▶️ Watch on 3Speak - Telegram - YouTube - Summary - Hive Notes - Sections playlist - MES Links --- In this video, I derive the real spherical harmonics by excluding the imaginary complex i terms fr...
Transcript
Let's derive the real spherical harmonics. So the real ones basically remove all of these imaginary terms out of there. So you don't have any of those imaginary terms. So the real spherical harmonics are defined such that the functions are real valued everywhere with no complex term I, with a typical definition shown as follows. So it's typically written like this, or defined, I guess. So we're going to call this. y real n m, and then this is a function of theta and phi, this equals to, it's going to be a giant thing, and again, we went over to this in the overview slash introduction, just to get an overview of what we're trying to achieve with this spherical armonics, but now this is the definition like this, we'll go over some of the terms in a bit, so it's going to be negative 1 to the power of m, and then divided by square root 2 times it by, So times it by YNM, and this is the complex one, this is the real one, this is the complex, which is just basically this part right here. YNM plus negative 1 to the power of M, I'll make this regular bracket like that, the square bracket, and then YN minus M like that, and this is 4M is greater than 0. And then this is equal, so basically it's equal to that, but if m is equal to 0, then we just get y and zero like that, this is just m equals zero. And then lastly, for m is negative, we get this term, negative 1, power of m plus 1 divided by i square root 2, and then we have the same setup, put a square bracket. Y, N, absolute value of M minus, this is going to be minus negative 1 to power of M, or absolute value of M, and then Y, N, and then this is the term negative absolute value of M, and then this is 4, M is less than 0. And here I quickly just move this over to the other side, out of this square bracket. So, and now before we go further, so note that the complex spherical harmonics are chosen using the pure complex exponentials, i.e. the basis solutions. So instead of using this full linear combinations right there, yeah, so rather than the linear combinations, going back here, yeah, so rather than using this giant thingy there, we're only going to, only just going to use these right here, like that. And because again, those are still solutions. Basically, we're going to use the basis solutions rather than the linear combinations, and this term right here, Y, and then a negative M over there. So this one right here. So inside this, when we get the real one, we're just going to use the basis one. That'll explain in a bit. And then for this term, we're going to define that, basically, defined via the following complex. conjugate form i.e. flip the sign of the complex rotation angle term, because this makes important formulas and operations and quantum mechanics cleaner, simpler, and more elegant. Again, there's a link to the GROC AI explanation for that, so that is for another day, but apparently that makes it just much, much easier. So, and using the proportional forms, basically ignoring, ignoring the normalization constant discussed earlier we obtained. So for these y n, m and this y n negative m, what we get is we get y n, and the complex form is going to be proportional to, because yeah, we're ignoring the proportionality constant. So PNM, the associated legron function P and M of cosine theta, and then multiplied by just the basis solution, EIM. 5. So that's the plus one. Yes, in other words, we just do this one here, times by this, and there's going to be proportionality, constant, etc., which ensures that the probability integral thing equals to 1. So that is that, and then we have y and negative m is going to be proportional 2, and now this one's going to be written like this, negative 1 to the power of m, and then p n m. Yes, it still has the m there, and I'll explain why it's still there. cosine Theta, and then this is going to be E negative I m thigh. So now we have this, and again, just want to elaborate more on this. So note the complex conjugation in the above definition, again, that's just using this negative there. and if you go to the wiki, go back to the wiki right here. So if you go right here, complex conjugate right here, for rectangular one, just goes z equals x plus i y, then you just flipped it to negative. It's negative i y. This one is theta there. This one's going to be basically negative theta. So basically the complex conjugate of a plus b i is a minus b i, but in polar form, because we're really dealing with spherical harmonics, this is going to be the angle there. So R, E, I, Phi is just going to be R E negative phi. Yeah, I Phi. Yeah, so in other words, what we have, I'm just going to write this, how this is usually written. So we have Y, N, negative, M of Theta Phi, this is equal to negative 1, or proportional to it, or actually this just equals to this one. just equals