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 start doing some example problems on the Schrӧdinger equation, particle in a box,
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particle in a 2 dimensional box, 3 dimensional box, things like that.
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We are going to be doing a lot of problems.
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The only real way to wrap your mind around any of this is doing a ton of problems.
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The only warning that I have is as you already figured out quantum mechanics can be notationally intensive.
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The best advice I can give is keep calm, cool, and collect it, and work slowly.
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Let us jump right on in.
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Our first example problem is let ψ be a wave function for a particle in a 3 dimensional box.
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Find del² of the ψ and show that the ψ is an Eigen function of Laplacian operator.
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Let us see what we have got.
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First of all, let us start off with our equations.
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Again, I think is always a good idea to just write down the equations that you know simply
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because repetition locks it in your mind, makes you feel more comfortable with it.
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Let us go ahead and write down our equations.
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Ψ for a 3 dimensional box, N sub X N sub Y N sub Z = 8/ ABC ^½ × sin of the N sub X π/ AX × sin of the N sub Y π/
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B × Y × sin of N sub Z × π/ C × Z.
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That is our wave function.
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The Laplacian operator, del² was this thing.
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The del² that was that D² / DX² + D² DY² + D² DZ².
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What we have to do is we have to do is to apply the Laplace operator to this function.
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I know what you are thinking, I'm thinking the same thing.
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Our del² ψ = del² DX² + del² DY² + del² DZ² of ψ.
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Again, operators they can distribute so we are going to do this, we are going to do this, and we are going to add.
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Let us just jump right on in.
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Let us go ahead and do the second partial derivative of the X with respect to X,
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Which means we are going to hold this constant and this constant.
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The only thing we are going to differentiate is this.
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Let me actually do this in red.
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This first term, we are just going to differentiate this function because we are holding all the other variables constant.
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Let me go ahead and write this out.
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When I take the first derivative of this, I’m going to write it this way.
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N sub X π/ A × X, when I take the first derivative, just that D means take the derivatives.
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I end up with the following.
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I end up with NX π/ A × cos of NX π/ A × X that is the first derivative.
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If I take the second derivative, if I differentiate again, I end up with N sub X² π²/ A² and the derivative of the cos is –sin,
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I get the negative sin there and I get N sub X π/ A × X.
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This is my second derivative.
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Since, I’m holding everything else constant, this is the only thing that I need to concern myself with.
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My D² DX² of ψ is going to equal to -8/ ABC ^½.
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I will take this thing , -sin N sub X² π²/ A².
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I have this × the rest of this.
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I have included this, that is here.
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The negative sin from here, this thing the N sub X² π²/ A² that is here.
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Now, I have this × this and this.
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What I get is a sin of N sub X π/ A × X × sin of N sub Y π/ A.
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BY × sin of N sub Z × π/ C × Z.
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This is the first thing that I want.
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Let me go ahead and circle it, that is the first thing that I want.
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That is my first distribution.
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I have done 1/3 of my operation so far.
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It is going to be same, these functions are the same.
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It is the exact same thing except this time with respect to Y and with respect to Z.
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When I operate on ψ with the second partial with respect to Y and operate with respect to Z,
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I end up with the following.
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I'm just working very carefully.
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Let me go to red.
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I get D² DY² that is going to equal the same thing as before.
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Everything is the same except now the variable is different.
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You get -H/ ABC ^½ and this time you have N sub Y² π²/ B² and everything else is the same.
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Sin of N sub X π/ A × X, you will get sin of N sub Y π / B × Y × sin N sub Z π/ C × Z.
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The last one, we get the del²/ del Z² it looks exactly the same.
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8/ ABC¹/2 except now we differentiate it with respect to Z.
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We have N sub Z², let me make this a little bit more clear here.
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N sub Z² × π/ C² and everything else is the same.
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Sin of N sub X π/ A × sin of N sub Y π/ B.
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There is an X here, there is a Y here.
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Sin of N sub Z π/ C × Z.
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You can see that it is really easy for a problem in quantum mechanics to go south on U for no other reason than for arithmetic or missing some letter.
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It is going to make you crazy. It is going to make you want pull your hair out but that is the nature of the game.
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We have our 3 partial derivatives.
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We have operated on the wave function.
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We have this and this and the previous one, so now we just add them up.
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That was the last part of the operator, we have to add them.
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When we add them up, adding these 3 expressions together, notice the sins are all the same.
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The -8/ ABC¹/2 that is the same, you have the π² that is the same.
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When you put all the terms together, what you end up with is the following.
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Let me write out the operator here.
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Del² of ψ N sub E= - 8/ ABC ^½ × we have π² that is going to be × N sub X²/ A² + N sub Y²/
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B² + N sub Z²/ C² × this thing.
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I have the sin of N sub X π/ A × X.
