Insert Latex formulas into the threads?

1. Nov 10, 2004

jet10

hi there
I am new here. Could some one tell me how to insert Latex formulas into the threads?

we get imanary no. in these matrices.

Let's say we have S= n1.sx + n2.sy +n3.sz showing us the magnitude in each component (sx, sy, sz are the spin matrices/vectors). What does it show us? How should we interprete the imaginary no.?

2. Nov 10, 2004

dextercioby

1.Go to a thread where u can find latex formulas written.Click on one of them and u'll be opened a window in which u find the "sourse code" for that formula.Below it there's a link to the webpage (pdf file which can be downloaded) in which,basically on 4 pages,u are explained how to write tex.

2.What complex number??

3. Nov 10, 2004

humanino

You seem not to take the difference between matrices and vector not too seriously. This is a big issue. Consider say a 3-vector (ordinary vector). How do you define such an object ? Usually by giving its transformation law under the rotations. This is equivalent to saying you deal with a certain representation of $$SO(3)$$.

Alternatively, one can represent such a three vector using a hermitean, traceless 2x2 matrix. This is related to the fact that $$SU(2)$$ is the covering group of $$SO(3)$$. The usual rotation of such a matrix is a unitary transformation. Now the reason for the imaginary numbers is the following : when passing from a Lie group to its algebra, the usual convention is $$g=\exp^{\imath G}=\exp^{\imath \alpha^iG_i}$$ where $$g$$ is in the group and $$G$$ is in the algebra. The $$\alpha^i$$s are the coordinates in the $$G_i$$ basis. The fact that we actually obtain an algebra can be seen as $$\exp^{A}\exp^{B}=\exp^{A+B+\frac{1}{2}[A,B]+\cdots}$$ where $$\cdots$$ contain commutators of commutators, commutators of commutators of commutators....
so by the very definition of what generators are, we can infer the fact that the commutator of two generators is a linear combination of the generators : the algebra "closes". In the case of $$SO(3)$$ we have antisymmetric structure constants : $$[G_i,G_j]=-\imath\epsilon_{ij}^kG_k$$. Now the punchline : we actually can make the $$\imath$$ disappear by absorbing it into the generator :
If $$G\rightarrow -\imath G$$ then $$g=\exp^{G}$$ and $$[G_i,G_j]=\epsilon_{ij}^kG_k$$. But then the generators are antihermitean !

For instance, for translation, we can set $$P_i=\partial_i$$ and for rotations $$J_{ij}=x_{[i}\partial_{j]}$$ since anyway the $$\imath$$ is doomed to be canceled eventually.

I think there is not much more into the $$\imath$$ than convenience for manipulating expressions. In the case of spacetime symmetries it is legitimate to eliminate the $$\imath$$ from the definition and use antihermitean generators. In the case of internal symmetries, it seems more convenient to keep hermitean generators. Also, complex Lie algebra are nice to use, whereas real Lie algebras require more care (just as real polynomials are not as nice to factorize as complex polynomials).

Again about the vector/matrix "duality" : the $$\gamma_\mu$$ matrices of Dirac are both spin matrices and (one single) Lorentz vector (the set of Dirac matrices is a vector. Each component of the vector is a matrix). They are the link between Lorentz and spin indices.
Also in the case of Pauli matrices for the fundamental $$SU(2)$$, the 3-vector "duality" previously mentionned is explicitely

$$(V)_{\alpha\dot{\beta}}=\left( \begin{array}{cc}V_+,V_1^*\\V_1,V_-\end{array} \right) = \frac{1}{\sqrt{2}} \left( \begin{array}{cc}V_0+V_1,V_2-\imath V_3\\V_2+\imath V_3,V_0-V_1\end{array} \right) =V_a (\sigma^a)_{\alpha\dot{\beta}}$$
and in that case also, the Pauli $$\sigma$$ matrices are (the three components of) a 3-vector (even though each component is a matrix).

EDIT : in the last formula, one should set $$V_0 = 0$$ for a three vector. This form displayed also works with 4-vectors.

Last edited: Nov 10, 2004
4. Nov 12, 2004

jet10

Thanks. I think I am not far enough to understand what it really means. I am just beginning to understand the formalism of QM. I am not familiar with terms $$SO(3)$$ and Lie groups. I will read your answer again later, when I am more acquainted with them.