WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com, welcome back to Physical Chemistry.
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We spent some time talking about the hydrogen atom and now we are going
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to do the example problems for the hydrogen atoms.
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There are going to be a lot of them.
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No worries, we would spend plenty of time working in example problems because this is very important.
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Let us jump right on in.
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Example number 1, let S² equal to this expression.
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This of course is the square of the angular momentum operator.
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For L = 0 and L = 1, find this expression.
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L² of S sub L super M θ φ, so these are the spherical harmonics.
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We want you to find, use this operator to operate on the spherical harmonics for L = 0 and L = 1.
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Let us see, let us go ahead and start.
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Let me see, should I do black or blue, or red.
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I will stick with black today, at least for this one.
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For L = 0, we have that M is equal to 0.
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You remember that, you know M is equal to, it goes all the way from -L all the way to +L, passing through 0.
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In the case of L = 0, M just equal 0.
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We have S0, 0 and it is equal to, let me go ahead actually use my θ and φ.
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S0, 0 of θ φ is equal to 1/ 4 π ^½.
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Of course, you will get all of these from your book.
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There are list of the spherical harmonics are listed.
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The Legendre polynomials are listed.
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The Laguerre polynomials are listed.
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All the full wave functions are listed.
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And if for any reason you are not using a book, I will just go ahead and look them up online.
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There they are listed. These are spherical harmonics.
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This is our S0, 0, now we want to find L² S0, 0.
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In this case, let us go ahead N equal to 0.
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This one is an easy one because S0, 0 is not a function.
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It is a function of θ and φ but it does not depend on θ and φ.
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Notice, this is just some constant so θ and φ do not show up.
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I had nothing to actually take the derivative of that.
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When I take the derivative of a constant, it is just going to equal 0.
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Because S0, 0 does not depend on θ or φ.
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Let us go ahead and do for L = 1.
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For L = 1, N is going to equal -1, 0 and + 1.
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We have three of these to actually take care of.
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Let us go ahead and do S1, 0 first.
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Let us take care of this 0 and 1 first.
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S1, 0 that is equal to 3/ 4 π to ½ cos θ.
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-H ̅², we are going to use this operator and we are going to operate on this right here.
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Let me go ahead and write it all out.
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Again, as you know quantum mechanics is mostly, it is not particularly difficult,
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it is just computationally intensive, notationally intensive.
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That is the real issue and it can be intimidating.
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Taking a look at an expression like this and realizing what is going on.
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Nothing, it is just the second derivative here, first derivative, first derivative, that is all that is happening here.
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No worries.
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Let us see, sin θ DD θ + 1/ sin² θ D² D φ.
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We are going to operate on the function 3/ 4π ^½ cos of θ.
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Let us just take this one at a time.
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We are going to go this on that and this on that.
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Essentially, this distributes over that, this distributes over that.
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I’m going to start here, I’m going to do this one first.
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When we take DD θ of this cos θ, I’m going to go ahead and put the constant here.
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I’m just going to deal with what the functions of θ and φ.
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Stick with constants at the end.
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DD θ of cos θ, this thing is going to end up being sin θ × - sin θ.
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It is going to equal - sin² θ and then we are going to do this one, the DD θ of that.
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We are going to end up getting -2 sin θ cos sin θ.
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We are going to multiply that by 1/sin θ.
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This is a nice way of actually doing it, in barrels with each operation.
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We end up getting - 2 cos θ.
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We end up getting that when we operate on that.
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The second term, the second term does not matter.
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The reason it does not matter is because this D² D φ of this is just 0.
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There is no φ here, so when we take the derivative of it, it just goes to 0.
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This 1 / sin² θ does not matter because we are just multiplying by 0.
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I’m going to go ahead and write that in blue here.
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The second term of the L² operator does not matter because S1, 0 does not involve the φ.
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We have L² of S1, 0 is equal to, I will go ahead and put all the constants back in.
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-H² 3/ 4 π ^½ × - 2 cos θ.
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That is what we got when we get the operation.
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That is going to equal -2 H ̅² 3/4 π ^½ cos θ.
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Notice this part, it is the same as that is the S1, 0.