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I should have made a little bit more room here.
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Sin of N sub Y π/ B × Y × sin of N sub Z π/ A × Z.
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This is my final solution, the first part of the problem, that is it right there.
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I wanted to apply the del operator ⁺to the wave function or particle in a 3 dimensional box that it is right there.
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On the second part I have to show that it is actually an Eigen function of the operator.
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For a function to be an Eigen function of an operator.
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Let me write this down.
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For a function F, speak in general terms, for a function F to be an Eigen function of an operator A, remember operators have a little hat on top of them.
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The following must hold, this is the definition.
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The following must hold applying the operator to F gives me some constant × F.
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In other words, when I apply an operator to a function, what I end up getting is just some constant × that function.
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If that holds, then the function is an Eigen function of the operator.
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In other words, an F is an Eigen function of the operator A and L happens to be the Eigen value associated with that particular Eigen function.
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That is what is happening.
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Notice what we have, our wave function.
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I will do this in black here.
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Our original wave function ψ, in other words, what we want is this.
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We want Del² ψ to equal sum constant.
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I will just use λ again.
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To be some λ × ψ.
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I want to get ψ back so I operated on it with the del² operator and I got this.
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My function, my original ψ is this one right here.
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It includes this, that, and this.
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That right there is ψ.
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I operated on it, I got this thing to actually to equal.
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I can write it this way, that is my function.
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I end up with del² of ψ = I'm left with - π² × N sub X²/ A² + N sub Y²/ B² + N sub Z²/ C² × ψ.
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Thing is just the actual function ψ.
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Sure enough, when I operated on it, I got this thing.
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This thing is nothing more than a constant which is that × the original function.
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Therefore, ψ my wave function for a particle in a 3 dimensional box is an Eigen function of the Laplacian operator.
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This thing happens to be the associated Eigen value for the associated function.
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Let me write that down.
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Do I have an extra page here?
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I do, so let me go ahead and use it.
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Our ψ N sub X N sub Y N sub Z is an Eigen function of the Laplacian operator.
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The corresponding Eigen value is - π² × N sub X²/ A² + N sub Y²/ B² + N sub Z²/ C².
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That is it so again, I have an operator and I operate on some function F.
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If what I end up getting when I do that is just the constant × the function back, that means that F is an Eigen function of this operator.
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That means it is very intimately connected and λ is the corresponding Eigen value for this particular Eigen function.
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There are going to be many Eigen functions for a given operator.
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It one of those Eigen functions has an Eigen value.
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What you are doing is this, it is saying when you take the derivative, when you operate on a function you are just getting that function back.
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All you are doing is multiplying it by some constant.
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That is very extraordinary and we will say more about it as we go on in the course.
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Let us go on to another problem here.
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Example 2, what are the most likely positions of a particle in a 1 dimensional box of length A when it is in the N = 2 state?
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What is the most likely position of a particle?
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In other words, what position in a 1 dimensional box?
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This is 0 and this is A, where along this interval is the highest probability that
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I’m going to find the particle when the wave function is in the 2 state?
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That is all the question is asking.
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Let us go ahead and write down our equations.
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We know that the equation for a 1 dimensional box ψ sub N = 2/ A.
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I hope you do not mind.
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I’m going to make one slight change here.
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It is a capital here, I am accustomed to using the small a so I’m going to change this, of length A.
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I’m just accustomed to writing the small a.
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We have 2/ A ^½ × sin of N π/ A × X.
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This are actual wave function for a particle in a 1 dimensional box.
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We said that it is in the N = 2 state.
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Ψ of 2 = 2/ A ^½ × sin of 2 π/ AX.
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We just actually pick the value.
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We are talking about probabilities here.
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We said that ψ conjugate × ψ or in the case of a real function just ψ² is the probability density or in other words, the probability.
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When we speak of probability density, we can go ahead and say probability.
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We said that when you multiply the conjugate of ψ × ψ or in the case of a real function just ψ² is a probability density.
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This is the function that we have to maximize.
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This is the function we have to maximize.
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Let me write a little bit better.
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Ψ² is the function we must maximize because we wanted the most likely positions.
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Once we have the function which we do, ψ² for the probability, we want to maximize the probability.
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That is the function we have to maximize.
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You remember from calculus, maximize means that the DDX of this ψ² function = 0.
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We are going to form ψ², we are going to differentiate it with respect to X.
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We are going to set it equal to 0 and we are going to find the X values that actually maximize it.
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What you are going to get are some of that maximize and minimize it.
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We are going to have to pick the ones that maximize it.
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That is what we are doing.
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This is a straight maximum problem from single variable calculus.
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Let us go ahead and write down,
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The ψ * × ψ which is the same as ψ² because we are dealing with a real function.
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That is just going to equal this function × itself.
