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In the lecture, my professor wrote this on the board

$$ \begin{equation} \begin{split} (\vec{\sigma}_{1}\cdot\vec{\sigma}_{2})|++\rangle &= |++\rangle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\blacktriangledown)\\ (\vec{\sigma}_{1}\cdot\vec{\sigma}_{2})(|+-\rangle+|-+\rangle) &= (|+-\rangle+|-+\rangle)\\ (\vec{\sigma}_{1}\cdot\vec{\sigma}_{2})(|+-\rangle-|-+\rangle) &= -3(|+-\rangle+|-+\rangle) \end{split} \end{equation} $$

but I don't get how these are correct. I know that

$$ \begin{equation} \begin{split} |1\;1\rangle &= |++\rangle \\ |1\;0\rangle &= \frac{1}{\sqrt{2}}(|+-\rangle+|-+\rangle) \\ |0\;0\rangle &= \frac{1}{\sqrt{2}}(|+-\rangle-|-+\rangle) \end{split} \end{equation} $$

I will work out equation $(\blacktriangledown)$ in the usual matrix representation of the eigenstates of $S_z$ basis: $$ |+\rangle=\begin{pmatrix}1\\ 0 \end{pmatrix},\;\;\;\;\;\;\;\;\;\;\;\;\;\;|-\rangle=\begin{pmatrix}0\\ 1 \end{pmatrix}, $$ So we have $$ \begin{equation} \begin{split} (\vec{\sigma}_{1}\cdot\vec{\sigma}_{2})|+\rangle_{1}\otimes|+\rangle_{2}&=&\vec{\sigma}_{1}|+\rangle_{1}\otimes\vec{\sigma}_{2}|+\rangle_{2}\\&=&\begin{pmatrix}1 & 1-i\\ 1+i & -1 \end{pmatrix}_{1}\begin{pmatrix}1\\ 0 \end{pmatrix}_{1}\otimes\begin{pmatrix}1 & 1-i\\ 1+i & -1 \end{pmatrix}_{2}\begin{pmatrix}1\\ 0 \end{pmatrix}_{2}\\&=&\begin{pmatrix}1\\ 1+i \end{pmatrix}_{1}\otimes\begin{pmatrix}1\\ 1+i \end{pmatrix}_{2} \end{split} \end{equation} $$ but this is not $|++\rangle=|+\rangle\otimes|+\rangle$. What did I do wrong here? What have I misunderstood?

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Your expression for: $$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\vec \sigma_1 |+\rangle\otimes \vec \sigma_2 |+\rangle_2$$ Is wrong. It sould read: $$(\vec \sigma_1 \cdot \vec \sigma_2) |+\rangle_1 \otimes |+\rangle_2=\sigma_{1x}|+\rangle_1\otimes \sigma_{2x}|+\rangle_2+\sigma_{1y}|+\rangle_1\otimes \sigma_{2y}|+\rangle_2+$$ $$\sigma_{1z}|+\rangle_1\otimes \sigma_{2z}|+\rangle_2$$ $$=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}$$ $$+\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}0&-i\\i&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}$$ $$+\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}$$ $$=\begin{pmatrix} 0 \\1 \end{pmatrix}\otimes\begin{pmatrix} 0 \\1 \end{pmatrix}$$ $$+\begin{pmatrix} 0 \\i \end{pmatrix}\otimes \begin{pmatrix} 0 \\i\end{pmatrix}$$ $$+\begin{pmatrix} 1 \\0 \end{pmatrix}\otimes \begin{pmatrix} 1\\0\end{pmatrix}$$ $$=\begin{pmatrix} 0 \\1 \end{pmatrix}\otimes\begin{pmatrix} 0 \\1 \end{pmatrix}$$ $$-\begin{pmatrix} 0 \\1 \end{pmatrix}\otimes \begin{pmatrix} 0 \\1\end{pmatrix}$$ $$+\begin{pmatrix} 1 \\0 \end{pmatrix}\otimes \begin{pmatrix} 1\\0\end{pmatrix}$$ $$=\begin{pmatrix} 1 \\0 \end{pmatrix}\otimes \begin{pmatrix} 1\\0\end{pmatrix}$$ i.e. I think you have to do the dot product between the Pauli matrices vectors first then put them through the tensor product.

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  • $\begingroup$ @PeterShor serves me right for copying and pasting :) $\endgroup$ – Quantum spaghettification Jun 6 '16 at 6:51

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