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Therefore, we have L ̅² of S1, 0 is equal to -2, the minus sign go away so we are left with.
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I’m sorry this is a + here.
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We have 2 H ̅ S1, this is the S1, 0 so I can just call it that.
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LR² S1, 0 is equal to some constant × S1, 0.
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Therefore, S1, 0 is an Eigen function of the square of the angular momentum operator.
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L1 with Eigen value 2 H ̅, that is it.
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We are just confirming something that we already know but it is nice to go through the process to do the mathematics.
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We have taken care of S1, 0, let us go ahead and take care of S1, 1.
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Let us do S1, 1, it is equal to 3 / 8 π to ½ and it is going to be sin θ E ⁺I φ.
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This time, both θ and φ are involved.
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We are going to operate, this is going to be –H ̅².
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I hope you will forgive me if I keep repeating the operator over and over again.
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To write down as much as possible and I’m not sure if that is bad.
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I think that is probably pretty good.
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Sin θ DD θ + 1/ sin² θ D² with respect to φ² this.
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Of course, we are operating on for B/ 8 π ^½ sin θ E ⁺I φ.
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We just have to do keep things straight, do our best to keep everything in line.
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We are going to ignore the constant and I’m just going to do this one first.
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This one is going to be DD θ of this, it is going to be the derivative of that is sin θ.
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The derivative of that is cos θ so we are going to have sin θ × cos θ.
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When we apply now this one, so we are going to apply the DD θ part.
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We are going to get - sin² θ and I hope that I actually done my differentiation correctly.
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- sin² θ + cos² θ and I'm going to multiply by 1/ sin θ.
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That is going to end up giving me cos² θ.
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Let us see if I can keep track of everything that is going on here.
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We got cos² θ, let me drop it down here.
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I'm going to end up getting this, I’m doing everything here.
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Cos² θ - sin² θ/ sin θ.
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Of course, we cannot forget about that.
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E ⁺I φ that is going to be + sin θ or sin² θ I² E ⁺I φ.
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That is why we operated this one on that.
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It is going to end up being, cos² θ – sin² θ/ sin θ -1 ×,
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this is going to be -1/ sin θ × E ⁺I φ.
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I can go ahead and do on the next page here.
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It is going to be -2 sin² θ, I’m just using basic identities.
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Θ/ sin θ × E ⁺I φ which is equal to - 2 sin θ E ⁺I φ, there we go.
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This is from our final answer after operating on that.
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L ̅² of S1, 1 is equal to -H ̅² × 3/ 8 π ^½ × - 2 sin θ E ⁺I φ, which is equal to 2 H ̅ S1, 1.
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S1, 1 is an Eigen function of L ̅².
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The Eigen value is going to be 2 H ̅.
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We see the pattern here and let us go ahead and do one more thing.
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Now, we have to do the S1,-1.
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That is going to equal, the constant is the same.
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It is 3/ 8 π, this is listed in your books or on the net, sin θ.
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Except, it is going to be E ⁻I φ.
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When I go ahead and apply the L² operator to this function, I actually end up getting the same thing as I did before.
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It is going to end up being L ̅² operating on S1, -1.
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It is actually going to end up giving me 2 H ̅ S1, -1.
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It is going to be the same as for the S1, 1.
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S1, -1 is an Eigen function of the square of the angular momentum operator.
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We already saw in the previous lesson, we already saw the general result in a previous lesson.
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L ̅² of SLM is equal to H ̅² × L × L + 1 × SLM.
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Here is the Eigen value and the spherical harmonic is an Eigen function of the square of the angular momentum operator.
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Of course, the magnitude of the angular momentum vector is going to equal H ̅ × L × L + 1.
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That is the general result.
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Let us go ahead and go to example problem 2.
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For the Legendre polynomials, P1 of X, P2 of X, and P3 of X, they are orthogonal.
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We look them up either online or in our books, these are listed at least for the first few, for the first 4 and 5.
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P1 of X is just equal to X, P2 of X is equal to ½ 3X² -1 and P3 of X is equal to ½ 5X³ -3X.
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Recall that X is equal to cos Θ, a little change of variable.
