:59.000
Put them together, you form the product of the functions, you get the overall wave equation for a particle in a 3 dimensional box.
00:36:59.000 --> 00:37:05.200
You add up the energies to get the energy of the particle in a 3 dimensional box.
00:37:05.200 --> 00:37:12.000
Once again, the N sub I they vary independently.
00:37:12.000 --> 00:37:16.300
Do not get stuck on the idea that it all has to be 111, 222, 333.
00:37:16.300 --> 00:37:19.800
It does not.
00:37:19.800 --> 00:37:24.100
Let us talk about something interesting that happens.
00:37:24.100 --> 00:37:27.900
Something happens when the sides of the box are equal.
00:37:27.900 --> 00:37:30.100
A can be anything, B can be anything, C can be anything.
00:37:30.100 --> 00:37:33.900
When they are equal, here is what happens.
00:37:33.900 --> 00:37:42.600
The energy N sub X, let us specify what we mean by this sides being equal.
00:37:42.600 --> 00:37:49.900
We mean when A = B = C, when you have a cubed.
00:37:49.900 --> 00:37:51.300
Here is what happens to the energy.
00:37:51.300 --> 00:37:59.300
The energy of N sub X, N sub Y, N sub Z, that is equal to H²/ 8M.
00:37:59.300 --> 00:38:07.600
Now, it is going to be N sub X²/ A² + N sub Y²/ A²,
00:38:07.600 --> 00:38:11.700
Because B = A so I can just put in A².
00:38:11.700 --> 00:38:19.000
They are all the same length, + the N sub Z²/ A².
00:38:19.000 --> 00:38:22.000
The A² is a constant so you can just pull it out.
00:38:22.000 --> 00:38:35.800
The energy = H²/ 8 M A² × N sub X² + N sub Y² + N sub Z².
00:38:35.800 --> 00:38:37.700
This is just a constant.
00:38:37.700 --> 00:38:41.200
What is important is this thing.
00:38:41.200 --> 00:38:46.300
The N depending on what N is, the energy of that level is going to be different.
00:38:46.300 --> 00:38:51.200
If N is 1, if N sub X is 1, N sub Y is 1, N sub Z is 1.
00:38:51.200 --> 00:39:00.700
You are going to end up with 1 + 1 + 1 it is going to be 3 A²/ 8 MA² that is the energy of the 111 level.
00:39:00.700 --> 00:39:05.500
Now, for a particle in a 3 dimensional box.
00:39:05.500 --> 00:39:10.700
The state of the system is expressed by the N sub X N sub Y N sub Z.
00:39:10.700 --> 00:39:18.000
Let me say that again.
00:39:18.000 --> 00:39:35.800
For a particle in a 3 dimensional box, the state of the system is expressed by these quantum numbers.
00:39:35.800 --> 00:39:41.800
It is expressed by N sub X, N sub Y, and N sub Z.
00:39:41.800 --> 00:39:47.700
Let us go ahead and actually work out some energy for some different states.
00:39:47.700 --> 00:39:59.600
When we have N sub X = 1, N sub Y = 1, and N sub Z = 1, this is going to be the energy of the 111 state,
00:39:59.600 --> 00:40:09.600
That is going to equal H² / 8 MA² 1² + 1² + 1².
00:40:09.600 --> 00:40:12.000
I’m just putting it into the equation.
00:40:12.000 --> 00:40:24.700
I end up with 3 H²/ 8 MA², that is the energy of the 111 state of a particle in a 3 dimensional box.
00:40:24.700 --> 00:40:33.100
If I said I have a particle in this 3 dimensional box, the sides are length A.
00:40:33.100 --> 00:40:41.200
What is the energy of the 111 state?
00:40:41.200 --> 00:40:44.200
That particular wave function, what is the energy of that particle?
00:40:44.200 --> 00:40:50.200
There it is, planks constant², multiply by 3, divide by 8, divide by the mass of whatever the particle is,
00:40:50.200 --> 00:40:58.900
electron, proton, whatever and then divide by the square of the sum of the side length.
00:40:58.900 --> 00:41:00.800
Let us do some more.
00:41:00.800 --> 00:41:10.200
This time let us do N sub X = 2 and N sub Y = 1 and N sub Z = 1.
00:41:10.200 --> 00:41:13.200
This is the energy of the 211 state.
00:41:13.200 --> 00:41:26.200
That is going to equal A²/ 8 M, A² this time it is going to be 2² + 1² + 1².
00:41:26.200 --> 00:41:29.400
N sub X, N sub Y, N sub Z.
00:41:29.400 --> 00:41:40.800
You will end up with 6 H²/ 8 MA².
00:41:40.800 --> 00:41:42.700
Let us do another level.
00:41:42.700 --> 00:41:51.100
For N sub X = 1, this time N sub Y = 2, N sub Z = 1.
00:41:51.100 --> 00:41:56.800
This is not the same state 211, this is 121, this is an entirely different wave function.
00:41:56.800 --> 00:42:04.400
An entirely different wave, an entirely different set of probabilities, an entirely different of state of the system.
00:42:04.400 --> 00:42:13.300
We know the energy, the energy of the 121 state = H²/ 8 MA².
00:42:13.300 --> 00:42:17.900
This time it is going to be 1² + 2² + 1².
00:42:17.900 --> 00:42:22.800
Again, we end up with 6 H²/ 8 MA².
00:42:22.800 --> 00:42:29.300
It ends up being the same energy as the 211 state.
00:42:29.300 --> 00:42:31.300
You can imagine what we are going to do next.
00:42:31.300 --> 00:42:37.400
This time we are going to do X = 1 and Y = 1 and this time we are going to take the N sub Z = 2.