to it by definition, Y and M, and then this is going to be Theta, Phi, and then this is the complex conjugate, you just put a star. This is indicating complex conjugate. Complex conjugate. Conjugate. All right, and now here quickly, just to quickly recap on this complex conjugate slash just imaginary numbers in general. So if we have, this is, let's say we have set up like this. This is the real part. This is how we typically written, and imaginaries the y-axis there. So then let's say we had this up to here, this is R. So over here is going to be R, and there's an angle, angles phi. and this height is going to be Y. So it's going to be Y is on the imaginary side and X is on the real side like that. So this is going to be R, E, this is going to be I, phi. That's how you could, that's written like that, and again, it matches Euler's identity. This just equals to, well, equals to X, the real one plus the imaginary times by Y, the height of the imaginary, like that. And where, well, cosine number using just basic Pythagoras, so. I mean basic trigonometry. So, sine phi, remember, is this equals opposite over adjacent, over R. Likewise for the, yeah, likewise for cosine will be adjacent, it would be the X over R. So in other words, then move the R to the other side. So this equals two, but then just take the factor out the R. This is going to be cosine, phi, plus I, sine, and again, this is just Euler's identity. this is Euler's identity that we covered earlier. I'm Euler's, yeah, Euler's formula. Yeah, I just wrote formula there, and if we scroll all the way back up to Euler, I'll see where Euler is, was way back over here. Yeah, so this is Euler's formula right here, we covered earlier, so EIX equals cosine X plus I sine X, and then Euler's identity is EI-Pi equals negative 1, yeah, so this is the Euler's formula, and you can also write as a matrix and so on, which is pretty fascinating. That's for, yeah, that's for my earlier video, so definitely watch that, to see how you can do complex stuff in terms of the matrices. All right, anyways, let's go back over here, so now the complex conjugate of this would just be, well, the bottom up there, so this would be you would go down R, so R is always positive just the length to this point. So that's R, e, negative, i, phi, like that, and this goes like that. And this is, again, height is going to be y, and then this is the other, so you could say height is negative y, and so on. So negative y, like that, and x is still positive, so this equals 2, well, x, minus, iY, which equals to r, take the bracket out, cosine, phi, minus i sine, phi. Anyways, I just want to recap on this setup here. I'm going to circle this because it's not really part of our main theme, but just to give you a recap on that. All right, so now that we have this, the next setup is, well, we take this over here. So now we want to just show you what this is, this complex conjugate, just illustrate that. So we take this YNM, the complex spherical harmonic. So YNM is proportional to PNM, the associated Legendre function of cosine theta times EIM phi. So thus the complex conjugate of this, YN, m star, do a better star, is going to equal, or it's going to be proportional to, let's do the proportional setup like that, of, do like that, okay, this is going to be PN, so the same thing as the one on the left, cosine theta, e negative I, m, phi. So we just take, this set up, make a negative, exactly this set. That's a negative, that's a complex conjugate, and this is positive, goes to negative. That's all it is. Just take that. Same thing as plus iY, same thing as minus iY, like that. All right, so as we have, so thus, we have it by this definition, so y, n, negative m. Yes, we're going to have y, n, negative m. is going to be, well, proportional to, and I'll just put it fully like this. This is going to be a, actually, we'll just, yeah, put it like this, it's going to be a function of theta and phi. Portional two, put it off from the top, then down. All right, like that. It's going to be proportional to, well, this one here, yeah, it's going to be proportional to this one here, negative, yeah, negative 1, the power of m. And then this is going to be this, this one now, it's going to be changed this. I know we're using proportional, because it's this proportional there. So we're using that. So this is going to be PNM of cosine theta E negative I.M. Phi, like that. So that's what we have there. All right, now, going further now, so also note that for this Y and negative M term right here, this is the spherical harmonic with the negative M there, we use PNM instead of PN negative M, so notice that even though there's negative M there, we're still using this positive, this plus M there, and yeah, we use that instead, Yeah, and we use that, yeah, because the associated Legendra equation depends on M squared, not M. So what this means is, these are going to be proportional to each other. So these two are actually proportional to them. They're not going to be