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And that θ ran from 0 to π.
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Cos of 0 is equal to 1, cos π is equal to -1.
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Therefore, the value of X actually goes from -1 all the way to 1.
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This is change of variable, when we put X in here, we actually get the actual function of θ.
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When we do the change of variable and we call the cos θ X, just to make it look a little simpler, that is all.
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Let us see what we can do for P1 and P2.
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We have the P1 P2, P1 P3, and P2 P3.
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We are going to check that they are orthogonal.
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In other words, what we are going to check is the following.
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We want to check to see that this integral P1 P2, we want that to equal 0, that is what we want.
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-1 to 1, P1 is X, X conjugate is just X.
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It is a real number so it is just going to be X × ½ 3X² -1 DX.
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You can solve the integral directly if you want, it is not a problem.
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Just go ahead and multiply out, integrate it, just like you would in any other function.
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You can do this one by hand very quickly.
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Integral directly or just like the Hermite polynomials of the harmonic oscillator.
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We will try to make our lives a little bit easier if we can by dealing with integrals
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that we are going to see over and over again.
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Instead of just actually integrating over and over again.
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P sub L, the Legendre polynomial sub L is even when L is even.
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It is an even function.
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It is even when L is even and it is odd when L is odd.
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This integral -1 to 1 of this thing X × ½ 3X² -1 DX, it is actually equivalent to the integral for -1 to 1, an odd function.
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X is an odd function × an even function, this ½ 3X² -1 is actually an even function.
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We know from previous work that an odd function × an even function is an odd function.
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The whole thing when we multiply that out, it is actually going to be an odd function.
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The integral of an odd function / a symmetric interval, symmetric about 0 -1, -π, something like that is going to equal 0.
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The integral of an odd function/ a symmetric interval, in this case from -1 to 1.
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Symmetric interval is 0.
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The integral from -1 to 1 of P1* P2 is equal to 0.
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P1 and P2 are orthogonal.
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Let us do P1 and P3.
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For P1 and P3, we are going to have the integral from -1 to 1.
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P1 conjugate P3, that is going to equal the integral from -1 to 1 of X × ½.
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This is going to be 5 X³ -3 X DX.
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X is an odd function and this one also happens to be an odd function.
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An odd × an odd function is actually equal to an even function.
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This one we have to integrate directly.
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We integrate directly and it is not a problem at all.
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This is going to equal 1/2 integral from -1 to 1.
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Let us go ahead and pull the ½ out, distribute the X.
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We have 5 X⁴ -3 X² DX = ½.
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Of course, this one is going to be X⁵ – X³ from -1 to 1 = ½ is going to be 1 -1 - -1.
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--1 that is going to be the hard part to keep track of all of these.
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It is going to end up being equal to 00, it is going to end up being 1/2 is 0 = 0.
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These are orthogonal.
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For our last one, for P2 and P3.
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P2 is an even function, P3 is an odd function.
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We know that they are even × an odd is going to equal an odd function.
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We know that the integral / symmetric interval of an odd function is equal to 0.
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These are also orthogonal.
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There we go, and of course this is true for all the Legendre Polynomials.
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They are all going to be pair wise orthogonal.
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Example 3, for the Legendre polynomials P2 and P3 show that they satisfy the general normalization condition
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for the Legendre polynomials which is the integral of its² DX is equal to 2/ 2L + 1.
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This is the general normalization condition for the Legendre polynomials.
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This is how we get R, if we needed to, to get our particular constants that make the integral equal to 1.
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We just need to put in a P2 and put in a P3 to show that it actually satisfies this.
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Let us see what we have got.
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P2 is equal to ½ 3X² -1.
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In this particular case, L is equal to 2.
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When we do the integral, we should get 2/ 2 × 2 + 1.
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L is 2 so we should get 2/5.
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That is our target, let us go ahead and see if that is true.
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P₂² that is going to give us ¼ × 9 X² , we should do our algebra correctly, -6X² + 1.
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We have ¼ of the integral from -1 to 1 of 9X⁴ -6 X² + 1 DX.
