WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and 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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Make sure we are on this page.
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I’m going to go to blue.
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In the last lesson, we finished off by a little discussion of the angular momentum then
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we talked about how the angular component of the hydrogen wave function the S of θ φ
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is actually an Eigen function of the square of the angular momentum operator.
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We would like to continue along that track today.
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Angular momentum is a vector quantity, in other words there is some angular momentum vector and
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it is going to have three components.
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It is going to have an X component, Y component, and a Z component.
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It is going to be L sub X in the I direction, X direction + L sub Y in the J direction, the Y direction and L sub Z in the K direction.
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This is just the standard unit vector representation of the angular moment vector.
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By the operators corresponding to the three components, the LX, LY, and LZ, we saw them before.
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The operator is corresponding to the three components of L.
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The three components of the angular momentum operator, we are talking about R.
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We had L sub X is going to equal, I will write out everything.
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It is Y PZ – Z PY.
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The PY and PZ are the linear momentum components.
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It = - I H ̅ Y DDZ - Z DDY.
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I’m just recalling what these are.
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The Y component is going to end up being Y P sub X - X P sub Z - I H ̅ Z DDX - X DDZ.
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These are the operators, I’m going to list them out.
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Of course, we have the final one which is L sub Z and that is going to equal the X sub Y.
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It is going to be X P sub Y – Y P sub X.
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It is going to be - I H ̅ X DDY - Y DDX.
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I’m sorry, is my order correct?
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These are Cartesian coordinate representations, we are working at spherical coordinates.
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Let us go ahead and see what these look like in spherical coordinates.
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Let me do this in red and hopefully you do not have to actually go through
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the process of converting it because it is a very tedious exercise.
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It is part of your exercises in your book, I know it is but hopefully you do not have to do it yourself.
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In spherical coordinates, it looks like this.
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The X component of the angular momentum is going to equal - I H ̅ that is going to be – sin φ.
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Let me sure I will get all of this correct here.
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DD θ, I’m telling you try to keep track of all the symbols is the real challenge with all the quantum mechanics, - the cot of θ.
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The cos of φ DD φ and the Y component is going to equal - I H ̅ is going to be cos of φ DD θ - cot of θ
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and this is going to be the sin φ and it should be DD φ.
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We have L sub Z, it is going to look very different.
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This one looks really simple.
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It is going to be -I H ̅ DD φ.
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This looks very different from the Y and the X.
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Let us go ahead and apply this last one, the Z component of the angular momentum.
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Let us apply it to our angular component of the rigid rotator function of S of θ φ.
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Let me go back to blue.
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Let us apply L of Z to the S sub LM of θ φ.
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Let us see what we get.
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I will write this one very carefully when I go through it.
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L sub Z of S LM, I wonder if I should use my θ and φ.
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It is going to equal, we have the normalization component LN and then we have the L sub Z of S sub L sub M,
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that is going to equal the associated Legendre function P sub L absolute of M of cos θ × E ⁺IM φ.
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That is going to equal N LM P sub L absolute M of cos θ × –I H ̅ DD φ applied to E ⁺IM φ.
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The L sub Z is you are taking the partial derivative with respect to φ.
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There is no θ involved so I can just go ahead and skip this all together.
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It is like a constant, I just apply it to the function that has a φ in it which is the exponential portion of it.
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We end up with N LM P sub L absolute of M of cos of θ × IM × – I H ̅ × E ⁺IM φ.
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When I actually differentiate DD φ with respect to this, this IM comes here.
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This is going to equal, I'm going to bring this so I and I is -1, negative and negative is positive.
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I’m going to bring this H and this M out front so I get H ̅ × M × N sub LM P sub L absolute M cos θ × E ⁺IM φ.
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This is equal to, this part right here, this is just S LM of θ φ.
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That is just S, it is the same thing.
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Now, I have an operator, something operating on a function gives me the function back × a constant.
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That is equal to H ̅ M × S sub LM of θ φ.
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Our spherical harmonics, S sub L M of θ φ are Eigen functions of the Z component angular momentum operator with Eigen value H ̅ M.
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Let us go back to blue.
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In other words, when something is an Eigen function, when some function is an Eigen function of an operator.
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In other words, when we measure values for the Z component of the angular momentum we get integral multiples of H ̅.
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H ̅ is a basic unit.
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Notice H ̅ is a basic unit of measuring angular momentum for a quantum mechanical system.
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H ̅ is a basic unit of measure for angular momentum in the quantum mechanical system.
