Spherical Harmonics: Atomic Orbitals, Laplacian in Polar and Spherical Coordinates, Laplace Equation

Transcript

Hi, it's time for another math, easy solution. Today I'm going to go over a pretty interesting topic, and that's going to be on spherical harmonics, and a whole lot of other stuff related to. It's going to be quite epic. All right, and before I go further, so note, if you don't have time to watch this whole video, you could, well, in this case, it's played at faster speed, top secret left, your brain gets used to faster speed. Here's a link to my tutorial, and also you could download and read the notes, it'll be in PDF format as well as on the Hive, blockchain, and article format, and you could also watch a video in parts, times times of all the parts will be in the description, as well as links of everything else, as well as the video sections playlist. After the video, I'll be making short clips of... of the whole video, so it's gonna be pretty epic. Alright, so let's jump right in, so topics to cover, so we'll go over introduction to spherical harmonics, look at atomic orbitals and magnetic fields, and then I'm gonna go over a mathematical review. and then basically over the Laplace operator and polar coordinates, Laplace operator in spherical coordinates, derive these equations, and then look at the solid spherical harmonic solutions to Laplace's equation and spherical coordinates, and then discuss the associated legendary equation, or... or, I mean, it's a Legendra equation, something like that. And then go over spherical harmonics, a solution to the angular Laplastian, and then deriving the real spherical harmonics, and then graphing spherical harmonics in Desmos calculator. Or more specifically, graphing real spherical harmonics in Desmos calculator. All right, so let's continue further. So, introduction to spherical harmonics. Yeah, so I'll be going over the introduction and overview of the equations that are derived, and then go over basically atomic orbitals and magnetic fields. Those are basically the driving force behind why I am covering spherical harmonics now. All right, so spherical harmonics are special functions defined on the surface of a sphere and have many applications in physics. From Wikipedia, spherical harmonics originate from solving Laplace's equation in the spherical domains. And functions that are solutions to Laplace's equation are called harmonics. and yeah, we'll discuss that in a bit. So, the Laplace equation in rectangular coordinates, or in other words, XYZ, the typical ones here is rectangular coordinates, XYZ, those ones, yeah, the Laplace equation is shown below. So here's a Laplace equation. It's basically the acceleration, or the, this is the second partial derivative of views. Basically, it's a derivative. of u in this case whatever the function is uh typically i use f uh uh uh uh f but in this case using u so this is a function u of x y z etc and this is a point p on there but anyways um Yeah, so the derivative with respect, yeah, of the function with respect to X only, so everything else is constant, we're just looking at the rate of change of this one. And this is the second derivative. So in other words, this essentially what it's saying is that it's the rate change of X on its own, on its own, Y is on its own, and Z on its own, basic acceleration on each axis. while everything else is fixed. Yeah, for a function in 3D space. So, and we'll go over that when we do the mathematical review. So, anyways, this is the partial, the second partial derivative of U in terms of Y, and then plus the second partial derivative of U in terms of Z, and then this equals a zero. And note that the above equation describes steady state conditions, no sources or sinks, i.e., where things are in, quote, harmony, hence the solutions are called harmonic functions. the acceleration on each axis is going to equal, when you sum all up, it's going to equal zero, so there's no net change in that respect. All right, so this is Laplace equations in rectangular coordinates. So now... The above equation can be rewritten to obtain the Laplace equation in spherical coordinates, and it uses R, which is the distance from the origin like that, or the radius, and then you have theta, which is here, this is the polar angle, and you have phi, this is the angle about like that, but the x-axis, I think that's called azimithal angle, or as-mythal angle. Yeah, so this is the physics convention like that, spherical coordinates, R-theta. and phi as commonly used radial distance, slant distance of the origin, polar angle theta, angle with respect to positive polar axis, and as methyl angle, phi, angle of rotation from the initial meridian plane. And yeah, that meridian plane is this is what it's referring to, that's the initial one there, and then you're just rotating it about. this phi there, all right? Yeah, I'm pretty sure that is. Anyway, so going further so, this is, yeah, this one here is in rectangular coordinates. That's the Laplace