Hi everyone! Im trying to get the eigenvalues of L^2 operator but without using the usual method of solving Laplace equation in spherical coordinates. I mean only using the matrixes for Lx, Ly and Lz. But I dont get j(j+1) as it should be! Any ideas? I should have named this threat SO(3) by the way... Please Register or Log in to view the hidden image!
Hi arfa brane! I was just realizing if it is posible to use these operators in their matrice form and use only linear algebra to get the eigenvalues as it usually done (in linear algebra). The matrices im using are the 3x3 "version" of pauli matrices. So I dont know if there is any mathematical reason that doesnt allow me to use the sum of the squares of Lx Ly and Lz ( (Lz)^2 + (Ly)^2 + (Lx)^2 = L^2 ) and try to get the eigen values for L^2 as usually ( det ( L^2 - I \lamda ) = 0 ).
The Pauli matrices are all elements of SU(2). I don't understand what you mean by "But I dont get j(j+1) as it should be!" And, have you looked at the quaternions?
Ok we can put it other way: the orbital momentum operaror has the eigenvalues j(j+1) (that is demonstrated when solving Laplace equation in spherical coordinates). Then I undertand that there is an homomorphism between SO(3) and SU(2). In my lectures of group theory (long ago) I have that in SU(2) the Lie algebra generators are some matrices 3x3 someway related to Pauli matrices. After that we apply Lz, L+ and L- to the state vector and get the eigenvalues for these operators and through the Kazimir operator we get that for L^2 it is j(j+1). My question is: can I use these generators and use them as mtrices and get the same eigenvalues (as it is done in linear algebra (through the determinant) ?
Great! That's a really useful article I just didnt know (nobody told me ever) that the representation used is related to a certain eigenvalue. Thank you so much!
Ok, it would be instructive, I think, to go over what exactly the problem was here. oscmh3 says in the OP he's "trying to get the eigenvalues of L^2 operator but without using the usual method of solving Laplace equation in spherical coordinates." The \( L^2 \) operator is the Casimir operator, and I think in the representation (spherical coordinates) it's the angular (spin) momentum operator. The link I referenced above about representations of SO(3) says: And at https://en.wikipedia.org/wiki/Casimir_element it says: . But, you have the Pauli operators "out of phase" by a factor of \( \sqrt {-1} \), relative to the SO(3) operators . . .?
I think the interesting part is to demonstrate these formulas for the different representations: Please Register or Log in to view the hidden image! Maybe someone knows how is that done)
That sounds like a question someone who knows about Lie algebras could answer. When j = 1/2 the above gives: \( (J_z)_{ba} = (\frac {3} {2} - a)\delta_{ab,a} \), and \( 1 \le a,b \le 2\). Whether \( (J_z)_{ba} \) is nonzero depends on the Kronecker delta, and so it depends on ab = a. Just thinking out loud.
Sorry, I must be misinterpreting what \( \delta_{ab,a} \) means. It must mean you take the Kronecker delta of ab first, then repeat with a; otherwise it doesn't seem to describe a matrix that looks like (up to a constant) \( \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \). It's not familiar notation.
Well that isnt a familiar notation for me either, but I also interpreted it the way you did for spin 1 and got the correct result. For spin 1/2 you are getting one of Paulies matrices... I couldnt understand why is the spin related to the dimention of the representations if they are all in fact representations of the SU(2) group...But the same article in wikipedia says: So, above, the 3×3 generators L displayed act on the triplet (spin 1) representation, while the 2×2 ones (t) act on the doublet (spin-½) representation. By taking Kronecker products of D1/2 with itself repeatedly, one may construct all higher irreducible representations Dj. That is, the resulting generators for higher spin systems in three spatial dimensions, for arbitrarily large j, can be calculated using these spin operators and ladder operators. So that explains a lot Please Register or Log in to view the hidden image!
SU(2) can be described in the following way: take an electron and its antineutrino partner. Then the two bosons, the photon and W boson define transformations, the photon is the identity transformation, and the W boson transforms electrons into antineutrinos and antineutrinos into electrons. This works with positrons and neutrinos too. The transformation symmetries are expressed by Feynman diagrams. Actually, the above is a bit inaccurate, the photon describes U(1) symmetry, you have SU(2) x U(1) where SU(2) is defined over the spin of electrons and their antineutrinos because both are left-handed. In that case, the 2 x 2 transformation matrix representation of SU(2) has \( W^0 \) on the diagonal, and the charged \( W^{\pm} \) on the off diagonal. U(1) is 'spinless'. The photon appears in SU(2) x U(1) acting only on electrons (as the identity), neutrinos don't interact with the EM field. Neutrinos and antineutrinos are leptons, but with zero charge and a very small rest mass; you can't stop a neutrino and reverse its direction as you can with an electron, so neutrinos have invariant spin. Here, it's like SU(2) is the reason you can only measure electrons, measuring neutrinos means transforming them into photons via W bosons.
