# Matrix Element of (1/2, 0) Representation of the Lorentz Group

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Let us consider the Lorentz group $SO(1, 3)$. There are two copies of $SU(2)$ in it. We specify a matrix representation of $SO(1, 3)$ by a doublet $(j, j')$; where $j$ corresponds to the SU(2) generated by $N_i^+$ and $j'$ corresponds to the SU(2) generated by $N_i^-$; $N_i^\pm$ are generators of the $SU(2)$s.

Say, I want to write a matrix element of the $(0, \frac{1}{2})$ representation of $SO(1, 3)$. How can I mathematically construct such an element? A detailed discussion would be highly appreciated.

recategorized Oct 22, 2016

Represent the first SU(2) trivially and the second in its fundamental representation. Represent the desired infinitesimal  generator of SO(1,3) as a linear combination of the generators of the two SU(2)s. Then you can take the matrix elements without problems.

@ArnoldNeumaier  The problem is that the generators and the matrix elements of the $j = 0$ representation of the $SU(2)$ are $1 \times 1$ matrices. But the generators and the matrix elements of the $j' = 1$ representation of the $SU(2)$ are $2 \times 2$ matrices. How can I take their linear combination, when we can't add $1 \times 1$ matrices with the $2 \times 2$ matrices?

You need to take the tensor product of the representations, not (as your comment implies that you currently do) the direct sum. Note that $1 \otimes 2 = 2$. This means that your Hilbert space is $1\times 2=2$-dimensional and the action of the generators of the 0-representation is identically zero.

@ArnoldNeumaier: Would you mind if I request you write an equation?

A general element of $so(1,3)$ is $X= a\cdot N_1+b\cdot N_2$ with 3-vectors $a,b$, and $X(1\otimes \psi_2)=1\otimes b\cdot N_2 \psi_2$.

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