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That is going to equal ¼ 9X⁵/ 5 – 2 X³ + X from -1 to 1.
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This is going to equal ¼, that is what it is going to give you for actually going through all the tedium of these integrations.
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I would not do to all the time but at least for now, I think it is fine.
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We are going to have 9/5 -2 + 1, a – and --9/5 + 2 -1 = 1/4 4/5 –and - 4/5 is going to equal 2/5, when you do the arithmetic.
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Yes, it satisfied that relation for P2.
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Let us go ahead and see if it actually works for P3.
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P3 of X is equal to ½ 5X³ -3X, in this particular case L is equal to 3 so we should get the 2/2 × 3 + 1.
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We should get our final answer of 2/7 when we do the integral.
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P3² is going to equal ¼ and is going to be 25 X⁶ -30 X⁴ + 9 X².
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We have a 1/4 of integral from -1 to 1 of this thing which is 25 X⁶ -30 X⁴ + 9 X² DX.
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This is going to end up equaling, let me go ahead and put is here.
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The ¼ and it is going to be 25 X⁷/ 7 -6 X⁵ + 3X³, all the way from -1 to 1.
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This is going to equal ¼ 25/7 -6 + 3 -8 -25/ 7 + 6 -3.
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It is going to be ¼ × 4/7 -and -4/7.
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When you do the arithmetic, it is going to equal 2/7.
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Yes, it also worked.
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That is your general formula.
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Let us see what we have got next.
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These are the Legendre polynomials, add them aside.
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Your book is going to list the first 5 or 6 Legendre polynomials.
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But I'm guessing that your book probably does not have a general formula for generating the Legendre polynomials.
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They are actually several them available, I'm going to go ahead and give you the one that is the most easiest.
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As inside, there are several formulas for the nth degree Legendre polynomial P sub N of X.
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The easiest is probably Rodriguez’s formula.
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It is the nth degree Legendre polynomial is equal to 1/ N! 2 ⁺nth power.
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The nth derivative, let me make my L a little bit clearer here.
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The nth derivative of X² -1 ⁺N.
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If you actually want to generate a particular polynomial, let us say the 15th degree Legendre polynomial, this is the formula that you would use.
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Let us see what we have got.
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Example number 4, the general formula for generating the associated Legendre function.
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This is not the Legendre polynomial, this is the associated Legendre function.
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It actually has 2 indices, a subscript and a superscript is this right here.
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Use this relation to generate P2, 0, P2, 1, and P2, 2.
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Express in terms of θ not X.
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Recall the X is equal to cos θ, when you find the function of X, you are just going to put sin θ for that and simplify if you can.
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I will use the basic trigonometric identities that you remember from pre calculus.
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Let us see what we have here.
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P2 of 0, we are just going to put them into this.
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That is it, that is all we are going to do.
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It = 1 - X², this is L the subscript, the superscript is M.
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This is L = 2, N = 0, so we put it up here and it is going to be 0/ 2 D0 DX is 0 of P2.
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In this particular case, I can go to notational state here.
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L is the associated Legendre function.
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This should be M, there we go.
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In this case D00 P2, this just is 1 and this 0 or the derivative is just P2.
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It = ½ 3X² -1.
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It is LL P2.
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X = cos θ, therefore, our P2 0 as a function of θ is equal to ½ × 3 cos² θ – 1.
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In other words, we are generating the full function, the associated Legendre function from the Legendre polynomial.
00:35:25.400 --> 00:35:28.200
The one with just the subscript is the Legendre polynomial.
00:35:28.200 --> 00:35:35.400
The one with subscript and superscript is the associated Legendre function.
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Let us go ahead and do P2, 1.
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P2, 1 that is going to equal 1 - X² ^½ D1 D X1 of P sub L P₂ which is ½ 3 X² -1.
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That is going to equal 1 -X² ^½.
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When I take the derivative of this, it is going to equal 6 X/ 2.
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It is just going to equal 3 X so I end up getting 3 X × 1 -X²¹/2.
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I have X equal cos θ, this is my function associated Legendre function in X.
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X = cos θ, remember we want to express it in terms of θ.
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We are going to have 3 × cos θ × 1 - cos² θ ^½.