00:42:37.400 --> 00:42:42.000
This is going to be the 112 state.
00:42:42.000 --> 00:42:50.400
Again, a completely different state 112.
00:42:50.400 --> 00:42:55.300
Sorry to elaborate the points here, I think it is actually good to go through it like this.
00:42:55.300 --> 00:42:57.700
+1² + 2²,
00:42:57.700 --> 00:43:04.700
Once again we end up with 6 H²/ 8 MA².
00:43:04.700 --> 00:43:18.800
For 3 different states, the 211, 121, and 112, the energies are the same.
00:43:18.800 --> 00:43:35.500
For 3 different states, the energies are the same, this is what we call degeneracy.
00:43:35.500 --> 00:43:41.800
Let me go ahead and do this in red.
00:43:41.800 --> 00:43:56.900
We say this particular energy level, this energy level in this particular case, the 6 H², this particular energy level.
00:43:56.900 --> 00:44:02.000
Let us say the 6 H²/ 8 MA², whatever the energy happens to be.
00:44:02.000 --> 00:44:11.200
That level is 3 fold degenerate or has degeneracy of order 3.
00:44:11.200 --> 00:44:16.900
In other words, there are 3 states that have that energy.
00:44:16.900 --> 00:44:20.100
That should be the case.
00:44:20.100 --> 00:44:25.000
A particle in one state is given by a particular wave function should have a different energy.
00:44:25.000 --> 00:44:33.600
All of a sudden, simply by making the side of the box equal, whether it is a perfect square or a perfect cubed,
00:44:33.600 --> 00:44:39.900
all of a sudden you have these particles that are completely different states, completely different wave functions,
00:44:39.900 --> 00:44:43.100
if they have the same energy that is what we call degeneracy.
00:44:43.100 --> 00:44:52.800
We say that this energy level is 3 fold degenerate.
00:44:52.800 --> 00:45:26.600
That is the level has 3 states that have that particular energy.
00:45:26.600 --> 00:45:29.600
Let me go ahead and go back to blue here.
00:45:29.600 --> 00:45:47.200
I will just write in general, degeneracy is the number of different states having the same energy.
00:45:47.200 --> 00:45:53.400
That is it that all degeneracy is.
00:45:53.400 --> 00:45:59.100
It is a very important concept.
00:45:59.100 --> 00:46:02.100
Degeneracy emerges when the system becomes symmetric.
00:46:02.100 --> 00:46:07.000
Remove the symmetry and you actually end up removing the degeneracy.
00:46:07.000 --> 00:46:12.400
Once again we have some rectangular box A, B, C are different.
00:46:12.400 --> 00:46:15.900
You are going to get different values.
00:46:15.900 --> 00:46:23.000
With them, you actually end up making the size of the box equal, you actually introduced degeneracy into the system.
00:46:23.000 --> 00:46:28.200
You actually solve by introducing symmetry into the system, the cube is perfectly symmetric.
00:46:28.200 --> 00:46:29.800
The square is perfectly symmetric.
00:46:29.800 --> 00:46:32.600
You actually introduced degeneracy.
00:46:32.600 --> 00:46:35.300
That was actually a fundamental principle of quantum mechanics.
00:46:35.300 --> 00:46:42.800
It says that degeneracy is up here, they emerge, they show up when you introduce symmetry into the system.
00:46:42.800 --> 00:46:47.200
If you remove the symmetry, you end up removing the degeneracy.
00:46:47.200 --> 00:46:50.700
Degeneracy is not necessarily like an open door and a closed door.
00:46:50.700 --> 00:46:54.600
It is not like yes it symmetrical, not it is not symmetrical.
00:46:54.600 --> 00:46:59.700
There are actually degrees of symmetry and we are going to see some of that when we actually did example problems.
00:46:59.700 --> 00:47:05.700
But again, degeneracy emerges as a result of introducing symmetry.
00:47:05.700 --> 00:47:08.700
Take away the symmetry, you reduce the degeneracy.
00:47:08.700 --> 00:47:10.300
You take away the degeneracy.
00:47:10.300 --> 00:47:13.000
This is actually a very profound quantum mechanical principle.
00:47:13.000 --> 00:47:14.900
Not just quantum mechanical.
00:47:14.900 --> 00:47:22.200
The idea of symmetry is what permeates nature, permeates all of mathematics and physics.
00:47:22.200 --> 00:47:29.700
I’m going to go ahead off with what I said earlier, when I said I would discuss more formally the idea of the particle in a 3 dimensional box
00:47:29.700 --> 00:47:37.000
being a product of the functions of the 1 dimensional box and the energies being the sum.
00:47:37.000 --> 00:47:39.700
Let me go back to black here.
00:47:39.700 --> 00:48:00.500
We said that the ψ of N sub X N sub Y N sub Z = ψ of N sub X × ψ of N sub Y × ψ of N sub Z,
00:48:00.500 --> 00:48:08.000
and that the energy of N sub X N sub Y N sub Z state is just the energy of the N sub X state +
00:48:08.000 --> 00:48:26.300
the energy of N sub Y state + the energy of the N sub Z state.
00:48:26.300 --> 00:48:30.300
I think I’m going to leave this for another time, I apologize.
00:48:30.300 --> 00:48:34.900
I think I can go ahead and close out this lesson right here and I will discuss this formality a little bit later.
00:48:34.900 --> 00:48:37.600
Perhaps, I will actually do some problems.
00:48:37.600 --> 00:48:39.500
That should not be a problem at all.
00:48:39.500 --> 00:48:41.600
Thank you so much for joining us here at www.educator.com.
00:48:41.600 --> 00:48:43.000
We will see you next time, bye.