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We get 2/ A × the sin² of 2 π/ A × X.
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This is our ψ².
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We need to go ahead and differentiate that function.
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When we do DDX of our ψ², I will leave the 2/ A.
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I’m going to differentiate this, this is going to be × 2 × sin of 2 π/ A × X.
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Chain rule so I differentiate this, I get the cos of 2 π/ A × X and I differentiate what is inside.
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It is going to be the 2 π/ A.
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That is the derivative with respect to X of this function.
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When I simplify it, let me go ahead and put some things together here.
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I end up with 4 π/ A² × 2 × sin of 2 π/ A × X × cos of 2 π/ A × X.
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I’m going to use the identity, 2 sin θ cos θ = sin of 2 θ cos θ.
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I'm going to rewrite this using that identity and I end up with the following.
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I end up with the DDX of our ψ² is going to equal 4 π/ A² × sin of 4 π/ A × X.
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That is our derivative.
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That is the derivative that we now need to set equal to 0.
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That means that sin of these 4 π/ A × X = 0 because this is just a constant.
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Now sin of angle = 0 whenever this 4 π/ X = 0 π 2 π 3 π 4 π 5 π 6 π 7 π.
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Therefore, this is true.
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Whenever 4 π/ A × X = some integer × π, where K = 0, 1, 2 and so one.
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That is what makes this possible.
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Rearrange for X, go ahead and cancel the π, and I get X = K × A/ 4.
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For K= 0, 1, 2, 3, 4.
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Try different values of K.
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Let me go ahead and rewrite my function ψ² just so I have it.
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My ψ² which we have actually maximized.
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You have to go back to the original function not the derivative.
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Ψ² = 2/ A × the sin² of 2 π/ A × X this is the function that we have maximized.
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We differentiated it, set it equal to 0, these are the values.
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When K =0, what you will end up with is.
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Let me put K and 0 into here, and we put this value of X that we get into here.
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What you will end up with is the following.
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You end up with 2/ A × the sin² of 2 π/ A × K 0.
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This is 0 that means X is 0 × 0 = 0.
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Let us try K = 1.
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When I do K= 2 in here, I get 2A = 4, X = 2A/ 4.
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I put that into this X so I end up with 2/ A × the sin² of 2 π/ A × 2A/ 4.
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The sin of π is 0, so again this = 0.
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Let us try K=4.
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4 is as far as we can go, I will tell you why in just a minute.
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When we use K =4, we get 4A/ 4 that is X.
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We put that into this and we end up with,
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Again, we are doing the original function sin² of 2 π/ A × 4A / 4, that cancels.
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A and A what you get is a sin of 2 π², sin² of 2 π.
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The sin of 2 π is 0 so we end up with 0.
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These are the minimum values.
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4 is as far as we go for K and here is why.
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When K = 4, 4/ 4 will give us A.
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We got as far as we are going to go.
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Again, we are going from 0 to A that is our box.
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We cannot go farther than that.
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We cannot go for example 2, 6A/ 4 which is 3/ 2.
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That is out here.
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We are sticking here is the case, from 0 to 4.
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If K is equal 0, 2 and 4 give us minimum values that means that the 1 and the 3, when K = 1 and K = 3,
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that is going to give us the maximum values.
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When K = 1 and K = 3 give us the Max values of this function.
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With K = 1, X = ¼ and when K = 3 X = 1 A/ 4.
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It is A/ 4.
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This is going to be 3A/ 4.
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Therefore, let us do an extra page.
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A/ 4 and 3A/ 4 are the most likely positions of finding a particle in a 1 dimensional box for the N = 2 state.
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What is it that we have done?
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We have the wave function, let me go back to black.
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We have the wave function ψ.
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We know the probability or the probability density.
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In this particular case, because it is a real function, under normal circumstances is this one.
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This, because it is a real function it is just ψ².
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We form the ψ² and we differentiated it.
00:28:27.800 --> 00:28:29.600
We took the derivative of it.
00:28:29.600 --> 00:28:33.700
We set the derivative equal to 0.
00:28:33.700 --> 00:28:45.300
We found the values of X that make that true and we took those values of X for different values of K 0, 1, 2, 3, 4.
00:28:45.300 --> 00:28:49.400
We plug them back in the original equation to see which one give us a minimum value,
00:28:49.400 --> 00:28:58.100
which in this case was 0 and which one give us a maximum value which was in this case, you end up with a sin =1.
00:28:58.100 --> 00:28:59.000
That is what we did.
00:28:59.000 --> 00:29:06.500
We found out that when K = 1 and K = 3, that gives us the maximum values.
00:29:06.500 --> 00:29:15.200
Therefore, when we put that back into the X = KA/ 4 which gives us our max and min values,
00:29:15.200 --> 00:29:22.200
we end up with A/ 4 and 3/ 4 as the most likely positions to find a particle.