equation, and now it can convert the rectangle ones to spherical, you get this giant equation, which is 1 divided by r squared, and then you have partial derivative with respect to r of r squared, and multiplied by partial derivative of f in terms of r. and then this one here, and then plus 1 divided by r squared sine theta, and then now you have the partial derivative in terms of theta of sine theta times partial derivative of f in terms of theta. So there's a partial derivative of this whole term, and you have to do it of this one inside, so again, a lot of partial derivatives, plus 1 divided by r squared times sine squared theta. So 1 of r squared sine squared theta times by the second partial derivative in terms of phi of f or function f. This case we had U, this one's F, it's just convention. U or F is typically used. So now solving the above equation using separation of variables obtains the solid spherical harmonic. So if you solve this, by assuming that you have the function F equals, this right here, F is a function of R, theta, and phi, and then you could, and then you assume that you could separate them to have three different equations that are independent of the others, multiplied by each other. So basically, E, no, actually, yeah, yeah, basically has functions of this, like that. Yeah, so that they're independent of each other. So assume that you could do the function like that. What you end up having is this setup here, F of R Theta Phi, equals this one right here. This is the E. This one is the R function. And now this one is PNM cosine, Theta, where PNM is, we'll get that a bit, that is the associated Lejondra function. beyond the scope of this video, but I'll do be, I'll be going over that in a future one, deriving it. So anyways, so we're n, right here. So we have n, these terms are n is 0, 1, 2, 3, 4, it could be 0, 1, 2, 3, etc, it's a non-negative integer. And m is equal to negative n, negative n plus 1, dot, dot, all the way to n like that. that is what m is and it's also an integer as well and then we have p nm or p subscript and and then superscript should be a um a subscript how we do this i think it's like this yeah well minus oops All right, I just pause the video and just add the superscript, or the subscript there first, and then the superscript on top there. It's a PNM like that. I'm not sure if we could put them stack on top of each other. They'll be for another day, but anyways, so PNM is the of T. So now here where T is equal to cosine theta. So this one, I said T equals to that, just so we have this part there. so the formula here so anyway so where p n m t is the associated la genre function with t is equal to cosine theta so in other words t is going to be an element of all real numbers between negative 1 and 1 so basically that's the close interval and that's because cosine is like this so you get number of cosine is going to be a maximum 1 minimum negative 1 so t is in between there And now here is the associated Lezondra function, and I will not be deriving this in this video, but I probably will be doing it in another video. But anyways, that becomes this, so PN, m of t is equal to negative m to the power of m, times it by 1 minus t squared to power of m divided by 2, and then times it by the mth derivative, in terms of t, the PN of T, where PNFT is the PNFT is the ordinary legerna polynomials. Oh, yeah, polynomial or polynomials. And this is using the broad regress formula. And again, this would be for another video. And where basically this thing is this. Where that equals to PN of T is equal to 1 divided by 2N, yeah, 2n squared. I mean 1 divided by 2 to the power of n times n factorial, times D, I mean, this is going to be the nth derivative in terms of t of t squared minus one to power of n. All right. All right, guys, this is absolutely ridiculous. This is quite amazing, though. So now, so basically compacted together, the associated leisandra function equals to here, or come back together, I basically put this. inside here, and now we throw this all in here, and then we get this. We get this giant formula, PNM of T is equal to negative 1 to the power of m times by 1 divided by 2 to the power of n factorial, times by 1 minus t squared to power of m divided by 2, and then take the n plus into the n plus mth derivative of t squared minus 1 to power of n. And this is in terms of t. Fascinating, fascinating stuff. So again, basically throw this entire thing inside here. Yeah, so that one there. And then T replaced T with cosine theta. So this replaced T with cosine theta, and then throw this all inside to the other part. It's fascinating stuff. So anyways, so the general solutions to the Laplace equation are called solid spherical harmonics, as they contain both the radial term R and angular terms theta and phi. So in other words, solving this full thingy, you're going to get the terms of R, and it's the solid spherical marks, and you have the angular ones. Here's the angle ones, and then you also have