# Why is (1/2,1/2) the vector representation?

1. Jul 6, 2013

### Mesmerized

Hello,
I'm reading Zee's book 'Quantum Field Theory in a Nutshell', the chapter about Lorentz group representations at the moment. In the end of the chapter there is suggested an exercise - "Show by explicit computation that (1/2,1/2) is indeed the Lorentz vector". And I just can't figure it out.

Here' s a little background. Lorentz group has 6 generators - 3 rotations $J_i$ and 3 boosts $K_i$ with commutation relations
$[J_i,J_j]=i\epsilon_{ijk}J_k, [J_i,K_j]=i\epsilon_{ijk}K_k, [K_i,K_j]=-i\epsilon_{ijk}J_k,$. We define $J_{+}=1/2(J_i+i K_i), J_{-}=1/2(J_i-i K_i)$, which now satisfy the SU2 groups commutation relations $[J_{+i},J_{+j}]=i\epsilon_{ijk}J_{+k},[J_{-i},J_{-j}]=i\epsilon_{ijk}J_{-k},[J_{+i},J_{-j}]=0$, which means that Lorentz group is isomorphic to SU2$\otimes$SU2.
Now coming back to (1/2,1/2) representation, it means that $J_+$ as well as $J_-$ are represented by 1/2 representation of SU2, i.e. by $1/2\sigma_i$, the Pauli matrices. Here comes the first misunderstanding, Pauli matrices are 2 by 2 matrices, Lorentz generators (therefore $J_{+}, J_{-}$ too, as they are just the sum of the Lorentz matrices) are 4 by 4.

As I couldn't solve it I decided to go backwards. J's and K's can be represented by
$$J_1 = \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0\\ \end{bmatrix}, etc, K_1 = \begin{bmatrix} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}, etc \\$$
therefore, for example $J_{+1}, J_{-1}$ have the following forms
$$J_{+1} =1/2(J_1+i K_1)=1/2 \begin{bmatrix} 0 & i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0\\ \end{bmatrix}, J_{-1} =1/2 \begin{bmatrix} 0 & -i & 0 & 0\\ -i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0\\ \end{bmatrix}. \\$$
I just can't see how these matrices can be related 1/2 representation of SU2 group, i.e. to Pauli sigma matrices. While the words 'Show by explicit computation' in the exercise tell me, that they somehow should. Any help, even some general sketch of that computation would be much appreciated.

2. Jul 6, 2013

### Bill_K

In the (1/2, 1/2) representation, J+ is represented by the tensor product σi ⊗ I, and J- by the tensor product I ⊗ σi which are 4x4 matrices.

3. Jul 6, 2013

4. Jul 6, 2013

### dextercioby

The representation (1/2,1/2) of SL(2,C) is isomorphic to the representation of weight 1 of the (restricted homogenous) Lorentz group due to the existence of the Infeld-van der Waerden symbol, $\left(\sigma^{\mu}\right)_{\alpha\dot{\beta}}$. The direct proof by calculation is done by Wiedemann and Mueller Kirsten in their book on Supersymmetry, Proposition iii), Page 74.

5. Jul 6, 2013

### Mesmerized

Sam,
this is what I understood from your responses in those threads. (1/2,1/2) is a representation with two indices (dotted and undotted), and each of those two 1/2 representations acts on the corresponding index. I understand this part. Then, as you write, "It (the representation with two indices) can be identified with vector field via"
$$\Psi_{r\dot{r}}=V_μ(x)(σ^μ)_{r\dot{r}}.$$
I guess this equation somehow contains the answer to my question, but I can't see how. The right hand side of this equation reminds me of the discussion, which leads to equivalence of Lorentz transfomations and SL(2,C) group. But how will I get some Lorentz transformation from Pauli matrices of those two 1/2 representations still puzzles me.

6. Jul 6, 2013

### dextercioby

Hi, check the source I gave you above and note one more thing. The direct product of 2 copies of SU(2) is a double-cover of the compact SO(4) group (which contains only rotations in a 4D flat space), not of the Lorentz group which is non-compact due to the boosts (sub)manifold being homeomorphic to R^3.

7. Jul 6, 2013

### Mesmerized

hi dex, thanks for the hint, I will check.

8. Jul 6, 2013

### Bill_K

Mesmerized, Put together the responses you've gotten so far, and they seem to be saying it will be necessary to master the representations of the Lorentz group in order to solve this simple exercise. Well if that's what you'd like, and you have the time for it, by all means do so.

But I had the impression that your main interest is reading along further in Zee, and you should stop and ask yourself what his aim was in posing this exercise, and how Zee expects you to approach it. It says "explicit calculation", and to do this you do not need the Infeld-van der Waerden symbol, or the two-fold covering, or any of the other stuff.

You were initially on the right track in the OP, so you just need to continue with what you began. Zee wants you to write out the six 4 x 4 generators in terms of Pauli matrices. Then compare them to the 4 x 4 generators acting on the four components of a vector. And see that the two sets are the same!

9. Jul 6, 2013

### samalkhaiat

The problem of associating fields (geometrical objects on space-time) with the abstract representations (of any Lie group) is a bit complicated (mathematical) problem. So, at least for now, you will be better off taking the following as a rule.
Since the dimension of $( j_{ 1 } , j_{ 2 } )$ is $[ n ] = ( 2 j_{ 1 } + 1 ) ( 2 j_{ 2 } + 1 )$. You can identify the representation $( j_{ 1 } , j_{ 2 } )$ with n-component field (on space-time) transforming under $SO(1,3)$.
To each $SL( 2 , C )$ transformation of the hermitian spinor-tensor $\psi_{ r \dot{ r }} \in ( 1/2 , 1/2 )$:
$$\psi_{ r \dot{ r } } \rightarrow A_{ r }{}^{ s } \bar{ A }_{ \dot{ r } }{}^{ \dot{ s } } \psi_{ s \dot{ s } }, \ \ \ ( A, \bar{ A } ) \in SL( 2 , C ) ,$$
there corresponds a Lorentz transformation of a real 4-component vector, $V_{ \mu } ( x ) \in M^{ ( 1 , 3 ) }$, in the Minkowski space-time:
$$V_{ \mu } ( x ) \rightarrow \Lambda ( A )_{ \mu }{}^{ \nu } \ V_{ \nu } ( x ) , \ \ \ \Lambda ( A ) \in SO( 1 , 3 ) .$$
It is easy to verify this, if you know that the “connection” $( \sigma^{ \mu } )_{ r \dot{ r } }$ behaves as a spinor-tensor under $SL( 2 , C )$, and as a 4-vector under $SO( 1 , 3 )$. Indeed, we have the following identity
$$A_{ r }{}^{ s } \bar{ A }_{ \dot{ r } }{}^{ \dot{ s } } ( \sigma^{ \mu } )_{ s \dot{ s } } = \Lambda_{ \nu }{}^{ \mu } ( \sigma^{ \nu } )_{ r \dot{ r } } .$$

Sam