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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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marked as duplicate by Qmechanic quantum-mechanics May 29 at 11:50

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    $\begingroup$ Should this be on the mathematics stack? $\endgroup$ – user207455 May 29 at 10:37
  • $\begingroup$ 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. $\endgroup$ – jacob1729 May 29 at 10:45
  • $\begingroup$ $\langle a|$ 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. $\endgroup$ – gented May 29 at 11:33
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Because it is the bra that is associated with dual vector, not the ket. If it were bra that was used as the element of original Hilbert space, then it would be like linear algebra. Blame Dirac for wanting operators to act from the left AND on the"flat side", just because it looks better than acting from the left and on the pointy side.

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