1. May 29, 2015

### dyn

Hi. For the angular momentum l=2 case are the L2 and Lz matrices both 5 x 5 matrices with the following eigenvectors ?
$\begin{pmatrix} 1\\0\\0\\0\\0 \end{pmatrix}$ , $\begin{pmatrix} 0\\1\\0\\0\\0 \end{pmatrix}$ , $\begin{pmatrix} 0\\0\\1\\0\\0 \end{pmatrix}$,$\begin{pmatrix} 0\\0\\0\\1\\0 \end{pmatrix}$,$\begin{pmatrix} 0\\0\\0\\0\\1 \end{pmatrix}$

I am definitely confused about the addition of angular momentum using J = L + S. These are vectors but represented by square matrices ? And also the L and S matrices are usually not even of the same size so how can they be added ? Any help would be appreciated.

2. May 29, 2015

### blue_leaf77

The more proper way of writing that sum is $\mathbf{J} = \mathbf{L}\otimes I + I \otimes \mathbf{S}$ where $I$ in the first and second term correspond to the identity matrix in the spin space and orbital angular momentum space, respectively. $\otimes$ denotes kronecker/direct product. The resulting matrices in both terms in that sum will have $(2l+1)(2s+1) \times (2l+1)(2s+1)$ dimension, which determines the dimensionality of the $J$ space.

Last edited: May 29, 2015
3. May 31, 2015

### dyn

How do you get to that dimensionality ? I would have thought the 1st term has the dimensionality of the L matrix and the 2nd term the dimensionality of the S matrix which are usually different.
Ps am I right about the l=2 case ?

4. May 31, 2015

### blue_leaf77

The basis vector $|l,m \rangle$ and $|s_z \rangle$ live in different space, as you have mentioned, therefore if you want to construct a system in which both spaces interact, the new composite basis vector would be given by the direct product from the two constituent spaces, that is $|l,m \rangle \otimes |s_z\rangle$ and the linear transformations that act on this new space is also constructed by the direct product.

If you are only working within the l=2 subspace, yes $L^2$ and $L_z$ are 5x5.

Eigenvectors of which operator? If for example you have your $L_z$ diagonal, then they are the eigenvectors of $L_z$, if it's $L_x$ which is diagonal then they are the eigenvectors of $L_x$. But the general convention is indeed that, we use the eigenvectors of $L_z$ (and $L^2$) to span the subspace of a given $l$.

Last edited: May 31, 2015
5. Jun 1, 2015

### my2cts

Yes these are eigenfunctions of Lz and of L^2.

6. Jun 1, 2015

### blue_leaf77

Not so quick for taking a conclusion. You have to be careful if you are only given a set of eigenvectors in matrix form such as those above. To know which operator (in matrix form), a given set of standard basis vectors (vectors with one element being unity while the rest are zeros such as those above) are eigenvectors of, one has to find one or more matrices which are diagonal. It will be of different issue if you are given already in ket form, namely $|l,m \rangle$'s, they are certainly the eigenvectors of $L_z$ and $L^2$ because our convention has reserved the number $m$'s to be the eigenvalues of $L_z$.