00:29:22.200 --> 00:29:26.300
I hope that makes sense.
00:29:26.300 --> 00:29:32.000
Let us try our next example here.
00:29:32.000 --> 00:29:40.700
For an electron in a 3 dimensional box which sides are 1 nm × 2 nm × 4 nm and X, Y, Z respectively,
00:29:40.700 --> 00:29:53.100
X is 1, Y is 2, Z is 4, calculate the frequency of the radiation required to stimulate a transition from the state 123 to the state 234.
00:29:53.100 --> 00:29:57.800
In other words, N sub X is 1, N sub Y is 2, N sub Z is 3.
00:29:57.800 --> 00:30:00.100
For a particle in a box, we have 3 quantum numbers.
00:30:00.100 --> 00:30:02.000
In this case, 123 state.
00:30:02.000 --> 00:30:07.900
I want to stimulate it to the 234 state.
00:30:07.900 --> 00:30:09.800
How much energy do we need?
00:30:09.800 --> 00:30:13.400
Or what is the frequency of the radiation that I need?
00:30:13.400 --> 00:30:18.200
Let us go ahead and see if we can work this one out.
00:30:18.200 --> 00:30:20.200
Let us start off with our equations.
00:30:20.200 --> 00:30:45.100
I know the energy N sub X, N sub Y, N sub Z, is equal to planks constant/ 8M × N sub X²/ A² + N sub Y²/ B² + N sub Z²/ C².
00:30:45.100 --> 00:30:57.500
The change in energy from the state is going to be the energy of the 234 state - the energy of the 123 state.
00:30:57.500 --> 00:31:04.800
Energy of the 123 is the difference between them.
00:31:04.800 --> 00:31:08.400
Let us find the energy of the 234 and the energy of the 123.
00:31:08.400 --> 00:31:14.300
The energy of the 234 state, we just plug the values in.
00:31:14.300 --> 00:31:26.500
That is fine, let us go ahead and do this way.
00:31:26.500 --> 00:31:28.000
I will write down what these values are.
00:31:28.000 --> 00:31:34.500
A²/ 8 × the mass of the electron ×,
00:31:34.500 --> 00:31:42.300
Well, NX² NX is 2 so this is going to be 2²/ A is 1 nm.
00:31:42.300 --> 00:31:47.200
1 × 10⁻⁹².
00:31:47.200 --> 00:31:52.100
You want to work in meters, you want to work in kg, you have to watch the units.
00:31:52.100 --> 00:31:56.100
Time is going to be in seconds, mass is going to be in kg.
00:31:56.100 --> 00:32:01.800
Length is going to be in meters as energy is going to be in Joules.
00:32:01.800 --> 00:32:04.200
+ N sub Y,
00:32:04.200 --> 00:32:07.800
N sub Y is 3²/ B².
00:32:07.800 --> 00:32:23.700
B is going to be 2 × 10⁻⁹² + 4²/ 4 × 10⁻⁹² because that is 4 nm long.
00:32:23.700 --> 00:32:43.200
Planks constant H = 6.626 × 10⁻³⁴ J/ s and the mass of the electron, the rest mass of the electron is 9.109 × 10⁻³¹ kg.
00:32:43.200 --> 00:32:59.400
When I put these values in to here and run the calculation on my calculator, I get an energy for state 234 = 4.37 × 10⁻¹⁹ J.
00:32:59.400 --> 00:33:01.800
Pretty standard range.
00:33:01.800 --> 00:33:05.400
Let us go ahead and calculate the energy of 123.
00:33:05.400 --> 00:33:13.500
The energy of the 123 state is again, the same thing A²/ A × the mass of the electron.
00:33:13.500 --> 00:33:34.500
This time, we have 1²/ 1 × 10⁻⁹² + 2²/ 2 × 10⁻⁹² + 3²/ 4 × 10⁻⁹².
00:33:34.500 --> 00:33:41.600
And when I go ahead and calculate this, I end up with 1.54 × 10⁻¹⁹.
00:33:41.600 --> 00:33:44.300
Again, I hope that you are going to be checking my arithmetic.
00:33:44.300 --> 00:33:49.400
I am notorious for arithmetic mistakes.
00:33:49.400 --> 00:33:55.900
Therefore, the change in energy is equal to, like we said the energy of the 234 state –
00:33:55.900 --> 00:34:06.700
the energy of the 123 state is going to end up equaling 2.83 × 10⁻¹⁹ J.
00:34:06.700 --> 00:34:13.500
We know that the change in energy = planks constant × the frequency of the radiation
00:34:13.500 --> 00:34:21.300
which implies that the frequency of the radiation = the change in energy ÷ planks constant.
