WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, and welcome to www.educator.com, and welcome back to Physical Chemistry.
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Today, we are going to start our discussion of the hydrogen atom.
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We are ready to put everything together and we are ready to talk about the orbitals,
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the energy levels of the hydrogen atom.
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One little bit of warning before we start here.
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This is going to consist of multiple lessons and is going to be very heavily mathematical,
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at least in the sense of all the symbols floating around on the page.
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Ultimately what is important is, we want to be able to find the solutions for ψ,
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the wave function for the electron that is moving around the proton in the hydrogen atom.
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That is what is most important.
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It is easy to lose our way, I'm going to present it in the normal fashion.
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I’m going to present most of the mathematics, not all of it.
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It is really important that you separate, do not get intimidated and just alienated from all the symbols on the page.
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I will try to keep it clear as to what is important and what is just cultural information.
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What is just part of your scientific literacy.
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I could just present the solutions and say this is what it is and jump one to some problems
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but we also want you to understand where does that these things come from because that is where real understanding takes place.
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Ultimately, you are only responsible for the solutions
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but we want to show you where the solutions come from and why they take the form it takes.
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Just a little bit of warning, do not be intimidated, it is just symbols.
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Let us get started.
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In discussing the rigid rotator, we held R, the radius of the turn constant.
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We allowed θ and φ to vary.
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Something rotating this way, this way, is basically just something spinning around some fixed center.
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You had the θ which is this way and the φ which is this way.
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Those are the things that vary so it gave you all of the points on a sphere.
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We are going to allow R to actually change.
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We are going to allow not just move it about a sphere, we are going to allow it to be close.
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We are going to allow it to be far, so all of space is going to be considered now.
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We allow R to vary also.
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For the hydrogen atom, we take as our model a fixed proton of the center,
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a fixed proton at the origin so that will be our 0 position or origin position.
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An electron of mass M sub E interacting with the proton via the Coulombic potential.
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Basically, this is just a fancy way of saying we have a positive charge and a negative charge and
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they are going to interact with each other based on the fact that positive and negative charge attract each other.
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That is it, that is all we are saying.
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The Coulomb’s, the potential is the potential energy is 1 charge of positive charge and negative charge,
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the distance between them being R.
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The potential energy is going to equal – E² / 4 π ε₀ R.
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E here, it is the charge on 1 proton, the charge an electron in Coulomb’s.
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It is going to be 1.602 × 10⁻¹⁹ C.
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This ε₀, that is a permittivity of free space and its value is 8.854 × 10⁻¹² C² / J-m.
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It is called the permittivity of free space.
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R is the distance between the proton and the electron.
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Basically, you just have this proton, center, and you already know it is already electron is moving around all over.
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Not necessarily in circular orbits but now it is moving around this way, it is getting close, it is getting far, it can be anywhere.
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Let us see what we have got.
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Basically, we dealt with this rigid rotator which is a perfect sphere and all we have really done is allow the electron to move in and out.
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Get close to the proton, get far from the proton.
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We still want to work in spherical coordinates because the symmetry of the situation naturally lends itself to spherical coordinates.
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That is why these various coordinates systems exist
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because certain problems lend themselves to a particular coordinate system so the math actually becomes easier.
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In spherical coordinates, let me go ahead and erase this.
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Let me see, should I go ahead and put on this page.
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A spherical symmetry of the model compels us to use spherical coordinates.
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We are going to end up with something like this.
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The Schrӧdinger equation is going to be H of ψ.
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Ψ is a function of 3 variables, R, θ, and φ.
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R, θ, and φ is equal to E × ψ of R, θ, and φ.
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The Eigen function, Eigen value representation of that Schrӧdinger equation.
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H is going to be, the Hamiltonian operator here is going to be – H ̅² / 2 × the mass of the electron Del² - the potential energy.
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The potential energy was the Coulombic potential here, - E² / 4 π ε NR.
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We know what del² looks like in Cartesian coordinates.
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A spherical coordinates, it looks like this.
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Let me go ahead and rewrite this.
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H R is equal to - R² / 2 ME Del² -E².
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Where the del² operator in spherical coordinates, when we actually make
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the change of variable from cartesian to spherical, it looks like this.
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It looks really complicated.
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Do not let it scare you, it is just symbols.
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We just want you to see it.
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It is in your books, we want you to see it.
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We want you to at least recognize it, that is all we ask.
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Whatever R², DDR, R², DDR, so this is the del² operator in a spherical coordinate.
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I'm sure that I’m forgetting something here, R sin θ DD θ sin θ DD θ +,
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let me actually write it here, so I’m not squeezing it anywhere, + 1/ R² sin θ × E² D φ².