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You remember basic trigonometry identity 1 – cos² θ is equal to sin² θ.
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We have 3 cos θ time sin² θ ^½.
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We end up with 3 cos θ sin θ, there you go.
00:37:12.700 --> 00:37:30.200
This is the associated Legendre function P2, 1 expressed as a function of θ.
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Note, L = 2 which mean that N is equal to -2 -1, 0, 1, and 2.
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We have taken care of this one, the 0 one.
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I will be taking care of the 1.
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The formula actually has the absolute value of M.
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There is also a P2 absolute value of -1 but the absolute value of -1 is 1.
00:38:14.000 --> 00:38:18.500
It is actually the same as P2, 1 so there is another one.
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There is a P2 absolute value of -1 which happens to be the same as this.
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We are going to end up with three of them but we actually have 5 of them.
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One for each value of M because M runs from -L all the way to +L, in increments of 1 passing through 0.
00:38:37.800 --> 00:39:03.300
There is also at which is the same as P2, 1 because the absolute value of -1 is equal to 1.
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Let us go ahead and do P2, 2 here.
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We have P2, 2, it is going to be the formula 1 - X²²/ 2.
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We have D2² DX² of ½ 3X² -1.
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We are going to differentiate this twice.
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Let us go ahead and just do this part.
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We have ½ 3 X² -1.
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Now, we are going to differentiate it the first time and we would end up with 3X.
00:39:53.100 --> 00:40:00.400
We take the second derivative, we are going to end up with 3.
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Our P2, 2 is going to equal 1 - X²¹ × 3.
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It is going to equal three × 1 - cos² θ.
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1 – cos² is sin² θ.
00:40:20.600 --> 00:40:24.900
We have three sin² θ.
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This is our P2, 2 expressed as a function of θ.
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And we also have a P2 absolute value of -2, that is going to be the same.
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It is going to be 3 sin² θ.
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Recall what these are, this is going to be the important part here.
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Recall what these associated Legendre functions are, in case you have forgotten.
00:41:20.200 --> 00:41:37.500
They are the T θ portion of the spherical harmonics.
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The angular portion of the wave function for the hydrogen atom.
00:41:45.500 --> 00:41:51.000
It is also the wave function of the rigid rotator.
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They are the T θ portion of the spherical harmonics.
00:41:57.700 --> 00:42:00.000
Let me go ahead and do one more page here.
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Basically, what we have is we have this ψ which is a function of R θ and φ.
00:42:07.200 --> 00:42:12.800
As a function of three variables, spherical coordinates, the radius changes, the φ changes, the θ changes.
00:42:12.800 --> 00:42:19.300
That is how we get the wave function for the electron of the hydrogen atom.
00:42:19.300 --> 00:42:32.800
We separated that into a radial portion R N sub L which is just a function of R and S sub LM which was a function of θ and φ.
00:42:32.800 --> 00:42:36.800
These are the spherical harmonic, the radial function, the angular function.
00:42:36.800 --> 00:42:39.500
The angular function we call the spherical harmonics.
00:42:39.500 --> 00:42:43.300
When we are solving this equation, this is a function of 2 variables θ and φ.
00:42:43.300 --> 00:42:46.600
We broke out that up into 2.
00:42:46.600 --> 00:42:50.600
S L sub M of θ φ.
00:42:50.600 --> 00:42:59.600
We brought that up into a function T of θ and a function F of φ.
00:42:59.600 --> 00:43:12.600
This right here, this T of θ, those are the associated Legendre functions.
00:43:12.600 --> 00:43:15.100
A lots of functions flowing around now.
00:43:15.100 --> 00:43:19.600
It is very easy to get them all confused.
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Hermite polynomials, Legendre polynomials, Legendre functions, Laguerre polynomials,
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radial functions, having keep them all straight, this is what we did.
00:43:29.900 --> 00:43:33.400
We broken it up into 2 and we broke this up into further down.
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Right now, we are dealing with the associated Legendre functions.
00:43:37.100 --> 00:43:42.900
These are the associated Legendre functions.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time for a continuation of example problems.
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Take care, bye.