independent, so you're not going to have just a random different solution. Yeah, right here, they're going to be proportional. Yeah, these ones are going to be proportional. Let me see if I could write, yeah, this ones are to be proportional to each other. Yeah, so they're basically by a constant, so on, and then I'll explain that as well, up here again. Let me just go look at the Associated Legendra equation. So if we look at it right here, here's the associated Legendra equation. This is I-X squared and so on. And then instead of this one, instead of writing, actually, that's the normal one, this is the one. This is the associate, the other one's just a legendary, this is the one we want. And what I'll do is I'm going to replace this D, this second derivative, PLM, I'm just going to call this Y double prime. So that's, just make it more generic. So what I'll do is I'll copy this so we can write this down easier, all the way back up here. All right, so I'm going to put this right here just so we can see it. Okay, that is the, yeah, that was the associate Lazandra question. I'm going to go 1 minus, use the T setup there, so 1 over T squared, and then this is going to be Y double prime, like that. And then this next one is going to be 2T, then D over D x, that's just the first derivative, so Y prime, and then plus, and then change these other ones to the our usual N, N, set up N, N plus 1. minus m squared. So notice that the only m is there is over there, so 1 minus t squared, and then y equals to zero. So we got that, we get rid of it. So now this all depends on m squared. So, yeah, so what this means is if, if we have y equals to, I'll just say, yeah, yeah, let's say y of t. All right, so if y of t, yeah so if y of t is equal to the associated lezondra function p n of m like that of t and this is a solution so if is a solution yeah i'll say yeah if this is a solution also yeah is a solution is a solution yeah so as a solution then then basically the other one is automatically a solution too. Just put a bot there. So thus, yes, then this one's going to be, yeah, then y of t is also equal to solution. This one is negative m. Ft is also a solution. A solution. And again, this is all because, let's say, I'll write it down, I'll rather write it than type it. So that's because, well, m plus or minus m squared is still going to be m squared. Yeah, it's just equals to m squared. It doesn't matter what, it's going to be still positive. Yeah, so it does not matter. Basically, this is m squared, this is unchanged. So it doesn't change the actual thing. This is unchanged. Thus, yeah, this is pretty, a good clear set of it, actually. This is, thus, we'll have PN, negative M of T, is going to be, well, proportional to, it's going to be proportional to PN of T. Yeah, PN of T. In other words, this is going to be a constant multiple applied, so that's why we have this. Yeah, just make it like this. So basically, if we have it, for example, if we have something like, let's say, K, PNM, that's pretty much what is up to a constant. So it could be just equal to like a constant multiply by that. And then basically, when you plug in the constant, when you take out the constant here, all these derivatives, you're just going to get a K out of this whole thing either. And then divide by k just all cancels. So that's pretty much what it is. Yeah, so that is that. Yeah, so this is just going to be proportional to it. And, yeah, actually, I'm just going to write that set up there. So this actually equals to a constant. So this equals to, yeah, this equals to the constant, I'll put it over here actually, K, N, or actually, just to make it appear better, this equals to constant K, PN. m of t and what it actually turns out is that there's actually an identity for this where k is going to equal to well is actually going to equal to negative uh negative one power of m times it by and this is going to be times by n minus m uh factorial divided by n plus m factorial p and m of t so that's what it actually equals to i know what this here is apparently from rodriguez formula, I'll, I might drive this in a future video. So that equals that, and I'll box this whole thing. Here, I just fix that up, and those box this whole thing in. Yes, we have that box, and also is right here where k is a constant. K equals constant, like that. And here, just fix that up, so where k equals a constant, and we have it over here. So basically, K is this whole giant term right here, negative 1 times, Yeah, this is going to be this whole setup there. Yeah, this whole setup there, like that. So that is the, or not the M there in this one here. So that is our K. So note that this exact relation is obtained from Rodriguez formula for the associated Lezondra function in which we have the following, yeah, which we have, have, have, you have the following identity, which I may derive in a future video, so PL, negative, negative M of x equals to a negative 1, power of M. times by l minus m factorial divided by l plus m factorial