WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to continue our discussion of the rigid rotator.
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Let us jump right on in.
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In the last lesson, we said that because there is no potential energy term V of X,
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we said that the hamiltonian operator is just equal to the kinetic energy operator
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which is equal to -H ̅² / twice the reduced mass × the del² operator.
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We know or we have seen that the del² is equal to D² DX² + D² DY² + D² DZ²,
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as an operator in Cartesian coordinates.
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For an object rotating in space, Cartesian coordinates are not exactly the most,
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We can certainly use it, it just that one thing is rotating in a circle, polar coordinates,
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spherical coordinates, tend to be a more natural choice.
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They actually make the math easier which is why we actually have these various coordinate systems.
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Certain problems just by the virtue of their particular symmetry lend themselves to a particular coordinate system.
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However, for an object rotating about a fixed center in any spatial orientation, and when I say spatial orientation,
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I mean it can rotate like that or like this or like this.
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The molecule itself can be oriented in any way and can rotate however it wants.
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Rotating in any special orientation, spherical coordinates are more natural choice.
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This is going to take a couple of minutes to review, spherical coordinates and what they are.
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I’m not going to say too much about them.
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A point in 2 space requires 2 coordinates to represent.
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A point in 3 space requires 3 coordinates.
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A point in N space requires N coordinates.
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Whatever the dimension of the space is, you need at least that many numbers to represent that point.
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A point in 3 space requires 3 coordinates to describe its position.
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We know in Cartesian coordinates we use X, Y, and Z.
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In spherical coordinates, which is the 3 dimensional version of polar coordinates,
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the spherical coordinates we are going to use R θ and ϕ.
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Let us talk about what R θ and ϕ actually are.
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I will draw on the next page, that is not a problem.
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Let us go ahead and draw ourselves a little coordinates system here.
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A 3 dimensional coordinates system is something like this.
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We have our X axis, we have our Y axis, and we have our Z axis.
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I’m going to pick some random point in the first octet and this is going to be the vector.
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This is going to be our point R θ and ϕ, in terms of R θ and ϕ.
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Let me go ahead and draw a couple of things.
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We dropped a perpendicular projection down to the XY plane.
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That is going to give me this thing right here from there and that is it.
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This is the point.
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Here is where the R θ ϕ comes in.
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R is the length of the actual vector.
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It is a length from the origin to the point X, Y, Z.
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In this case, it is represented by R θ and ϕ.
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R θ is going to be this angle right here.
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That is angle of θ.
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That is the angle that makes, the R makes with the +Z axis.
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Φ is this angle right here.
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It is angle that it makes from the +X axis.
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R is going to be greater than 0, greater than or equal to 0.
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Θ is going to run from 0 π and ϕ is going to be greater than or equal to 0 and less that or equal to 2 π.
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Those are the possible values.
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R is the length of the vector.
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Do not worry if this picture is not making complete sense.
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Right now, I’m going to explain what it means.
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It is the length of the vector’s θ.
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It is the angle that a vector makes with the +Z axis.
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Θ runs from 0 to π.
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Φ is the angle at the projection of R onto the XY plane makes with the +X axis.
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We need a reference point, so we have something that we start with.
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Let me go this way.
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This is the Z axis right, if I move along the Z axis at distance R and this is the Z axis, this is going to be the X axis.
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Why you, yourself, from your perspective you do not see it.
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If I move along the Z axis and I pull away an angle θ from the Z axis and then from here, if I end up rotating this way.
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Now R is here and I’m going to rotate it this way.
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This third angle rotation that is the ϕ.
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R θ and then swing out the ϕ.
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If you move a distance R from the origin along the positive Z axis
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and then run θ from 0 all the way to π,
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then run ϕ from 0 all the way to 2 π, you are going to end up sweeping out the sphere of radius R.
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This is the origin of the name spherical coordinates.
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Let us see what this actually this looks.
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I'm guessing that you guys are familiar with this already but again for those of you that are not,
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we definitely want to make sure that you understand this.
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Let me draw this out.
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What I'm going to draw out here is I'm going to draw just the Z axis and the X axis.
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We said from the origin we want to move at distance R up to Z axis.
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Let us come up here.
