Hermitian conjugate of a Hermitian Conjugate

[teaJ]

I know that $(A\mp )\mp =A$ . Where A is an Hermitian operator How does one go about proving this through the standard integral to find Hermitian adjoint operators?

I should mention, I don't want anyone to just flat out show me step by step how to do it. I'd just like a solid starting place.

$\int (A\mp \psi*) \varphi dx = \int (\psi*)A\varphi dx$

 

[teaJ]

Edited with the question. Forums are difficult sometimes :)

 

[stevendaryl]

I know that $(A\mp )\mp =A$ . Where A is an Hermitian operator How does one go about proving this through the standard integral to find Hermitian adjoint operators?

I should mention, I don't want anyone to just flat out show me step by step how to do it. I'd just like a solid starting place.

$\int (A\mp \psi*) \varphi dx = \int (\psi*)A\varphi dx$

I prefer $\dagger$ for Hermitian conjugate. Anyway...

Writing $\langle \psi|\phi \rangle$ for $\int \psi^* \phi dx$, we have the definition of $A^\dagger$:

Equation 1: $\langle A^\dagger \psi | \phi \rangle = \langle \psi | A \phi \rangle$

But we also have a fact about inner products:

Equation 2: $\langle X | Y \rangle^* = \langle Y | X \rangle$.

Now, apply equation 2 to both sides of equation 1 to get another fact about Hermitian conjugates.

 

[DrDu]

I know that $(A\mp )\mp =A$ . Where A is an Hermitian operator

That's not true. An operator is hermitian or synonymously symetric, if $\langle A^\dagger \phi|\psi\rangle =\langle \phi| A\psi \rangle$ for $\phi,\; \psi \in \mathcal{D}(A)$. However, the domain of $A^\dagger$ may be larger than that of A, $\mathcal{D}(A)\subset \mathcal{D}(A^\dagger)$ and then also $\mathcal{D}(A)\subset \mathcal{D}(A^{\dagger \dagger})$. However if A is self adjoint, i.e. if additionally $\mathcal{D}(A)=\mathcal{D}(A^\dagger)$ then also $\mathcal{D}(A)=\mathcal{D}(A^{\dagger \dagger})$.