# Doubt about property of hermitian operator

For any hermitian operator M, prove that $$$$\langle Ma|b \rangle = \langle a|Mb \rangle$$$$

My attempt:
Let $$\begin{eqnarray} \langle a| = \sum_i a_i^*\langle i|\\ |b\rangle = \sum_j b_j|j\rangle \end{eqnarray}$$ Then $$\begin{eqnarray} \langle Ma| &=& \sum_{i} a_i^*M\langle i|\\ \therefore\langle Ma|b\rangle &=& \sum_{i,j} a_i^*b_jM\langle i|j\rangle\\ &=& \sum_{i,j} a_i^*b_jM\delta_{ij}\\ &=&\sum_{i} a_i^*b_iM \end{eqnarray}$$ Now similarly, $$\begin{eqnarray} \langle a|Mb\rangle &=& \sum_{i,j} a_i^*b_j\langle i|M|j \rangle\\ &=& \sum_{i,j} a_i^*b_j M_{ij} \end{eqnarray}$$ I don't know what to do next. I feel that I am very close to the answer

• How exactly are you defining a Hermitian operator? Commented Sep 21, 2021 at 16:27
• @BySymmetry I have not used the hermiticity of the operator but the theorem does mention operator M being hermitian. Commented Sep 21, 2021 at 16:29
• So my issue is that the statement you are trying to prove is what is normally taken as the definition of a Hermitian operator, so in order to prove this you must be using some other definition. Using an alternative equivalent definition is fine, but we need to know what it is in order to help you here Commented Sep 21, 2021 at 16:41
• @BySymmetry I am using the definition that if M is hermitian, then $M$ = $M^\dagger$ Commented Sep 21, 2021 at 16:44
• And how are you defining the $\dagger$ operation? Commented Sep 21, 2021 at 16:46

A is Hermitian if $$A=A^\dagger$$ or equivalently if $$\langle x,Ay\rangle =\langle Ax,y\rangle \ \ \forall\ \ x,y.$$
This can be seen as follows: $$\langle a|A^\dagger|b\rangle \equiv \langle b|A|a\rangle ^*$$ At this point, We can proceed in two equivalent way $$\langle a|A^\dagger |b\rangle=\langle b|aA\rangle^*=\langle Aa|b\rangle$$ Or $$\langle a|A^\dagger|b\rangle =\langle Ab|a\rangle^*=\langle a|bA\rangle$$ Hence$$\langle a|bA\rangle =\langle Aa|b\rangle$$