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We see we have R, we have θ, and we have the φ.
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We got the Laplacian operator in spherical coordinates.
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That is all it is.
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If we put this last expression into the equation 1, the original equation that we had
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which was this ψ of R θ and φ is equal to energy × ψ R θ and φ.
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In other words, if we apply this to this, we end up with something like this.
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We apply this last expression, meaning this thing right, the whole thing,
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that where this del square is equal to as, we apply this expression to, let us call this equation 1 and multiply by 2 M sub ER².
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We get this thing.
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Let us see if we can, that is getting crazy here.
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We get –H ̅² DDR² D of ψ DR –H ̅² 1/ sin θ × DD θ sin θ D of ψ D θ + 1/ sin².
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Here are the mistakes, I think there is an R² there and an R² there.
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1/ sin² θ D² ψ.
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Once again, I’m going to keep saying it this is not something that you have to know.
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We just want you to see it because I think it is actually need to see it.
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It is in your books but more than likely you are just sorting orders in your books, like the rest of us in school.
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-2 M sub E R² E/ 4 π E₀ R + the energy × ψ = 0.
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This is the equation that we end up solving.
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It is really nice to see it because I think it is one intellectual achievement back
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but we have managed to come up with this and would actually managed to solve it and be very successful in our solutions.
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It is actually quite extraordinary.
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It is really nice to know that you are in a position to appreciate this, see it, and understand what it means.
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It is not necessarily the mechanics of the actual solution of it so you really should be very proud of where it is that you are.
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This ψ, we said is actually a function of 3 variables.
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It is a function of R, θ, and φ, so in order to solve this, we are going to separate variables.
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We are going to write this ψ R θ and φ, we are going to write it in terms of some function just of R and some function of θ and φ.
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The wave function for the hydrogen atom, we are going to separate them into two functions, a product of two functions.
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One of them is just a function of the radius.
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One of them is just a function of the angles θ and φ.
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This one right here, this is called the radial component of the hydrogen wave function.
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This is called the angular component.
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We are going to deal these one at a time.
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Since we have separated variables and written ψ as a product of two functions, we recovered two separate differential equations.
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We are going to have a radial equation and we are going to have an angular equation.
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We are going to deal with the angular one first.
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We will deal with the angular equation first.
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Let us go ahead and start working in blue.
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When we do this, we are going to take these angular differential equation that we get
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and we are going to subject it to further manipulation.
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I’m not going to go through all of the manipulations but when we do that,
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I will just say with further manipulation which is nothing more than simplifying it so we are going to deal with it better.
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With further manipulation, we get sin of θ DD θ sin of θ DS D θ.
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S is the function of θ and φ, the angular component that consist of the two variables.
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Θ + D² S D φ² + β sin² θ × S is equal to 0.
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This is for the angular component, the S θ of φ, this is the differential equation that we have solved in order to find S.
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We are looking for S.
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Notice that this equation is the same as the Schrӧdinger equation for the rigid rotator.
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It is the same as the Schrӧdinger equation for the rigid rotator.
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The angular portion of the hydrogen atom wave function is the same as the rigid rotator wave functions.
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Let us see here, let us go ahead and do this.
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We are looking for this S of θ and φ.
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That is what we are looking for, the angular components.
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We are dealing with this so we subject this to another separation of variables.
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We are going to take this S of θ and φ because we are dealing only with that and deal with it by separating variables.
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We are going to express this as a product of two functions, 1 just of θ, 1 just of φ.
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I will call this T of θ and F of φ.
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We are just making our life easier, this is what we do.
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We start with 3 variables, we separated into 1 and 2 variables.
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We deal with the 1 and 2 variables, we separate that into 1 and 1, that is what we are doing here.
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The solutions for F of φ are F of φ is equal to this normalization constant A sub M E ⁺IM φ,
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where M is another quantum number and it is equal to 0 + or -1, + or -2, and so on.
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This is the normalization constant.
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Let us determine what the normalization constant is so we can actually write a full function for this.
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Let us determine the A sub M.
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The normalization condition, we know that the normalization condition is that
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the integral of the function conjugate × the function has to equal 1.
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In this particular case, φ itself is greater than or equal to 0 and less than or equal to 2 π.
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Those are our limits of integration, 0 to 2 π.
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We have the integral from 0 to 2 π of S * sub M × FM = 1.
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This is going to be the integral from 0 to 2 π.
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We are going to take F sub M*, we are going to star this, we are going to conjugate this.
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It is going to be A sub M, when we conjugate, we conjugate everything.
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It is going to be the constant, conjugate of the constant E and – IM φ
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because the conjugate of the IM φ is – IM φ × the function itself F sub M which is A sub M E ⁺IM φ D φ.
