WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com.
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Welcome back to Physical Chemistry.
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Today, we are going to continue our discussion of the hydrogen atom.
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Let us jump right on in.
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Let me go ahead and take a blue here.
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In the last lesson, we found the Legendre polynomials, those P sub L of X where M is equal to 0.
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Remember, we were solving for that function of θ and we have found these Legendre polynomial
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and we decided that was only for the value of N = 0.
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Now, we are going to look at the polynomials 4M not equal to 0, + 1 -1, + 2 -2, and so on.
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These are called the associated Legendre functions.
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They are defined in terms of the Legendre polynomials that we saw, the P sub L of X.
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They are defined terms of P sub L of X which are the actual Legendre polynomial.
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Here is what they look like.
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The symbol for P sub L absolute value of N of X, this is the symbolism.
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It is equal to 1 - X² × V M derivative D absolute value of M DX, absolute value of M.
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DDX will be the first derivative, that would be the first associated polynomial for M = 1.
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If it was M was or equal to 2, it would be second derivative D² DX².
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If M was = 3, it will be D3/ DX3, the third derivative, and so on.
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That is what this M means, let me make this one little bit more clear, the P sub L of X.
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All this is saying is that, for example, if you had L = 3, you would pick the P sub L of 3.
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You would take the third meal, this would be P3.
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Remember what we said, we said that L is going to equal 0, 1, 2, 3, and so on.
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In the case of L = 3, we said that M takes on the value 0, + or -1, + or -2, and + or - all the way to L, so + or – 3.
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In this case, we have 1, 2, 3, 4, 5, 6, 7, different values of M.
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For those different values of M, we take various derivatives and we multiply by 1 - X², in order to get this function.
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That is all we are doing here.
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For every Legendre polynomial, we are going to end up with multiple associated Legendre functions.
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That is was happening, when M does not equal 0.
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Let me write out the specifications here for M a little bit more clearly.
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This is for the value of L is going to be 0, 1, 2, and so on.
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The value of M is going to equal 0, + or -1, + or – 2, all the way to + or – L, whatever that happens to be.
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Because the leading term of P sub L of X because the leading term of the actually Legendre polynomial.
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A polynomial has just a subscript.
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The associated function has the subscript of the function comes from and it has a different value of M for the function itself.
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Polynomial has just a subscript.
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The associated Legendre function has both the sub and a superscript.
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The subscript being L, basic polynomial that all of those associated functions come from.
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Because the leading term of the polynomial P sub L of X is X ⁺L, for example the P₂ was X².
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The P₃ had an X³, the P₄ had an X⁴, that is all of these means.
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P sub L absolute value of M of X becomes 0.
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0 is the absolute value of M is greater than L.
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All that means is that, let us say we take L = 3, if L = 3 the leading term of the polynomial is going to be X³.
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The coefficient does not matter, it is going to be some X³.
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When I take the third derivative of that, the coefficients is going to be some constant.
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If I take the fourth derivative, it is going to be the derivative of a constant which is going to be 0.
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Whenever absolute value of M is bigger than L, these just go to 0.
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It is just a property of the polynomial itself.
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We decided to just throw that in there.
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Let us go ahead and list the first few Legendre associated functions.
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I know this is I want to take in, do not worry about that.
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That is the nature of the beast, we have to run for the material.
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All of this is going to make more sense when we do the problems, when we see over and over and over again.
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It will start to make sense.
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The nature of the beast is such that we have to present the material in such a way.
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We have to present it in a rather large amount, before we can actually work problems that make sense.
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The P₀ is 0 of X is equal to 1.
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Here L = 0 and M is going to be 0 but you can only go as high as L.
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There is only one of those.
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We have P₁ is 0 of X is going to equal cos θ.
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I will do both, it is equal to X which is equal to cos of θ.
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Because, again X is equal to cos of θ, X is not the coordinate.
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It is the change of variable.
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Here, L is 1 and M is 0.
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P of 1, now N is going to go from 0 and is also do 1 and -1.
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P1 1 is equal to 1 - X² ^ ½ and that is going to end up equaling sin θ.
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I'm just putting all the values into the definition from the previous slide.
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Let me write it now.
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Let me do it in red.
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There is also a P₁, M is 0, + or -1, + or -2, + or -3, all the way to + or – L.
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In this case L =1.
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We have done the one M = 0.
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You have done the M, and we have done the one M =1.
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There is also a 1 -1, we will go ahead and put in absolutes.
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That ends up because this is an absolute value sign, the absolute value of -1 is just 1.
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It actually ends up being the same as this.
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For the L = 1, we actually have 3 associated Legendre functions.
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We have this one and we have this one, and we have the one for 1 -1,
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which is actually the same as this because absolute value of 1 is 1 so it ends up being the same as this function.
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That is also 1 - X² ^½ which is equal to sin θ.
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Let me go back to blue here.
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P2 0 is going to equal ½ 3X² -1 = ½ 3 cos² θ -1.
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We have P2 1 that is going to equal 3X × 1 - X² ^½.
