I want to prove the following identity:
$$\langle v|\Omega^{\dagger}|u\rangle = \langle u|\Omega | v \rangle^*$$
How should I go about this? I believe I can prove it when $\Omega$ is hermitian, but I do not know how to prove it in general.
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$\begingroup$ What's the definition of $\Omega^\dagger$? $\endgroup$– AccidentalFourierTransformApr 27, 2018 at 20:01
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$\begingroup$ Related: physics.stackexchange.com/q/43069/2451 , physics.stackexchange.com/q/216846/2451 and links therein. $\endgroup$– Qmechanic ♦Apr 27, 2018 at 20:05
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2 Answers
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By the definition. It is $$(\Omega^\dagger v, u) = (v, \Omega u) = (\Omega u, v) ^* $$
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The complex conjugate operation distributes in an inverse order. Let a, b, c be some numbers that the conjugate action can act. Then, $$ (abc)^* = c^* b^* a^* $$ For a general case where a, b, c are not numbers but could be state, operator, etc., such that the product is a number, then one needs to use Hermitian conjugate, of course. So, $$\langle u | \Omega | v \rangle^* = | v\rangle^\dagger \Omega^\dagger \langle u |^\dagger $$ and of course, due to the fact that the conjugate of a bra is a ket and vice versa, one can conclude the proof.
