Six Generators of Group SO(4)

[YOBDC]

Recently,I have read a article which referred "the six generators of group SO(4)".And who can tell me what are these generators and what are their matrix forms?

 

[tom.stoer]

There a three rotations L^a (generating the SO(3) subgroup of SO(3,1) and three boosts K^a.

In addition one can complexify the algebra to SO(4,C) and define the new linear combinations

L_±^a = L^a ± iK^a.

These new sets of generators define two mutually commuting SU(2,C) algebras.

The matrix form can e.g. be found in "Jackson: Electrodynamics".

 

[DrDu]

The Hamiltonian of the Coulomb problem is invariant under SO(4) and is discussed e.g. in Landau Lifshetz, Quantum Mechanics. In that case the six generators are the three generators of rotations and the three so called Runge Lenz vectors which describe the orientation of the big axis of the ellipses of the particles.

 

[arkajad]

They are:
$$L_{12}=\begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$$

and five similar $L_{ij},\quad i<j$

 

[YOBDC]

They are:
$$L_{12}=\begin{pmatrix}0&1&0&0\\-1&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$$

and five similar $L_{ij},\quad i<j$

Did you mean that for each L_ij(i<j),there are only two non-zero elements:l_ij=1 and l_ji=-1?

 

[arkajad]

Did you mean that for each L_ij(i<j),there are only two non-zero elements:l_ij=1 and l_ji=-1?

Yes. But, of course, one can always choose a different basis of generators by taking independent linear combinations of the above ones,

 

[YOBDC]

Yes. But, of course, one can always choose a different basis of generators by taking independent linear combinations of the above ones,

Oh, I see. These generators must be anti-symmetric. Thank you very much!

 

[tom.stoer]

Note that the counting is different; arkajad uses ij with i<j (ij being spacetime indices) whereas I am using a=1..3 (NOT being spacetime indices; suppressed). Of course in both cases you have six generators in total and of course they are one-to-one.

 

[samalkhaiat]

These generators must be anti-symmetric.

You can represent the n(n-1)/2 generators of any SO(n) in some orthonormal base, by the operator

$$L_{nm} = |n\rangle \langle m| \ - |m \rangle \langle n|$$

where n,m = 1, 2, ...,n and;

$$\langle n|m\rangle = \delta_{nm}$$

From that you can find the location of the non-zero matrix elements by calculating the martix element;

$$\langle r|L_{nm}|s \rangle$$