the length R, it can vary, and so on. So it's not just on a sphere, you can have a varying, multiplying by this varying R value. All right. So if we instead remove the radial term so that we obtain only the angular terms, then such solutions are termed spherical harmonics. Now this is solid spherical, now here's a spherical one, which is what we're interested in for this case. So we have, this is going to be... This symbol right here, you could also use this right here, this triangle. This is called a nabla or something like that. This is where this nabla squared, or is upside-down triangle squared, is the Laplace operator or Laplacian, like that. And I'll be discussing that later. But basically, if you have the function like this, the Laplacian is equal to this in spherical coordinates, and then you set it to equal to zero. That's the Laplas equation. So we have, this is the full thing in here. This was here. yeah this is just this part here but this this all thing equals to delta f the that's a laplocci and all you can do the other one like that f equals that all right just put that in there um Anyways, so that is that. So multiplying, so here's the way to get rid of it is you will multiply R on both sides. I'm just doing this just to a high level, and then I'll derive this later in this video. But I want to do a high level, just to wrap your head around, what in the world do people refer to when they say spherical harmonics? Because even though this is the solid spherical harmonics right here, but typically it's grafted without this right here, for just pure spherical harmonics. All right, anyway, so multiplying R on both sides, so then if you multiply R in both sides, this cancels, whoops, yeah, if you multiply by R squared on both sides, this cancels, this all cancels, and then multiply by zero, it's just still zero. So you get rid of all that, and you end up with this right here. So partial root of R, of this function in terms of R, of R squared, and then this is, partial F, partial R, plus 1 divided by sine theta, times by partial derivative of this thing in terms of theta, sine theta, and then partial derivative of F divided by, yeah, basically partial derivative of f, in terms of theta. And then plus 1 divided by sine, sine square theta, equals this, and then multiplied by the second partial derivative of F in terms of phi. All right, so we have that, and then that whole thing equals to zero. So, in other words, this, And you can also put it using this thing here to call the angular Laplasian. So get rid of that and put that this right here. So you can just write it as this. So partial derivative of this function in terms of R of R, of R squared, times by partial derivative of F in terms of R, plus the Laplace, the angular Laplacian of F equals zero. So where this symbol is the angular Laplaceian or Laplace. bell trami operator on the sphere, all right, and that equals to this thingy, which we just did right here. All right, so we have that. So now, rewriting F, so basically the next step's in a derivation, you will rewrite F, so this is the original one we had, R, in terms of the separated variables, R of R, P of Theta, E of Phi, this equals to R, capital R of R, times it by, now we're going to define these two terms, to get the angular parts only as y. So y of theta phi. So where y of theta phi is equal to p of theta times e of phi. So there it is. And now when you solve this, so solving y, we obtain the spherical, complex spherical harmonics. I'll put harmonics. So now if you solve this, this giant formula, I'm just doing a high level overview. Okay, you solve this right here. by using separating variables, but for y like this, what you end up having is, you know, what you end up having is, we have YNM of Theta Phi equals to n times by PNM, that's the associated Lejondra function of cosine theta times it by E to the IM phi. So that's still including the R1, because we have this R there. So we have that phi. Actually, no, no, we don't have that, actually. This is the Y one. know that that's separated. I mean, this one you solve for this part specifically. We're ignoring this, this one here. That is, yeah, we're just carrying about the Y one right here. And then this equals to this E, M, phi, like that. That's the phi term. This is the theta term. And this capital N is a normalization constant, usually equals to this giant thing, n equals to negative 1 to power of m times by square root. 2n plus 1 times, yeah, divided by 4 pi times by n minus 1 factor, n minus m factorial, divided by n plus m factorial. And the reason they have this giant, this normalization constant, is so that, to ensure that this following integral right here, used in quantum mechanics for probability is equal to 1. And that's this right here. So integral from theta is equal to 0 up to pi of this right here, the y, n, m of the spherical harmonic of theta and phi squared times it by sine theta, D, D, this is D theta, then D phi, where this integral from 0, from phi to 