 Solution

Write $K=\mathbb{Q}(\sqrt{d})$ and let $\sigma$ be its nontrivial automorphism. On the rational vector space $E(K)\otimes \mathbb{Q}$, the involution $\sigma$ gives the eigenspace decompositionThe positive eigenspace is $E(\mathbb{Q})\otimes \mathbb{Q}$. Choose the quadratic twist $E^{\prime }=E^d$ and an isomorphism $\iota :E^{\prime }\to E$ over $K$ for which ${\iota }^{\sigma }=-\iota$. It maps $E^{\prime }(\mathbb{Q})\otimes \mathbb{Q}$ isomorphically onto the negative eigenspace. Taking dimensions gives