If I reduce the Lorentz group to the representation $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$, I can write left and right-handed Weyl spinors respectively as $\left( \frac{1}{2},0 \right)$ and $\left(0, \frac{1}{2} \right)$. I can get a Dirac spinor by summing them, i.e. $\left( \frac{1}{2},0 \right) \oplus \left(0, \frac{1}{2} \right)$.

In my lecture notes, it is said that one can obtain higher representations by taking the tensor product. As an example, my prof gives $\left( \frac{1}{2},0 \right) \otimes \left(0, \frac{1}{2} \right) = \left(\frac{1}{2}, \frac{1}{2} \right)$, and he says that this corresponds to spin $1$ with $4$ components. I don't really understand what are the numbers on the right-hand side, and how I can relate this object to spin $1$. What does this notation mean, and how does the calculation work?

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    $\begingroup$ possible dup.? physics.stackexchange.com/q/192349/84967 $\endgroup$ – AccidentalFourierTransform Jul 15 '19 at 18:27
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    $\begingroup$ First, due diligence. $\endgroup$ – Cosmas Zachos Jul 15 '19 at 18:32
  • $\begingroup$ @CosmasZachos Yes, I was there of course, but I couldn't find a tensor product similar to what I wrote in the OP. Or did I miss it? $\endgroup$ – Jxx Jul 15 '19 at 18:34
  • $\begingroup$ The Kronecker product $1/2 \otimes 0 =1/2$ you already know from addition of spins: adding spin 1/2 to spin 0, i.e. combination of a doublet with a singlet yields a doublet. This happens to both direct sum subspaces. Try a matrix example. $\endgroup$ – Cosmas Zachos Jul 15 '19 at 18:39
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    $\begingroup$ Might find this useful. $\endgroup$ – Cosmas Zachos Jul 15 '19 at 21:40