times p l m layout so this is this one i might derive it in a future video and here's ours using our t and n setups instead of the x and so on all right now that we have gone over this and basically explained why we're using these two setups here yes so uh so this this one here and that the y nm and y n minus m so now we're going to look at the definition so this is the real spherical harmonics nm of theta phi equals this for when m is greater than zero so we'll look at this one first right there so i'll just uh i'll just copy this all right we'll copy that and let's go right here yeah so for m is greater than zero the definition is so move it like that This is going to be, we'll write this as Y, N, M, real. The real spiral harmonic, and this is M is greater than zero. All right, so now the next setup is, let's just plug in our setups there, so we have our proportionality is stuff, so YNM is over here. Yes, we plug these in, Y, NM is just PNM, cosine, and then E to this one. The other one's same thing, but negative E to the I, M, phi, and there's a negative 1M set up there. So let's put that all inside. This becomes, again, this can be proportional to because we're using a proportionality set up. So negative 1, 2, divided by square root 2, and then YNM is going to be, well, P, N, N, Theta, then E to the I, m, phi. All right, and then the other one's going to be, again, we're doing proportional, because we're not including all those random normalization constants. So this is going to be plus, and now it's going to be, well, this is the, this part right here. It's going to be negative 1m square, I mean, to power of, yeah, negative 1 to power of m. and then, well, now we're going to have a second one, so we're going to times it by negative 1 to power of m. And now this is going to be, well, the proportionality constant, same thing, p, and m cosine theta, of cosine theta, and then this is going to be times it by e negative i. m5, so the complex conjugate of that one, the negative side of it. And all right here, I just quickly shrunk it to fit over there, but I'm also going to write here as YN M real. Like that. All right, so then we have this. Next setup, we can go over here and basically simplify. all this, but we're going to take out this PNM, so first we have this negative 1, power of M like that, square root 2, take out, just factor out the associated Lezondra function, so we're going to have a PNM of cosine Theta. All right, so take that one out, now we're going to have a, we're going to have this setup here in E. Actually, before we even get to that, so we factor that out. Move it down here. Note that right here, this negative 1 to power of m. So, negative 1 power of m times negative 1 power of m, well, that's just equals to 2m. Equals, yeah, negative 1 to power of 2m, but now there's a 2 there. So if we have a negative, well, this is going to be an even number. So if you have, for example, negative, if you have three, two times three, it would be six. So you're always going to have a, basic factor of two. In other words, even number. So this is even. So this whole thing equals the 1. So in other words, we could just say, yeah, get rid of that, so it equals 1. Now, this one here, PN, cosine, theta, move that out as well, that's over here. So we're going to get a E-I -M-fi, and then plus E-negative I -M-fi. And again, using Euler's formula, what we get here is, Alright, so what this becomes is, well, this equals to, and again, if you scroll back up to our Euler's formula, so we had R-E-I-5 equals to X plus I-Y, and then the other there times by the Euler's formula, cosine 5 plus I -sign-5. And then for the complex conjugate, the negative side, this E-R-I-5 is this term there. And again, this is, take the factor out the R. Yeah, we have X minus I, y, that's R, and then factored out, cosine, phi, minus i-sign phi. But in our case, we have M-fi, so we replace this phi with M-fi. And then it's just minus, that's the other ones plus. So we get, this whole top part's going to be, well, this is going to be a cosine. Yeah, cosine phi, and then plus, I mean, cosine, M-fi. m phi, put a bracket like that, plus i. sine m phi. All right. And then plus the other ones will be the same thing, cosine m phi, minus now, minus i sine. Yes, minus i sine phi. and the sine m phi, like that. Now right here, I just shifted it over to the left. So what this means is, well, this i -sign-fi, and this negative i-sign-fi cancels. Now we're just going to left with this is 2-cosine 5. This equals to 2-cosine m-fi. M-fi. So in other words, this whole setup equals to now negative 1, the power of m times by 2, divided by, I'm going to put the square two over here, PNM, cosine, uh, cosine, uh, Theta, times it by cosine, and fie, like that. But again, and also note that here, this is going to be, well, 2, this right here, just make it for completeness. Calcalspoken, most, a lot of areas always try to remove, simplify this. So 2 of the power of, this is equal to 2 the power of 2, divided by, it's…