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Spherical harmonics are Eigen functions of the square of the angular momentum operator L².
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They are also Eigen functions of the Z component of the angular momentum operator.
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The spherical harmonics S sub LM are not Eigen functions of the X and Y components of the angular momentum operators.
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They are not Eigen functions of L sub Y and L sub X.
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The spherical harmonics of sub LM are Eigen functions of L² and L sub Z.
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Because the S sub LM are Eigen functions of both L² and LZ, these two operators commute.
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These two operators, they commute.
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Remember, when we said when we talked about operators that commute.
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For two operators that do commute, their commutator = 0.
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It think we can apply the operators on either order.
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When that is the case, that means that we can measure each one to any degree of precision that we want.
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When operators do not commute, like position and linear momentum,
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that is when Heisenberg uncertainty principle comes into play.
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It says that if you more accurately measure one, the less accurately you know the other.
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There is a certain compromise that I have to come to.
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But for operators that do commute, I can measure them both to any degree of accuracy that I want.
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In this case, I can measure the square of the angular momentum or the angular momentum and
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I can measure the Z component of the angular momentum to any degree of accuracy I want
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but I cannot say anything about the X and Y components.
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I can tell you in which direction the X and Y components are.
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I can tell you in which direction the actual angular momentum vector is.
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I can tell you the direction of the Z component but I cannot tell you where X and Y are pointed.
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That is what this is saying.
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Because they are Eigen functions of both L² and L sub Z, these two operators commute.
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This means that we can measure values for L² and LZ simultaneously to an arbitrary degree of precision or accuracy.
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I’m a little loose with those words but it is not going to hurt us here.
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I was going to actually prove this as an exercise but it will be the extra mathematics here that is not altogether necessary.
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We may instead be doing it in the part of the problem sets, if I decide to do that.
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For now, I think we will actually save ourselves a little bit of time and little bit of mathematical grief
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and we will just take it on fate if this is the case.
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The angular momentum operator and the Z component of the angular momentum operator, they commute.
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Let us go ahead and move forward here.
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We already know that L takes on the values 0, 1, 2, . . . And M
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which is dependent on L takes on the values of 0, + or -1, + or -2, all the way to + or –L.
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Those are the highest values.
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For every value of L, there are 2L + 1 values of M.
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There are 2L + 1 values of M for each value of L.
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Let us recall what we have here.
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We have L ̅² of S L sub M.
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I going to skip this and say of φ.
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We said that was equal to H ̅² × L × L + 1 × S LM.
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This just is a repetition of the fact that the spherical harmonics are Eigen functions of the angular momentum square operator.
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The square of the angular momentum operator.
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This is exhausting.
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We also have that, they are Eigen functions of the Z component so L sub Z applied to S LM is that is equal to, we found H ̅ M S LM.
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If we choose the state L = 1, we are going to end up with the following.
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We are going to get L ̅² S1 M is equal to H ̅² × 1 × 1 + 1 S1 M.
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We are going to end up with L sub Z S1 M is equal to H ̅ M S1 M.
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Let me erase this and let me we just put and here.
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For M = 0, + 1 -1.
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L is 1 so M is going to be 0 + 1 - 1.
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Over here, because this is the case when we measure the L²,
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we are going to end up with these values because these are Eigen values for the Eigen function.
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This is just 1 + or -2, we are going to end up with L² as a value for the angular momentum.
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Now, it is going to equal to H ̅² which means that the magnitude of the angular momentum is actually going to be H ̅ × √ 2.
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That is the magnitude of the angular momentum that I measure.
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Over on this side, I have L sub Z.
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When I measure these, these are the Eigen values so those are the values that I get.
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L sub Z is going to equal H ̅ M.
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Well, M is equal to 0 + 1 -1.
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Therefore, LZ is going to take on the values –H ̅ 0 and +H ̅.
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I have to put the positive there.
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My angular momentum for the state L = 1, my angular momentum itself is going to have a value of √ 2 × H ̅.
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The Z component of the angular momentum is going to have the value of H or 0 or –H ̅.
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H ̅ is 0 or – H ̅.
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The absolute value of L is the magnitude of the angular momentum vector.
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In other words, this electron is flying around this proton.
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It was moving a circular orbit.
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It is going in and out, there is a circular component to it.
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Anything that moves in a curve around a fix center is going to have an angular momentum.
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This actually gives me the magnitude of the angular momentum of electron for wherever it is.
