diagonalization vs partial waves

Diagonalization vs partial waves PhL 1.11.11 The word "diagonalization" refers to a certain matrix Knm in the discussion below. This matrix is "the matrix elements" of an operator K which here is an integral operator. The term "partial waves" here refers to expanding a function on sines and cosines (or corresponding exponentials) which then are "partial waves" of the SO(2) group. At the end we mention partial waves of SO(3). A "partial wave" refers in general to the projection of a function onto some basis functions which are in fact EF's of some other operator(s) [ Jz for SO(2), and J2, Jz for SO(3) ] . We show that SO(2) diagonalization of a convolution form integral equation causes Knm to be diagonal and this happens when all the functions are expanded on their SO(2) partial waves. Sometimes I forget that in "group diagonalization", there is in fact a certain matrix that becomes diagonal (or at least achieves block form diagonal). See elsewhere for more details. Overview (2/3 page, written 1.16.11) 1 A. Discussion 1 B. Summary of what happened above. 5 __________________________________________________________________________________ Overview (2/3 page, written 1.16.11) Given an arbitrary complete set of states |n> and eigenfunctions φn(x) on (a,b), a 1D integral equation f(x) = !Syntax Error, I dy/C k(x,y) g(y) can be converted to a matrix equation fm = Σn=0∞ Kmn gn where fm is the projection of f(y) etc and where the matrix Knm = <n|K|m> is in general fully populated and not diagonal. In the special case that k(x,y) = r(x-y), this conclusion still obtains. However, if this r(x) is periodic on the interval (a,b), then we can select for our set φn(x) the functions exp(iknx) where kn=2πn/(b-a) which happen to be the 1D representations of the group SO(2). With this selection for φn(x), the matrix Knm becomes diagonal and the matrix equation becomes fm = Kmmgm which is trival to solve for gm. This basis φn(x) is in this case the set of eigenfunctions of K, and the rn = Knn are the eigenvalues. Sections A and B just fill out this paragraph. A. Discussion Section 1 gives the QM notation for how this all works, and talks about a certain constant C. Section 2 notes the fairly obvious fact that if the φn(x) are EF's of K, then Knm is diagonal. But of course you don't in general know the EF's of K, so this is just an academic statement. Section 3 notes that k(x,y)=r(x-y) in general does not make Knm be diagonal. Section 4 notes that if k(x,y)= r(x-y) AND r(x) is periodic in (a,b), for general φn(x) again nothing has been simplified. Knm is shown in equation (*) and is in general not diagonal. Section 5 notes that if k(x,y)= r(x-y) AND r(x) is periodic in (a,b) AND you take φn(x) = exp(iknx) where kn=2πn/(b-a), then Knm becomes diagonal. We don't consider the infinite interval case in this doc, but we know exactly how to handle it. B. Summary of what happened above. This long section just reviews with numbered items what we showed above. ___________________________________________________________________________________ A. Discussion 1. Suppose we have a 1D integral equation on some interval (a,b) f(y) = !Syntax Error, I dx/C k(y,x) g(x) where C is some arbitrary constant. Suppose we have a set of orthonormal eigenfunctions φn(x) = <n|x> on this interval (a,b). Assume that the eigenfunctions can be labeled by n = 0,1,2,.... This means that completeness orthogonality !Syntax Error, Idx/C |x><x| = 1 <x|x'> = C δ(x-x') Σn=0∞ |n><n| = 1 <n|m> = δnm In what follows, we will always have the |n> states be orthonormal, so <n|m> = δnm . But we are free to normalize the coordinate representation states |x> any way we want, and we have chosen to add the constant C which is the same as the C appearing in the integral equation. Normally people use C = 1, but we shall use C = C. Note then that we write 1 = <n|n> = <n|!Syntax Error, Idx/C |x><x|n> = !Syntax Error, Idx/C <n|x><x|n> = !Syntax Error, Idx/C φn(x)* φn(x) Then we do some expansions like so (we show the corresponding projections on the far right) f(x) = Σn=0∞ fnφn(x) ie <x|f> = Σn=0∞ <x|n><n|f> fn = !Syntax Error, Idx/C φn(x)* f(x) g(x) = Σn=0∞ gnφn(x) ie <x|g> = Σn=0∞ <x|n><n|g> gn = !Syntax Error, Idx/C φn(x)* g(x) Now use QM notation to write the integral equation as <y|f> = !Syntax Error, Idx/C <y|K|x><x|g> = <y|K|g> => |f> = K|g> Then process this last result: |f> = K|g> = Σn=0∞ K|n><n| g> <m|f> = <m|K|g> = Σn=0∞ <m|K|n><n| g> fm = Σn=0∞ Kmn gn In this way we have converted our equation f = Kg from coordinate space to eigenfunction space. For some random kernel k(y,x) and some random set φn(x) on (a,b), Knm is not going to be a diagonal matrix. We can write Kmn = <m|K|n> = <m!Syntax Error, Idx/C |x><x|K|!Syntax Error, Idy/C |y><y|n> = !Syntax Error, Idx/C!Syntax Error, Idy/C φm(x)* <x|K|y> φn(y) = !Syntax Error, Idx/C!Syntax Error, Idy/C φm(x)* K(x,y) φn(y) 2. But suppose we can find eigenvalues and eigenfunctions of the operator K which form a complete set, K|n> = λn|n>, and suppose we take these (orthonormalized) to be our set φn(x). Then we have Kmn = <m|K|n> = <m|λn|n> = λn<m|n> = λn δm,n so with this special set of φn(x), Kmn is a diagonal matrix, and our integral equation becomes fm = Σn=0∞ Kmn gn = Σn=0∞ λn δm,n gn = λm gm In a sense, we could say I think that we have "diagonalized" our integral equation. The solution is then gn = fn/λn g(x) = Σn=0∞ gnφn(x) = Σn=0∞ (fn/λn)φn(x) so barring that some λm = 0, we have the problem completely solved. The Catch 22 here is of course that it may be extremely difficult to find and enumerate the eigenfunctions of K, and if K is not Hilbert-Schmidt, the eigenfunctions may not form a complete set. Stakgold has lots to say about this integral equation example and the solution outlined here. 3. How is the above discussion affected if k(x,y) = r(x-y) so the integral equation then is of "convolution form" ? Since this is just a special case, everything above still applies. We would then have for some general φn(x) set Kmn = <m|K|n> = !Syntax Error, Idx/C!Syntax Error, Idy/C φm(x)* r(x-y) φn(y) Nothing is really any simpler than it was before. Kmn is still in general non-diagonal for some general set φn(x) you pick. 4. Suppose we now add the extra condition that r(x) is periodic on the interval (a,b). Does that make things any simpler? To make things easier, assume that a = 0 so we then have Kmn = !Syntax Error, Idx/C!Syntax Error, Idy/C φm(x)* r(x-y) φn(y) r(x +b) = r(x) If some f(x) is periodic in this way, then we know that for any real α, !Syntax Error, Idx f(x) = !Syntax Error, Idx f(x) Suppose in our Kmn integral we define x' = x-y, does that help? Kmn = !Syntax Error, Idy/C !Syntax Error, Idx/C φm(x)* r(x-y) φn(y) = !Syntax Error, Idy/C φn(y) !Syntax Error, Idx'/C φm(x'+y)* r(x') Officially the functions φn(x) are defined only on the interval (a,b), but we can replicate what we see in this interval across the entire x axis, and thus cause this extended φn(x) to be periodic with period b. Then the dx' integrand shown above would be periodic with period b and we could say = !Syntax Error, Idy/C φn(y) !Syntax Error, Idx'/C φm(x'+y)* r(x') = !Syntax Error, Idx/C !Syntax Error, Idy/C φm(x+y)* r(x) φn(y) (*) but this does not seem to be a simplification for arbitrary φn(x). 