Equivalence between two definitions of hermitian adjoint Given the two definitions of hermitian adjoint:
$(1): <\Psi|A|\Phi>=(<\Psi|A^+|\Phi>)^*$
$(2): <\Psi|A\Phi>=<A^+\Psi|\Phi>$
I want to show that they are equivalent
However I fail to prove it as once I apply wither definition to prove the other I get stuck because by taking the complex conjugate the functions $\Phi$ and $\Psi$ get flipped.
For instance, in $(1) \Rightarrow (2)$:
$<\Psi|A\Phi>=<\Psi|A|\Phi>=(<\Psi|A^+|\Phi>)^*=<\Phi|A^+|\Psi>=<\Phi|A^+\Psi>$
Where the order of the function is wrong
I'd appreciate if someone gave me a hint on how to prove this
Thanks in advance
 A: A big problem with the bra-ket notation is that the expression $\langle\Psi|A|\Phi\rangle$ is ambiguous if $A$ is not selfadjoint. For instance if you use the usual inner product on $\mathbb C^2$, then $\langle \Psi|\Phi\rangle$ means $\Psi^*\Phi$, where the $*$ here means the conjugate transpose. Then, is
$
\langle \Psi|A|\Phi\rangle
$ equal to $\Psi^*A\Phi$, or to $(\Psi A)^*\Phi$? These are not the same if $A$ is not selfadjoint as can already be seen with $\Phi=\Psi$ and $A=i\,I$.
Even then, the first equality does not define the adjoint as written. This can be seen if you take $A$ selfadjoint and $\Phi=i\,\Psi$; the two sides are not equal.
Assuming that $
\langle \Psi|A|\Phi\rangle
$ means $\langle \Psi|A\Phi\rangle$, the right equality for $(1)$ is
$$\tag{$*$}
\langle \Psi|A|\Phi\rangle=\langle \Phi|A^\dagger|\Psi\rangle^*. 
$$
Now the two equalities are the same, as the left-hand-side on $(*)$ is $\langle \Psi,A\Phi\rangle$, and the right-hand-side is
$$
\langle \Phi|A^\dagger|\Psi\rangle^*=\langle \Phi,A^\dagger\Psi\rangle^*=\langle A^\dagger\Psi,\Phi\rangle. 
$$