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Now the next thing we are going to do is we are going to swing this R, we are going to take θ from 0 to π.
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We are going to swing this 180°.
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I’m going to take this and I'm going to swing it around this way, a full circle 2 π.
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When I do that, when I take this and I sweep it around,
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Remember, rotations in calculus class, I'm going to end up actually sweeping out a sphere.
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What I end up getting is something that looks like this.
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I’m going to start from R, the origin then I go up the Z axis, that is my length R.
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I’m going to end up swinging this down.
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I already used up so much space, it does not have to be so big you will still understand it.
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I'm going to end up swinging it up R and I’m going to swing it down this way.
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I will draw a sphere and make it a lot better.
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This is the center along the Z axis.
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I swing it down this way, this right there is my θ.
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My θ goes from 0 all the way down to π.
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I’m going to take this and I'm going to swing around full circle.
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It is going to go all the way around that is my ϕ.
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I’m going to end up sweeping out a sphere and that is where this comes from.
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We do not want to say any more about that.
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Hopefully that is pretty clear.
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We are going through a lot of these mathematical stuff because it is part and
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parcel of your mathematical education, your scientific literacy.
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You need to see these techniques.
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You do not have to know it, you do not have to do it yourself, you do not have to do the derivations.
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They are so many partial differential equations but you want to be able to see it.
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It is important, we do not want to leave you with a feeling that they just dropped out of the sky.
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At least it would make things possible for you, it is easier to wrap your mind around it.
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That is why we are doing this.
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We do not want to get caught up in the derivation.
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We want to be able to follow derivation.
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We want to spend most of our time thinking about the final equations that we get.
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The energy functions and wave functions, but we do need to go through the derivation.
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It is part of your education.
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In spherical coordinates, our del² is actually equal to this.
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It looks like a very complicated expression and it is, but we can simplify it.
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It is going to be 1/ R² DDR.
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I, myself, do not memorize this.
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That is fine, I will go ahead and move θ and ϕ here anytime we have some partial derivative.
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Out here as a subscript, that means holding this constant while we take the partial derivative.
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It is implicit, it is part of the notation.
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I will just go put it this one time and not going to keep repeating it.
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+1/ R² sin θ × DD θ of sin θ DD θ.
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This is going to be R and ϕ + our final term which is 1/ R² sin² θ and D², this time D ϕ².
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This is going to be RN θ.
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For the rigid rotator, this is where the R, θ and ϕ, all three of these things varying.
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R changes length, θ changes length, ϕ changes length.
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It could be anywhere in space.
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For the rigid rotator, R is fixed, R is constant.
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R is the length in between the two masses.
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R is constant so del² ends up actually simplifying a little bit.
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We end up with 1/ R² sin θ DD θ sin θ DD θ + 1/ R² sin² θ D² D ϕ².
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I know it does not look well altogether but it is, believe me.
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We had that the rotational inertia is equal to the reduced mass × the R².
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Let me go ahead and solve for the reduced mass.
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M= I / R².
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The hamiltonian operator is equal to – H ̅² / twice the reduce mass × Del².
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I'm going to put this M here and I end up with.
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I get –H ̅² / 2 I / R² del² which is equal to - H ̅² all / 2 I.
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-H ̅² R² / del².
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Previously, we had that 1/ R² 1/ R².
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R² is on the numerator here so for the del², the R² on top cancels the R² on the bottom.
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The R² in the numerator cancels the R² terms in the denominator our del² operator
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leaving the hamiltonian equaling –H ̅² all / two I × 1/ sin θ DD θ sin θ DD θ + 1/ sin² θ × D² with respect to ϕ.
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That ends up being our hamiltonian operator.
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We said that the hamiltonian operator is also equal to the kinetic energy operator.
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The kinetic energy operator expressed in terms of angular momentum is L²/ 2I.
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This L² as an operator/ 2I happens to equal the Hamiltonian.
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I’m going to go ahead and call, everything that is in this bracket, I'm just going to go ahead and
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call it Z just to make my life a little bit easier.
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That is equal to - H ̅² / 2 I × Z which implies that L bar² is equal to -H ̅² × Z.
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We ended up with finding an expression for the angular momentum operator
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or actually the square of the anger momentum operator.