Ok, that's the electroweak 'sector'; but SU(2) as we've seen above covers the spin degrees of freedom of fermions and bosons; photons have SU(2) symmetry too. The close relation between SU(2) and SO(3) and the relations between fermion and boson statistics implies this "computational" structure. Think about what happens at the LHC when it's running, then (if possible) reversing the direction of time without breaking any laws of physics--all the gathered data and all the discarded data generate opposing beams of protons at each detector. It's a kind of statement about what we can compute and why.
As a freshman I learned enough linear algebra to barely squeek by. As a telecom engineer, I learned enough about Galois fields to apply it to error correction technology. And then I met a savant who could best the answers of anyone so instructed, apparently effortlessly, and without using anything like group theory in order to do so. Needless to say, this was not sufficient cause to make me wish I had bothered to learn any more about eigenvectors.
Sorry, I've gotten the parity wrong above: you take electroweak doublets with the same spin direction, so their weak charges sum to zero. So with the electron and electron neutrino you have an element of SU(2) in the gauge theory 'representation'; the W boson "gauges" the weak interactions between leptons. In some sense, the neutrino spin fixes the spin direction of its weak doublet partner, the electron.
Well that is a kind of mediocre point of view. Of course if solving something easy often logic is enough, but try to create grand unification theory without using SU(5) or groups of even higher dimentions. Group theory is an incredible useful field that predicted lots of particles and it is very beautiful too.
I am a bit confused by this thread - is it about spin? For my benefit only I will assume so. I note first that "spin" in quantum physics refers to a certain sort of symmetry, rather than that which a wheel, for example might do. The Lie groups describe continuous symmetries, so with a massive leap of faith assume the group we want is the special unitary group \(SU(2)\). (This means that for any element \(A \in SU(2)\) - a matrix with entries from the complex field \(\mathbb{C}\) -that \(A^{\dagger}= A^{-1}\) where the \dagger denotes the conjugate transpose and the "special" means that {tex]A[/tex] has determinant one). Let's assume that is all we know for now. Now to every Lie group we associate an algebra, in the present case call it \(\mathfrak{su(2)}\), Let's assume the Lie theory assertion that {tex]SU(2)[/tex] is endowed with the structure of a differential manifold and that 1. We can identify \(\mathfrak{su(2)}\) with the tangent space at the identity\(e\) - \(T_eG\) - and that we will require the "commutator" \([\, ,\,] \in \mathfrak{su(2}\). Let's further assume that the group] \(SU(2)\) has the topology of the 3-sphere. It follows that we seek a basis for the vector space \(T_eG\) which is represented by a set of 3 traceless 2 x 2 matrices. (There is another bit of Lie theory that states that if \(\text{det}A \in SU(2) = 1\) then its adjoint representation in \(\mathfrak{su(2)}\) is traceless). A final bit theory tells us that we require every basis vector \(X \in \mathfrak{su(2)}\) be anti-Hermitian, that is \(X^{\dagger}=-X\). With that windy preamble, and taking it all on board, we are left with few choices, Here is one. For the basis I suggest \(X_1= \begin{pmatrix}0&i\\i&0 \end {pmatrix}\) \(X_2 =\begin{pmatrix}0&1\\-1&0 \end {pmatrix}\) \(X_3 =\begin{pmatrix}i&0\\0&-t \end {pmatrix}\) Working out the commutators I find this give \([X_i,X_j] = -2X_k\). Since these are twice as "big" as they need to be, I end up with \(X_1= \frac{1}{2}\begin{pmatrix}0&i\\i&0 \end {pmatrix}\) \(X_2 =\frac{1}{2}\begin{pmatrix}0&1\\-1&0 \end {pmatrix}\) \(X_3 =\begin{pmatrix}i&0\\0&-i \end {pmatrix}\) Umm. This is already to long. In a while ;ll show you how to construct the group from its algebra, and what, in the present case, is has to do with spin
It is not only about spin, it represents lots of things that have the same symmetry. But assuming it represents spin you dont lose generality. But I didnt really understand your question. Please Register or Log in to view the hidden image!
