# How do get from $|0\rangle=\alpha|a\rangle+\beta|b\rangle$, $|1\rangle=\gamma|a\rangle+\delta|b\rangle$ to an expression for $|01\rangle-|10\rangle$?

My question is linked to the Nielsen Chuang book. Particularly equation 2.216 on basis change from $$|0\rangle$$, $$|1\rangle$$ to orthonormal $$|a\rangle$$ and $$|b\rangle$$. How do we get the equation from the following?

I understand everything except why we get the negative sign in the coefficient.

This is just algebra because $$(a-b)|u\rangle + (c-d)|v\rangle = (a-b)|u\rangle - (d-c)|v\rangle$$.
\begin{align} |01\rangle = \big( \alpha |a\rangle + \beta |b\rangle\big) \otimes \big(\gamma |a\rangle + \delta |b\rangle \big) = \alpha \gamma |aa\rangle + \alpha \delta |ab\rangle + \beta \gamma |ba\rangle + \beta \delta |bb\rangle \end{align} and similarly \begin{align} |10\rangle = \big( \gamma |a\rangle + \delta |b\rangle \big) \otimes \big( \alpha |a\rangle + \beta |b\rangle\big) = \gamma \delta|aa\rangle + \gamma \beta |ab\rangle + \delta \alpha |ba\rangle + \delta \beta |bb\rangle \end{align}
\begin{align} \dfrac{|01\rangle - |10\rangle}{\sqrt{2}} &= \dfrac{\alpha \gamma |aa\rangle + \alpha \delta |ab\rangle + \beta \gamma |ba\rangle + \beta \delta |bb\rangle}{\sqrt{2}} \\ &- \\ &\hspace{0.6 cm}\frac{\gamma \delta|aa\rangle + \gamma \beta |ab\rangle + \delta \alpha |ba\rangle + \delta \beta |bb\rangle}{\sqrt{2}} \\ &= 0 + \dfrac{\big(\alpha \delta - \gamma \beta \big)|ab\rangle}{\sqrt{2}} + \dfrac{\big( \beta\gamma - \delta \alpha\big) |ba\rangle}{\sqrt{2}} + 0 \\ &= \dfrac{\big(\alpha \delta - \gamma \beta \big)|ab\rangle - \big( \delta \alpha -\beta\gamma \big) |ba\rangle}{\sqrt{2}} \\ &= \dfrac{\big(\alpha \delta - \gamma \beta \big)|ab\rangle - \big( \alpha \delta -\gamma \beta\big) |ba\rangle}{\sqrt{2}} \\ &= \big(\alpha \delta - \gamma \beta \big) \dfrac{|ab\rangle - |ba\rangle}{\sqrt{2}} \end{align}