5. Let's now further assume that b = 2π, and write coordinates as if they were angles. At the same time, we shall now set C = 2π, since we can set it to whatever we want. Then we have Kmn = !Syntax Error, Idθ'/2π !Syntax Error, Idθ/2π φm(θ)* r(θ-θ') φn(θ') For general a and b, we could scale things to obtain the above form. We now continue our list of assumptions by assuming now that the eigenfunctions φm(θ) are those associated with the group SO(2), so that φm(θ) = e-inθ now for n = all integers (making a slight change to what we assumed above about n). This is well known to be a complete set of basis functions. These eigenfunctions have "the group property" which is this φm(θ)φm(θ') = φm(θ+θ') Now let's look at our Knm integral in the form (*) above Kmn = !Syntax Error, Idθ/2π !Syntax Error, Idθ'/2π φm(θ+θ')* r(θ) φn(θ') = !Syntax Error, Idθ/2π !Syntax Error, Idθ'/2π φm(θ)*φm(θ')* r(θ) φn(θ') = !Syntax Error, Idθ/2π φm(θ)* r(θ)!Syntax Error, Idθ'/2π φm(θ')* φn(θ') = !Syntax Error, Idθ/2π φm(θ)* r(θ) δm,n = δm,n !Syntax Error, Idθ/2π φm(θ)* r(θ) = δm,n <m|r> = δm,n rm where we now define rm in the obvious manner (just as we did for f and g above) rm = <m|r> = <m| !Syntax Error, Idθ/2π |θ><θ|r> = !Syntax Error, Idθ/2π <m|θ><θ|r> = !Syntax Error, Idθ/2π φm(θ)* r(θ) Suddenly Kmn is diagonal! This means that the rm and φm(θ)are the eigenvalues and eigenfunctions of the operator K. ( Above we referred with the eigenvalues as λm). Our φn(θ)-space integral equation is then fm = Σn=0∞ Kmn gn = Σn=0∞ δm,n rm gn = rmgm which we write again fn = rn gn The solution to our problem is then gn = fn/ rn g(θ) = Σn gnφn(θ) = Σn (fn/rn)e-inθ so barring that one of the rn = 0, we have a complete solution to our problem. B. Summary of what happened above. 1. We first considered an integral equation with some arbitrary kernel f(y) = !Syntax Error, I k(y,x) g(x) dx/C We would like to solve this integral equation for g(x), given k and f. 2. There are an infinite number of complete sets of eigenfunctions φn(x) on the interval (a,b) -- each would correspond to some self-adjoint differential operator Lx with some boundary conditions at a and b, a so-called Sturm-Liouville problem. If we re-express the integral equation in the |n> basis instead of the |x> basis, where <x|n> = φn(x), we obtain fm = Σn=0∞n Kmn gn where Kmn = <m|K|n> = !Syntax Error, Idx/C!Syntax Error, Idy/C φm(x)* K(x,y) φn(y) fn = !Syntax Error, Idx/C φn(x)* f(x) f(x) = Σn=0∞ fnφn(x) gn = !Syntax Error, Idx/C φn(x)* g(x) g(x) = Σn=0∞ gnφn(x) We have assumed that the eigenfunctions φn(x) can be labeled by n = 0,1,2... but some other infinite numbering system might be appropriate depending on the Sturm-Liouville problem. It might come out that we want n = all integers from -∞ to +∞ including 0, for example. 3. For some general k(y,x) kernel and some randomly chosen complete set φn(x) on (a,b), the matrix Kmn will not be diagonal. The integral equation has now become a matrix/vector equation which we could write as f = K g so we have a linear algebra situation, except the vectors have an infinite number of components and K is an ∞ x ∞ matrix. Perhaps if we keep only the first 50 components and the upper left 50 x 50 part of the matrix K, we can use linear algebra to solve the above equation for g = K-1f and our problem will then be approximately solved if we can show that higher components make very small contributions. 4. For certain kernels k(y,x) the corresponding Hilbert Space operator K will have eigenvectors and eigenfunctions which form a complete set which we could then use as our φn(x). This is of course a very special set of basis functions, and it might be very difficult to find out what the φn(x) are. If k(x,y) is symmetric k(x,y) = k(y,x), and if the double integral of k2 over the (a,b) x (a,b) square is finite, then K is a Hilbert-Schmidt integral operator and such a complete set of φn(x) is known to exist. There will be some eigenvalues λn such that K|n> = λn|n>. In this very special basis, the matrix Knm is trivially diagonal, having the form Knm = λn δn,m. The matrix equation f = K g then becomes fn = λn gn and our problem is completely solved, barring technical difficulties, since we then have gn = fn/λn and we then put these coefficients into our expansion for g(x). The big problem here is that, even assuming k(y,x) has the required properties, it is usually very hard to find the λn and φn(x) and one can only do that by some numeric approximation method. 