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That is what is happening here, that is what we are doing.
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L is the magnitude of the angular moment vector.
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Notice that the magnitude of L is not equal to the magnitude of LZ.
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The magnitude of L is H ̅ × √ 2.
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The magnitude of LZ is just H ̅, they are not the same.
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The angular momentum vector and the Z component of the angular momentum vector, they do not point to the same direction.
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Otherwise, they would have the same value.
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They do not point to the same direction.
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We have the magnitude to the angular momentum vector.
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We have the magnitude and the direction of the Z component of the angular momentum,
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it is moving in the Z direction and has certain magnitude, in this case H ̅.
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In order to specify the direction of the angular momentum vector L itself,
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we need to know which direction X and Y are pointed at also.
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We do not know that, all we know is the Z component.
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We have the magnitude of the total angular momentum but we do not know its direction.
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In order to know the direction of L, we must also be able to specify LX and LY.
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We need to know which direction they are in to but this is not possible.
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Let me erase these.
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We know the directions, the direction of X and the direction of Y is the direction of Y,
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we do not know their values but this is not possible.
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This commutes with this, so those commute.
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The square of the angular momentum operator commutes with LZ.
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It will commute with LY.
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It will commute with LX.
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One of them, it does not commute with all of them simultaneously.
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LZ or LY, in other words there is nothing particular special about Z.
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It just depends on how we fit for coordinate system.
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The Z ended up being the one that ended up being the easiest to deal with.
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The truth is, L² commutes with LZ or it commutes with LY, or it commutes with LX.
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But the issue is the LX, LY, and LZ, do not commute among each other.
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What that means is if I can specify the Z component of the angular momentum, I cannot say anything about the X or the Y.
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If I specify the Y component of the angular momentum, I cannot say anything precisely about the Z or the X.
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That is what is going on here, I cannot specify all of them simultaneously.
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I have to choose one of them but only one, that is what this means by commute among each other.
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We get the following picture.
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The picture that we end up getting looks like this.
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For a particular point in space, this is going to be the origin.
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Here is what is going to happen, this is the Z axis, this is the X axis, and this is the Y axis.
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I’m going to go ahead and draw my angular momentum vector.
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Let me go ahead and draw this, let me make a circle here.
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Let me work in red.
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This vector right here, that is the Z component.
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This is L sub Z that is equal to H ̅.
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This right here, this is equal to H ̅ √ 2.
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This is equal to the actual magnitude of the angular momentum.
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This right here is the magnitude of the angular momentum.
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We know the Z component, we know the actual angular momentum, the problem is we do not know X or Y.
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We do not know how much and what direction.
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What ends up happening is this is the angular momentum vector, this has to be this thing right here.
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We know the Z component straight up, the problem is we do not know about the X or the Y.
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What ends up happening is anything is possible, any direction is possible.
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This angular momentum vector actually ends up recessing, going in a circle around the Z axis
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and actually ends up sweeping out of cone.
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It is actually going to sweep out this cone, that is what is happening.
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We do not know about X and Y, all we know is Z.
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And because we do not, anything is possible in the X or Y direction.
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We end up getting this angular momentum vector, the actual L sweeping out this cone.
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I hope that makes sense.
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Let us go ahead and write this down.
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Because we cannot specify LX or LY, this is the magnitude, the vector L, this thing right here, it sweeps out the cone.
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This is a classical interpretation of what it is that actually going on.
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We are just interpreting it in terms of what it is we know from classical mechanics.
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The angular momentum be a vector value thing.
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This is a classical interpretation because we need to make sense of it somehow for all purposes.
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We need to know what is it that we are looking at.
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With respect to something that we know in classical mechanics angular momentum, we can go ahead and talk about what is happening.
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We can go ahead and assign values to different vectors, that is what we are doing.
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This is a classical interpretation of what is happening.
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The angular momentum vector processes around the Z axis to sweep out a cone.
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LX and LY do not have values that can be specified precisely because they cannot commute with LZ.
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Because they do not commute, there is no way of actually specifying the LX and LY,
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or LX and LY precisely because of the uncertainty principle.
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LX do not have values that can be specified precisely but they have the average values.
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We can always find average values.
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If we cannot specify precisely what value is, we can always get an average value.
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Let us stop for a second and talk about what we mean by precise and average value.
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We said that these spherical harmonics are Eigen functions of the square of the angular momentum operator
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and they are also Eigen functions of the Z component of the angular momentum operator.