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This is this, that ends up becoming this and this, ends up becoming the absolute value of A sub M².
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The integral from 0 to 2 π, this and this may cancel out, - IN φ + IN φ is 0, which is equal to 1.
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All you are left with is D φ, that is equal to 1.
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Therefore, what we have is absolute value of A sub M² × 2 π.
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This integral from 0 to 2 π of D φ is equal to 2 π is equal to 1.
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That implies that A sub M is equal to 1 / 2 π M =0 + or -1, + or -2, and so on.
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Our final function F of φ, this thing right here, what it looks like is the following.
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F of φ is equal to 1/ 2 π, I will write it as, the exponent of ½ E ⁺IM φ.
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Again, subject to these values for M.
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This is the φ component of the angular component of the hydrogen wave function, fully normalized.
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This is definitely one of the equations, one of the functions that we are interested in.
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We are making good progress here.
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We will see what we have got.
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We had S of θ and φ.
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When we separate variables for this, we end up with two differential equations.
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We did mention the differential equation for the φ component, we just gave you the solutions.
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For the other part, the T of θ.
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This one I’m going to write down the differential equation.
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We go to blue here.
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We had this S of θ and φ, we separated into a function just of θ, and the function just of φ.
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Let me go back to red.
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We just found the F of φ, now we want to find the T of θ so we can multiply them together to get our full angular solution.
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The differential equation for T of θ is the following.
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It is going to be sin of θ T of θ DD θ sin of θ DT D θ + this β sin² θ = M².
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The solution to this differential equation involves something called the change of variable.
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We are going to change the variable and we are going to actually solve this equation.
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The solutions will get our act going to be polynomials.
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The variable inside those polynomials is actually going to be a cos θ.
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It is actually is going to be a function of θ.
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Solving this involves using a change of variable.
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You do not have to know this, you do not have to reproduce this.
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If you can follow it, that is great.
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If not, this is just part of your general scientific literacy.
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The change of variable we are going to use is going to be X = cos θ.
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Θ is running from 0 to π so T of θ becomes a polynomial in X comes which will designate as P of X.
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Under this change of variable, what ends up happening to the differential equation is the following.
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I will go ahead and stay with red.
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Under this change of variable, the differential equation becomes the following.
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It becomes 1 -X² D² ψ DX² -2 X DP DX + β – M²/ 1 - X² × P is equal to 0.
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Again, M is equal to 0 + or -1, + or -2, and so on.
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This is a very important equation in physics, it is called the Legendre equation.
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This is called the Legendre equation.
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It shows up a lot.
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Let me go ahead and go back to blue here.
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What is the equation we solve?
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When we solve this equation, in order for these polynomials, the solutions P of X to stay finite, we need it to stay finite.
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This β, it must equal λ × L.
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In order for the solutions to stay finite, this constant β has to equal this L × L + 1.
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The value of L is another quantum number.
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It is actually equal to 0, 1, 2, and so on but with a restriction that the absolute value of M has to be less than or equal to L.
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There is a connection, M is actually dependent on L.
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We write, we are just going to put this into β.
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We write 1 - X² D² P DX² -2X DP DX + L × L + 1 - M²/ 1 - X² P = 0.
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Where, L is going to take on the value 0, 1, 2, and so on.
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M is going to take on the values + or -1, + or -2, all the way to + or – L.
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The absolute value is less than or equal to L, which means that M is going to take on the values when we write them out explicitly 0, 1, -1, 2, -2, 3, -3.
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If L is 4, then is going to be 4, -4 all the way up to that value.
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Now, we have two quantum numbers that are going to arise here.
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There are different sets of solutions for the various values of L.
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Let us actually start writing this so we can read it.
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For the various values of M, we are going to look first, let us first look at M = 0.
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When M is equal to 0, the solutions which we symbolize as
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P sub L of X are called appropriately Legendre polynomials because they are coming from Legendre equation.
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The Legendre polynomials are very important.
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I’m going to go ahead and list the first few Legendre polynomials.
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We are taking M = 0.
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The first few Legendre polynomials, P₀ of X is equal to 1.
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P₁ of X now M is equal to 0, this subscript right here, that is the L value.
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P₁ of X is going to equal X.
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P₂ of X is going to equal ½ × 3 X² – 1.
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P₃ of X is going to equal ½ 5X³ -3X.
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We will go ahead and stop with P₄, that is going to equal 1/8 35 X⁴ - 30 X² + 3.
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It is very important to remember.
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Recall that X is not a variable here, they are the change of variable.
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X is actually equal to cos of θ.
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If I do P2 of X that is actually P2 of cos θ.
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Wherever X is, I'm just going to stick in to cos θ.
