- #1

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As far as I understand, an Hermitian adjoint ##\hat{A}^{\dagger}## of an operator ##\hat{A}## is defined by $$\left(\hat{A}\lvert\psi\rangle ,\lvert\phi\rangle\right)=\left(\lvert\psi\rangle ,\hat{A}^{\dagger}\lvert\phi\rangle\right)$$ and the inner product of two kets defined such that $$\left(\lvert\psi\rangle ,\lvert\phi\rangle\right)\equiv\langle\psi\lvert\phi\rangle ,\qquad\langle\psi\lvert\phi\rangle =\langle\phi\lvert\psi\rangle^{\ast}$$ but using these two definitions I fail to arrive at the expression that I'm trying to prove, as it would appear from this that $$\langle\psi\lvert \hat{A}\rvert\phi\rangle =\langle\psi\lvert \hat{A}^{\dagger}\rvert\phi\rangle$$ If someone could explain my misunderstanding and the correct way to do it, I'd be grateful.