00:34:21.300 --> 00:34:34.600
We get 2.83 × 10⁻¹⁹ J ÷ 6.626 × 10⁻³⁴ J/ s.
00:34:34.600 --> 00:34:48.600
Joules cancel and we end up with a frequency of 4.27 × 10¹⁴ inverse seconds which is as you now is a Hertz.
00:34:48.600 --> 00:34:51.600
There you go, nice and simple.
00:34:51.600 --> 00:34:57.300
You just plug the values in, that is all you got to do.
00:34:57.300 --> 00:35:06.500
Let us see, what do we got next?
00:35:06.500 --> 00:35:17.600
For a 1 dimensional box of length A, calculate the probability of finding a particle between A/ 2 and 3A/ 4.
00:35:17.600 --> 00:35:19.700
Let us do this one in blue.
00:35:19.700 --> 00:35:34.400
We have 0 and we have A, they are saying calculate the probability of finding a particle between A/ 2 and 3A/ 4.
00:35:34.400 --> 00:35:36.300
This is 3/4 of A.
00:35:36.300 --> 00:35:39.000
What is the probability?
00:35:39.000 --> 00:35:42.700
What are the chances of me actually finding a particle right there?
00:35:42.700 --> 00:35:44.600
Let us do it.
00:35:44.600 --> 00:35:50.100
This is a 1 dimensional box so let us start off with our 1 dimensional wave equation
00:35:50.100 --> 00:36:00.600
which is equal to 2/ A ^½ power × sin of N π/ A × X, that is our equation.
00:36:00.600 --> 00:36:12.800
We also know that the probability of finding a particle between, when X is between some left and right point,
00:36:12.800 --> 00:36:25.000
that is our L, this is our R, the probability of that is equal to the integral from L to R of ψ × ψ DX.
00:36:25.000 --> 00:36:32.300
This is the probability density when we integrate over the entire interval that we are interested in,
00:36:32.300 --> 00:36:36.100
we end up getting the probability of finding a particle there.
00:36:36.100 --> 00:36:38.800
This is the definition of probability.
00:36:38.800 --> 00:36:43.500
You just take the complex conjugate × the function, or in this case it is a real function.
00:36:43.500 --> 00:36:48.100
Ψ² and you integrate it, that is all you are doing.
00:36:48.100 --> 00:36:58.900
We know what ψ² is, ψ² is just equal to 2/ A × sin² of N π/ A × X.
00:36:58.900 --> 00:37:03.300
It is just this thing squared.
00:37:03.300 --> 00:37:24.900
Our probability is actually equal to, in this particular case it is going to be the integral from A/ 2 to 3A/ 4 of 2/ A sin² N π/ A × X DX.
00:37:24.900 --> 00:37:26.900
Again, you can use mathematical software for this.
00:37:26.900 --> 00:37:29.800
You can use a table of integrals for this.
00:37:29.800 --> 00:37:32.500
It just depends on what it is that I happen to be doing.
00:37:32.500 --> 00:37:36.800
In this particular case, I’m going to go ahead and use a table of integrals to look up this integral,
00:37:36.800 --> 00:37:39.800
to see what it is, put the values in so I did this manually.
00:37:39.800 --> 00:37:43.900
There are going to be other times when you just want some good straight numerical value and you just plug it into your software.
00:37:43.900 --> 00:37:47.700
It does not really matter.
00:37:47.700 --> 00:37:51.200
From a table of integrals I found the following.
00:37:51.200 --> 00:37:56.300
Let us go ahead and do this in red.
00:37:56.300 --> 00:38:13.400
The integral of sin² B X DX = X/ 2 - sin of 2 BX/ 4B.
00:38:13.400 --> 00:38:20.900
In this particular case, our B is this N π/ A.
00:38:20.900 --> 00:38:29.600
I'm going to just plug in N π/ A whenever and wherever I have a B and I’m going to evaluate this.
00:38:29.600 --> 00:38:30.900
We end up with the following.
00:38:30.900 --> 00:38:41.400
We end up with this integral, this constant comes out of course.
00:38:41.400 --> 00:38:49.000
I have the integral of sin² N π/ X DX and again this is going to be our B here.
00:38:49.000 --> 00:38:51.200
This is what I’m going to use.
00:38:51.200 --> 00:38:53.100
I end up with the following.
00:38:53.100 --> 00:38:58.300
I end up with the probability of being equal to 2/ A, that came out.
00:38:58.300 --> 00:39:20.500
And when I integrated this thing, it is going to be X/ 2 - sin 2 N π/ A × X ÷ 4 N π/A.
00:39:20.500 --> 00:39:28.300
We are evaluating it from A/ 2 to 3 A/ 4.
00:39:28.300 --> 00:39:31.300
When I put these values in, here is what I get.
00:39:31.300 --> 00:39:34.000
I’m going to write all this out.
