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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?

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  • $\begingroup$ Which variable is conjugated? Mathematics and physics use the opposite conventions for that. Your "bra" and "ket" notation $\langle a \rvert$ is only used in physics, so you probably should use that convention. $\endgroup$ – GEdgar May 29 at 11:31
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In linear algebra, $u\cdot v$ is $u^* v$, the first vector gets conjugated.

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