# Complex conjugate in inner products [duplicate]

When we solve for inner product of $$\rvert a \rangle \cdot \rvert b \rangle$$ we solve for $$\langle a \rvert b \rangle$$ where $$\langle a \rvert$$ is complex conjugate of $$\rvert a \rangle$$. However this confuses me because in linear algebra, $$u \cdot v$$ is $$uv^*$$. The latter vector is conjugated. Why does braket notation conjugate prior vector and linear algebra conjugate latter vector?

• Should this be on the mathematics stack? – user207455 May 29 '19 at 10:37
• I'm pretty sure the physicist convention is $u^{\dagger}v$. The mathematician's convention is different, I think they might consider the bra-vectors $\langle \psi \rvert$ fundamental. – jacob1729 May 29 '19 at 10:45
• $\langle a|$ is NOT the complex conjugate of $|b\rangle$, rather it is the functional in the dual space associated with the corresponding vector. Also, your "linear algebra" notation isn't necessarily always true. – gented May 29 '19 at 11:33