00:39:34.000 --> 00:39:40.200
2/ A × 3A/ 4 /2.
00:39:40.200 --> 00:40:09.900
I get 3A/ 8 - sin of 2 N π/ A × X which is 3 A/ 4 all of that / 4 N π/ A -, now we will do A/ 2 in here.
00:40:09.900 --> 00:40:31.000
I get A/ 4 - sin of 2 N π/ A × A/ 2 all over 4 N π/ A.
00:40:31.000 --> 00:40:33.200
I just put this in here and this in here.
00:40:33.200 --> 00:40:44.000
I’m just evaluating the integral when I start combining things 3A/ 8 -2A/ 8, I get A/ 8.
00:40:44.000 --> 00:40:46.200
Here I get some cancellations.
00:40:46.200 --> 00:40:55.100
The A and A cancel, the 2 and 4 cancel here, the 2 and 2 cancel, the A and A cancel.
00:40:55.100 --> 00:40:57.100
I end up with the following.
00:40:57.100 --> 00:40:58.000
Let us see here.
00:40:58.000 --> 00:41:03.100
This one, I end up with sin of N π.
00:41:03.100 --> 00:41:07.900
The sin of N π, regardless of when N is going to be 0, this term actually goes to 0.
00:41:07.900 --> 00:41:09.800
It drops out.
00:41:09.800 --> 00:41:14.900
When I take 3 A/ 8 – 2 A/ 8, I get A/ 8.
00:41:14.900 --> 00:41:30.900
What I’m left with here, 2 and 3 is 6.
00:41:30.900 --> 00:41:34.800
Here is what I end up getting.
00:41:34.800 --> 00:41:37.000
I end up with the following.
00:41:37.000 --> 00:41:55.300
I end up with 2/ A × A/ 8 - A × sin of 3 N π/ 2 all over 4 N π.
00:41:55.300 --> 00:41:59.400
When I multiply, when I distribute this I end up with the following.
00:41:59.400 --> 00:42:06.400
I end up with the probability equaling.
00:42:06.400 --> 00:42:22.400
The A cancels so I end up with ¼ - 1/ 2 N π × sin of 3 N π/ 2.
00:42:22.400 --> 00:42:24.500
This is my probability.
00:42:24.500 --> 00:42:31.600
Whatever N happens to be, I will put it in and this is going to be a probability that we are seeking.
00:42:31.600 --> 00:42:43.200
This is the probability that I will find a particle between A/ 2 and 3 A/ 4.
00:42:43.200 --> 00:42:46.200
Notice something really interesting here.
00:42:46.200 --> 00:42:54.300
As N goes to infinity, 1, 2, 3, 10, 20, 30, 40, 50, 100,
00:42:54.300 --> 00:43:08.900
The term 1/2 N π goes to 0.
00:43:08.900 --> 00:43:20.800
The probability ends up equaling ¼.
00:43:20.800 --> 00:43:23.000
Here is how you calculate the probability.
00:43:23.000 --> 00:43:27.300
It is going to be ¼ - something, ¼ + something depends.
00:43:27.300 --> 00:43:34.600
But as you take N higher and higher and higher, as quantum numbers become larger and larger and larger,
00:43:34.600 --> 00:43:41.600
the probability just comes down to ¼.
00:43:41.600 --> 00:43:47.600
In other words, this particular area accounts for ¼ of the interval.
00:43:47.600 --> 00:43:55.400
Therefore, my chances of actually finding the particle there is ¼ or 25%.
00:43:55.400 --> 00:44:01.600
This is classical behavior and here is the correspondence principle in action.
00:44:01.600 --> 00:44:04.300
The probability is going to change based on what N is.
00:44:04.300 --> 00:44:05.900
It is not going to be ¼.
00:44:05.900 --> 00:44:12.400
For N even it is going to be 1/4 but for N odd it is going to be different values ¼ - something, ¼ + something.
00:44:12.400 --> 00:44:19.100
But as N gets higher and higher and higher, this term goes to 0 because this term goes to 0,
00:44:19.100 --> 00:44:25.600
which means that the probability is going to be fixed at ¼.
00:44:25.600 --> 00:44:30.500
If I do a million experiments, I'm going to find 1/4 of the time that the particle is going to be
00:44:30.500 --> 00:44:33.800
in this particular area because it accounts for 1/4 of the interval.
00:44:33.800 --> 00:44:39.200
Again, the correspondence principle says that as quantum numbers become larger and larger and larger,
00:44:39.200 --> 00:44:53.500
quantum mechanical results which is this thing, they end up approaching classical mechanical results which is just the ¼ in general.
00:44:53.500 --> 00:45:02.200
Let us see our last problem here.
00:45:02.200 --> 00:45:06.800
Let A = B = 2C for the sides of a 3 dimensional box.