5. If it happens that the kernel has the "convolution form" k(x,y) = r(x-y), in general there is no simplification of the problem outlined above. For a general set φn(x), Knm won't be diagonal. And K will have some eigenfunctions which will likely still be hard to find. The Love Equation is an example. 6. However, if k(x,y) = r(x-y) and if r(x) is periodic with a period equal to the interval width b-a, then all of a sudden the problem becomes simple. Assuming a and b are finite, the fact that r(x) is periodic means it can be expanded in a complex Fourier Series, r(x) = Σn rn φn(x) // expansion rn = !Syntax Error, Idx/C r(x) φn(x)* // inversion C = (b-a) where φn(x) = e-inx[2π/(b-a)] and we note in passing that <n|n> = !Syntax Error, Idx/C <n|x><x|n> = !Syntax Error, Idx/C e+inx[2π/(b-a)] e-inx[2π/(b-a)] = !Syntax Error, Idx/C = 1 It then follows, due to the obvious "group property" of these basis functions, φn(x+x') = φn(x)φn(x'), that the basis functions are in fact eigenfunctions of K with eigenvalues rn. There are various ways to show this, here is one way. First, we have k(y,x) = r(y-x) = Σn rn φn(y-x) = Σn rn φn(y)φn(-x) = Σn rn φn(y)φn(x)* Second, we write φm(y) ≡ <y|K|m> = <y|K !Syntax Error, Idx/C|x><x |m> = !Syntax Error, Idx/C <y|K|x><x|m> = !Syntax Error, Idx/C k(y,x) φm(x) = !Syntax Error, Idx/C r(y-x)φm(x) so we can insert the first result into the second to get φm(y) = !Syntax Error, Idx/C[ Σn rn φn(y) φn(x)*]φm(x) = Σn rn φn(y) !Syntax Error, Idx/C φn(x)* φm(x) = Σn rn φn(y)<n|m> = Σn rn φn(y)δn,m = rm φm(y) which we can write also <y|K|m> = rm <y|m> or just K|m> = rm|m>. Thus, K is diagonal Knm = <n| K |m> = rm <n|m> = rm δn,m and then our matrix equation f = K g becomes just fm = rm gm and the solution to our integral equation problem is then g(x) = Σn gn φn(x) where gn = fn/rn If a and/or b are infinite, we must modify the above analysis replacing Fourier Series by Fourier Transforms, but the general nature of the result comes out the same. 7. Let's now go back to the original integral equation f(y) = !Syntax Error, I dx/C k(y,x) g(x) and let's go ahead and assume the convolution form for the kernel, f(y) = !Syntax Error, I dx/C r(y-x) g(x) y in (a,b) The whole problem is defined on the interval (a,b) of width L = b-a. Notice that, depending on the value of y and the location of the integration point x, the function r(x) is accessed from x = -L to x = +L, which is a width of 2L. If we wanted, we could take this 2L wide chunk of r(x) and replicate it along the entire x axis. Replacing the original r(x) with this replicated r(x) makes no difference since we only access (-L,L) of this function. But this does make r(x) be periodic, but alas, the period is 2L, not L, so we don't get the massive simplification described above for some general r(x). Nevertheless, we could expand all the functions on a set of basis functions over the interval (-L,L) which has width 2L. We could even perhaps use our same complex Fourier Series basis functions, but geared to 2L. This will then lead to a matrix equation of the form f = K g but K won't be diagonal and we will have to use some kind of approximation method. This is done in the Love numerical paper I downloaded. We only get the massive simplification (diagonalization) when r(x) is periodic with period L = b-a. When we say that we can "diagonalize" a convolution equation using some φn(x), we have shown above that we are in fact "diagonalizing" matrix Kmn so the projected equation is not f = Kg, but fm = rm gm.The matrix in question is <m|K|n> where |n> are the SO(2) representation states. If we were to diagonalize using SO(3) instead of SO(2) as discussed here, the matrix K is then Kjm,j'm' = <jm| K |j'm'> = δj,j'<jm| K |jm'> In this case the matrix Kjm,j'm' is not completely diagonal, but has a block diagonal form where the size of each block is (2j+1)2.