# Addition for generators of rotation

1. Aug 14, 2011

### jfy4

Hi,

I have here the following rules for representations of the Lorentz algebra including spinors
• The representations of the Lorentz algebra can be labeled by two half-integers $(j_{-},j_{+})$.
• The dimension of the representation $(j_{-},j_{+})$ is $(2j_{-}+1)(2j_{+}+1)$.
• The generator of rotations $\mathbf{J}$ is related to $\mathbf{J}^{+}$ and $\mathbf{J}^{-}$ by $\mathbf{J}=\mathbf{J}^{+}+\mathbf{J}^{-}$; therefore, by the usual addition of angular momenta in quantum mechanics, in the representation $(j_{-},j_{+})$ we have states with all possible spin $j$ in integer steps between the values $|j_{+}-j_{-}|$ and $j_{+}+j_{-}$.

My question is that given a representation $(1,\frac{1}{2})$, the dimension for the plus and minus generators are different (3 and 2), yet I am suppose to add these matrices together for the generators $\mathbf{J}$. Would it be a direct product $A\oplus B$? I don't see how you can add matrices together with different dimension.

Thanks,

2. Aug 14, 2011

### kith

I'm not very familiar with group theory, but direct products and direct sums do not require the matrices to be of the same dimension. Else it would not be possible to describe composite systems of two parts whose Hilbert spaces are of different dimensions.
http://en.wikipedia.org/wiki/Tensor_product#Kronecker_product_of_two_matrices
http://en.wikipedia.org/wiki/Direct_sum_of_matrices#Direct_sum

If you add two operators which act in different spaces, usually the following is meant:
$A + B = A \otimes 1 + 1 \otimes B$

Last edited: Aug 14, 2011
3. Aug 14, 2011

### jfy4

Thanks kith,

That's my tentative answer too right now. I think it is the direct sum (I wrote direct product, oops, ). That would get me a vector that has one 5x5 matrix for each component of the vector.

4. Aug 14, 2011

### Bill_K

No, actually you want the direct product. Saying that a representation of the Lorentz group is (1, ½) means that J+ acts on a 3-dimensional representation and J- acts independently on a 2-dimensional representation. The Lorentz group representation has (2j-+1)(2j++1) components, which in this case is six. You need to construct a six-dimensional matrix as a direct product, and then presumably reduce it.

5. Aug 14, 2011

### jfy4

Really!? How should I interpret
$$\mathbf{K}=-i(\mathbf{J}^{+}-\mathbf{J}^{-})$$
Is this the direct product of the generators such that one of them is negative? Before I thought I understood these, I had thought before this was the direct sum with $\mathbf{J}^{-}$ negative. Help!

6. Aug 14, 2011

### Bill_K

The Lorentz group has six generators: three rotations J and three boosts K. You form linear combinations J+ and J-, and discover that J+ commutes with J- and each of J+, J- generates a 3-D rotation group, SU(2).

States in a representation of SU(2) are labeled as |j, m>. A representation of the Lorentz group will be a direct product of two such SU(2) representations and have states labeled as |j+, m+>|j-, m->.

As kith pointed out above, for an operator like K which is a linear combination of J+ and J-, each term acts separately on its own space, and the action can be written (J+ ⊗ 1) ⊕ (1 ⊗ J-).

7. Aug 14, 2011

Thanks guys.