00:45:06.800 --> 00:45:12.700
What are the degeneracy of the first five energy levels of a particle in this box?
00:45:12.700 --> 00:45:19.200
The first five energy levels are the lowest, the next highest, next highest, the next highest, different numbers.
00:45:19.200 --> 00:45:21.700
Let A = B = 2C.
00:45:21.700 --> 00:45:25.700
When we talk about degeneracy, we set A = B =C, a perfect cubed.
00:45:25.700 --> 00:45:29.500
In this case, two of the sides are equal but one of the sides is not.
00:45:29.500 --> 00:45:30.500
What is going to happen?
00:45:30.500 --> 00:45:31.800
Are there going to be degeneracy?
00:45:31.800 --> 00:45:33.200
Are there not going to be degeneracy?
00:45:33.200 --> 00:45:36.400
Let us find out.
00:45:36.400 --> 00:45:38.600
Let us go ahead and do this in blue.
00:45:38.600 --> 00:45:44.500
Here we have A= B and we have A = 2C.
00:45:44.500 --> 00:45:49.700
Therefore, C = A/ 2.
00:45:49.700 --> 00:45:55.300
When I take C² I'm going to get A²/ 4.
00:45:55.300 --> 00:46:08.300
The energy in a 3 dimensional box, the energy of N sub X, N sub Y, N sub Z = planks constant²/ 8 M ×
00:46:08.300 --> 00:46:20.100
N sub X²/ A² + N sub Y² / B² + N sub Z²/ C².
00:46:20.100 --> 00:46:26.400
But in this particular case, A = B and C happens to equal A/ 2.
00:46:26.400 --> 00:46:44.300
I can plug in, I can fix this so the energy actually ends up becoming H²/ 8M × N sub X²/ A² +
00:46:44.300 --> 00:46:57.600
N sub Y²/ A² because B = A + 4 N sub Z²/ A²,
00:46:57.600 --> 00:47:00.800
Because C² = A²/ 4.
00:47:00.800 --> 00:47:04.300
When I flip it, the 4 comes on top and I’m left with A².
00:47:04.300 --> 00:47:09.200
Now A² is everywhere in the denominator, I can pull it out as a constant.
00:47:09.200 --> 00:47:23.300
I'm left with H²/ 8 M A² × N sub X² + N sub Y² + N sub Z².
00:47:23.300 --> 00:47:26.000
This right here, this is a constant.
00:47:26.000 --> 00:47:28.200
This is what is going to change.
00:47:28.200 --> 00:47:30.200
N is going to change, it is going to vary independently.
00:47:30.200 --> 00:47:34.000
111, 121, 131, 657 whatever.
00:47:34.000 --> 00:47:39.100
This is what we are going to analyze right here.
00:47:39.100 --> 00:47:51.000
This term is what we will analyze and which will give us the different values of the energy, what we will analyze.
00:47:51.000 --> 00:47:53.200
Let us go ahead and start.
00:47:53.200 --> 00:47:58.300
I think I would actually do this on the next page.
00:47:58.300 --> 00:48:01.600
Let me rewrite the expression.
00:48:01.600 --> 00:48:20.200
Our energy = H²/ 8 × M × A² × N sub X² + N sub Y² + 4 N sub Z².
00:48:20.200 --> 00:48:23.500
We will go ahead and take the 111 state.
00:48:23.500 --> 00:48:32.900
For 111, I’m just going to plug the 1 here, the 1 here, the 1 here, I end up with,
00:48:32.900 --> 00:48:43.700
1² + 1² + 4 × 1² that is going to equal 1 + 1 + 4 that is going to equal 6.
00:48:43.700 --> 00:48:52.100
6 × H²/ 8 MA² that is the energy of the 111 state.
00:48:52.100 --> 00:48:57.000
I hope that makes sense.
00:48:57.000 --> 00:49:02.700
I’m going to write it this way.
00:49:02.700 --> 00:49:10.300
I will put E of 111 is equal to, this is what we are analyzing.
00:49:10.300 --> 00:49:18.900
It is going to be 1 + 1 + 4 = 6 that is level 1.
00:49:18.900 --> 00:49:20.200
We are going to try different values in here.
00:49:20.200 --> 00:49:26.400
We are going to try 121, 211, 212, and you end up with different numbers.
00:49:26.400 --> 00:49:31.400
You can try it at any random order but you are going to arrange the energy levels increasingly.
00:49:31.400 --> 00:49:34.100
Here is what you come up with when you do this.
00:49:34.100 --> 00:49:49.000
The energy of the 211 state, that is going to be 4 + 1 + 4 that is going to equal to 9.
00:49:49.000 --> 00:49:57.700
For this particular state, the energy is going to be 9 H²/ 8 MA².
