Explicit construction of $(\frac{1}{2}, \frac{1}{2})$ representation of Lorentz group

Question

For the vector representation of the Lorentz group (actually the algebra), the $J^1$ generator is

$$J_1 = i \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 &0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}, \tag1$$

while what I get applying trying to work it out from the $(\frac{1}{2}, \frac{1}{2})$ representation is

$$J_1 = \begin{pmatrix} 0 & 1/2 & 1/2 & 0 \\ 1/2 & 0 & 0 & 1/2 \\ 1/2 & 0 & 0 & 1/2 \\ 0 & 1/2 & 1/2 & 0 \end{pmatrix}\tag2$$

Obviously there is a mistake in my calculation, the question is where and how to rectify it. Here is how I arrived at the above calculation.

To construct the irreducible representations of the homogeneous Lorentz group, starting with the $J^{\mu\nu}$ generators, one defines $\mathbf{J}=\{J^1,J^2,J^3\}$ as $J^1 = J^{23}$ and so on by cyclic permutations and $\mathbf{K} = \{K^1, K^2, K^3\}$ as $K^i = J^{0i}$. Then by defining

$$ \mathbf{A}=\frac{1}{2}(\mathbf{J}+i\mathbf{K}), \mathbf{B}=\frac{1}{2}(\mathbf{J}-i\mathbf{K})\tag3$$

one gets two independent $SU(2)$ algebras, since

$$[A_i,A_j]=i\epsilon_{ijk}A_k,\tag4$$

$$[B_i, B_j] = i\epsilon_{ijk} B_k,\tag5$$

$$[A_i, B_j]=0.\tag6$$

Thus $\mathbf{A}$ can be represented by the usual angular momentum matrices $\mathbf{J}^{(A)}_{a,a^\prime}$, where $A$ is the maximum value (integer or half integer) and $a,a^\prime = -A,...,+A$. Similarly for $\mathbf{B}$. Combining the two for the Lorentz group, in term of indices we have

$$(\mathbf{A})_{ab,a^\prime b^\prime} = \mathbf{J}^{(A)}_{aa^\prime} \delta_{bb^\prime}\tag7$$

$$(\mathbf{B})_{ab,a^\prime b^\prime} = \delta_{aa^\prime} \mathbf{J}^{(B)}_{bb^\prime}\tag8$$

Now here starts my trouble. As I understand it, in matrix form these two are

$$\mathbf{A} = \mathbf{J}^{(A)} \otimes \mathbf{1}_{2b+1}\tag9$$

and

$$\mathbf{B} = \mathbf{1}_{2a+1} \otimes \mathbf{J}^{(B)}.\tag{10}$$

$\mathbf{1}$ stands of course for the unit matrix of the appropriate dimensions and $\otimes$ is the matrix direct product.

Using these equations for $A=\frac{1}{2}=B$, the $J$s are $\frac{1}{2}\sigma^i$, half the Pauli matrices and what I get is

$$A_1 =\frac{1}{2}\sigma_1 \otimes \mathbf{1}_2 = \begin{pmatrix} 0 & 1/2 \\ 1/2 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \tag{11}$$

$$= \begin{pmatrix} 0 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & \frac{1}{2} \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \frac{1}{2} \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & 0 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1/2 & 0 \\ 0 & 0 & 0 & 1/2 \\ 1/2 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 0 \end{pmatrix}\tag{12}$$

and similarly

$$B_1 = \mathbf{1}_2 \otimes \frac{1}{2}\sigma_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1/2 \\ 1/2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1/2 & 0 & 0 \\ 1/2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/2 \\ 0 & 0 & 1/2 & 0 \end{pmatrix}\tag{13}$$

From the definitions of $\mathbf{A}$ and $\mathbf{B}$, eq. (3), $\mathbf{J} = \mathbf{A} + \mathbf{B}$, therefore $J_1 = A_1 + B_1 $ and

$$J_1 = \begin{pmatrix} 0 & 1/2 & 1/2 & 0 \\ 1/2 & 0 & 0 & 1/2 \\ 1/2 & 0 & 0 & 1/2 \\ 0 & 1/2 & 1/2 & 0 \end{pmatrix}\tag{14}$$

which hardly looks like the $J_1$ for the vector representation.

Somewhere along the line there is a mistake, yet I fail to see it. Any help and advise will be greatly appreciated.

Answer 1

You, in fact, did not make any mistake. You just did not pursue the problem to the end, changing bases to reduce your reducible representation. In fact, regardless of your Lorentz embedding usage, this is just the Kronecker product composition of two spin 1/2 representations into Kronecker sum of a spin 1 and a spin 0 representation... undergraduate addition of angular momentum. You never actually used any boosts!

The reducible representation of angular momentum you found, (14), $$ \Delta (J_1) = \frac{1}{2}\begin{pmatrix} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{pmatrix}, $$ $\Delta (J_3)=$diag (1,0,0,-1), etc, can be reduced to the direct sum of a triplet (spin one) and a singlet (spin 0, so generators are realized trivially by 0!) rep, by the orthogonal Clebsch–Gordan matrix, $$ C=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. $$ That is to say, $$ C^{-1} \Delta(J_1) C= \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}. $$ The 3rd component of the 4-vectors has decoupled (it is the singlet!), while the other 3 components constitute just $$ j_1=\frac{1}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, $$ the triplet generator in the spherical basis. The footnote in that article, or this answer, gives you the similarity transformation to the Cartesian basis expression you wrote, (1), after the 0th row and columns are likewise excised. Of course, $\Delta (J_3)=$diag (1,0,0,-1) remained unaffected by the above Clebsching to yield just $j_3=$diag(1,0,-1).

You may convince yourself of the analogous reduction with the same Clebsch-Gordan matrix for $\Delta(J_2)$. You may then fit this rep subspace into a 3-by-3 block, (the lower right one) of the 4-vector rep you are using in 1. As a last check to appreciate the structure, work out the full quadratic Casimir invariant and its reduction to a 3d identity and a zero entry for the singlet.