# Representation of angular momentum matrix in Cartesian and spherical basis

1. Jul 27, 2010

### kof9595995

The two sets of matrices:
$${G_1} = i\hbar \left( {\begin{array}{*{20}{c}} 0 & 0 & 0 \\ 0 & 0 & { - 1} \\ 0 & 1 & 0 \\ \end{array}} \right){\rm{ }}{G_2} = i\hbar \left( {\begin{array}{*{20}{c}} 0 & 0 & 1 \\ 0 & 0 & 0 \\ { - 1} & 0 & 0 \\ \end{array}} \right){\rm{ }}{G_3} = i\hbar \left( {\begin{array}{*{20}{c}} 0 & { - 1} & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array}} \right)$$
and
$${J_1} = \frac{\hbar }{{\sqrt 2 }}\left( {\begin{array}{*{20}{c}} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array}} \right){\rm{ }}{J_2} = \frac{{i\hbar }}{{\sqrt 2 }}\left( {\begin{array}{*{20}{c}} 0 & { - 1} & 0 \\ 1 & 0 & { - 1} \\ 0 & 1 & 0 \\ \end{array}} \right){\rm{ }}{J_3} = \hbar \left( {\begin{array}{*{20}{c}} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & { - 1} \\ \end{array}} \right)$$
They both satisfy the common commutation relation
$$[{L_i},{L_j}] = i\hbar {\varepsilon _{ijk}}{L_k}$$ with L substituted by G or J
So both can be used to describe angular momentum. G is called the Cartesian basis representation and J is called spherical basis representation. This is absolutely new to me, so I want to acquire some general knowledge about it. For example, why are G and J named like that (Cartesian and spherical....), how are they related and so on. And any source of reference will be helpful.

2. Jul 27, 2010

### xepma

This is not something terribly deep, but it's good to know.

The two representations are related through a basis transformation. So it's just a different way of looking at the same physical system.

To be specific: take any of the G's (usually one takes G3), diagonalize it and you end up with J3. So there is a (unitary) basis transformation such that

$$J_3 = U^{-1} G_3 U$$

This same basis transformation transforms G2 and G1 into J2 and J1. More importantly: the basis transformation preserves the algebraic structure: the Lie bracket. Since a basis transformation is invertible you are dealing with an algebra isomorphism -- a bijective mapping which preserves the algebra structure.

This is the definition of similarity of two representations: the two representations are equivalent. Usually, you do not distinguish between two representations which are equivalent -- it's just a different look at the same physics. One way might be more convienent to use than the other, but it's important to realize that they do not describe different physical systems.

Sidenote: that what you call the spherical basis is called the Chevally basis in the (more general) context of Lie algebras. It refers to the basis in which you diagonalize as many generators of the Lie algebra as possible. The subspace spanned by these generators is called the Cartan subalgebra. The remaining, non-diagonalized operators are then formed to so-called step operators. Very interesting stuff, although quite an overkill for su(2) (the Cartan subalgebra is one dimensional).

You should really consult a proper book on Lie groups, Lie algebras and representation theory -- I recommend the book by Jones as an easy start.

I'll stop my rant now.

Last edited: Jul 27, 2010
3. Jul 27, 2010

### kof9595995

Thanks, the information is quite helpful. Yet I still don't understand why G has anything to do with Cartesian coordinates and J has anything to do with spherical coordinate. And what's the name of Jones' book?

4. Jul 27, 2010

### inempty

the eigenfunctions of the J's are spherical harmonic functions in position representation, J's generate the rotation matrices that act on state vectors when spherical functions are chosen to be bases . On the other hand, G's are the infinitesimal generators of space rotating matrices of cartesian vectors.

5. Jul 27, 2010

### kof9595995

Thanks inempty, nice to see a familiar face here:rofl:

6. Jul 28, 2010

### inempty

haha

7. Jul 28, 2010

### weejee

As a sidenote, quantum states that make a basis for G is p_x, p_y and p_z states.

8. Jul 28, 2010

### inempty

Why? I couldn't understand it clearly. After simple calculation it turns out that the basis functions are corresponding to x,y and z. They're not momentum operator's eigenfunctions.

9. Jul 28, 2010

### weejee

Hi inempty,

p_x, p_y and p_z don't mean momentum eigenstates. They are p orbitals, where p_x and p_y are linear combinations of |l=1,m=1> and |l=1,m=-1> states, while p_z corresponds to the |l=1,m=0> state. Their wavefunctions are proportional to x, y and z, respectively, as you have pointed out.

Last edited: Jul 28, 2010
10. Jul 28, 2010

### inempty

Oh, I got it :) you are right.