00:34:42.900 --> 00:35:00.500
In the case of the P2, I'm going to have ½ 3 cos² θ -1.
00:35:00.500 --> 00:35:03.000
Θ is still the variable that we are concerned with.
00:35:03.000 --> 00:35:06.900
Remember ψ is a function of R, θ, and φ.
00:35:06.900 --> 00:35:11.200
All we have done is change the variable in order for us to actually solve this differential equation
00:35:11.200 --> 00:35:14.000
and make it a little bit easier to deal with, that is all we have done here.
00:35:14.000 --> 00:35:17.300
X is not the variable, XYZ is not a coordinate.
00:35:17.300 --> 00:35:22.900
It is just a variable that we are using a hold the place of cos θ, the change of variable.
00:35:22.900 --> 00:35:28.000
The function is really this thing right here because again we need a function of θ
00:35:28.000 --> 00:35:31.100
but we are expressing it in terms of these polynomials.
00:35:31.100 --> 00:35:32.000
This is very important to remember.
00:35:32.000 --> 00:35:37.800
I will remind you of that several times because it is easy to lose ones way.
00:35:37.800 --> 00:35:40.300
We all do, I do all the time.
00:35:40.300 --> 00:35:44.800
Let us go back to blue here.
00:35:44.800 --> 00:35:50.000
We said that L takes all the values from 0, 1, 2, 3, 4, and so on.
00:35:50.000 --> 00:35:52.800
A couple of properties here of these polynomials.
00:35:52.800 --> 00:36:09.300
When L is odd, the Legendre polynomial P of X is an odd function.
00:36:09.300 --> 00:36:21.300
When L is even, the Legendre polynomial is an even function.
00:36:21.300 --> 00:36:44.000
It is also true that the integral from -1 to 1 of P sub Q of X, P sub R of X, is equal to 0.
00:36:44.000 --> 00:36:47.800
In other words, these Legendre polynomials are actually orthogonal.
00:36:47.800 --> 00:36:52.600
They are mutually orthogonal.
00:36:52.600 --> 00:37:09.900
That is the Legendre polynomials are mutually orthogonal.
00:37:09.900 --> 00:37:17.500
If you are wondering where I get this -1 to 1, let me go to the next page and tell you.
00:37:17.500 --> 00:37:34.900
The limits here are -1 to 1 because again X is equal to cos of θ.
00:37:34.900 --> 00:37:40.800
Θ is running from 0 all the way to π.
00:37:40.800 --> 00:37:56.900
X is cos θ so X goes from cos of 0 to cos of π.
00:37:56.900 --> 00:38:00.900
Cos of 0 is 1 to -1.
00:38:00.900 --> 00:38:04.900
It is θ is the variable that we are interested, θ runs from 0 to π.
00:38:04.900 --> 00:38:10.200
When we take the cos of 0 like this one, we take the cos of π that gives us -1.
00:38:10.200 --> 00:38:17.500
Our limit of integration, when we are integrating the polynomial is going to be -1 to 1 or 1 to -1.
00:38:17.500 --> 00:38:20.900
The order does not matter, you are just going to switch signs.
00:38:20.900 --> 00:38:24.600
When we are integrating with respect to θ, that is when we put 0 to π.
00:38:24.600 --> 00:38:30.200
That is why -1 to 1, that is where the limits come from.
00:38:30.200 --> 00:38:40.600
The last property, the coefficients in front of the polynomials, the ½, the 1/8, and so on,
00:38:40.600 --> 00:39:01.600
the coefficients in front of the P sub L of X are chosen so that when we do the P sub L of 1 it always ends up equaling 1.
00:39:01.600 --> 00:39:04.400
It is just for convenience, we have chosen it that way.
00:39:04.400 --> 00:39:19.000
For example, P₂ of X is equal to ½ 3 X² -1.
00:39:19.000 --> 00:39:31.600
P₂ of 1 is equal to ½ 3 × 1² - 1 is equal to ½.
00:39:31.600 --> 00:39:36.900
3 × 3 – 2, we actually end up getting a value of 2.
00:39:36.900 --> 00:39:41.700
We stick a ½ in front of it, in order for us to end up with a final value of 1.
00:39:41.700 --> 00:39:51.500
This is the actual polynomial, we have adjusted it by putting the coefficient in front of it to make sure that P of 1 is equal to 1.
00:39:51.500 --> 00:39:55.700
That is what gives us our final form of Legendre polynomials.
00:39:55.700 --> 00:39:56.800
We will go ahead and stop this lesson here.
00:39:56.800 --> 00:39:59.300
Thank you so much for joining us here at www.educator.com.
00:39:59.300 --> 00:40:00.000
We will see you next time for a continuation.