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That means any time that we measure the Z component, we are going to end up getting the same value over and over and over again.
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It is either H + or H-, they are very specific.
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When we take measurements of the LX and LY, we get a whole bunch of different values all over the place.
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That is what it means for them not to be Eigen functions, they are going to be different values.
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They are not going to be specified as this or this precisely.
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There is going to be lots of deviation.
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There is going to be a standard deviation for the X component and Y component.
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On average, when I take all of those values that 200, 500, 1000 values that measure,
00:31:49.800 --> 00:31:54.200
I averaged them out, I can find an average value for it.
00:31:54.200 --> 00:31:55.900
But they have average values.
00:31:55.900 --> 00:31:57.800
The average values are as follows.
00:31:57.800 --> 00:32:02.600
The average value for the X component of the angular momentum is going to be 0.
00:32:02.600 --> 00:32:07.900
And the average value for Y component is also going to be 0.
00:32:07.900 --> 00:32:10.400
The reason is this is the following.
00:32:10.400 --> 00:32:12.000
I will draw it out in just a minute.
00:32:12.000 --> 00:32:26.000
In the picture above, we do not know in which direction the LX and LY are going.
00:32:26.000 --> 00:32:29.300
Because we do not know which direction they are going, they can go in any direction.
00:32:29.300 --> 00:32:36.300
Because they can go in any direction, this way and that way, on average everything cancels out to 0.
00:32:36.300 --> 00:33:01.200
We do not know in which direction the LX and LY point because we know that
00:33:01.200 --> 00:33:04.500
the unit vectors has to point to the X and the Y direction but they are going to be certain magnitudes.
00:33:04.500 --> 00:33:11.500
Because of that, the vector can be this way or this way or this way or this way.
00:33:11.500 --> 00:33:13.400
I will draw it out in just a minute here.
00:33:13.400 --> 00:33:16.000
The picture that we get is the following.
00:33:16.000 --> 00:33:32.200
If I take this thing, if I turn around and look at it from the top, looking down along the Z axis on the XY plane, here is a picture that I get.
00:33:32.200 --> 00:33:38.100
This is the X axis and this is the Y axis, the Z axis is right here, it is coming straight up.
00:33:38.100 --> 00:33:44.000
If I'm looking down on this, the vector this is the Z axis.
00:33:44.000 --> 00:33:47.600
This angular momentum vector is sweeping out a cone.
00:33:47.600 --> 00:33:50.100
I'm looking at it from up above.
00:33:50.100 --> 00:33:54.300
From your perspective, what is happening is this thing.
00:33:54.300 --> 00:33:56.300
It is actually sweeping a circle.
00:33:56.300 --> 00:34:02.400
Looking at it from above, the vector actually sweeps out a circle as it swings this way,
00:34:02.400 --> 00:34:20.300
the projection of the precessing motion of the angular momentum vector.
00:34:20.300 --> 00:34:36.600
This vector right here looking at it from above, the projection of a precise motion of that along to the XY plane is a circle.
00:34:36.600 --> 00:34:47.200
Now, I cannot specify the X and Y components which means that the X and Y vector.
00:34:47.200 --> 00:34:52.200
Let me write what we have.
00:34:52.200 --> 00:35:01.200
L we said is equal to LX I + LY J + LZ K.
00:35:01.200 --> 00:35:09.600
I’m able to specify this, I can specify this and this so that means it can be this vector or this vector.
00:35:09.600 --> 00:35:11.800
It could be any of the vectors I do not know.
00:35:11.800 --> 00:35:15.500
Because it can be any of them, if I take a bunch of measurements,
00:35:15.500 --> 00:35:19.100
I'm actually going to end up with every single measurement possible.
00:35:19.100 --> 00:35:25.300
On average, it is going to end up being 0 because for every one of these, I'm going to have one of these.
00:35:25.300 --> 00:35:27.500
For every one of these, I'm going to have one of these.
00:35:27.500 --> 00:35:33.400
It is all going to average out to 0, that is why the average values.
00:35:33.400 --> 00:36:06.400
At any given instant, L sub XI + L sub YJ, let me make sure I got B, any vector along the circle on average,
00:36:06.400 --> 00:36:14.500
you are going to end up with 0 LY = 0.
00:36:14.500 --> 00:36:15.600
I hope that makes sense.
00:36:15.600 --> 00:36:17.600
Thank you for joining us here at www.educator.com.
00:36:17.600 --> 00:36:18.000
We will see you next time, bye.