00:49:57.700 --> 00:50:02.500
This is the constant so I’m just working out the numbers in the parentheses.
00:50:02.500 --> 00:50:18.100
Now the energy of the 121 state that equals, if I put the 121 into this expression, I end up with 1 + 4 + 4 = 9.
00:50:18.100 --> 00:50:22.200
Sure enough, it ends up having the same energy.
00:50:22.200 --> 00:50:25.100
Let us try 131, energy of 131.
00:50:25.100 --> 00:50:35.100
It is going to be 1 + 9 + 4 is going to equal 14.
00:50:35.100 --> 00:50:40.000
Let us try the energy of the 311 state.
00:50:40.000 --> 00:50:44.100
That is going to be 9 + 1 + 4.
00:50:44.100 --> 00:50:53.500
It is going to equal 14 so we see this.
00:50:53.500 --> 00:51:13.000
When we do the energy of the 321 state, I will put the 321 into here and I get 3² is 9, 2² is 4 + 4, that is going to equal 17.
00:51:13.000 --> 00:51:27.300
The energy of the 231 state is going to be 4 + 9 + 4 so I end up with 17.
00:51:27.300 --> 00:51:44.100
Let us try the 112 state, energy of 112 = 1 + 1 + 16 = 18.
00:51:44.100 --> 00:51:48.700
This is level 1, it has 1 degeneracy.
00:51:48.700 --> 00:51:56.000
This is level 2, it is level 2 because it is the next highest energy based on whatever N can be.
00:51:56.000 --> 00:52:00.600
It is the next highest but it is the lowest of the bunch.
00:52:00.600 --> 00:52:04.400
I can put in any N I want, I’m going to get a bunch of numbers.
00:52:04.400 --> 00:52:07.100
I'm going to end up with a bunch of energies.
00:52:07.100 --> 00:52:11.600
It is going to be 6 × this thing, 9 × this thing, 9 × this thing, 14 × this thing.
00:52:11.600 --> 00:52:16.400
What I'm looking for are the number of degeneracy for that particular energy level.
00:52:16.400 --> 00:52:23.400
For the energy level of 9 H²/ 8 MA², 2 states have that same energy.
00:52:23.400 --> 00:52:26.200
The 211 state, the 121 state.
00:52:26.200 --> 00:52:29.700
This particular energy level is 2 fold degenerate.
00:52:29.700 --> 00:52:31.500
It has a degeneracy of order 2.
00:52:31.500 --> 00:52:33.500
That is what I'm doing.
00:52:33.500 --> 00:52:43.200
The question asked, find the degeneracy of the first five energy levels for a particle in a 3 dimensional box.
00:52:43.200 --> 00:52:51.100
Here is level 1, here is level 2, here is level 3, I see it has 2 degeneracy.
00:52:51.100 --> 00:52:54.600
Here is level 4, it has 2 degeneracies.
00:52:54.600 --> 00:52:58.000
Here is level 5, it has no degeneracy.
00:52:58.000 --> 00:53:00.100
It is just one thing.
00:53:00.100 --> 00:53:02.000
That is what we have done.
00:53:02.000 --> 00:53:10.300
The degeneracy of order 1, its next level up has degeneracy of order 2, the third level is 2 fold degenerate,
00:53:10.300 --> 00:53:17.600
the fourth level is 2 fold degenerate, and the fifth level is not degenerate at all.
00:53:17.600 --> 00:53:20.100
It has 1 energy level.
00:53:20.100 --> 00:53:23.600
It might be, I continue on to see, that is the whole point.
00:53:23.600 --> 00:53:26.900
I have to try each different state.
00:53:26.900 --> 00:53:30.400
I hope that actually makes sense.
00:53:30.400 --> 00:53:39.600
We calculated the formula for the energy and we just tried different values of N by varying the NX, NY, NZ separately
00:53:39.600 --> 00:53:45.000
to end up with some increasing energies of the particles of different states.
00:53:45.000 --> 00:53:53.100
These are 2 different states, 2 entirely different wave functions that they have the same energy 9 H²/ 8 MA².
00:53:53.100 --> 00:54:00.900
For this particular setup, where A = B = 2C, the 131 and the 311 state also happen to have the same energy.
00:54:00.900 --> 00:54:04.700
The 321 and 231 state happen to have the same energy.
00:54:04.700 --> 00:54:08.000
That is all we are doing here, that is all we have done.
00:54:08.000 --> 00:54:10.800
I hope that makes sense.
00:54:10.800 --> 00:54:14.200
Do not worry about it, this is just the first set of example problems.
00:54:14.200 --> 00:54:17.200
We are going to be doing a lot of example problems here.
00:54:17.200 --> 00:54:20.500
I’m going to be doing them in the next several lessons.
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In the meantime, thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.