Spin up is described by a vector $\uparrow=(1\;0)^T$ and spin down by $\downarrow=(0\;1)^T$. When we write e.g. $\uparrow \uparrow$ what kind of product does it mean?
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Two spin up particles? – Connor Behan Dec 06 '22 at 16:10
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1It is a tensor product. See the "From bases" construction near the top of the page. The rest you don't need. Really, your quantum book should give you enough to figure out how to manipulate these objects and what they mean. – march Dec 06 '22 at 16:13
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1Related : Total spin of two spin- 1/2 particles. – Frobenius Dec 06 '22 at 19:15
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$\newcommand{\bra}[1]{\langle #1 \rvert}$ $\newcommand{\ket}[1]{\lvert #1 \rangle}$ $\newcommand{\avec}[2]{\left(\begin{matrix}#1 \\ #2 \end{matrix}\right)}$ $\newcommand{\bvec}[4]{\left(\begin{matrix}#1 \\ #2 \\ #3 \\ #4 \end{matrix}\right)}$
Spin up is described by a vector $\uparrow=(1\;0)^T$ and spin down by $\downarrow=(0\;1)^T$. When we write e.g. $\uparrow \uparrow$ what kind of product does it mean?
As stated in the comments, this is a tensor product.
Just like the original vectors can be represented by matrices (2-by-1 matrices), so too the tensor product vectors can be represented by matrices (4-by-1 matrices).
For example: $$ \uparrow\uparrow = \avec{1}{0}\times\avec{1}{0} = \avec{1\avec{1}{0}}{0\avec{1}{0}} = \bvec{1}{0}{0}{0} $$
For example: $$ \uparrow\downarrow = \avec{1}{0}\times\avec{0}{1} = \avec{1\avec{0}{1}}{0\avec{0}{1}} = \bvec{0}{1}{0}{0} $$
And so on.