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It is going to be 3 cos θ sin θ, when we actually do the change of variable and wherever we see X we put in a cos θ.
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There is also going to be a P2 2, write 2 0, 2 1, 2 2.
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There is also going to be at 2 -1, there is going to be a 2 -2, they happen to be same as these.
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It was a total of 5 of these, = 3 × 1 –X² = 3 sin² θ.
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For L = 0, we have P0 0.
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Let me go ahead and write this up here.
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Notice again, X is not the variable.
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Θ is the variable of interest.
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We just had under change of variable X = cos of θ.
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This is very important to remember.
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Let us go back to blue.
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For the L = 1, we have P1 of 0, we have P1 of 1, and we have P1 of -1.
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The definition uses the absolute value sign so we can leave it like this.
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Because this absolute value sign, the P1 -1 actually ends up being the same function as P1 1.
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It is important to remember that there are 3 of them because L 0, 1, 2, and so on.
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M takes some of value 0, + or -1, + or – 2, all the way to + or – L.
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M depends on L.
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For L = 2, we have P2 of 0, P2 of 1, we have P2 of -1, we have P2 of 2, and we have P2 of -2.
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This should be starting to look familiar.
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When L = 0, that is the S orbital, M = 0.
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When L = 1, we call that the P orbital, when M = 0, + 1, -1.
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When L = 2, that is the D orbital, M = 0, + 1, -1, + 2, -2, 1.
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PX PY PZ, DXY DXZ, DXY DZ².
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There are 5 D orbitals.
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There are 3 P orbitals.
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That is what is going on here.
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The P orbital will end up being PX PY PZ.
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When L = 2, that is the D orbital.
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M takes on the value 0, + 1 -1, + 2 -2, we end up with D sub XY, D sub YZ, D sub XZ, D sub Z², D sub X² – Y².
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That is what is going on here, this is where all this comes from.
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The normalization condition is the integral from -1 to 1, P sub S absolute value of M conjugate.
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P sub T absolute value of M DX is going to be S 0 to π P sub F of cos θ.
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The value of M conjugate × P sub T absolute value of M cos θ.
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This DX term is not just D θ, it is sin θ D θ.
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It is that because we said X is equal to cos θ, this is the change of variable.
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Therefore, DX is equal to sin θ D θ.
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We do it in increments here.
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DX D θ is equal to DX = sin θ D θ.
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This normalization condition, this thing, when we actually solve it we end up with the following.
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We end up with 2/ 2L + 1 × L + the absolute value of M!/ L - the absolute value of M!.
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Therefore, all of that is going to end up equaling 1.
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The integral of this thing is equal to this thing.
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This normalization condition is equal to 1.
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Therefore, when we actually solve this we end up with the following normalization constant.
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The normalization constant LM is equal to 20 + 1/ 2 × the absolute value of L – that !/ L - the absolute value of M.
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I hope you got them keeping all this straight, this is absolutely crazy, ½.
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That is the normalization constant for the particular associated Legendre polynomial
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P sub L absolute value of M, LM each one has a differential normalization constant based on this.
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It looks very complicated, it is not.
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L and M are really going to be just 1, 2, 3, that is about it 0, 1, 2, 3.
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Let us make sure we know where it is that we are.
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We started all this by looking for the angular component of the hydrogen wave function.
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We were looking for S of θ of φ which is equal to some T of θ and some F of φ.
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We have found this then we found that.
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We put them together now, right.
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We found F of φ, we found T of θ, now we are going to multiply them because S of θ φ is equal to T of θ × F of φ.
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So now we are going to write out the entire wave function symbolized.
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The angular component L M of θ φ is going to equal 2L + 1.
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It is going to look a lot more complicated than actually is in practice.
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× L - the absolute value of M!/ the absolute value of L + the absolute value of M! ½.
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The takes care of the normalization constant for that.
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Now, we are going to multiply by the normalization constant for the F of φ which is 1/ 2 π ^½ × P sub L absolute value of M cos θ × E ⁺IM φ.
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Where, L takes on the values 0, 1, 2, and M takes on the values 0, + or -1, + or – 2, all the way to + or – L.
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This is what we wanted, we wanted the angular component of the hydrogen atom wave function is this.
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That is the normalized wave function for the angular component of the hydrogen atomic orbital.
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This is what it looks like.
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Let us go ahead and combine the two normalization constants.
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Make it a little bit easier to look at here.
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We have S sub L M of θ of φ is going to equal 2 O + 1, when we combine the two constants
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it is going to end up being that, / 4 π × L - absolute of M!/ L + absolute of M! ^½ P sub LM cos θ E ⁺IM φ.
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Where L is 0, 1, 1…
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M = 0, + or -1, + or – 2, all the way to + or – L.
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These are the wave functions for the angular component of
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the hydrogen atom wave function ψ of R θ φ, which is R of R, S of θ φ.
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This is just the part right there.
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They are also the wave functions for the quantum mechanical rigid rotator that we promised earlier, when we are just dealing with the energies.
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These normalized wave functions are here.