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We see that the square of the angular momentum
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is a naturally occurring operator for quantum mechanical rotating systems.
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Anything that rotates, the angular momentum is the most important aspect of that system.
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Let us go ahead and continue on here.
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Let us go back to blue.
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Since R is fixed, the orientation of the rotation in space depends only on the two other variables, θ and ϕ.
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It does not depend on R, R is fixed.
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With the rigid rotator, R does not change.
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It is only θ and ϕ change so you end up with a function of two variables only.
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Since R is fixed, the orientation of the rotation is space depends on θ and ϕ.
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The wave functions for the rigid rotator are functions of θ and ϕ only.
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I’m going to call them S for spherical.
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S, θ, and ϕ.
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The wave function of a rigid rotator S is a function of two variables θ and ϕ.
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The Schrӧdinger equation for the quantum mechanical rigid rotator which is our model for a rotating diatomic molecule
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is the hamiltonian of S which is a function of θ and ϕ is equal to the energy × the function S θ and ϕ.
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This is the equation that we have to solve, where the hamiltonian is –H ̅² / 2I × 1/ sin θ DD θ of sin θ.
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DD θ is an operator, +1/ sin² θ D² D ϕ² × S of θ and ϕ = E × the function S which is the function of θ and ϕ.
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I’m going to end up multiplying this whole thing by sin ^θ and I’m going to end up rearranging it.
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I'm going to get the following.
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I'm going to get sin θ DD θ sin θ DD θ + D² D ϕ² of S + sin² θ × 2 I E/ H ̅² × S = 0.
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I’m going to actually perform the operation by distributing this / the S and multiplying out
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so we are going to get sin θ DD θ × sin θ DSD θ + I’m going to do this one.
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D² of S D ϕ² + this is going to be M sin² θ × S =0.
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I have where M is actually equal to 2 I E/ H ̅² to make my life simpler by putting everything all into one.
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For right now, I’m going to discuss the energies of the rigid rotator.
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I’m going to save the discussion of the wave functions.
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The S I’m going to wait until we talk about the hydrogen atom so we can talk about the wave function.
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It is going to end up working a little more, I do not want to complicate things too much right now.
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Right now, let us worry about the energy functions and later we will talk about the actual wave function.
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Let me go back to blue here.
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When we solve this equation, the M must obey the following constraint.
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The constraint is M has to equal J × J + 1, where J =0, 1, 2, and so on, the whole numbers.
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M= 2I E/ H ̅², that has to equal to J × J + 1.
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Let me make this J a little bit more clear so it do not look like a 6.
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J + 1, when I rearrange to solve for E, which is what we always do we end up with the following relation E sub J.
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J is the level of energy = H ̅²/ 2 I × J × J + 1 or again J =0, 1, 2, and so on.
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These are the energies of the rigid rotator.
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Anytime I notice J is in our quantum number, the 0 level of energy, the energy is 0.
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The first level of energy is going to be 1 × 1 + 1=2.
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It is going to be 2 H ̅² / 2 I.
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It is going to be H ̅² / I.
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I put in this 2, the 3, these are the different energy levels of the rotating diatomic molecule for rigid rotator.
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Notice, the quantize energy levels cannot just take on any values at all.
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There are very specific values that it can take, the quantize energy levels.
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Now each energy level J has a degeneracy.
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Degeneracy D sub J of 2J + 1.
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All that means is that when I have a particular energy level, let us say the second energy level,
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when the particle is in the wave functions for that particular state, for the J = 2 state, there are going to be 2 × 2 + 1.
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There are going to be 5 wave functions that have that same energy level.
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Remember what degeneracy was.
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Degeneracy was if you have a degeneracy of 10 that means there are 10 wave functions that actually share that same energy level.
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For a particular energy level J, that level has 2 J + 1 degeneracy.
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It has that many wave functions that share that same energy level, that is all that means.
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These are the two important relations so far.
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Again, we are going to save our discussion of the wave functions which is our S of θ and ϕ for our future lesson.
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We are going to talk about that when we discuss the hydrogen atom and the Schrӧdinger equation for the hydrogen atom.
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We will go ahead and stop it here for now.
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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, bye.