2 pi. So it's a double integral, and that'll be again for another topic there, but then that whole thing is just to ensure that this equals to 1, because in quantum mechanics, they use it for probability, so you need to have this normalized so that the total probability is equal to 1 when you sum up everything. Again, integrals are sum of pretty much every single thing there. So they kind of use that whole cervical harmonics, the equations on the sphere, and then normalized it to 1. so that they can use it as a probability distribution for atomic orbitals and electron shells, as I will discuss soon. And this normalization setup here is also written like this. So integral from, yeah, just the integral of this spherical harmonic squared, the basically absolute value, you just take that squared, and of this D omega equals to one, where D omega is equal to sine, theta, D, theta. So yeah, that's something for a future possible video to just see how they get these exact terms. Anyways, the real sphere, no, yes, and this is a complex one, and the reason it's complex, it's called complex, is because you're going to have this right here. There's an I there, there's an imaginary number there, but typically the ones that you graph in, like, when you use a different calculator, just any calculator, they don't graph complex, it's not easy to graph, or just not, doesn't appear as well. So basically the real ones, so the real spherical harmonics are, I'll put this in, bold, so keep that in mind. So the real spherical harmonics are defined such that the functions are real valued everywhere with no complex term I. The usual definition is given below and is almost always used when displaying spherical harmonics visually, since they exclude the complex phase factor E to the IM5, which is rotating about the Z-axis. Remember, this is the azimethyl angle there. this phi, so basically you're going to have a rotation set up there. This right here, a rotation there, and that's for the complex, but for the real one is to ignore that, so we basically get like a scalar value, assume, and you can actually visually see it, it makes sense when you graph it. So anyways, so this is how it looks like, so I was defined, I'm not going to go over exactly how they get this, but I just do, I want to emphasize, this is what they do. But actually, I may or may not cover this later in this video, because I haven't fully written up the notes at the end. I'll be going that as I film this video and see what I come across. Because again, this is a lot of, yeah, I cannot imagine how much just complexity they have in this sphericalomarks. Anyways, so the spherical harmonic, the real one is Y and M of Phi of Theta and Phi is equal 2. So if M is greater than 0, you have this set up negative 1 to power of m divided by 2 squared times by YNM. plus and then you have this negative 1 to power of m times y n of uh yeah and now you have a negative n so yeah they do this so that you get there's a negative term there so that's interesting so they kind of throw that back in there and again this is the positive negative one and it's just so it somehow cancels out all the um complex terms and now if m equals zero you just you just replace m yeah just put that in there so it's just y and and um m equals zero And then if m is less than zero, you're going to have this setup, negative 1, m plus 1, and then now you have the i there, because you're going to be able to cancel later in this setup here. So these ones i cancels within, this one cancels, it gets one i term that cancels here. So i divided by i is i times two, square root 2. And then you have y, n, and then now m is absolute value, minus negative 1, the power of absolute value of m. times it by y n of negative absolute value of m so again fascinating stuff there yeah um i may cover this near the end of the video just to see what happens anyways uh plugging in all the values into the above equation obtains the simplified form below in which the proper proper i mean proportionality Yeah, the proportional symbol or the proportionality symbol, this right here, is used just to show the functional shape rather than the normalization constant. So then basically, once you plug this in and basically simplify this, so you have this YNM, and you know what YNM is, that is, this PNM there, and that's the Legendra, the associated Lezondra function. So throw that all in there and simplify you get this set up. So it's proportional to this, excluding the proportionality constant, n. So we've got m is greater than 0. You'll have p n, m is a absolute value of m, and then cosine theta, and then cosine times it by cosine m-fi. And then the second part of m equals 0 is plug in the 0 here for the associated legera function. And then if m is less than 0, it just equals to the PN. and then absolute value of m, you plug that into associated losanitor function, cosine theta, sine, and then absolute value of m, and then phi. Or…

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