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The normalized wave functions a formal orthogonal set.
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They are mutually orthogonal functions form and orthonormal set.
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Let me go back to blue here.
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These functions are called the spherical harmonics.
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The first few spherical harmonics, let us list them out.
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The first few spherical harmonics S0 of 0 is equal to 1/ 4 π ^½, it is a constant.
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We are just putting for every value of L with the value of M and we list them out.
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We just put in the equation.
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The equation looks big but L and M are such tiny numbers but everything ends up condensing becoming really easy to deal with.
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S1 of 0 is equal to 3/ 4 π ^½ × the cos of θ.
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S₁ 1 is equal to 3/ 8 π ^½ sin θ E ⁺I φ.
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S1 -1, 1 0, 1 1, 1 -1, it = 3/ 8 π ^½ sin θ E ⁻I φ, because now N is -1.
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Remember, it is E ⁺IM φ.
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We are going to have E ⁻I φ, that is always going to be like that for the + or -1, + 2 -2.
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Let us do, where are we next?
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We are at S20 that is going to equal 5/ 16 π ^½ × 3 + sin² θ -1.
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Let me write then, I can continue on this page, not a problem.
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We will do S21 is going to equal 15/ 8 π.
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It looks like I forgot about my notes, it has a ½ there.
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We have sin of θ cos θ E ⁺I φ.
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Let us come up here and do S2 -1 is equal 15/8 π.
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Everything else is absolutely the same.
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Sin θ cos θ E ⁻I φ -1 + 1.
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Let us do S22, it = 15/ 32 π, just running for the numbers.
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L0 and M0, L1 M0 1 -1, L2 M0 1 -1 2 – 2.
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That is all that we are doing.
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32 π ^½ this one is going to be sin² θ E ⁺I φ.
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S2 -2 = 15/ 32 π ^½ sin² θ.
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What do you think?
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E ⁻I2 φ, that is all that is happening here.
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Let us talk a little bit about angular momentum.
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When we talk about the rigid rotator, when we discussed the rigid rotator,
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we found that the square of the angular momentum operator
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was actually equal to – H ̅² × 1/ sin θ DD θ sin θ DD θ + 1/ sin² θ D² D φ².
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This is the operator for the square of the angular momentum.
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I do not want to keep repeating myself over and over again so I’m just going to call of that Z.
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We had for S of θ φ, we had -1/ S × 1/ sin θ DD θ sin θ DS D θ + 1/ sin² θ D² S D φ²,
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all of that was equal to θ which is equal to L × L + 1.
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We said θ is equal to L × L + 1.
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If we multiply both sides by H ̅² S of θ φ, we end up with – H ̅² Z is equal to H ̅² × S × L × L + 1.
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We have L ̅² S of θ φ is equal to H ̅² × L × L × L + 1 × S.
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I just rearrange this thing.
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Notice A of ψ = λ of ψ, Eigen value, Eigen function kind of thing.
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We have L ̅² of S is equal to H ̅² × L × L + 1 × S.
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Here is the operator, here is the function.
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It is equal to this scalar × the function which means that the S sub L of M of θ φ, they are Eigen functions.
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In other words, when we apply the angular momentum operator to S, when we apply it to S we end up getting this thing.
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We end up getting some constant × S back.
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This is the condition for Eigen value Eigen function.
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The angular component, the wave functions for the rigid rotator happens to be Eigen functions of DL² operator.
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Let me rewrite this so that we have it on one page.
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L ̅² of S, I will get rid of θ and φ, is equal to H ̅² × L × L + 1 × S.
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The square of the angular momentum, when I actually take measurements
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of the angular momentum because this is an Eigen function of the operator.
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Because when we operate what we actually get the function back × the constant,
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that means when we take measurements, those are the only values that we actually get.
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Square of the angular momentum can only have the values² = H ̅² × L × L + 1.
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Or again L = 0, 1, 2, and so on.
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Those are the only possible values for the square of the angular momentum.
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If I measure the angular momentum, the square of the value that I end up getting
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is going to be one of these values depending on what L is.
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It was the only values that can take.
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The angular momentum is quantized.
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Recall that the Hamiltonian operator is equal to the square of the angular momentum operator
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÷ twice the rotational inertia for the rigid rotator.
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If I apply this, I’m going to do H ̅ of S LM, it is equal to L ̅²/2I S LM is equal to H ̅² × L × L + 1/ 2I S LM.
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This thing should look familiar.
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All I have done is, I already had this L² of S = something, = this thing.
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H is just L²/ 2I.
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I took this thing and just divide it by 2I.
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They should be familiar.
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These are just the energies of the quantum mechanical rigid rotator.
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Remember, the Hamiltonian is just the energy function.
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It is the energy operator, the total energy operator.
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They should look familiar.
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They are the energies of the rigid rotator.
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Back when we discuss the rigid rotator, we used J E sub J = H ̅²/ 2I J × J + 1.
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Now, we are replacing J with this L that actually shows up.
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That is all that is going on here.
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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 the hydrogen atom, bye.